Fundamentals of Mechanical Engineering Unit 2 Notes- AKTU B.Tech 1st Year

University: Dr. A.P.J. Abdul Kalam Technical University (AKTU)

Programme: B.Tech First Year, Semester I & II

Course Codes: AME101A / AME201A

Branches: Common to All

Unit: II — Thermodynamics and Heat Transfer Basics

Introduction

Thermodynamics and Heat Transfer Basics explains stored energy, thermal transfer and the limits on energy conversion. Its applications connect these principles to processor coolers, server-room air conditioning and EV battery thermal management.

Thermodynamics establishes how energy is stored and transferred across a selected boundary. Heat transfer explains the mechanisms and rates through which thermal energy moves. Together, they help determine why a component warms, how a cooling arrangement transports energy and what information is needed to predict temperature.

A complete solution begins by identifying the system. A processor package, a processor-and-cooler assembly, a room and a complete refrigeration cycle have different boundaries. A transfer that crosses one boundary can become an internal interaction when the boundary is enlarged.

The equations must also distinguish energy from power. Internal energy, heat and work are measured in joules. Heat-transfer rate and electrical power are measured in watts. Temperature is a state property and is not interchangeable with either energy or its rate.

Topic Map

Symbols and Units

Sign Convention

For a stationary closed system:

• Q is positive into the system.
• W is positive when done by the system.
• Work supplied to the system is therefore negative W.
• With negligible kinetic and potential energy changes, ΔU = Q − W.

Cooling-cycle balances use separately named positive magnitudes, such as heat removed from a room, input power and rejected heat. The sign convention must remain consistent within each calculation.

How to Approach the Unit

• Draw the boundary before balancing energy.
• Identify matter, heat and work crossing that boundary.
• Specify whether the problem concerns an interval or a steady rate.
• Distinguish steady temperature from thermal equilibrium.
• Name each segment of the thermal path.
• Use resistance values with matching endpoints.
• Convert to kelvin before temperature ratios or fourth powers.
• Keep air, coolant and refrigerant streams separate.
• State the assumptions behind a temperature-rise or flow calculation.
• Check the answer’s units, direction and expected trend.

1. System, Surroundings and Boundary

Thermodynamics studies energy, its storage and transfer, and the restrictions on energy conversion.

The system is the selected matter or region being studied.

Everything outside the selected system is its surroundings.

The boundary separates the two. It may be a real wall or an imaginary surface. It can be fixed or moving.

A processor package can be selected as a system receiving electrical energy and transferring thermal energy to a cooler. If the boundary is enlarged to include the cooler, package-to-cooler conduction becomes an internal interaction.

Heat rejection from the combined package-and-cooler system to room air still crosses the enlarged boundary.

The physical equipment has not changed, but the bookkeeping has. This is why a boundary must be chosen before energy terms are added.

Boundary Selection Method

• Name the object or region.
• Draw a closed boundary around it.
• Mark every mass stream crossing it.
• Mark heat and work transfers.
• Specify an interval or steady condition.
• Identify stored-energy change.

Avoiding Double Counting

In a steady series path, processor-to-sink conduction and sink-to-air heat rejection are successive transfers of the same energy.

They should not be added as two independent heat sources in the overall room budget.

Thermodynamics gives the energy balance. Heat-transfer models determine the temperature differences and rates associated with the path.

Worked Problem 2.1: Electrical Input Exceeds Heat Rejection

Problem: A stationary assembly receives 100 W electrical input and rejects 80 W as heat for 60 s. No mass crosses the boundary and other transfers are negligible. Find stored-energy change.

Step 1: Calculate transferred energies.

Electrical input = 100 × 60 = 6000 J

Heat leaving = 80 × 60 = 4800 J

Step 2: Apply input minus output.

ΔU = (100 − 80) × 60 = 1200 J

Step 3: Internal energy increases by 1200 J.

Step 4: Check using signed transfers.

Q = −4800 J

W = −6000 J

ΔU = Q − W = −4800 − (−6000) = 1200 J

Electrical work enters the selected system, so its work-by-system value is negative.

Worked Problem 2.2: Balanced Input and Output

Problem: An assembly receives 150 W and rejects 150 W for 120 s.

Step 1:

Electrical input = 150 × 120 = 18000 J

Heat leaving = 18000 J

Step 2:

ΔU = (150 − 150) × 120 = 0 J

Step 3: Stored energy remains unchanged.

Step 4:

Q = −18000 J

W = −18000 J

The signed first law also gives zero.

Zero storage change does not mean there was no heat transfer. Energy entered and left continuously.

Worked Problem 2.3: Heat Rejection Exceeds Input

Problem: An assembly receives 90 W and rejects 110 W for 30 s.

Step 1:

Input = 90 × 30 = 2700 J

Output = 110 × 30 = 3300 J

Step 2:

ΔU = (90 − 110) × 30 = −600 J

Step 3: Internal energy decreases by 600 J.

Step 4:

Q = −3300 J

W = −2700 J

Q − W = −600 J

The system supplies part of the rejected energy by reducing its stored energy.

2. Closed, Open and Isolated Systems

A closed system contains a fixed amount of matter. Energy can cross its boundary as heat or work, but matter cannot.

A sealed heated vessel is a simple closed-system model.

An open system or control volume permits mass to enter and leave. The flowing matter also carries energy.

A cooling-water passage is an example.

An isolated system exchanges neither matter nor energy with its surroundings.

Insulation alone does not establish isolation. A closed insulated vessel can still receive shaft work from stirring or electrical work through wires.

System Comparison

A control surface around a running fan is open if air crosses it.

Selecting only the solid blade as a material body gives a different model.

The object’s name does not determine the system type; its boundary does.

Steady Flow and Mass Accumulation

A steady open system can have equal inflow and outflow with constant mass inside.

No accumulation is different from no mass transfer.

The mass balance over an interval is:

mfinal = minitial + (ṁin − ṁout)t

for constant supplied flow rates.

Worked Problem 2.4: Equal Inlet and Outlet Flows

Problem: A control volume initially contains 2 kg. Inlet and outlet mass flows are each 0.05 kg/s for 10 s.

Step 1:

Mass entering = 0.05 × 10 = 0.5 kg

Mass leaving = 0.5 kg

Step 2:

Accumulation = 0 kg

Step 3:

mfinal = 2 kg

Step 4: The system is open because matter crosses the boundary. Equal flow rates do not make it closed.

Worked Problem 2.5: Mass Accumulation

Problem: Initial mass is 3 kg, inlet flow 0.08 kg/s and outlet flow 0.06 kg/s for 10 s.

Step 1:

Mass entering = 0.8 kg

Mass leaving = 0.6 kg

Step 2:

Accumulation = 0.2 kg

Step 3:

mfinal = 3 + 0.2 = 3.2 kg

Step 4: The control volume remains open. The flow imbalance increases the mass inside.

Worked Problem 2.6: Mass Depletion

Problem: Initial mass is 1 kg, inlet flow 0.02 kg/s and outlet flow 0.03 kg/s for 10 s.

Step 1:

Mass entering = 0.2 kg

Mass leaving = 0.3 kg

Step 2:

Accumulation = −0.1 kg

Step 3:

mfinal = 0.9 kg

Step 4: The model is open and loses stored mass. The supplied interval gives a physically positive remaining mass.

3. Properties, State and Thermal Equilibrium

A thermodynamic property describes the state of a system.

Temperature, pressure, volume, mass and internal energy are examples.

A state is specified by enough independent properties for the selected material model.

A process changes the system from one state to another. A cycle returns it to its starting state.

Extensive and Intensive Properties

An extensive property depends on the amount of matter.

Examples include:

• Mass.
• Total volume.
• Total internal energy.

An intensive property does not scale directly with total amount.

Examples include:

• Temperature.
• Pressure.
• Density.
• Specific internal energy.

Specific internal energy is:

u = U/m

Its unit is J/kg or kJ/kg.

Thermal Equilibrium and Steady Temperature

Thermal equilibrium means no net thermal transfer between bodies in thermal contact.

A processor can maintain a steady temperature while rejecting heat continuously to cooler air.

It is then steady at the stated location, but it is not in thermal equilibrium with that cooler air.

A stable reading also does not prove uniform temperature throughout an assembly.

State and Path

For a closed system, ΔU depends on the end states.

Heat and work depend on the process connecting those states.

A cycle has ΔU = 0, but its heat and work transfers can each be nonzero.

Worked Problem 2.7: Total and Specific Internal Energy

Problem: A uniform material has mass 2 kg and supplied u = 180 kJ/kg. Find total internal energy and the energy of two equal portions at the same state.

Step 1:

U = mu = 2 × 180 = 360 kJ

Step 2:

Mass per portion = 1 kg

Step 3:

Energy per portion = 1 × 180 = 180 kJ

Step 4: Total U scales with mass, but u remains 180 kJ/kg at the stated state.

No temperature follows without an appropriate material relation.

Worked Problem 2.8: Three Equal Portions

Problem: Mass is 3 kg and u = 250 kJ/kg. Divide it into three equal portions at the same state.

Step 1:

U = 3 × 250 = 750 kJ

Step 2:

Mass per portion = 1 kg

Step 3:

Uportion = 250 kJ

Step 4: The portion energies sum to 750 kJ. Specific internal energy remains unchanged because the state is unchanged.

Worked Problem 2.9: Two Portions of a Smaller Mass

Problem: Mass is 1.5 kg and u = 120 kJ/kg. Find total energy and energy of each of two equal portions.

Step 1:

U = 1.5 × 120 = 180 kJ

Step 2:

Mass per portion = 0.75 kg

Step 3:

Uportion = 0.75 × 120 = 90 kJ

Step 4: The extensive total changes when the amount changes, while the supplied specific property remains the same.

4. Temperature Scales and Differences

Temperature determines the direction of spontaneous net thermal transfer.

Heat is energy in transit because of a temperature difference. Temperature itself is a state property.

Kelvin is the absolute scale used in thermodynamic temperature ratios and radiation equations.

T(K) = t(°C) + 273.15

A rise of 1 K has the same size as a rise of 1°C.

However, an absolute temperature of 1 K is not 1°C.

Choosing the Correct Form

For a simple subtraction, the 273.15 offset cancels.

For a ratio or individual fourth power, the offset does not cancel.

A CPU at 70°C and air at 25°C have a 45 K difference. The ratio 70/25 is not an absolute thermodynamic temperature ratio.

Worked Problem 2.10: 80°C and 25°C

Problem: Convert both temperatures, find their difference and identify natural net heat-flow direction.

Step 1:

Th = 80 + 273.15 = 353.15 K

Step 2:

Tc = 25 + 273.15 = 298.15 K

Step 3:

ΔT = 80 − 25 = 55 K

Step 4: Natural net transfer is from the 80°C body toward the 25°C body.

A numerical rate requires a heat-transfer model and more data.

Worked Problem 2.11: 60°C and 20°C

Step 1:

Th = 333.15 K

Step 2:

Tc = 293.15 K

Step 3:

ΔT = 40 K

Step 4: Natural net transfer is from 60°C to 20°C.

Knowing the temperatures establishes direction but not the complete transfer rate.

Worked Problem 2.12: Two Negative Celsius Temperatures

Problem: Bodies are at −10°C and −25°C.

Step 1:

Th = −10 + 273.15 = 263.15 K

Step 2:

Tc = −25 + 273.15 = 248.15 K

Step 3:

ΔT = −10 − (−25) = 15 K

Step 4: The −10°C body is hotter. Natural net transfer is toward the −25°C body.

Negative Celsius readings do not imply negative absolute temperature in this ordinary model.

5. Internal Energy, Heat and Work

Internal energy U is stored microscopic energy associated with the material’s state.

Heat Q crosses a boundary because of temperature difference.

Work W is a non-thermal transfer mode, such as electrical input or shaft work.

A body contains internal energy. It does not contain a stored substance called heat.

For the stated stationary closed system:

ΔU = Q − W

Work supplied to the system is negative W.

Energy and Rate

A watt is a joule per second:

1 W = 1 J/s

For constant power:

Energy = Power × Time

A 100 W transfer for 60 s corresponds to 6000 J.

The 100 W value is a rate. The 6000 J value is an amount.

Heater Boundary

Electrical input to a heater is work at the electrical boundary.

Energy leaving its hot surface because of temperature difference is heat.

If the heater also warms during the interval, part of the electrical input changes its stored energy.

Worked Problem 2.13: An 80 W Heater

Problem: A heater receives 80 W for 90 s. Under the supplied ideal model, all input reaches a separate receiving body, with no heater storage or other loss.

Step 1:

Ein = 80 × 90 = 7200 J

Step 2: The zero-storage assumption makes transferred heat equal to input.

Q = 7200 J

Step 3:

Q = 7.2 kJ

Step 4: The rate remains 80 W. Energy must not be labelled in watts.

Worked Problem 2.14: A 250 W Heater

Problem: A 250 W heater operates for 120 s under the same ideal assumptions.

Step 1:

Ein = 250 × 120 = 30000 J

Step 2:

Q = 30000 J

Step 3:

Q = 30 kJ

Step 4: Changing heater storage would require another energy term. The result relies on the supplied zero-storage approximation.

Worked Problem 2.15: A 1200 W Heater

Problem: A heater receives 1200 W for 30 s with all input transferred to a receiving body.

Step 1:

Ein = 1200 × 30 = 36000 J

Step 2:

Q = 36000 J

Step 3:

Q = 36 kJ

Step 4: The transfer rate is 1200 W, while the transferred energy is 36 kJ.

The two values describe different aspects of the same interval.

6. Zeroth Law and Temperature Measurement

The zeroth law states that if two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

This supports the use of a thermometer to compare temperatures.

A thermometer is placed in suitable thermal contact with the region being measured and allowed to respond.

At equilibrium, its temperature matches that region within the measurement conditions.

A real sensor has response time, calibration uncertainty and a location.

A cooler surface temperature is not automatically CPU junction temperature. Thermal resistance creates differences between the heat-generating region, package, sink and air.

Three Different Conditions

• Thermal equilibrium: No net thermal exchange between bodies in contact.
• Steady temperature: Temperature at a location does not change with time.
• Uniform temperature: Temperature does not vary spatially within the selected body.

A powered heat sink can be steady and hotter than air. Balanced input and output do not make it isothermal with the surroundings.

Thermal Inspection Context

Thermal observations need calibration and correct interpretation.

Surface temperature, stored energy and transfer rate remain different quantities. An image does not supply every boundary condition.

Worked Problem 2.16: Negative Sensor Offset

Problem: A thermometer indicates 65°C. Offset is defined as indicated minus true temperature and equals −1.5°C.

Step 1:

Offset = Tindicated − Ttrue

Therefore:

Ttrue = Tindicated − Offset

Step 2:

Ttrue = 65 − (−1.5) = 66.5°C

Step 3:

Ttrue = 339.65 K

Step 4: The corrected value belongs to the surface in equilibrium with the sensor. It does not automatically represent an internal junction.

Worked Problem 2.17: Positive Sensor Offset

Problem: The thermometer indicates 25°C and has an indicated-minus-true offset of +0.8°C.

Step 1:

Ttrue = 25 − 0.8

Step 2:

Ttrue = 24.2°C

Step 3:

Ttrue = 297.35 K

Step 4: The offset is a temperature difference and has the same numerical size in kelvin.

Its sign follows the supplied definition, rather than an assumed correction rule.

7. First Law for a Stationary Closed System

Energy is conserved. It can change form, be stored and cross boundaries, but it is not created or destroyed.

For a stationary closed system with negligible kinetic and potential energy changes:

ΔU = Q − W

A positive ΔU indicates increasing internal energy.

A negative ΔU indicates decreasing internal energy.

For example, 800 J heat input and 250 J work output give a 550 J increase.

If 250 J of work is instead supplied, W = −250 J and the increase becomes 1050 J.

Four-Step Method

• Select the system and sign convention.
• Write heat and work with signs.
• Substitute into ΔU = Q − W.
• Explain whether energy rises or falls.

It is also valid to use positive input and output magnitudes, provided every transfer is placed on the correct side.

For an electronic assembly with input P and rejection q:

Stored-energy rate = P − q

under the stated simplified model.

Worked Problem 2.18: Heat In and Work Out

Problem: Q = 800 J and W = 250 J.

Step 1: Heat entering is 800 J.

Step 2: Work done by the system is 250 J.

Step 3:

ΔU = 800 − 250 = 550 J

Step 4: Internal energy increases by 550 J.

The input-minus-output form gives the same result.

Worked Problem 2.19: Heat and Work Both Enter

Problem: Q = 600 J and W = −200 J.

Step 1: Heat input is 600 J.

Step 2: The negative work means 200 J is supplied to the system.

Step 3:

ΔU = 600 − (−200) = 800 J

Step 4: Internal energy increases by 800 J.

Adding the two physical inputs is consistent with the signed equation.

Worked Problem 2.20: Heat and Work Both Leave

Problem: Q = −500 J and W = 150 J.

Step 1: Heat leaving is 500 J.

Step 2: Work output is 150 J.

Step 3:

ΔU = −500 − 150 = −650 J

Step 4: Internal energy decreases by 650 J.

The negative result identifies depletion of stored energy.

Worked Problem 2.21: Work Input Exceeds Heat Loss

Problem: Q = −200 J and W = −700 J.

Step 1: Heat leaving is 200 J.

Step 2: Work entering is 700 J.

Step 3:

ΔU = −200 − (−700) = 500 J

Step 4: Internal energy increases by 500 J.

The sign convention is unchanged even though the physical direction of work differs from the positive reference.

8. Adiabatic, Rigid and Isothermal Models

An adiabatic process has no heat crossing the selected boundary:

Q = 0

It can still involve work and temperature change.

A rigid vessel has fixed volume and therefore no moving-boundary expansion work.

Other work, including electrical input or stirring, can still cross the boundary.

An isothermal process maintains constant temperature.

Constant temperature does not universally mean zero heat or zero work.

For a supplied ideal-gas model, internal energy depends only on temperature. An isothermal ideal-gas process therefore has ΔU = 0 and Q = W.

That conclusion cannot be assigned to every material without a constitutive model.

Comparison

An insulated outer enclosure can contain internal heat exchange. The outer-boundary heat can be zero while components exchange energy internally.

Worked Problem 2.22: Adiabatic Work Input

Problem: A stationary closed system is adiabatic. Q = 0 and W = −400 J.

Step 1:

ΔU = 0 − (−400) = 400 J

Step 2: Adiabatic eliminates heat transfer, not work input.

Step 3: The supplied work increases stored energy.

Step 4: Temperature change requires a material energy relation or heat capacity. The process label alone does not give that change.

Worked Problem 2.23: Rigid Electrically Powered System

Problem: A rigid system has Q = −200 J and W = −900 J.

Step 1:

ΔU = −200 − (−900) = 700 J

Step 2: Rigid geometry eliminates volume-change work.

Step 3: It does not prohibit the specified electrical work input.

Step 4: The resulting temperature requires additional material information. Fixed volume alone does not mean fixed temperature.

Worked Problem 2.24: Isothermal Ideal Gas

Problem: A supplied isothermal ideal-gas model has Q = 300 J and W = 300 J.

Step 1:

ΔU = 300 − 300 = 0 J

Step 2: Constant ideal-gas temperature gives no internal-energy change under the supplied model.

Step 3: Heat input balances work output.

Step 4: This result depends on the ideal-gas energy relation. Isothermal is not a universal synonym for no heat transfer.

9. Cycles and Energy-Conversion Bookkeeping

A cycle returns the working system to its original state.

Because internal energy is a property:

ΔUcycle = 0

For the stationary closed cyclic model:

Qnet = Wnet

with consistent signs.

Using positive magnitudes for a heat engine:

QH = Wout + QC

Thermal efficiency is:

η = Wout/QH

The denominator is heat received, rather than rejected heat.

Conservation and Feasibility

An energy balance is necessary but not sufficient to establish physical possibility.

A proposed cyclic engine can conserve energy yet violate the second law if its sole effect is complete conversion of heat from one reservoir into work.

A boundary around the complete cycle eliminates internal transfers from the overall balance.

Detailed property charts are not needed for these supplied overall energy calculations.

Worked Problem 2.25: Engine Efficiency of 35%

Problem: A cyclic model receives 1000 kJ and rejects 650 kJ per cycle.

Step 1:

QH = Wout + QC

Step 2:

Wout = 1000 − 650 = 350 kJ

Step 3:

η = 350/1000 = 0.35 = 35%

Step 4: Reservoir conditions and the second law would also be needed for a complete feasibility assessment.

Worked Problem 2.26: Engine Efficiency of 25%

Problem: Heat input is 2400 kJ and rejection is 1800 kJ per cycle.

Step 1: Cyclic storage change is zero.

Step 2:

Wout = 2400 − 1800 = 600 kJ

Step 3:

η = 600/2400 = 25%

Step 4: Rejected heat is part of the energy balance but is not the efficiency denominator.

Worked Problem 2.27: Engine Efficiency of 30%

Problem: Heat input is 5000 kJ and rejection is 3500 kJ per cycle.

Step 1:

ΔUcycle = 0

Step 2:

Wout = 1500 kJ

Step 3:

η = 1500/5000 = 0.3 = 30%

Step 4: The balance establishes energy bookkeeping under the supplied model, rather than unrestricted performance at unspecified temperatures.

10. Second Law: Direction and Practical Limits

The first law counts energy.

The second law restricts process direction and possible performance.

Heat flows spontaneously from higher to lower temperature.

Moving heat from a cold region to a warmer region requires an accompanying effect, such as work input.

Kelvin–Planck Statement

A cyclic device cannot have the sole effect of taking heat from a single reservoir and converting it completely into work.

Clausius Statement

A cyclic device cannot have the sole effect of transferring heat from a colder body to a hotter body.

These statements do not prohibit an air conditioner. The AC receives work and rejects both the removed heat and input energy to the warmer sink.

A fan also receives electrical input. It changes airflow and convection rather than creating cold energy.

Irreversibility

Friction, mixing and heat transfer across finite temperature differences are common irreversible processes.

Energy remains conserved, but its usefulness for producing work is affected.

Calling rejected heat wasted does not remove it from the first-law equation.

Worked Problem 2.28: Cooling Device with Work Input

Problem: A cyclic device removes 600 J from a cold region and receives 200 J work per cycle.

Step 1:

QH = QL + Win

Step 2:

QH = 600 + 200 = 800 J

Step 3: Work input is the accompanying effect required for cold-to-hot transport.

Step 4: The balance does not establish performance at arbitrary unspecified reservoir temperatures. It shows why reversed transfer alone is not a violation.

Worked Problem 2.29: Larger Cooling Transfer

Problem: The device removes 1200 J and receives 300 J work.

Step 1: Cyclic storage change is zero.

Step 2:

QH = 1200 + 300 = 1500 J

Step 3: The work input distinguishes the process from forbidden sole-effect cold-to-hot transfer.

Step 4: Reservoir temperatures remain relevant to a complete performance limit.

Worked Problem 2.30: Another Heat-Pump Balance

Problem: A device removes 1500 J and receives 500 J work.

Step 1:

QH = QL + Win

Step 2:

QH = 2000 J

Step 3: Warm-side rejection includes both cold-side heat and input work.

Step 4: Energy conservation and second-law assessment are complementary checks. One does not replace the other.

11. Entropy: An Introductory Interpretation

Entropy S is a thermodynamic property used to express the second law quantitatively.

Its SI unit is J/K.

It is different from energy and is not a force or heat source.

For reversible heat transfer at constant absolute boundary temperature:

ΔS = Qrev/T

For an isolated system, total entropy cannot decrease.

It remains unchanged for an ideal reversible process and increases for an irreversible process.

A non-isolated component can have decreasing entropy if entropy leaves its boundary. The surroundings must be included in the total assessment.

For heat Q transferred from TH to TC:

ΔSH = −Q/TH

ΔSC = Q/TC

If TH > TC, the combined change is positive.

Formula Conditions

Use kelvin.

The reservoir expressions assume constant temperatures.

Do not apply Q/T as a universal real-process entropy formula to a body whose temperature changes arbitrarily.

The simple models here support introductory interpretation rather than advanced property-table analysis.

Worked Problem 2.31: Transfer from 400 K to 300 K

Problem: Heat of 1000 J passes between ideal reservoirs at 400 K and 300 K.

Step 1:

ΔSH = −1000/400 = −2.5 J/K

Step 2:

ΔSC = 1000/300 = 3.33333 J/K

Step 3:

ΔStotal = 0.833333 J/K

Step 4: The cold reservoir gains more entropy than the hot reservoir loses because the same energy is divided by a lower absolute temperature.

Energy remains conserved.

Worked Problem 2.32: Transfer from 350 K to 280 K

Problem: Q = 600 J.

Step 1:

ΔSH = −600/350 = −1.71429 J/K

Step 2:

ΔSC = 600/280 = 2.14286 J/K

Step 3:

ΔStotal = 0.428571 J/K

Step 4: The positive total is consistent with spontaneous hot-to-cold transfer under the reservoir model.

Worked Problem 2.33: Transfer from 500 K to 320 K

Problem: Q = 2000 J.

Step 1:

ΔSH = −2000/500 = −4 J/K

Step 2:

ΔSC = 2000/320 = 6.25 J/K

Step 3:

ΔStotal = 2.25 J/K

Step 4: Entropy and energy have different balances. The positive entropy change does not mean energy was created.

12. Third Law and Absolute Zero

The third law gives the zero-temperature entropy reference.

The entropy of a perfect crystal approaches zero as temperature approaches absolute zero, provided its ground state is unique.

The perfect-crystal condition matters because residual disorder can be relevant in real materials.

Absolute zero = 0 K = −273.15°C

The unattainability principle states that absolute zero cannot be reached through a finite sequence of cooling operations.

This does not prohibit low-temperature research. It identifies a limiting state.

The third law does not mean ordinary internal-energy references must all become zero at 0 K.

It also does not make 0°C an entropy-zero condition.

Electronic-System Connection

Temperature comparison supports sensing.

Energy conservation governs input and rejection.

The second law governs process direction.

Low-temperature operation still follows absolute-temperature principles.

Worked Problem 2.34: Cooling from −100°C to −200°C

Step 1:

T1 = 173.15 K

Step 2:

T2 = 73.15 K

Step 3:

Temperature drop = 100 K

Step 4: The final state is above 0 K. Removed heat requires a material energy relation; the temperature drop alone does not establish it.

Worked Problem 2.35: Cooling from −200°C to −250°C

Step 1:

T1 = 73.15 K

Step 2:

T2 = 23.15 K

Step 3:

Temperature drop = 50 K

Step 4: The sample has not reached absolute zero. A negative Celsius value is not an absolute-zero condition.

Worked Problem 2.36: Cooling from −250°C to −270°C

Step 1:

T1 = 23.15 K

Step 2:

T2 = 3.15 K

Step 3:

Temperature drop = 20 K

Step 4: The sample remains above absolute zero. The perfect-crystal entropy limit concerns approach to 0 K.

13. Sensible Heating, Cooling and Thermal Storage

For approximately constant specific heat, no phase change and the stated energy model:

Q = mcΔT

Use mass in kg, c in J/(kg·K) and temperature difference in K to obtain joules.

The whole-body heat capacity is:

C = mc

with unit J/K.

A larger C slows temperature rise for the same net input.

It does not remove the need to reject heat during sustained operation.

For a suitable uniform-temperature approximation:

C(dT/dt) = Pnet

For constant Pnet and C:

ΔT = Pnet t/C

Rate and Total Energy

A net 50 W over 20 s gives 1000 J.

If C = 200 J/K:

ΔT = 1000/200 = 5 K

A 50 J input would instead give 0.25 K.

Phase change requires latent energy and cannot be represented merely by extending mcΔT through the transition.

Worked Problem 2.37: Heating a 0.5 kg Body

Problem: m = 0.5 kg, c = 900 J/(kg·K), temperature changes from 25°C to 65°C and net heating is 100 W.

Step 1:

ΔT = 40 K

Step 2:

Q = 0.5 × 900 × 40 = 18000 J

Step 3:

t = 18000/100 = 180 s

Step 4: Energy increases. The model assumes uniform temperature, constant specific heat and no phase change.

Worked Problem 2.38: Heating a 2 kg Body

Problem: m = 2 kg, c = 4200 J/(kg·K), temperature changes from 20°C to 35°C and net input is 500 W.

Step 1:

ΔT = 15 K

Step 2:

Q = 2 × 4200 × 15 = 126000 J

Step 3:

t = 126000/500 = 252 s

Step 4: The positive temperature change gives positive energy change under the supplied sensible-heating model.

Worked Problem 2.39: Cooling a Body

Problem: m = 1.5 kg, c = 800 J/(kg·K), temperature falls from 70°C to 30°C and net cooling magnitude is 200 W.

Step 1:

ΔT = −40 K

Step 2:

Q = 1.5 × 800 × (−40) = −48000 J

Step 3:

t = 48000/200 = 240 s

Step 4: Energy decreases, while elapsed time is positive. Signed energy and positive duration must not be confused.

14. The Three Modes of Heat Transfer

Conduction

Conduction transfers thermal energy through molecular, atomic and electronic interactions without requiring bulk material motion.

Convection

Convection couples a surface to a moving fluid.

Near-wall conduction and bulk motion both affect the transfer.

Radiation

Radiation transfers energy through electromagnetic emission and absorption.

It can occur across a vacuum.

The modes are mechanisms rather than mutually exclusive device categories.

A processor cooler combines conduction through solids and interfaces, convection to air and radiation from exposed surfaces.

A fan primarily changes fluid flow. A metal sink primarily supplies a conductive route and exposed area.

Worked Problem 2.40: Package, Base and Air

Problem: Identify the principal mode for package-to-metal-base transfer and base-to-fan-driven-air transfer.

Step 1: Package to base is principally conduction.

Step 2: Base to moving air is principally convection.

Step 3: Introductory equations are kAΔT/L for a stated plane layer and hAΔT for the supplied surface-fluid model.

Step 4: Geometry, temperatures and the appropriate k or h are needed. Naming a mechanism alone does not give a numerical rate.

Worked Problem 2.41: Cold Plate and Surroundings

Problem: Identify the mode from a cold-plate wall to coolant and from a hot surface to a large enclosure.

Step 1: Wall-to-coolant transfer is principally convection.

Step 2: Surface-to-enclosure exchange is radiation in the stated description.

Step 3: Use hAΔT or εσA(Ts⁴ − Tsur⁴), with their assumptions.

Step 4: The fluid and surface references must be named. The radiation model additionally requires emissivity and absolute temperatures.

Worked Problem 2.42: Interface and Evacuated Gap

Problem: Identify the mode in a solid interface and across an evacuated gap by electromagnetic waves.

Step 1: The solid interface is a conduction segment.

Step 2: Electromagnetic transfer across the gap is radiation.

Step 3: Use the relevant conduction or radiation equation only after supplying geometry and temperatures.

Step 4: A vacuum prevents ordinary convection across the gap but does not eliminate radiative exchange.

15. Conduction Through a Plane Wall

For steady one-dimensional conduction through a plane layer with constant k and no internal generation:

q = kA(Th − Tc)/L

This gives the magnitude from hot face to cold face.

Use:

• k in W/(m·K).
• A in m².
• L in m.
• Temperature difference in K.

The local Fourier relation is:

qx = −kA(dT/dx)

The minus sign indicates transfer toward decreasing temperature.

A larger area, conductivity or temperature difference increases q at fixed remaining quantities. Greater thickness reduces q.

Heat Flux and Resistance

Heat flux is:

q″ = q/A

Its unit is W/m².

Layer resistance is:

Rcond = L/(kA)

Therefore:

q = ΔT/Rcond

Model Boundaries

The expression does not directly represent cylindrical geometry, strong three-dimensional spreading or internal generation.

Contact resistance is separate unless included in a supplied effective value.

High material conductivity alone does not establish complete cooler performance.

Worked Problem 2.43: Conductive Metal-Like Layer

Problem: k = 200 W/(m·K), A = 0.01 m², L = 0.005 m and face temperatures are 70°C and 40°C.

Step 1:

ΔT = 30 K

Step 2:

q = 200 × 0.01 × 30/0.005 = 12000 W

Step 3:

q″ = 12000/0.01 = 1.2 × 10⁶ W/m²

Step 4:

R = 0.005/(200 × 0.01) = 0.0025 K/W

Check: 12000 × 0.0025 = 30 K.

The large rate follows the specified fixed face temperatures and geometry.

Worked Problem 2.44: Lower-Conductivity Layer

Problem: k = 0.5 W/(m·K), A = 0.02 m², L = 0.004 m, Th = 50°C and Tc = 30°C.

Step 1:

ΔT = 20 K

Step 2:

q = 0.5 × 0.02 × 20/0.004 = 50 W

Step 3:

q″ = 50/0.02 = 2500 W/m²

Step 4:

R = 0.004/(0.5 × 0.02) = 0.4 K/W

The product qR recovers 20 K.

Worked Problem 2.45: Thicker Plane Layer

Problem: k = 1.5 W/(m·K), A = 0.03 m², L = 0.02 m, Th = 80°C and Tc = 25°C.

Step 1:

ΔT = 55 K

Step 2:

q = 1.5 × 0.03 × 55/0.02 = 123.75 W

Step 3:

q″ = 123.75/0.03 = 4125 W/m²

Step 4:

R = 0.02/(1.5 × 0.03) = 0.444444 K/W

Check: qR = 55 K.

Worked Problem 2.46: Thin Conductive Layer

Problem: k = 150 W/(m·K), A = 0.004 m², L = 0.003 m and temperatures are 60°C and 45°C.

Step 1:

ΔT = 15 K

Step 2:

q = 150 × 0.004 × 15/0.003 = 3000 W

Step 3:

q″ = 3000/0.004 = 750000 W/m²

Step 4:

R = 0.003/(150 × 0.004) = 0.005 K/W

Again, qR recovers the imposed temperature difference.

16. Layers in Series and Temperature Drops

Several plane layers can be represented by resistances:

Ri = Li/(kiA)

If the path is steady and interfaces have no storage or generation, the same heat-transfer rate passes through every layer.

Rtotal = R1 + R2 + …

q = (Th − Tc)/Rtotal

Each temperature drop is:

ΔTi = qRi

The largest resistance produces the largest drop at a common rate.

It is not necessarily the thickest layer, because conductivity and area also matter.

Contact and Double Counting

An interface may need extra contact resistance.

If a supplied resistance already includes that interface, do not add it again.

Adding series rates would also count the same steady transfer several times.

The layer drops must sum to the overall temperature difference.

Worked Problem 2.47: Two Series Layers

Problem: R1 = 0.1 K/W, R2 = 0.4 K/W and overall difference is 50 K.

Step 1:

Rtotal = 0.5 K/W

Step 2:

q = 50/0.5 = 100 W

Step 3:

ΔT1 = 100 × 0.1 = 10 K

ΔT2 = 100 × 0.4 = 40 K

Step 4: The drops sum to 50 K. The rate is common, rather than split between the layers.

Worked Problem 2.48: Different Resistance Distribution

Problem: R1 = 0.25 K/W, R2 = 0.15 K/W and ΔT = 60 K.

Step 1:

Rtotal = 0.4 K/W

Step 2:

q = 60/0.4 = 150 W

Step 3:

ΔT1 = 37.5 K

ΔT2 = 22.5 K

Step 4: The sum is 60 K. The larger resistance has the larger temperature drop.

Worked Problem 2.49: Low Total Resistance

Problem: R1 = 0.02 K/W, R2 = 0.08 K/W and ΔT = 25 K.

Step 1:

Rtotal = 0.1 K/W

Step 2:

q = 25/0.1 = 250 W

Step 3:

ΔT1 = 5 K

ΔT2 = 20 K

Step 4: Both layers carry 250 W. Their individual drops add to the imposed difference.

17. Convection and the Heat-Transfer Coefficient

A common convection model is:

q = hA(Ts − T∞)

Ts is surface temperature and T∞ is the specified bulk fluid temperature.

When the surface is hotter, the result is outward heat transfer.

The coefficient h has unit W/(m²·K).

It depends on geometry, fluid, flow and operating condition.

Natural and Forced Convection

Natural convection involves buoyancy-driven motion caused by density differences.

Forced convection uses a fan, pump or another imposed motion.

For supplied constant h:

Rconv = 1/(hA)

A larger h generally reduces the required surface-fluid rise for a fixed rate and area.

Effective Area

Real fins have nonuniform temperature.

Using every visible fin surface at base temperature can overpredict transfer unless effective area or a suitable efficiency is supplied.

If effective area is already given, an additional efficiency factor should not be applied without instruction.

Fan Interpretation

A fan brings fluid through the intended route and carries warmed fluid away.

It does not necessarily reduce the incoming air temperature.

Hot recirculation can make local inlet air warmer than a remote room sensor indicates.

Worked Problem 2.50: Convection from 65°C Surface

Problem: Effective area is 0.2 m², Ts = 65°C, air is 25°C and h = 25 W/(m²·K).

Step 1:

ΔT = 40 K

Step 2:

q = 25 × 0.2 × 40 = 200 W

Step 3:

Rconv = 1/(25 × 0.2) = 0.2 K/W

Step 4: The positive outward transfer agrees with the hotter surface. h represents this condition rather than a universal solid property.

Worked Problem 2.51: Higher Convection Coefficient

Problem: A = 0.15 m², Ts = 55°C, air is 30°C and h = 60 W/(m²·K).

Step 1:

ΔT = 25 K

Step 2:

q = 60 × 0.15 × 25 = 225 W

Step 3:

Rconv = 1/(60 × 0.15) = 0.111111 K/W

Step 4: qR recovers the 25 K difference.

Worked Problem 2.52: Lower Convection Coefficient

Problem: A = 0.3 m², Ts = 50°C, air is 20°C and h = 8 W/(m²·K).

Step 1:

ΔT = 30 K

Step 2:

q = 8 × 0.3 × 30 = 72 W

Step 3:

Rconv = 1/(8 × 0.3) = 0.416667 K/W

Step 4: The result uses the supplied effective coefficient. A different flow condition would require an updated model.

18. Thermal Radiation and Absolute Temperature

For a grey surface exchanging with large isothermal surroundings:

qrad = εσA(Ts⁴ − Tsur⁴)

where:

σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴)

Every temperature in a fourth power must be in kelvin.

Radiation needs no material medium.

The surroundings also emit, so the net expression subtracts opposing irradiation under the stated geometry.

Emissivity depends on the surface and modelling conditions. Visible colour is not a complete numerical emissivity specification.

Fourth-Power Error

For 60°C and 25°C, use:

333.15⁴ − 298.15⁴

Do not use:

• 60⁴ − 25⁴.
• (60 − 25)⁴.

The large-surroundings model does not replace a full enclosure analysis for arbitrary geometry.

Worked Problem 2.53: Surface at 80°C

Problem: ε = 0.8, A = 0.05 m², Ts = 80°C and surroundings are 25°C.

Step 1:

Ts = 353.15 K

Tsur = 298.15 K

Step 2:

Surface emission term = 35.2784 W

Opposing term = 17.923 W

Step 3:

qrad = 17.3554 W outward

Step 4: Equal absolute temperatures would give zero net exchange, although both bodies would still emit.

Worked Problem 2.54: Larger Surface

Problem: ε = 0.9, A = 0.2 m², Ts = 60°C and surroundings are 20°C.

Step 1:

Ts = 333.15 K

Tsur = 293.15 K

Step 2:

Emission term = 125.731 W

Opposing term = 75.3779 W

Step 3:

qrad = 50.3535 W

Step 4: The positive result agrees with the surface being hotter than its surroundings.

Worked Problem 2.55: Lower Emissivity

Problem: ε = 0.3, A = 0.1 m², Ts = 100°C and surroundings are 30°C.

Step 1:

Ts = 373.15 K

Tsur = 303.15 K

Step 2:

Emission term = 32.9812 W

Opposing term = 14.3669 W

Step 3:

qrad = 18.6143 W

Step 4: The rate follows the specified emissivity, area and absolute temperatures.

19. Complete Thermal-Resistance Paths

For a supplied steady series path:

Thot = Tamb + qRtotal

Resistances can represent package conduction, interfaces, contact, a base and convection.

Only resistances belonging to the selected series route should be added.

At fixed q, reducing resistance lowers the temperature rise.

For a specified limit:

Rmax = (Tmax − Tamb)/q

Endpoint Consistency

Junction-to-case resistance differs from junction-to-ambient resistance.

A cool sink does not establish an equally cool internal junction if the package or interface drop is large.

A complete path must not be combined with its own included sub-resistances again.

If parallel routes exist, the single-series-route assumption must be stated.

Worked Problem 2.56: Three-Series-Resistance Source

Problem: q = 100 W, source-to-case R = 0.1 K/W, interface R = 0.05 K/W, sink-to-air R = 0.25 K/W, air = 25°C and source limit = 85°C.

Step 1:

Rtotal = 0.4 K/W

Step 2:

Rise = 100 × 0.4 = 40 K

Tsource = 65°C

Step 3:

Margin = 85 − 65 = 20 K

Step 4:

Rmax = (85 − 25)/100 = 0.6 K/W

The supplied total is below the permitted model value.

Worked Problem 2.57: Source at 150 W

Problem: q = 150 W, resistances are 0.08, 0.04 and 0.2 K/W, air is 30°C and limit is 90°C.

Step 1:

Rtotal = 0.32 K/W

Step 2:

Rise = 48 K

Tsource = 78°C

Step 3:

Margin = 12 K

Step 4:

Rmax = (90 − 30)/150 = 0.4 K/W

The model remains below the supplied temperature limit.

Worked Problem 2.58: Smaller Margin

Problem: q = 80 W, resistances are 0.12, 0.08 and 0.3 K/W, air is 35°C and limit is 80°C.

Step 1:

Rtotal = 0.5 K/W

Step 2:

Rise = 40 K

Tsource = 75°C

Step 3:

Margin = 5 K

Step 4:

Rmax = (80 − 35)/80 = 0.5625 K/W

The small margin belongs to the supplied steady model and conditions.

Worked Problem 2.59: Source at 120 W

Problem: q = 120 W, resistances are 0.1, 0.05 and 0.2 K/W, air is 28°C and limit is 85°C.

Step 1:

Rtotal = 0.35 K/W

Step 2:

Rise = 42 K

Tsource = 70°C

Step 3:

Margin = 15 K

Step 4:

Rmax = (85 − 28)/120 = 0.475 K/W

The named resistance endpoints must remain unchanged when comparing these values.

20. Thermal Interfaces and Contact Quality

Real solid surfaces are not perfectly smooth.

Microscopic gaps and imperfect contact can create substantial resistance.

A thermal-interface material fills suitable gaps and improves the effective path between package and cooler.

For an ideal uniform layer:

R = L/(kA)

At fixed conductivity and area, increasing thickness increases bulk resistance.

More paste is not automatically better. Too much can increase thickness, while too little or poor installation can leave gaps.

Real Contact Factors

• Contact pressure.
• Flatness.
• Contamination.
• Mechanical tolerances.
• Material compatibility.
• Required electrical insulation.
• Intended gap size.

Thermal conductivity is not electrical conductivity.

A supplied thermal k does not establish electrical safety or material compatibility.

Thermal-Interface Material

Thickness, conductivity, contact area and contact quality are distinct influences.

The analytical layer model does not prescribe mounting force or assembly procedure.

Worked Problem 2.60: A 0.5 mm Interface

Problem: Thickness is 0.5 mm, k = 5 W/(m·K), area is 6 cm² and q = 120 W.

Step 1: Convert dimensions.

L = 0.0005 m

A = 0.0006 m²

Step 2:

R = 0.0005/(5 × 0.0006) = 0.166667 K/W

Step 3:

ΔT = 120 × 0.166667 = 20 K

Step 4: Doubling ideal thickness doubles this bulk resistance and drop, but real gap filling and contact remain separate considerations.

Worked Problem 2.61: A 0.1 mm Interface

Problem: Thickness is 0.1 mm, k = 2 W/(m·K), area is 3 cm² and q = 60 W.

Step 1:

L = 0.0001 m

A = 0.0003 m²

Step 2:

R = 0.0001/(2 × 0.0003) = 0.166667 K/W

Step 3:

ΔT = 60 × 0.166667 = 10 K

Step 4: Square-centimetre conversion uses 10⁻⁴ m² per cm². Using 10⁻² would give an incorrect area.

21. Processor Cooling: From Chip to Room Air

A processor converts much of its input into heat under the stated operating condition.

A typical path includes the die, package, spreader, interface, cooler base, fins and air.

The exact arrangement varies, so calculations must begin with the supplied model.

Transient and Steady Operation

During warm-up, input can exceed output and stored energy rises.

At steady state, storage is constant and rejection balances the supplied thermal input.

A larger thermal mass delays warming. Lower steady resistance reduces the eventual temperature rise at unchanged load and ambient.

These are different effects.

Reading a Cooling Calculation

• State the heat rate to be rejected.
• Name resistance endpoints.
• Use local incoming air temperature.
• Identify the temperature being predicted.
• Compare with the supplied limit.

Fan power inside a room boundary eventually contributes to room load. It is not automatically package heat generation in a package-only model.

Worked Problem 2.62: Cooler Improvement at 120 W

Problem: q = 120 W, local air = 30°C and complete resistance changes from 0.5 to 0.3 K/W.

Step 1:

Told = 30 + 120 × 0.5 = 90°C

Step 2:

Tnew = 30 + 120 × 0.3 = 66°C

Step 3:

Reduction = 120(0.5 − 0.3) = 24 K

Step 4: The supplied resistance is complete. Additional included package or interface values should not be added again.

Worked Problem 2.63: Cooler Improvement at 150 W

Problem: q = 150 W, air = 25°C and resistance changes from 0.4 to 0.25 K/W.

Step 1:

Told = 25 + 150 × 0.4 = 85°C

Step 2:

Tnew = 25 + 150 × 0.25 = 62.5°C

Step 3:

Reduction = 22.5 K

Step 4: Changed workload or inlet temperature would require a new comparison rather than reusing the result unchanged.

Worked Problem 2.64: Cooler Improvement at 80 W

Problem: q = 80 W, air = 35°C and resistance changes from 0.6 to 0.4 K/W.

Step 1:

Told = 35 + 48 = 83°C

Step 2:

Tnew = 35 + 32 = 67°C

Step 3:

Reduction = 16 K

Step 4: The reduction follows qΔR. It does not mean the receiving air became colder.

22. Heat Sinks, Fins and Cooling Fans

A heat sink conducts heat away from the source and provides exposed area for exchange with fluid.

Fins enlarge that area.

Their complete surfaces are not generally at base temperature, so effective area or fin efficiency is needed when specified.

A fan establishes forced flow.

Its benefit depends on cooler geometry, enclosure resistance, inlet temperature, dust and recirculation.

A faster fan does not guarantee a proportional improvement.

For supplied effective hA:

q = hAeff(Tbase − Tair)

If Aeff already includes an efficiency correction, do not apply that correction again.

Diagnosing the Path

A poor interface can leave the processor hot while the sink is relatively cool.

A poor air path can make both sink and source hot.

Named-node temperatures are more useful than fan rotation alone.

Processor Fan and Fins

Fan operation changes transfer conditions rather than eliminating resistance.

The source-temperature calculation still needs the complete path.

Worked Problem 2.65: Effective Fin Area

Problem: q = 150 W, physical area = 0.15 m², effective-area factor = 0.8, h = 50 W/(m²·K) and air = 30°C.

Step 1:

Aeff = 0.8 × 0.15 = 0.12 m²

Step 2:

Rise = 150/(50 × 0.12) = 25 K

Step 3:

Tbase = 55°C

Step 4: The result is base temperature. Internal junction temperature may be higher because of additional path resistances.

Worked Problem 2.66: Another Effective-Area Model

Problem: q = 80 W, physical area = 0.16 m², factor = 0.7, h = 25 W/(m²·K) and air = 20°C.

Step 1:

Aeff = 0.112 m²

Step 2:

Rise = 80/(25 × 0.112) = 28.5714 K

Step 3:

Tbase = 48.5714°C

Step 4: The supplied factor and h are assumed valid. Changed geometry or flow can change both.

23. Heat Pipes and Phase-Change Transport

A heat pipe is sealed and transports energy through evaporation and condensation of an internal fluid.

The hotter region is the evaporator.

Vapour moves toward a cooler condenser, where latent energy is released.

A return mechanism, often a wick, carries liquid back.

A heat pipe does not create cold energy or eliminate final rejection.

It can transport heat from a concentrated source to a larger finned region.

Its capacity depends on fluid, wick, geometry, orientation and operating limits.

For an ideal latent-transport calculation:

q = ṁhfg

Heat Pipe and Refrigerator

A heat pipe generally has no separate compressor.

A vapour-compression refrigerator uses compressor work and an expansion device to pump heat from a cold region to a warmer sink.

The devices therefore have different functions and operating models.

Heat-Pipe Cooler Construction

Visible tubes identify hardware, not a numerical resistance.

The assembly combines interface conduction, phase-change transport and rejection from fins.

Worked Problem 2.67: Latent Transport at 150 W

Problem: An ideal heat pipe transports 150 W with supplied latent energy 180000 J/kg.

Step 1:

q = ṁhfg

Step 2:

ṁ = 150/180000 = 0.000833333 kg/s

Step 3:

ṁ = 0.833333 g/s

Step 4: This is internal circulation, not external coolant consumption.

The condenser must still reject approximately the transported heat in the same steady idealisation.

Worked Problem 2.68: Latent Transport at 60 W

Problem: q = 60 W and latent energy is 240000 J/kg.

Step 1: Apply q = ṁhfg.

Step 2:

ṁ = 60/240000 = 0.00025 kg/s

Step 3:

ṁ = 0.25 g/s

Step 4: The result estimates the required phase-change circulation. It does not determine capillary limits or complete heat-pipe capacity.

24. Server Rooms and Data-Centre Airflow

Servers receive electrical input and reject most of it as heat within the selected room boundary.

The cooling system must transport this load out while maintaining suitable equipment inlet conditions.

Only sources belonging to the stated room model should be included.

Cool air reaches equipment inlets.

Warm exhaust returns toward the cooling unit.

Mixing exhaust with supply raises local inlet temperature and can create hot spots even when a remote room sensor reports an acceptable average.

Sensible Airflow Balance

For supplied constant cp and negligible moisture change:

q = ṁcp(Tout − Tin)

Increasing mass flow reduces the air rise for fixed q.

It does not reduce the total heat the cooling plant must reject.

Cooling capacity is a thermal rate. Electrical consumption is a different quantity connected through COP.

Server Airflow Hardware

Mass flow, volume flow and density are different quantities.

Fan size alone does not determine the numerical cooling rate.

Worked Problem 2.69: Airflow for an 18 kW Load

Problem: Air picks up 18000 W. cp = 1005 J/(kg·K), inlet = 18°C and outlet = 30°C.

Step 1:

ΔT = 12 K

Step 2:

ṁ = q/(cpΔT)

Step 3:

ṁ = 18000/(1005 × 12) = 1.49254 kg/s

Step 4: Multiplying ṁcp by 12 K recovers 18000 W.

Volume flow would additionally require density.

Worked Problem 2.70: Airflow for an 8 kW Load

Problem: q = 8000 W, cp = 1000 J/(kg·K), inlet = 22°C and outlet = 30°C.

Step 1:

ΔT = 8 K

Step 2:

ṁ = q/(cpΔT)

Step 3:

ṁ = 8000/(1000 × 8) = 1 kg/s

Step 4: The result is mass flow. It must not be labelled m³/s without a density conversion.

25. Vapour-Compression AC: Four Main Stages

A vapour-compression AC moves energy from a cooled region to a warmer sink using compressor work.

The refrigerant circulates through:

• Evaporator.
• Compressor.
• Condenser.
• Expansion device.

Evaporation

Low-pressure refrigerant absorbs heat through the evaporator.

In the basic cycle description, it leaves as vapour.

Room air transfers energy through the exchanger wall and remains a separate stream.

Compression

The compressor receives work and raises vapour pressure.

Under typical operation, refrigerant temperature rises.

Condensation

The refrigerant rejects heat to the external sink and condenses at higher pressure.

Expansion

The expansion device reduces pressure before the refrigerant returns to the evaporator.

In this basic AC model, throttling is not a useful work-producing turbine stage.

Separate Streams

Room air, refrigerant and any water loop have distinct paths.

The condenser rejects removed room heat plus input work.

With positive work input, condenser rejection exceeds evaporator absorption.

Worked Problem 2.71: AC Rates over 60 s

Problem: Cooling rate is 18 kW and input is 6 kW. Find warm-side rate and energies over 60 s.

Step 1:

qH = 18 + 6 = 24 kW

Step 2:

QL = 18 × 60 = 1080 kJ

Step 3:

Win = 6 × 60 = 360 kJ

QH = 24 × 60 = 1440 kJ

Step 4: Check:

1080 + 360 = 1440 kJ

The complete-cycle balance does not require refrigerant enthalpies for these supplied overall values.

Worked Problem 2.72: AC Rates over 120 s

Problem: Cooling is 12 kW and input is 4 kW for 120 s.

Step 1:

qH = 16 kW

Step 2:

QL = 1440 kJ

Step 3:

Win = 480 kJ

QH = 1920 kJ

Step 4: Removed heat plus input work equals rejected heat.

Worked Problem 2.73: AC Rates over 30 s

Problem: Cooling is 25 kW and input is 5 kW for 30 s.

Step 1:

qH = 30 kW

Step 2:

QL = 750 kJ

Step 3:

Win = 150 kJ

QH = 900 kJ

Step 4: The balance is 750 + 150 = 900 kJ.

Remember that kW multiplied by seconds gives kJ.

26. AC Energy Balance and Coefficient of Performance

For a cyclic cooling system using positive magnitudes:

qH = qL + Pin

where:

• qL is heat removed from the cooled region.
• Pin is the specified input-power boundary.
• qH is warm-side rejection.

Cooling COP is:

COPcooling = qL/Pin

For the same complete boundary:

COPheating = qH/Pin = COPcooling + 1

Why COP Can Exceed One

An AC moves existing heat using work.

It does not convert electrical input into an equal newly created amount of cooling energy.

COP is therefore different from heat-engine thermal efficiency.

Boundary Consistency

Compressor-only power and complete-unit power define different COP values.

Auxiliary inputs must be treated consistently.

A like-for-like comparison requires the same boundary and operating conditions.

Worked Problem 2.74: COP of Three

Problem: An AC removes 12 kW at complete-unit COP = 3 for 8 h.

Step 1:

Pin = 12/3 = 4 kW

Step 2:

qH = 12 + 4 = 16 kW

Step 3:

Ein = 4 × 8 = 32 kWh

Step 4:

COPheating = 16/4 = 4

The cooling COP above one does not imply energy creation.

Room Cooling System

The refrigerator transfers heat while receiving work.

Its rejected heat includes both extracted heat and input energy under the selected balance.

Worked Problem 2.75: COP of 2.5

Problem: Cooling rate is 9 kW, complete-unit COP is 2.5 and operation lasts 10 h.

Step 1:

Pin = 9/2.5 = 3.6 kW

Step 2:

qH = 9 + 3.6 = 12.6 kW

Step 3:

Ein = 3.6 × 10 = 36 kWh

Step 4:

COPheating = 12.6/3.6 = 3.5

Both ratios use the same input boundary.

27. Cooling-Load Budgets and Energy Use

A cooling budget adds distinct heat sources entering the selected space.

Servers, lights, people, enclosure gains and appropriate auxiliary loads can contribute.

A supplied total may already include some of these sources. They must not be added twice.

For a steady model with negligible storage:

Required removal = Total included thermal load

A supplied capacity allowance can be applied as an exercise:

Target capacity = Base load × (1 + Allowance fraction)

This target is different from actual operating load.

Energy Use

For constant input:

Electrical energy = Input power × Time

1 kWh = 3.6 MJ

Removed thermal energy follows qL, while electrical energy follows Pin.

A larger supplied COP reduces input at the same load and comparable conditions.

Worked Problem 2.76: Ten Servers

Problem: Ten servers produce 300 W each. Separate other loads total 500 W. Apply 10% capacity allowance. Actual-load complete-unit COP is 3.

Step 1:

Server load = 10 × 300 = 3000 W

qbase = 3500 W

Step 2:

Target = 3500 × 1.1 = 3850 W

Step 3:

Actual Pin = 3500/3 = 1166.67 W

Step 4: The unused allowance is not automatically consumed as electrical input. It is a stated capacity target.

Worked Problem 2.77: Eight Servers

Problem: Eight servers produce 500 W each, other loads are 800 W, allowance is 15% and COP at base load is 4.

Step 1:

qbase = 8 × 500 + 800 = 4800 W

Step 2:

Target = 4800 × 1.15 = 5520 W

Step 3:

Pin = 4800/4 = 1200 W

Step 4: Actual power uses the actual base load and its supplied performance condition.

Worked Problem 2.78: Twelve Servers

Problem: Twelve servers produce 250 W each, other loads are 600 W, allowance is 20% and COP is 3.5.

Step 1:

qbase = 12 × 250 + 600 = 3600 W

Step 2:

Target = 3600 × 1.2 = 4320 W

Step 3:

Pin = 3600/3.5 = 1028.57 W

Step 4: The allowance is not a complete redundancy or site-load assessment.

A different actual load requires its corresponding operating data.

28. EV Battery Thermal Management

Battery thermal management controls temperature and temperature differences within the pack.

Cells generate heat during charging and discharging.

Excessive or uneven temperature can affect performance, life and operating limits.

Real limits must come from the actual cell and pack information.

Simplified Thermal Route

Cells → Interface → Cold plate → Coolant → Radiator or chiller → Surroundings

A pump circulates coolant.

A radiator rejects heat to a suitable cooler environment.

A chiller can provide heat pumping when required by operating conditions.

Heat Generation

A supplied resistive model is:

Pheat = I²R

It does not represent every electrochemical and entropic contribution.

Total electrical output power is not automatically generated heat.

Temperature Uniformity

An acceptable average does not establish the hottest cell temperature.

Contact quality, flow distribution and sensor placement matter.

A one-temperature lumped model cannot identify local hot spots by itself.

Worked Problem 2.79: Battery Bulk Rise at 100 A

Problem: I = 100 A, effective R = 0.02 Ω, C = 40000 J/K, rejection = 80 W and duration = 120 s.

Step 1:

Pheat = 100² × 0.02 = 200 W

Step 2:

Pnet = 200 − 80 = 120 W

Step 3:

Estored = 120 × 120 = 14400 J

ΔT = 14400/40000 = 0.36 K

Step 4: This is a uniform bulk rise. Starting temperature is needed for final absolute temperature, and local peaks require more detailed information.

Worked Problem 2.80: Battery Bulk Rise at 150 A

Problem: I = 150 A, R = 0.01 Ω, C = 50000 J/K, rejection = 100 W and duration = 180 s.

Step 1:

Pheat = 150² × 0.01 = 225 W

Step 2:

Pnet = 225 − 100 = 125 W

Step 3:

Estored = 125 × 180 = 22500 J

ΔT = 22500/50000 = 0.45 K

Step 4: The result does not locate the hottest cell or establish a universal safe pack temperature.

Worked Problem 2.81: Battery Bulk Rise at 80 A

Problem: I = 80 A, R = 0.03 Ω, C = 30000 J/K, rejection = 60 W and duration = 150 s.

Step 1:

Pheat = 80² × 0.03 = 192 W

Step 2:

Pnet = 192 − 60 = 132 W

Step 3:

Estored = 132 × 150 = 19800 J

ΔT = 19800/30000 = 0.66 K

Step 4: Only the stated net thermal input enters the bulk-temperature model. Pack electrical output must not be substituted as heat generation.

29. Coolant Flow and Pack Temperature Rise

For a steady single-phase coolant with approximately constant cp:

q = ṁcp(Tout − Tin)

Mass flow is in kg/s and cp in J/(kg·K), giving watts.

The coolant rise is not the cell-to-coolant contact drop.

Cells can be hotter than the local fluid because of interface and plate resistance.

Coolant outlet temperature therefore does not automatically equal peak cell temperature.

Mass and Volume Flow

ṁ = ρV̇

V̇ = ṁ/ρ

A litre equals 0.001 m³.

Use supplied fluid density and cp. A coolant mixture should not automatically be treated as pure water.

Sustained Rejection

Heat gained by coolant must eventually reach the surroundings.

At the appropriate overall steady boundary, final rejection includes pack generation and relevant pump or auxiliary contributions.

Worked Problem 2.82: Coolant for 1000 W

Problem: q = 1000 W, cp = 3500 J/(kg·K), density = 1050 kg/m³, inlet = 25°C and outlet = 30°C.

Step 1:

ΔT = 5 K

ṁ = 1000/(3500 × 5) = 0.0571429 kg/s

Step 2:

V̇ = 0.0571429/1050 = 5.44218 × 10⁻⁵ m³/s

Step 3:

V̇ = 5.44218 × 10⁻⁵ × 1000 × 60

V̇ = 3.26531 L/min

Step 4: Converting back and multiplying by density and cpΔT recovers 1000 W.

Worked Problem 2.83: Coolant for 1800 W

Problem: q = 1800 W, cp = 3600 J/(kg·K), density = 1040 kg/m³, inlet = 20°C and outlet = 26°C.

Step 1:

ΔT = 6 K

ṁ = 1800/(3600 × 6) = 0.0833333 kg/s

Step 2:

V̇ = 0.0833333/1040 = 8.01282 × 10⁻⁵ m³/s

Step 3:

V̇ = 4.80769 L/min

Step 4: This fluid rise differs from a cell-to-coolant interface drop. The latter needs a separate resistance or temperature model.

Battery Cooling Hardware

Pumping, heat exchange and thermal storage are separate stages.

The fluid model and boundary establish the meaning of the calculated rate.

30. Troubleshooting from the Thermal Path

A high component temperature can result from:

• Increased heat generation.
• Higher receiving-fluid temperature.
• Increased path resistance.
• A combination of these effects.

A useful diagnosis compares source, path and receiving environment.

These are possible explanations, not a diagnosis proved by one observation.

Workload changes, sensor error and changing flow also matter.

Comparing Remedies

At constant q and R, lowering ambient by 5 K lowers predicted source temperature by 5 K.

Reducing R instead changes qR.

These remedies affect different terms in:

Tsource = Tamb + qR

Worked Problem 2.84: Five-Kelvin Inlet Rise

Problem: q = 100 W, R = 0.4 K/W and inlet air is 25°C. Air becomes 5 K warmer. Find the original and new source temperatures and resistance needed to retain the original value.

Step 1:

Toriginal = 25 + 100 × 0.4 = 65°C

Step 2:

Tnew = 30 + 40 = 70°C

Step 3:

Rneeded = (65 − 30)/100 = 0.35 K/W

Step 4: Lower resistance offsets the reduced available temperature difference. It does not remove the heat source.

Worked Problem 2.85: Eight-Kelvin Inlet Rise

Problem: q = 150 W, R = 0.3 K/W and air is 30°C. Air becomes 8 K warmer.

Step 1:

Toriginal = 30 + 45 = 75°C

Step 2:

Tnew = 38 + 45 = 83°C

Step 3:

Rneeded = (75 − 38)/150 = 0.246667 K/W

Step 4: The required resistance is below the original value. The calculation assumes unchanged load and a valid linear path model.

Worked Problem 2.86: Ten-Kelvin Inlet Rise

Problem: q = 80 W, R = 0.5 K/W and air is 20°C. Air becomes 10 K warmer.

Step 1:

Toriginal = 20 + 40 = 60°C

Step 2:

Tnew = 30 + 40 = 70°C

Step 3:

Rneeded = (60 − 30)/80 = 0.375 K/W

Step 4: A negative required resistance in another case would indicate that the target cannot be achieved through an ordinary passive positive-resistance path to that warmer environment.

31. Formula Sheet and Reliable Solution Method

Solution Sequence

Step 1: State the system boundary, temperatures and transfer directions.

Step 2: Choose the appropriate energy or heat-transfer model.

Step 3: Convert units before substitution.

Step 4: Calculate with enough intermediate precision.

Step 5: Identify the final quantity and check units, direction and trend.

A predicted junction temperature must not be confused with a measured sink temperature.

A coolant mass flow must not be labelled as volume flow.

An electrical energy must not be presented as cooling capacity.

32. Short Questions with Clear Answers

1. What Is a Thermodynamic System?

It is the matter or region selected for study.

Its boundary separates it from surroundings and determines which mass and energy transfers are counted.

The same equipment can be represented by different system models when the boundary changes.

2. Does a Closed System Exchange Energy?

Yes.

A closed system prevents mass crossing, but heat or work can cross.

An isolated system prevents both mass and energy exchange.

Insulation alone does not establish isolation if electrical or shaft work remains possible.

3. Is Heat Stored in a Body?

Heat is a transfer caused by temperature difference.

Internal energy is a stored state property.

A body can gain internal energy through either heat input or work input under the appropriate balance.

4. How Can Steady Temperature Coexist with Heat Flow?

Continuous input and output can balance, leaving storage unchanged.

A powered heat sink can be hotter than air while maintaining a constant temperature.

Steady temperature is therefore different from thermal equilibrium with that air.

5. What Is the Zeroth Law Used For?

It supports temperature comparison through thermal equilibrium with a third body.

This provides the physical basis of thermometry.

Real measurement still requires suitable contact, response time and calibration.

6. What Sign Convention Is Used?

Q is positive into the stationary closed system.

W is positive when done by the system.

Therefore:

ΔU = Q − W

Work supplied to the system is negative W.

7. Is an Adiabatic Process Necessarily Isothermal?

No.

Adiabatic means no heat crosses the selected boundary.

Work can still change internal energy and temperature.

Isothermal separately specifies constant temperature.

8. Does a Cycle Have No Heat or Work Transfer?

No.

The state-property change is zero after a complete cycle.

Net heat and work can both be nonzero and equal under the stated stationary closed-cycle balance.

9. Why Does an AC Not Violate the Second Law?

It moves heat from the cooled region to a warmer sink while receiving work.

The total process includes that work input and warm-side rejection.

It is not a sole-effect cold-to-hot transfer.

10. Why Is Kelvin Essential in Radiation?

Radiation uses absolute fourth powers.

The Celsius offset does not cancel inside individual fourth powers.

Convert each absolute temperature before evaluating the equation.

11. What Does a Heat Sink Do?

It conducts heat from the source and provides useful area for transfer to the surroundings.

It still requires a receiving air or coolant path.

It does not make generated heat disappear.

12. What Does the Fan Primarily Change?

A fan imposes fluid motion and affects convection and flow distribution.

It does not create cold energy or guarantee a lower inlet temperature.

A blocked or recirculating path can remain ineffective despite fan rotation.

13. Why Can Excess Interface Thickness Be Harmful?

At the same k and A:

R = L/(kA)

increases with thickness.

Real assembly must also avoid gaps and meet its specified installation conditions.

The ideal formula does not justify arbitrary thinning.

14. Why Can Cooling COP Exceed One?

COP compares heat moved from the cooled region with input work.

It is not a heat-engine conversion efficiency.

Warm-side rejection still includes both removed heat and input energy.

15. Why Is Average Pack Temperature Insufficient?

An average can hide hot cells and gradients.

Local generation, interface quality, coolant distribution and sensor placement influence peak temperature.

A lumped average alone does not establish every local condition.

33. Complete Examination Answers

1. Explain Thermodynamic Systems Using Computing Examples

A system is the selected matter or region. Its boundary separates it from surroundings.

A processor package can be modelled as a closed material system because matter remains inside while electrical energy enters and thermal energy leaves.

An airflow control volume around a fan is open because air crosses the boundary and carries energy.

An isolated model permits neither mass nor energy transfer.

An insulated device is not automatically isolated if electrical input or shaft work crosses its boundary.

A package-only boundary counts transfer to the cooler as external heat transfer.

Enlarging the boundary to include the cooler makes package-to-cooler conduction internal. The external thermal boundary then lies between cooler and air.

This explains why a solution begins with a diagram and named boundary.

Steady flow is not closed flow. Equal inlet and outlet mass rates can keep stored mass constant while matter continues to pass through the region.

2. Explain the Four Laws and Their Significance

The zeroth law establishes the transitivity of thermal equilibrium and supports temperature measurement.

If A and B are separately in equilibrium with C, they are in equilibrium with each other.

The first law expresses energy conservation.

For the stated stationary closed system:

ΔU = Q − W

A powered component warms when net input exceeds output and remains steady when input and rejection balance.

The second law establishes direction and performance restrictions.

Natural net heat transfer is hot to cold.

A cyclic refrigerator needs an accompanying effect, such as work input, to move heat cold to hot.

A cyclic heat engine cannot have the sole effect of converting heat from one reservoir entirely into work.

The third law gives the entropy limit of a perfect crystal approaching absolute zero under its ground-state condition.

Absolute zero is 0 K, rather than 0°C, and cannot be reached through a finite sequence of cooling operations.

Together, the laws distinguish temperature comparison, energy quantity, process feasibility and the low-temperature reference.

3. Compare Conduction, Convection and Radiation with Equations

Conduction transfers energy through material interactions without requiring bulk movement.

For the stated steady plane layer:

q = kA(Th − Tc)/L

Rate rises with conductivity, area and temperature difference and falls with thickness.

Package and metal-base regions provide conduction examples.

Convection combines near-wall transfer with fluid motion.

q = hA(Ts − T∞)

Natural convection is buoyancy-driven. Forced convection uses imposed flow.

The coefficient depends on fluid, geometry and operating conditions.

Fin area requires an effective-area or efficiency treatment when temperatures are nonuniform.

Radiation transfers energy electromagnetically and can cross a vacuum.

For the stated grey-surface model:

qrad = εσA(Ts⁴ − Tsur⁴)

Use kelvin.

Both bodies emit, and the equation gives net exchange.

One cooling assembly can involve all three modes along separate path segments or at one exposed surface.

4. Explain a Processor Thermal-Resistance Model

Select the source node and receiving-air reference.

Identify the package, interface, base and sink-to-air segments.

A plane layer has:

R = L/(kA)

A supplied convection surface has:

R = 1/(hA)

Add values only when they lie in series carrying the same rate.

Then:

Tsource = Tamb + qRtotal

Each segment drop is qRi.

A high source temperature can result from greater generation, warmer local air or increased resistance.

The same sink temperature does not prove the same junction temperature if package or interface drops differ.

Junction-to-case and junction-to-ambient resistances have different endpoints.

A larger heat capacity delays transient warming but does not by itself lower the constant-load steady rise.

State the assumptions of constant resistance, steady operation and a single represented route.

Parallel paths, varying airflow and variable power require additional treatment.

5. Explain the Roles of Heat Sink, Fan and Heat Pipe

The heat sink conducts heat into its base and fins.

Fins provide larger exposed area, although temperature varies along them.

The fan drives air through an intended path and changes convection.

Its effectiveness depends on inlet conditions, flow resistance, fin condition and recirculation.

Fan rotation alone is not proof of adequate cooling.

A wick-assisted heat pipe evaporates working fluid at the hot region, sends vapour to a cooler condenser and returns liquid through the wick.

It transports latent energy toward a rejecting surface.

It does not remove the need for final rejection and differs from a compressor-driven refrigerator.

Interfaces join solid parts and can introduce major temperature drops.

A complete explanation follows the route from source to surroundings and identifies conduction, phase-change transport, convection and relevant radiation.

6. Describe AC Operation and Derive the Overall Balance

The evaporator absorbs heat from cooled air into low-pressure refrigerant.

The compressor receives work, raises pressure and sends hot vapour toward the condenser.

The condenser rejects heat to the warm sink and produces liquid.

The expansion device lowers pressure before the refrigerant returns to the evaporator.

Refrigerant and room air exchange heat across a wall and remain distinct streams.

Over a complete cycle, the working fluid returns to its initial state, so net storage change is zero.

Using positive external magnitudes:

qH = qL + Pin

Cooling COP is:

COPcooling = qL/Pin

For qL = 18 kW and supplied complete-input COP = 3:

Pin = 6 kW

qH = 24 kW

A COP above one is permitted because the system moves heat using work.

Compressor-only and complete-unit power define different boundaries and should not be compared as identical ratios.

7. Explain Server-Room Cooling with an Airflow Balance

Servers and included room sources transfer energy into the space.

Under a steady sensible model, cooling removes the combined load.

Supply air gains heat as it passes through equipment:

q = ṁcp(Treturn − Tsupply)

At the same heat load, increasing mass flow reduces air temperature rise under the supplied constant-cp model.

Air must reach equipment inlets without excessive hot-exhaust recirculation.

A remote average room temperature can miss local inlet hot spots.

Cooling capacity is thermal power. Electrical input is connected through the supplied COP.

Fan and auxiliary loads must be included once according to the boundary.

State whether wall gains, leakage, storage and moisture effects are omitted.

The airflow balance identifies a transport requirement, while the complete cooling-cycle balance identifies final rejection and input energy.

8. Explain EV Battery Thermal Management with a Heat Path

A liquid-cooled path can contain cells, interface material, a cold plate, coolant, a pump and a radiator or chiller.

The interface and plate conduct heat.

Flowing coolant gains sensible energy:

q = ṁcp(Tout − Tin)

A radiator needs a suitable temperature difference to the receiving environment.

A chiller uses heat pumping when the operating requirement calls for it.

A supplied I²R value represents a resistive approximation, not every battery heat-generation mechanism.

Total electrical power delivered by the pack is not all converted to heat.

A lumped heat-capacity model estimates bulk warm-up from net thermal input.

It cannot locate local hot spots without more detail.

Uniformity, contact, flow distribution and sensor placement matter alongside average temperature.

A correct answer states these roles without inventing a universal operating limit or service procedure.

9. Solve a Combined Processor and Room-Energy Example

Given: Processor load is 120 W, local air is 28°C and complete source-to-air resistance is 0.35 K/W.

The source temperature is:

Tsource = 28 + 120 × 0.35 = 70°C

If the supplied limit is 85°C:

Margin = 85 − 70 = 15 K

A 5 K inlet increase with unchanged q and R gives:

Tsource,new = 75°C

Now consider a room with a combined 3 kW load already including the processors and indoor fan heat.

For complete-unit cooling COP = 3:

Pin = 3/3 = 1 kW

qH = 3 + 1 = 4 kW

The processors’ internal transfer rates must not be added again after they are included in the 3 kW room load.

This example connects component-temperature prediction to room energy accounting while preserving distinct boundaries.

34. Integrated Revision: Energy Boundaries

Electrical Work and Thermal Transfer

An electronic component can receive electrical work and reject heat.

The input mechanism and output mechanism differ even when their steady rates are equal.

Calling all electrical input heat at every boundary can obscure which interaction crosses the selected surface.

A package-only model treats electrical power as input work and package-to-cooler transfer as heat output.

A room model can treat the equipment’s eventual rejection as a room thermal load.

Both descriptions can be valid because their boundaries differ.

Internal Transfers Cancel in a Larger Boundary

In a steady package-interface-sink path, the same energy passes through successive segments.

A larger boundary containing all three makes their internal exchanges disappear from the overall balance.

Only external input, external rejection and storage remain.

This is why package-to-sink and sink-to-air transfers should not be added as independent room heat sources.

The resistance model retains the internal segments because it predicts their temperature drops.

The energy balance and temperature model therefore answer related but different questions.

Closed-System Energy and Open-System Transport

A closed system keeps the same matter and can exchange heat and work.

An open system transports mass and associated energy across its boundary.

A flowing coolant passage requires inlet and outlet information.

A solid cold plate may be represented as a closed material body in another model.

The complete assembly can therefore contain both material-body and control-volume descriptions.

The correct choice depends on the question, rather than the equipment’s name alone.

Rate Balance and Interval Balance

A rate balance uses watts.

An interval balance uses joules or kilojoules.

At constant rates, multiply by the same duration to move from one form to the other.

A steady rate balance has zero storage rate, but energy can still pass through continuously.

Over ten seconds, the amount transferred is different from the amount transferred over one hour even if the power remains unchanged.

The time basis should always be visible in an energy-use answer.

35. Integrated Revision: Temperature and Thermal Resistance

A Temperature Difference Is a Driving Quantity

A temperature difference is not heat itself.

The same difference can produce different rates through different areas, materials or flow conditions.

For conduction, geometry and conductivity matter.

For convection, the supplied coefficient and effective area matter.

For radiation, absolute temperatures, emissivity and exchange geometry matter.

Temperature readings therefore need a physical model before they can establish watts.

A Steady Temperature Does Not Mean No Transfer

A component can receive and reject equal power while maintaining constant stored energy.

Its steady temperature can exceed ambient because a finite difference drives rejection through the path.

The expression:

Tsource − Tamb = qRtotal

makes that difference explicit.

Thermal equilibrium would instead remove net exchange between the bodies in contact.

Steady operation and equilibrium should not be used interchangeably.

Series Paths Preserve a Common Rate

A series route with no intermediate generation or storage carries the same q through every segment.

Each drop is qRi.

The largest resistance gives the largest drop at that common rate.

This does not imply that the segment physically creates more energy or carries a larger rate.

Adding the drops recovers the total difference.

Adding the rates would count one transfer multiple times.

Resistance Endpoints Determine Meaning

A thermal resistance is meaningful only with its endpoint temperatures.

A source-to-case value cannot replace source-to-air unless the missing route is accounted for.

A complete source-to-air value may already include package, interface and cooler effects.

Adding those included values again overstates the path.

Named nodes are therefore more reliable than a list of resistance numbers without context.

Passive Cooling and Ambient Limits

For a positive heat source and ordinary positive resistance, the source is warmer than the receiving environment.

Reducing resistance reduces the required rise but does not reverse its sign.

If a target lies below the receiving air temperature, the simple passive path cannot achieve it.

A suitable active cooling process would require a different model, including input power and warm-side rejection.

The algebra should be interpreted physically rather than accepting a negative required resistance as a real passive component.

36. Integrated Revision: Cooling Hardware

Heat Capacity and Heat Rejection

Thermal mass stores energy.

A larger heat capacity slows a transient rise for the same net input.

It does not supply unlimited sustained cooling.

If generation continues, the stored energy eventually has to be rejected or temperature continues to change.

Heat capacity and steady thermal resistance therefore serve different functions.

One governs storage response, while the other relates steady rate and temperature difference.

Heat Sink and Fan

A sink distributes heat through solid material and offers exposed area.

A fan changes the fluid condition and intended route.

Neither function alone establishes complete source temperature.

A highly conductive base with poor airflow can still become hot.

A strong airflow with a poor interface can leave the junction hot while the sink is less warm.

Diagnosis should follow the entire route rather than attributing every symptom to fan speed.

Heat Pipe and Refrigeration

A heat pipe uses internal evaporation and condensation to transport heat under its operating limits.

A refrigerator uses work input to move heat against the natural temperature direction.

Both can involve phase change, but they are not equivalent devices.

A heat pipe still needs a condenser region connected to a receiving environment.

An AC additionally rejects input work along with removed heat.

Air, Coolant and Refrigerant

Airflow through equipment carries sensible energy.

Coolant through a plate can also carry sensible energy under a single-phase model.

Refrigerant in a vapour-compression cycle undergoes different state changes, including phase change.

These streams can exchange heat through walls without mixing.

Using one stream’s inlet temperature as another stream’s outlet without a stated exchanger model confuses the thermal route.

37. Common Numerical Errors

Mixing Energy and Power

A value in watts is a rate.

A value in joules is an amount.

Use the specified duration before comparing power with transferred energy.

For constant operation, kW multiplied by hours gives kWh, while kW multiplied by seconds gives kJ.

Using Celsius in Fourth Powers

Convert each absolute temperature to kelvin before evaluating radiation.

A simple difference can use the same numerical Celsius or kelvin increment.

Individual fourth powers cannot.

Reversing the Work Sign

The convention here uses work done by the system as positive.

Electrical input is therefore negative W.

The physical input-output form can be used as a check.

Both forms must give the same ΔU.

Treating Rigid as No Work

Rigid means no volume-change boundary work.

It does not eliminate electrical input or stirring.

The specific transfer modes must still be identified.

Treating Adiabatic as Constant Temperature

Adiabatic means Q = 0.

Work can still change internal energy.

A temperature change requires the material model.

Summing Series Heat Rates

Successive series segments carry the same rate.

Add their resistances and temperature drops, not repeated copies of q.

Applying Fin Efficiency Twice

If effective area is already supplied, it already represents the stated correction.

Do not reduce it again unless the model specifically requires another factor.

Confusing Cooling Capacity with Electrical Input

Cooling capacity is qL.

Electrical input is qL/COP for the supplied boundary.

Warm-side rejection is qL + Pin.

These three values differ.

Using Capacity Allowance as Actual Load

A capacity target includes a stated allowance.

Unused capacity is not automatically consumed as input power.

Actual energy use requires actual operating load and corresponding performance.

Confusing Coolant Rise with Cell Temperature

The fluid inlet-to-outlet difference belongs to the same stream.

Cell-to-fluid resistance creates another difference.

Outlet coolant temperature is not automatically the hottest cell temperature.

Treating a Bulk Average as a Peak

A one-temperature model cannot establish every local cell or junction condition.

Sensor location, gradients and local path quality matter.

A useful answer identifies the bulk nature of the result.

38. Final Application Connections

Processor to Room

A processor’s thermal path determines its source temperature relative to local air.

Its rejected heat then contributes to the room budget.

The room cooling system removes the combined load and requires input power.

The condenser rejects that removed load plus cooling-system input.

These are nested boundaries.

A clear calculation counts each source once at each appropriate boundary while avoiding repeated addition of internal transfers.

Server Airflow to Cooling Plant

Airflow transports heat from equipment to the cooling unit.

Its required mass rate follows the permitted air temperature rise.

The cooling plant’s input follows the supplied COP.

Changing airflow can change local temperatures without changing the total equipment heat generation.

Changing COP can change electrical input at the same room load.

The two improvements act on different parts of the overall system.

Battery Cells to Surroundings

Battery generation reaches an interface, cold plate and coolant.

The coolant then passes energy to a radiator or chiller.

A pump sustains flow but also consumes power.

A chiller adds a refrigeration input and warm-side rejection.

At each stage, the selected boundary determines whether a transfer is internal or external.

The model must also distinguish a bulk temperature estimate from local cell conditions.

Diagnosis with Controlled Comparisons

Hold workload, ambient and model endpoints consistent when comparing a resistance change.

Hold heat load and cp consistent when comparing flow effects.

Use like-for-like input boundaries when comparing COP.

A change in one measured temperature can otherwise be caused by several simultaneous changes.

Quantitative checks support a diagnosis, but the assumptions and measurement locations determine how strong the conclusion can be.

Related B.Tech First Semester Study Materials

CSE 1st Year All Subject PYQs → CSE 1st Year All Subject Notes → CSE 1st Year All Subject Syllabus →