ME 322: Thermodynamics — Unit 3 Master Review & Comprehensive Quiz

Problem 1 • Subunit 3.1: Heat Engines & Refrigeration Systems

Second Law Vocabulary & Core Concepts

Preclass 3.1 & Unit 3 Review Conceptual

Match each core thermodynamic concept or statement with its correct physical definition:

1. Kelvin-Planck Statement:

2. Clausius Statement:

3. Thermal Energy Reservoir:

4. Perpetual Motion Machine of the Second Kind (PMM2):

💡 Hint: Formulas & Approach

Review the classical statements of the Second Law and definitions from Chapter 6:

Equation Sheet Section: Heat Engines
A heat engine must exchange heat with at least two thermal reservoirs (a source and a sink) to produce net work ($\oint dQ = \oint dW$).
A refrigerator requires net work input to transfer heat from cold to hot.
PMM1 violates the 1st Law ($\Delta E e Q - W$). PMM2 violates the 2nd Law ($\eta_{th} = 100\%$).

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Master Definitions:

Kelvin-Planck Statement: It is impossible for any device operating on a cycle to receive heat from a single reservoir and produce a net amount of work. (No heat engine can be 100% efficient; $Q_L > 0$ always).
Clausius Statement: It is impossible to construct a cyclical device that transfers heat from a lower-temperature body to a higher-temperature body without external work input ($W_{in} > 0$).
Thermal Energy Reservoir: A system with large thermal mass that absorbs or releases heat isothermally ($\Delta T = 0$). Examples: oceans, atmosphere, lakes, large furnaces.
Perpetual Motion Machine of the Second Kind (PMM2): Any cyclical heat engine claiming 100% thermal efficiency ($Q_L = 0, W_{net} = Q_H$), directly violating the Kelvin-Planck statement.

Problem 2 • Subunit 3.1: Heat Engines & Refrigeration Systems

Heat Engine Power Output & Heat Rejection

Class Activity 3.2 Q1 & Preclass 3.2 Numerical

A heat engine receives heat at a rate of $\dot{Q}_H = 30{,}000\text{ Btu/hr}$ and operates with a thermal efficiency of $\eta_{th} = 40\%$.

Part A: Calculate the net power produced by the engine in $\text{Btu/hr}$.

Btu/hr

Part B: Convert the power output to horsepower ($\text{hp}$). Use the conversion $1\text{ hp} = 2544.5\text{ Btu/hr}$.

hp

Part C: What is the rate of waste heat rejected to the environment $\dot{Q}_L$ in $\text{Btu/hr}$?

Btu/hr

Figure: Thermodynamic Heat Engine Operating Between Two Reservoirs ($\dot{W}_{net} = \dot{Q}_H - \dot{Q}_L$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Heat Engines): $$\eta_{th} = \frac{\dot{W}_{net}}{\dot{Q}_H}$$ $$\dot{W}_{net} = \dot{Q}_H - \dot{Q}_L$$ Conversion: $1\text{ hp} = 2544.5\text{ Btu/hr}$ (or $2545\text{ Btu/hr}$).

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Step-by-Step Mathematical Solution:

Part A: Power Output: $$\dot{W}_{net} = \eta_{th} \cdot \dot{Q}_H = 0.40 \times 30{,}000\text{ Btu/hr} = \mathbf{12{,}000\text{ Btu/hr}}$$
Part B: Conversion to Horsepower: $$\dot{W}_{net} = \frac{12{,}000\text{ Btu/hr}}{2544.5\text{ Btu/(hp}\cdot\text{hr)}} = \mathbf{4.72\text{ hp}}$$
Part C: Heat Rejection: $$\dot{Q}_L = \dot{Q}_H - \dot{W}_{net} = 30{,}000 - 12{,}000 = \mathbf{18{,}000\text{ Btu/hr}}$$

Problem 3 • Subunit 3.1: Heat Engines & Refrigeration Systems

Coal-Fired Steam Power Plant Logistics & Air Flow

Class Activity 3.2 Q2 Numerical

A coal-burning steam power plant produces a net electrical power output of $\dot{W}_{net} = 300\text{ MW}$ with an overall thermal efficiency of $\eta_{th} = 32\%$. The heating value of the coal is $HV = 28{,}000\text{ kJ/kg}$. A typical rail train car holds approximately $105{,}000\text{ kg}$ of coal, and the gravimetric air-fuel ratio in the furnace is $AF = 12\text{ kg air / kg fuel}$.

Part A: What is the total rate of heat input $\dot{Q}_H$ required to run the facility (in $\text{MW}$)?

MW

Part B: Determine the mass of coal consumed during a 24-hour day in $\text{kg/day}$.

kg/day

Part C: How many train cars of coal must arrive at the power plant each day? (Round to nearest whole car).

cars/day

Part D: What is the mass flow rate of air $\dot{m}_{air}$ flowing into the furnace in $\text{kg/s}$?

kg/s

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\eta_{th} = \frac{\dot{W}_{net}}{\dot{Q}_H} \implies \dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}}$$ $$\dot{Q}_H = \dot{m}_{coal} \cdot HV$$ $$\dot{m}_{air} = AF \cdot \dot{m}_{coal}$$ Remember: $1\text{ MW} = 10^3\text{ kW} = 10^3\text{ kJ/s}$. A 24-hour day contains $86{,}400\text{ s}$.

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Step-by-Step Mathematical Solution:

Part A: Rate of Heat Input: $$\dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}} = \frac{300\text{ MW}}{0.32} = \mathbf{937.5\text{ MW}}$$
Part B: Daily Coal Consumption: $$\dot{m}_{coal} = \frac{\dot{Q}_H}{HV} = \frac{937{,}500\text{ kJ/s}}{28{,}000\text{ kJ/kg}} = 33.482\text{ kg/s}$$ $$\text{Mass per day} = 33.482\text{ kg/s} \times 86{,}400\text{ s/day} = \mathbf{2.893 \times 10^6\text{ kg/day}} \quad (2{,}892{,}857\text{ kg/day})$$
Part C: Train Cars per Day: $$\text{Cars/day} = \frac{2{,}892{,}857\text{ kg/day}}{105{,}000\text{ kg/car}} = 27.55 \approx \mathbf{28\text{ cars/day}}$$
Part D: Air Mass Flow Rate: $$\dot{m}_{air} = AF \cdot \dot{m}_{coal} = 12\text{ kg air/kg fuel} \times 33.482\text{ kg coal/s} = \mathbf{401.8 \approx 402\text{ kg/s}}$$

Problem 4 • Subunit 3.1: Heat Engines & Refrigeration Systems

Banana Refrigeration Enthalpy & Cooling Capacity

Class Activity 3.2 Q3 Numerical

Fresh bananas are to be cooled from $24^\circ\text{C}$ to $13^\circ\text{C}$ at a steady rate of $215\text{ kg/hr}$ by a refrigeration system. The specific heat of bananas above freezing is $c_p = 3.35\text{ kJ/kg}\cdot\text{K}$ (from Property Tables Table A-3 / food properties). The electrical power input to the refrigerator compressor is $\dot{W}_{in} = 1.4\text{ kW}$.

Part A: What is the rate of heat removal from the bananas (cooling load $\dot{Q}_L$) in $\text{kW}$ (or $\text{kJ/s}$)?

kW

Part B: Determine the Coefficient of Performance ($\text{COP}_R$) of the refrigerator.

dimensionless

Part C: What is the rate of heat rejected to the ambient surroundings $\dot{Q}_H$ in $\text{kW}$?

kW

Figure: Refrigerator & Heat Pump Cycle ($\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}, ext{COP}_{HP} = ext{COP}_R + 1$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\Delta \dot{H} = \dot{m} c_p \Delta T$$ $$\text{COP}_R = \frac{\text{Desired Output}}{\text{Required Input}} = \frac{\dot{Q}_L}{\dot{W}_{net}}$$ $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{net}$$ Conversion: $\dot{m} = \frac{215\text{ kg/hr}}{3600\text{ s/hr}}$. Specific heat $c_p = 3.35\text{ kJ/kg}\cdot\text{K}$.

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Step-by-Step Mathematical Solution:

Part A: Cooling Rate (Heat Removed from Bananas): $$\dot{m} = \frac{215\text{ kg/hr}}{3600\text{ s/hr}} = 0.05972\text{ kg/s}$$ $$\dot{Q}_L = |\Delta \dot{H}| = \dot{m} c_p (T_1 - T_2) = 0.05972\text{ kg/s} \times 3.35\text{ kJ/kg}\cdot\text{K} \times (24 - 13)\text{ K} = \mathbf{2.20\text{ kW}} \quad (2.20\text{ kJ/s})$$
Part B: Refrigerator COP: $$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}} = \frac{2.20\text{ kW}}{1.4\text{ kW}} = \mathbf{1.57}$$
Part C: Heat Rejection to Environment: $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in} = 2.20 + 1.40 = \mathbf{3.60\text{ kW}}$$

Problem 5 • Subunit 3.1: Heat Engines & Refrigeration Systems

Building HVAC Life-Cycle Cost & COP Optimization

Class Activity 3.2 Q4 Numerical

A commercial building has an annual cooling load estimated at $Q_L = 120{,}000\text{ kWh/year}$ in a region where electricity costs $\$0.10/\text{kWh}$. Two different air conditioners are being evaluated over a 10-year expected service lifespan:

Air Conditioner A: Seasonal average $\text{COP} = 3.2$, Initial purchase cost = $\$5{,}000$.
Air Conditioner B: Seasonal average $\text{COP} = 5.0$, Initial purchase cost = $\$8{,}000$.

Part A: What is the annual operating electricity cost for Unit A (in $\$/\text{year}$)?

$/yr

Part B: What is the 10-year total cost (capital cost + 10 years of operating costs) for Unit A?

$

Part C: What is the annual operating electricity cost for Unit B (in $\$/\text{year}$)?

$/yr

Part D: What is the 10-year total cost for Unit B?

$

Part E: How much money (in $\$$) does the owner save over 10 years by choosing the higher efficiency Unit B?

$

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\text{COP} = \frac{Q_L}{W_{in}} \implies W_{in} = \frac{Q_L}{\text{COP}}$$ $$\text{Annual Cost} = W_{in} (\text{kWh/yr}) \times 0.10\text{ \$/kWh}$$ $$\text{10-Year Total Cost} = \text{Capital Cost} + 10 \times \text{Annual Cost}$$

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Step-by-Step Engineering Economic Analysis:

Unit A: $$W_{in,A} = \frac{120{,}000\text{ kWh}}{3.2} = 37{,}500\text{ kWh/yr}$$ $$\text{Annual Cost}_A = 37{,}500\text{ kWh/yr} \times \$0.10/\text{kWh} = \mathbf{\$3{,}750/\text{year}}$$ $$\text{10-Yr Total}_A = \$5{,}000 + 10 \times \$3{,}750 = \mathbf{\$42{,}500}$$
Unit B: $$W_{in,B} = \frac{120{,}000\text{ kWh}}{5.0} = 24{,}000\text{ kWh/yr}$$ $$\text{Annual Cost}_B = 24{,}000\text{ kWh/yr} \times \$0.10/\text{kWh} = \mathbf{\$2{,}400/\text{year}}$$ $$\text{10-Yr Total}_B = \$8{,}000 + 10 \times \$2{,}400 = \mathbf{\$32{,}000}$$
10-Year Net Savings: $$\text{Savings} = \$42{,}500 - \$32{,}000 = \mathbf{\$10{,}500}$$ Conclusion: Recommend purchasing Unit B because the lower annual operating energy cost pays off the $\$3{,}000$ initial premium in just $\frac{3000}{3750-2400} = 2.22\text{ years}$, yielding $\$10{,}500$ in total net savings!

Problem 6 • Subunit 3.1: Heat Engines & Refrigeration Systems

Heat Pump vs. Refrigerator COP Identity

Preclass 3.2 Q7 & Unit 3 Review Q8 Numerical

A residential reversible heat pump is designed to provide heating during winter and air conditioning during summer. When operating in heating mode (as a heat pump), the system achieves a coefficient of performance of $\text{COP}_{HP} = 4.0$.

Question: What is its Coefficient of Performance when operating in cooling mode (as an air conditioner, $\text{COP}_R$) between the exact same space and outdoor conditions?

dimensionless

Figure: Refrigerator & Heat Pump Cycle ($\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}, ext{COP}_{HP} = ext{COP}_R + 1$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Heat Engines): $$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{net}}, \quad \text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{net}}$$ Since $\dot{Q}_H = \dot{Q}_L + \dot{W}_{net}$, derive the algebraic relationship between $\text{COP}_{HP}$ and $\text{COP}_R$.

📝 Show Full Worked Solution

Mathematical Proof & Calculation:

From First Law energy balance on the cycle: $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{net}$$
Substitute into the heat pump COP definition: $$\text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{net}} = \frac{\dot{Q}_L + \dot{W}_{net}}{\dot{W}_{net}} = \frac{\dot{Q}_L}{\dot{W}_{net}} + 1 = \text{COP}_R + 1$$
Therefore: $$\text{COP}_R = \text{COP}_{HP} - 1 = 4.0 - 1.0 = \mathbf{3.0}$$ Physical Insight: A heat pump is always more efficient than an equivalent electric resistance heater ($ ext{COP} = 1$), and its COP is always exactly 1 unit greater than the refrigerator COP operating between the same two temperatures because the compressor work input $\dot{W}_{net}$ is also converted into heat delivered to the heated space!

Problem 7 • Subunit 3.2: Carnot Cycles & Theoretical Limits

Carnot Heat Engine Output & Source Temperature

Class Activity 3.3 Q1 & Homework 3A Prob 3 Numerical

A heat engine operates on a reversible Carnot cycle and has a thermal efficiency of $\eta_{th} = 75\%$. The waste heat from this engine is rejected to a nearby lake at $T_L = 60^\circ\text{F}$ at a rate of $\dot{Q}_L = 800\text{ Btu/min}$.

Part A: Determine the power output of the engine $\dot{W}_{net}$ in $\text{Btu/min}$.

Btu/min

Part B: What is the power output in horsepower ($\text{hp}$)? Note: $1\text{ hp} = 42.41\text{ Btu/min}$ (or $2544.5\text{ Btu/hr}$).

hp

Part C: Determine the absolute temperature of the heat source $T_H$ in Rankine ($\text{R}$).

R

Part D: What is the heat source temperature in Fahrenheit ($^\circ\text{F}$)?

°F

Figure: Thermodynamic Heat Engine Operating Between Two Reservoirs ($\dot{W}_{net} = \dot{Q}_H - \dot{Q}_L$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Carnot Relationships): $$\eta_{th,C} = 1 - \frac{T_L}{T_H} = \frac{\dot{W}_{net}}{\dot{Q}_H}$$ $$\dot{Q}_H = \frac{\dot{Q}_L}{1 - \eta_{th,C}}, \quad \dot{W}_{net} = \dot{Q}_H - \dot{Q}_L$$ Temperature conversion: $T_R = T_{^\circ\text{F}} + 460$. Absolute temperature MUST be used in all Carnot formulas!

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Step-by-Step Mathematical Solution:

Part A: Power Output: $$\dot{Q}_H = \frac{\dot{Q}_L}{1 - \eta_{th}} = \frac{800\text{ Btu/min}}{1 - 0.75} = \frac{800}{0.25} = 3{,}200\text{ Btu/min}$$ $$\dot{W}_{net} = \dot{Q}_H - \dot{Q}_L = 3{,}200 - 800 = \mathbf{2{,}400\text{ Btu/min}}$$
Part B: Conversion to Horsepower: $$\dot{W}_{net} = \frac{2{,}400\text{ Btu/min}}{42.41\text{ Btu/(hp}\cdot\text{min)}} = \mathbf{56.6\text{ hp}} \quad \left(\text{or } \frac{2400 \times 60}{2544.5} = 56.59\text{ hp}\right)$$
Part C: Absolute Source Temperature: $$T_L = 60^\circ\text{F} + 460 = 520\text{ R}$$ $$\eta_{th,C} = 1 - \frac{T_L}{T_H} \implies 0.75 = 1 - \frac{520\text{ R}}{T_H} \implies \frac{520}{T_H} = 0.25$$ $$T_H = \frac{520\text{ R}}{0.25} = \mathbf{2{,}080\text{ R}}$$
Part D: Source Temperature in °F: $$T_{H,^\circ\text{F}} = 2080 - 460 = \mathbf{1{,}620^\circ\text{F}}$$

Problem 8 • Subunit 3.2: Carnot Cycles & Theoretical Limits

Evaluation of an Inventor Refrigeration Claim

Class Activity 3.3 Q2 & Homework 3A Prob 4 Numerical

An inventor claims to have developed a commercial refrigeration system that extracts heat from a cold refrigerated space at $T_L = -12^\circ\text{C}$ and transfers it to ambient outdoor air at $T_H = 25^\circ\text{C}$ while maintaining a steady coefficient of performance of $\text{COP}_R = 6.5$.

Part A: What is the maximum theoretical Carnot coefficient of performance ($\text{COP}_{R,c}$) possible between these two temperature limits?

dimensionless

Part B: Is this claim physically reasonable and thermodynamically valid?

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Carnot Relationships): $$\text{COP}_{R,c} = \frac{1}{\frac{T_H}{T_L} - 1} = \frac{T_L}{T_H - T_L}$$ Convert both temperatures to Kelvin: $T_K = T_{^\circ\text{C}} + 273.15$. A claim violates the 2nd Law if $\text{COP}_{actual} > \text{COP}_{Carnot}$.

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Rigorous Thermodynamic Verification:

Convert to Absolute Temperatures: $$T_L = -12 + 273.15 = 261.15\text{ K}$$ $$T_H = 25 + 273.15 = 298.15\text{ K}$$
Maximum Theoretical Carnot COP: $$\text{COP}_{R,c} = \frac{T_L}{T_H - T_L} = \frac{261.15\text{ K}}{298.15 - 261.15\text{ K}} = \frac{261.15}{37.0} = \mathbf{7.06}$$
Thermodynamic Conclusion: Since the claimed performance is $\text{COP}_{claimed} = 6.5 < 7.06$, the claim does NOT violate the Second Law of Thermodynamics. It is thermodynamically possible in principle! However, achieving $\frac{6.5}{7.06} = 92.1\%$ of the Carnot limit in a real vapor-compression cycle with compressor friction, motor losses, and finite $\Delta T$ across heat exchangers is practically implausible for standard commercial hardware.

Problem 9 • Subunit 3.2: Carnot Cycles & Theoretical Limits

Split Air Conditioner vs. Carnot Refrigerator

Class Activity 3.3 Q3 & Homework 3A Prob 5 Numerical

An $18{,}000\text{ Btu/hr}$ split air conditioner operates at full load to maintain a server room at $T_L = 22^\circ\text{C}$ in an outdoor environment at $T_H = 33^\circ\text{C}$. The measured electrical power input to the compressor is $\dot{W}_{in} = 3.4\text{ kW}$.
Unit Conversion Note: $18{,}000\text{ Btu/hr} = 5.275\text{ kW}$ ($1\text{ kW} = 3412.14\text{ Btu/hr}$).

Part A: What is the rate of heat rejected at the outdoor condenser $\dot{Q}_H$ in $\text{kW}$?

kW

Part B: What is the actual Coefficient of Performance ($\text{COP}_R$) of the air conditioner?

dimensionless

Part C: If this identical unit were operated in reverse as a heat pump, what would be its $\text{COP}_{HP}$?

dimensionless

Part D: What is the ideal Carnot $\text{COP}_{R,c}$ for these reservoir temperatures?

dimensionless

Part E: What would the rate of cooling be (in $\text{kW}$) if the air conditioner operated as an ideal Carnot refrigerator for the exact same $3.4\text{ kW}$ power input?

kW

Figure: Refrigerator & Heat Pump Cycle ($\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}, ext{COP}_{HP} = ext{COP}_R + 1$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\dot{Q}_L = \frac{18{,}000\text{ Btu/hr}}{3412.14\text{ Btu/(kW}\cdot\text{hr)}} = 5.275\text{ kW}$$ $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}$$ $$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}}, \quad \text{COP}_{HP} = \text{COP}_R + 1$$ $$\text{COP}_{R,c} = \frac{T_L}{T_H - T_L}, \quad \dot{Q}_{L,Carnot} = \text{COP}_{R,c} \cdot \dot{W}_{in}$$ Remember: Convert temperatures to Kelvin: $T_L = 22 + 273.15 = 295.15\text{ K}$, $T_H = 33 + 273.15 = 306.15\text{ K}$.

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Step-by-Step Mathematical Solution:

Convert Cooling Capacity to kW: $$\dot{Q}_L = 18{,}000\text{ Btu/hr} \times \frac{1.055056\text{ kJ}}{1\text{ Btu}} \times \frac{1\text{ hr}}{3600\text{ s}} = 5.275\text{ kW}$$
Part A: Condenser Heat Rejection: $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in} = 5.275 + 3.40 = \mathbf{8.68\text{ kW}}$$
Part B: Actual COP: $$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}} = \frac{5.275}{3.40} = \mathbf{1.55}$$
Part C: Heat Pump COP: $$\text{COP}_{HP} = \text{COP}_R + 1 = 1.55 + 1.0 = \mathbf{2.55} \quad \left(\text{or } \frac{8.68}{3.40} = 2.55\right)$$
Part D: Carnot COP: $$T_L = 22 + 273.15 = 295.15\text{ K}, \quad T_H = 33 + 273.15 = 306.15\text{ K}$$ $$\Delta T = 306.15 - 295.15 = 11.0\text{ K}$$ $$\text{COP}_{R,c} = \frac{295.15}{11.0} = \mathbf{26.83 \approx 26.8}$$
Part E: Carnot Cooling Rate: $$\dot{Q}_{L,Carnot} = \text{COP}_{R,c} \cdot \dot{W}_{in} = 26.83 \times 3.40\text{ kW} = \mathbf{91.2\text{ kW}}$$

Physical Explanation for Discrepancy: The real cooling rate ($5.275\text{ kW}$) is far lower than the Carnot rate ($91.2\text{ kW}$) because real systems suffer from large temperature differences across the evaporator and condenser coils (irreversible heat transfer through finite $\Delta T$), throttling valve expansion irreversibilities, mechanical friction, and compressor isentropic inefficiencies.

Problem 10 • Subunit 3.2: Carnot Cycles & Theoretical Limits

Diesel Engine Bottoming Steam Generator Claim

Homework 3A Prob 1 Numerical

An aftermarket manufacturer of diesel engine accessories claims they can replace the radiator on a large power-generation diesel engine with a waste-heat steam generator. The diesel engine currently operates at $\eta_1 = 35\%$ thermal efficiency. The steam generator keeps the engine block at $T_H = 250^\circ\text{F}$ to generate steam, which expands through a turbine exhausting heat to outdoor ambient air at $T_L = 68^\circ\text{F}$. The manufacturer claims the overall combined engine-turbine thermal efficiency will increase from $35\%$ to $55\%$.

Part A: What is the maximum theoretical (Carnot) thermal efficiency of the bottoming steam cycle $\eta_{steam,max}$?

%

Part B: What is the absolute maximum combined thermal efficiency $\eta_{overall,max}$ of the combined system (in $\%$)?

%

Part C: Does the manufacturer's claim violate the laws of thermodynamics?

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Carnot Relationships): $$\eta_{th,C} = 1 - \frac{T_L}{T_H}$$ Convert temperatures to Rankine: $T = T_{^\circ\text{F}} + 460$. $$T_H = 250 + 460 = 710\text{ R}, \quad T_L = 68 + 460 = 528\text{ R}$$ The bottoming cycle can only use the waste heat rejected by the diesel engine: $Q_{in,steam} = (1 - \eta_1) Q_{fuel}$.

📝 Show Full Worked Solution

Step-by-Step Thermodynamic Proof:

Carnot Efficiency of Bottoming Cycle: $$T_H = 250^\circ\text{F} + 460 = 710\text{ R}$$ $$T_L = 68^\circ\text{F} + 460 = 528\text{ R}$$ $$\eta_{steam,max} = 1 - \frac{T_L}{T_H} = 1 - \frac{528}{710} = 1 - 0.7437 = 0.2563 \quad (\mathbf{25.6\%})$$
Maximum Combined Efficiency: For $100\text{ units}$ of fuel input: $$W_{diesel} = 35\text{ units}$$ $$Q_{waste} = 100 - 35 = 65\text{ units}$$ The maximum work the steam cycle can produce from this waste heat is: $$W_{steam,max} = 65 \times \eta_{steam,max} = 65 \times 0.2563 = 16.66\text{ units}$$ Total maximum work produced: $$W_{total,max} = 35 + 16.66 = 51.66\text{ units}$$ $$\eta_{overall,max} = \frac{51.66}{100} = \mathbf{51.7\%} \quad (51.66\%)$$
Conclusion: The manufacturer claims an overall efficiency of $55\%$. Because $55\% > 51.66\%$, the claim directly violates the Second Law of Thermodynamics (Carnot principle) and is physically impossible.

Problem 11 • Subunit 3.2: Carnot Cycles & Theoretical Limits

R-134a Heat Pump with Geothermal Source

Homework 3A Prob 2 Numerical

A heat pump operating with R-134a is used to heat an indoor living space at $25^\circ\text{C}$ by absorbing heat from geothermal water that enters the evaporator at $60^\circ\text{C}$ at $\dot{m}_w = 0.065\text{ kg/s}$ and leaves at $40^\circ\text{C}$. Refrigerant enters the evaporator at $12^\circ\text{C}$ with a quality of $x = 15\%$ and leaves as saturated vapor ($x = 1.0$) at the same pressure. The compressor consumes $\dot{W}_{in} = 1.6\text{ kW}$ of power.

Part A: What is the rate of heat absorption from the geothermal water in the evaporator $\dot{Q}_L$ (in $\text{kW}$)? Use $c_p = 4.18\text{ kJ/kg}\cdot\text{K}$ (Table A-3).

kW

Part B: Determine the mass flow rate of R-134a ($\dot{m}_R$) through the cycle in $\text{kg/s}$. (Look up Table A-11 at $12^\circ\text{C}$).

kg/s

Part C: What is the rate of heat supplied to the heated room $\dot{Q}_H$ in $\text{kW}$?

kW

Part D: What is the actual Coefficient of Performance ($\text{COP}_{HP}$) of the heat pump?

dimensionless

Part E: Calculate the ideal Carnot COP ($\text{COP}_{HP,c}$) between the geothermal supply at $60^\circ\text{C}$ and the room at $25^\circ\text{C}$.

dimensionless

Part F: What is the minimum theoretical electrical power input $\dot{W}_{min}$ (in $\text{kW}$) required to deliver the same heat supply?

kW

Figure: Refrigerator & Heat Pump Cycle ($\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}, ext{COP}_{HP} = ext{COP}_R + 1$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Property Tables: $$\dot{Q}_L = \dot{m}_w c_p (T_{w,in} - T_{w,out})$$ From Table A-11 at $12^\circ\text{C}$: $h_f = 68.18\text{ kJ/kg}$, $h_{fg} = 188.58\text{ kJ/kg}$, $h_g = 256.76\text{ kJ/kg}$. $$h_1 = h_f + x_1 h_{fg}, \quad h_2 = h_g$$ $$\dot{m}_R = \frac{\dot{Q}_L}{h_2 - h_1}, \quad \dot{Q}_H = \dot{Q}_L + \dot{W}_{in}$$ $$\text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{in}}, \quad \text{COP}_{HP,c} = \frac{T_H}{T_H - T_L}, \quad \dot{W}_{min} = \frac{\dot{Q}_H}{\text{COP}_{HP,c}}$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Part A: Evaporator Heat Absorption: $$\dot{Q}_L = \dot{m}_w c_p (60 - 40) = 0.065\text{ kg/s} \times 4.18\text{ kJ/kg}\cdot\text{K} \times 20\text{ K} = \mathbf{5.434 \approx 5.44\text{ kW}}$$
Part B: Mass Flow Rate of R-134a: From Table A-11 at $12^\circ\text{C}$: $$h_1 = 68.18 + 0.15 \times (188.58) = 96.47\text{ kJ/kg}$$ $$h_2 = h_g = 256.76\text{ kJ/kg}$$ $$\Delta h = h_2 - h_1 = 256.76 - 96.47 = 160.29\text{ kJ/kg}$$ $$\dot{m}_R = \frac{5.434\text{ kW}}{160.29\text{ kJ/kg}} = \mathbf{0.0339\text{ kg/s}} \quad (0.0338\text{ kg/s})$$
Part C: Heat Supply Rate: $$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in} = 5.434 + 1.60 = \mathbf{7.034 \approx 7.04\text{ kW}}$$
Part D: Actual COP: $$\text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{in}} = \frac{7.034\text{ kW}}{1.60\text{ kW}} = \mathbf{4.40}$$
Part E: Ideal Carnot COP: Using the given reservoir temperatures $60^\circ\text{C}$ ($333.15\text{ K}$) and $25^\circ\text{C}$ ($298.15\text{ K}$): $$\text{COP}_{HP,c} = \frac{T_H}{T_H - T_L} = \frac{333.15\text{ K}}{60 - 25\text{ K}} = \frac{333.15}{35.0} = \mathbf{9.51}$$
Part F: Minimum Compressor Power: $$\dot{W}_{min} = \frac{\dot{Q}_H}{\text{COP}_{HP,c}} = \frac{7.034\text{ kW}}{9.51} = \mathbf{0.74\text{ kW}}$$

Problem 12 • Subunit 3.2: Carnot Cycles & Theoretical Limits

Winter Residential Heat Pump Minimum Power

Unit 3 Review Prob 15 & Textbook Ex 6-7 Numerical

A heat pump is used to maintain a house at a constant indoor temperature of $T_H = 21^\circ\text{C}$ throughout winter. When the outdoor temperature drops to $T_L = -5^\circ\text{C}$, the house is estimated to lose heat to the outdoors at a rate of $\dot{Q}_H = 135{,}000\text{ kJ/h}$.

Part A: What is the house heating load in kilowatts ($\text{kW}$)?

kW

Part B: What is the maximum theoretical Carnot coefficient of performance ($\text{COP}_{HP,c}$)?

dimensionless

Part C: Determine the minimum electrical power input $\dot{W}_{in,min}$ (in $\text{kW}$) required to drive this heat pump.

kW

Figure: Refrigerator & Heat Pump Cycle ($\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}, ext{COP}_{HP} = ext{COP}_R + 1$)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Carnot Relationships): $$\text{COP}_{HP,c} = \frac{1}{1 - \frac{T_L}{T_H}} = \frac{T_H}{T_H - T_L}$$ $$\dot{W}_{min} = \frac{\dot{Q}_H}{\text{COP}_{HP,c}}$$ Convert temperatures to Kelvin: $T_K = T_{^\circ\text{C}} + 273.15$. Remember $1\text{ hr} = 3600\text{ s}$.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Part A: Heating Load in kW: $$\dot{Q}_H = \frac{135{,}000\text{ kJ/h}}{3600\text{ s/h}} = \mathbf{37.5\text{ kW}}$$
Part B: Maximum Theoretical COP: $$T_H = 21 + 273.15 = 294.15\text{ K}$$ $$T_L = -5 + 273.15 = 268.15\text{ K}$$ $$\Delta T = 294.15 - 268.15 = 26.0\text{ K}$$ $$\text{COP}_{HP,c} = \frac{T_H}{T_H - T_L} = \frac{294.15\text{ K}}{26.0\text{ K}} = \mathbf{11.31}$$
Part C: Minimum Required Electrical Power: $$\dot{W}_{min} = \frac{\dot{Q}_H}{\text{COP}_{HP,c}} = \frac{37.5\text{ kW}}{11.313} = \mathbf{3.32\text{ kW}}$$ Significance: While an electric resistance baseboard heater would require the full $37.5\text{ kW}$ of electricity, an ideal heat pump delivers that exact same heat using only $3.32\text{ kW}$—saving over $91\%$ of the electrical energy!

Problem 13 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Clausius Inequality & Entropy Balance Fundamentals

Preclass 3.4, Preclass 3.5 & Unit 3 Review Conceptual

Evaluate the fundamental principles of the Clausius Inequality and Entropy Generation:

Part A: For any cyclic process operating in a closed system, what does the cyclic integral $\oint \frac{\delta Q}{T}$ evaluate to?

$\oint \frac{\delta Q}{T} = 0$ for reversible cycles, and $\oint \frac{\delta Q}{T} < 0$ for irreversible cycles. $\oint \frac{\delta Q}{T} > 0$ for irreversible cycles, and $= 0$ for reversible cycles. It is identically zero for all physical cycles because entropy is a property.

Part B: Which of the following mechanisms can directly decrease the entropy of a closed thermodynamic system?

Heat transfer out of the system ($Q_{out} > 0$) Work done by the system ($W_{out} > 0$) Internal irreversibilities ($S_{gen} > 0$) Entropy can never decrease in any system under any circumstances.

Part C: Two identical rigid insulated compartments are separated by a membrane. One side contains gas at high pressure and the other is evacuated. When the membrane is punctured and the gas fills the entire volume (unrestrained expansion), what happens to the internal energy $U$ and total entropy $S$?

Internal energy remains constant ($\Delta U = 0$), and entropy increases ($\Delta S > 0, S_{gen} > 0$). Both internal energy and entropy increase. Both internal energy and entropy remain constant because the tank is insulated and rigid.

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (2nd Law): $$\Delta S = S_2 - S_1 = \int_1^2 \left(\frac{dQ}{T}\right)_{int\,rev}$$ $$S_{gen,sys} \equiv \Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr} \ge 0$$ Work is energy transfer that carries NO entropy ($S_{work} = 0$). Only heat transfer carries entropy ($S_{heat} = \int \delta Q/T$). For an isolated system (rigid, adiabatic tank), $Q = 0, W = 0 \implies \Delta U = 0$, but spontaneous unrestrained expansion is completely irreversible ($S_{gen} > 0$).

📝 Show Full Worked Solution

Rigorous Engineering Explanations:

Part A (Clausius Inequality): $$\oint \frac{\delta Q}{T} \le 0$$ The equality holds for internally reversible cycles, while the inequality ($< 0$) holds for irreversible (real) cycles. If $\oint \frac{\delta Q}{T} > 0$, the process violates the Second Law and is impossible.
Part B (Decreasing System Entropy): Entropy can transfer across a system boundary via heat transfer and mass flow. Since work carries zero entropy, the only way to decrease the entropy of a closed system is through heat transfer out of the system ($Q < 0$). Irreversibilities always generate positive entropy ($S_{gen} \ge 0$).
Part C (Unrestrained Expansion): First Law on the isolated tank: $Q = 0$, $W = 0 \implies \Delta U = 0$ (internal energy remains constant; for an ideal gas, $T_2 = T_1$). However, the gas spontaneously expands into a larger volume without doing work. This is a classic irreversible process: $$\Delta S_{sys} = m R \ln\left(\frac{V_2}{V_1}\right) > 0 \quad (S_{gen} = \Delta S_{sys} > 0)$$

Problem 14 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Carnot Isothermal Heat Rejection on T-s Diagram

Class Activity 3.5 Q1 Numerical

During the isothermal heat rejection process of a Carnot cycle, the gaseous working fluid undergoes an entropy change of $\Delta S_{sys} = -0.7\text{ Btu/R}$. The temperature of the heat sink is $T_{sink} = 95^\circ\text{F}$.

Part A: Determine the amount of heat transfer $Q_{out}$ (in $\text{Btu}$) rejected by the working fluid.

Btu

Part B: What is the entropy change of the heat sink $\Delta S_{sink}$ (in $\text{Btu/R}$)? Include correct sign.

Btu/R

Part C: What is the total entropy change $\Delta S_{tot}$ of the isolated universe during this reversible heat rejection?

Btu/R

Part D: If the process were instead rejecting the exact same $388.5\text{ Btu}$ of heat across a finite temperature difference to a colder sink at $T_{sink2} = 75^\circ\text{F}$, what would be the entropy generation $S_{gen}$ (in $\text{Btu/R}$)?

Btu/R

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (2nd Law): $$\Delta S = \frac{Q}{T_0} \implies Q = T_0 \Delta S$$ $$T_{sink} = 95^\circ\text{F} + 460 = 555\text{ R}$$ $$\Delta S_{sink} = \frac{Q_{in,sink}}{T_{sink}} = \frac{+Q_{out,sys}}{T_{sink}}$$ $$S_{gen} = \Delta S_{sys} + \Delta S_{sink}$$ When heat transfers across a finite temperature drop ($95^\circ\text{F} \to 75^\circ\text{F}$), the sink temperature is $535\text{ R}$.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Part A: Heat Transfer Amount: $$T_0 = 95^\circ\text{F} + 460 = 555\text{ R}$$ $$Q_{sys} = T_0 \Delta S_{sys} = 555\text{ R} \times (-0.7\text{ Btu/R}) = -388.5\text{ Btu}$$ $$\text{Heat rejected } Q_{out} = |Q_{sys}| = \mathbf{388.5\text{ Btu}}$$
Part B: Entropy Change of the Sink: $$\Delta S_{sink} = \frac{+Q_{out}}{T_{sink}} = \frac{+388.5\text{ Btu}}{555\text{ R}} = \mathbf{+0.7\text{ Btu/R}}$$
Part C: Total Entropy Change: $$\Delta S_{tot} = \Delta S_{sys} + \Delta S_{sink} = -0.7 + 0.7 = \mathbf{0\text{ Btu/R}}$$ Note: In a reversible Carnot cycle, the total entropy change is identically zero ($S_{gen} = 0$).
Part D: Entropy Generation with Colder Sink ($75^\circ ext{F} = 535 ext{ R}$): $$\Delta S_{sink2} = \frac{+388.5\text{ Btu}}{535\text{ R}} = +0.7262\text{ Btu/R}$$ $$S_{gen} = \Delta S_{sys} + \Delta S_{sink2} = -0.7000 + 0.7262 = \mathbf{+0.026\text{ Btu/R}}$$

Problem 15 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Constant-Pressure Water Liquid-Vapor Phase Change

Unit 3 Review Prob 19 & Textbook Ex 7-1 Numerical

A piston-cylinder device contains a saturated liquid-vapor mixture of water maintained at a constant temperature of $T_0 = 300\text{ K}$. During an internally reversible, constant-pressure process, $Q_{in} = 750\text{ kJ}$ of heat is transferred to the water, causing part of the liquid to vaporize.

Question: What is the entropy change of the water $\Delta S_{water}$ in $\text{kJ/K}$?

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (2nd Law): $$\Delta S = \frac{Q}{T_0}$$ For any internally reversible isothermal process (including a constant-pressure phase change where temperature remains fixed at $T_{sat}$), no integration is required!

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

During a phase change inside the saturation dome at constant pressure, the temperature remains strictly constant at the saturation temperature ($T = T_0 = 300\text{ K}$).
From the equation sheet for an internally reversible isothermal process: $$\Delta S = \int_1^2 \left(\frac{\delta Q}{T}\right)_{int\,rev} = \frac{1}{T_0} \int_1^2 \delta Q = \frac{Q_{in}}{T_0}$$
Substitute the given values: $$\Delta S_{water} = \frac{750\text{ kJ}}{300\text{ K}} = \mathbf{2.50\text{ kJ/K}}$$

Problem 16 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Rigid Tank Cooling of Refrigerant-134a

Unit 3 Review Prob 20 & Textbook Ex 7-3 Numerical

A rigid sealed container contains $m = 5\text{ kg}$ of R-134a initially at $T_1 = 20^\circ\text{C}$ and $P_1 = 140\text{ kPa}$ ($0.14\text{ MPa}$). The tank is cooled by transferring heat to the surroundings until the internal pressure drops to $P_2 = 100\text{ kPa}$ ($0.10\text{ MPa}$).

Part A: What is the initial specific entropy $s_1$ in $\text{kJ/kg}\cdot\text{K}$? (Look up Table A-13 at $0.14\text{ MPa}, 20^\circ\text{C}$).

kJ/kg-K

Part B: Determine the vapor quality $x_2$ at state 2 inside the rigid container ($v_2 = v_1$, using Table A-12 at $100\text{ kPa}$).

dimensionless

Part C: What is the specific entropy change $\Delta s = s_2 - s_1$ in $\text{kJ/kg}\cdot\text{K}$?

kJ/kg-K

Part D: Determine the total entropy change of the refrigerant $\Delta S = m(s_2 - s_1)$ in $\text{kJ/K}$. Include correct negative sign.

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Property Tables:

State 1: Look up superheated R-134a in Table A-13 at $P_1 = 0.14\text{ MPa}$ and $T_1 = 20^\circ\text{C}$ to find $v_1$ and $s_1$.
Rigid Tank: Volume and mass are constant, so $v_2 = v_1$.
State 2: Look up saturated R-134a in Table A-12 at $P_2 = 0.10\text{ MPa}$ ($100\text{ kPa}$). Find $v_f$ and $v_g$. Then: $$x_2 = \frac{v_2 - v_f}{v_g - v_f}$$ $$s_2 = s_f + x_2 s_{fg}$$
$$\Delta S = m(s_2 - s_1)$$

📝 Show Full Worked Solution

Step-by-Step Property Evaluation:

State 1 ($P_1 = 0.14\text{ MPa}, T_1 = 20^\circ\text{C}$): From Table A-13 (Superheated R-134a): $$v_1 = 0.16544\text{ m}^3/\text{kg}, \quad s_1 = \mathbf{1.0624\text{ kJ/kg}\cdot\text{K}}$$
State 2 ($P_2 = 0.10\text{ MPa}, v_2 = v_1 = 0.16544\text{ m}^3/\text{kg}$): From Table A-12 (Saturated R-134a, Pressure table): $$v_f = 0.0007259\text{ m}^3/\text{kg}, \quad v_g = 0.19254\text{ m}^3/\text{kg}$$ $$s_f = 0.15405\text{ kJ/kg}\cdot\text{K}, \quad s_{fg} = 0.78363\text{ kJ/kg}\cdot\text{K}$$ Calculate quality $x_2$: $$x_2 = \frac{v_2 - v_f}{v_g - v_f} = \frac{0.16544 - 0.0007259}{0.19254 - 0.0007259} = \frac{0.164714}{0.191814} = \mathbf{0.8587 \approx 0.859}$$
State 2 Specific Entropy: $$s_2 = s_f + x_2 s_{fg} = 0.15405 + (0.8587)(0.78363) = 0.15405 + 0.6729 = 0.8270\text{ kJ/kg}\cdot\text{K}$$ $$\Delta s = s_2 - s_1 = 0.8270 - 1.0624 = \mathbf{-0.2354\text{ kJ/kg}\cdot\text{K}}$$
Total Entropy Change: $$\Delta S = m(s_2 - s_1) = 5.0\text{ kg} \times (-0.2354\text{ kJ/kg}\cdot\text{K}) = \mathbf{-1.177\text{ kJ/K}} \quad (\text{or } -1.173\text{ kJ/K})$$

Problem 17 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Superheated Steam Constant-Pressure Heating

Class Activity 3.5 Q2 Numerical

A weighted piston-cylinder device contains $m = 2.0\text{ lbm}$ of water at $P_1 = 300\text{ psia}$ with an initial volume of $V_1 = 2.5\text{ ft}^3$. The water is heated at constant pressure until the temperature reaches $T_2 = 500^\circ\text{F}$.

Part A: What is the initial specific volume $v_1$ in $\text{ft}^3/\text{lbm}$?

ft³/lbm

Part B: What is the initial vapor quality $x_1$ at $300\text{ psia}$? (Look up Table A-5E).

dimensionless

Part C: What is the initial specific entropy $s_1$ in $\text{Btu/lbm}\cdot\text{R}$?

Btu/lbm-R

Part D: What is the final specific entropy $s_2$ at $300\text{ psia}$ and $500^\circ\text{F}$? (Look up Table A-6E).

Btu/lbm-R

Part E: Determine the total entropy change of the water $\Delta S = m(s_2 - s_1)$ in $\text{Btu/R}$.

Btu/R

Figure: T-s Diagram Showing Isentropic (2s, vertical) vs Actual Irreversible Expansion (2a, entropy increases)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Property Tables: $$v_1 = \frac{V_1}{m} = \frac{2.5\text{ ft}^3}{2.0\text{ lbm}} = 1.25\text{ ft}^3/\text{lbm}$$ From Table A-5E at $P = 300\text{ psia}$: $T_{sat} = 417.35^\circ\text{F}$, $v_f = 0.01890\text{ ft}^3/\text{lbm}$, $v_g = 1.5435\text{ ft}^3/\text{lbm}$. $$x_1 = \frac{v_1 - v_f}{v_{fg}}, \quad s_1 = s_f + x_1 s_{fg}$$ State 2 is at $300\text{ psia}$ and $500^\circ\text{F} > T_{sat}$ (superheated steam, Table A-6E). $$\Delta S = m(s_2 - s_1)$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

State 1 Properties ($P_1 = 300\text{ psia}, v_1 = 1.25\text{ ft}^3/\text{lbm}$): $$v_1 = \frac{2.5\text{ ft}^3}{2.0\text{ lbm}} = \mathbf{1.25\text{ ft}^3/\text{lbm}}$$ From Table A-5E at $300\text{ psia}$: $$v_f = 0.01890, \quad v_g = 1.5435\text{ ft}^3/\text{lbm}$$ $$s_f = 0.58818, \quad s_{fg} = 0.92289\text{ Btu/lbm}\cdot\text{R}$$ $$x_1 = \frac{1.25 - 0.01890}{1.5435 - 0.01890} = \frac{1.2311}{1.5246} = \mathbf{0.8075 \approx 0.808}$$ $$s_1 = 0.58818 + (0.8075)(0.92289) = \mathbf{1.3334\text{ Btu/lbm}\cdot\text{R}}$$
State 2 Properties ($P_2 = 300\text{ psia}, T_2 = 500^\circ\text{F}$): Since $500^\circ\text{F} > T_{sat} = 417.35^\circ\text{F}$, it is superheated vapor. From Table A-6E (Superheated Water, $300\text{ psia}$ column at $500^\circ\text{F}$): $$s_2 = \mathbf{1.5711\text{ Btu/lbm}\cdot\text{R}}$$
Total Entropy Change: $$\Delta S = m(s_2 - s_1) = 2.0\text{ lbm} \times (1.5711 - 1.3334\text{ Btu/lbm}\cdot\text{R})$$ $$\Delta S = 2.0 \times 0.2377 = \mathbf{0.4754 \approx 0.475\text{ Btu/R}}$$

Problem 18 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Two-Process Steam Sequence: Isothermal Addition & Isentropic Expansion

Homework 3B Prob 2 Numerical

A piston-cylinder device contains $m = 5\text{ kg}$ of steam initially at $T_1 = 100^\circ\text{C}$ with a quality of $x_1 = 50\%$ ($0.50$). The steam undergoes two consecutive thermodynamic processes:

Process 1–2: Heat is transferred reversibly while the temperature is held constant at $100^\circ\text{C}$ until the steam exists as a saturated vapor ($x_2 = 1.0$).
Process 2–3: The steam expands in an adiabatic, reversible (isentropic) process until the final pressure is $P_3 = 15\text{ kPa}$.

Part A: Determine the heat transferred to the steam during the isothermal process 1–2 ($Q_{1-2}$) in $\text{kJ}$. (Look up Table A-4 at $100^\circ\text{C}$).

kJ

Part B: What is the vapor quality $x_3$ at the end of the isentropic expansion at $15\text{ kPa}$? ($s_3 = s_2$, look up Table A-5 at $15\text{ kPa}$).

dimensionless

Part C: Find the work done by the steam in process 2–3 ($W_{2-3}$) in $\text{kJ}$.

kJ

Figure: T-s Diagram Showing Isentropic (2s, vertical) vs Actual Irreversible Expansion (2a, entropy increases)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Table A-4 / A-5:

Process 1–2 (Isothermal Phase Change at $100^\circ ext{C}$): $$Q_{1-2} = m T (s_2 - s_1) = m (1 - x_1) h_{fg@100^\circ ext{C}}$$ From Table A-4 at $100^\circ ext{C}$: $h_{fg} = 2256.4 ext{ kJ/kg}$, $s_g = 7.3542 ext{ kJ/kg}\cdot ext{K}$.
Process 2–3 (Isentropic, $s_3 = s_2 = s_g = 7.3542 ext{ kJ/kg}\cdot ext{K}$): From Table A-5 at $15 ext{ kPa}$: $s_f = 0.7549$, $s_{fg} = 7.2522 ext{ kJ/kg}\cdot ext{K}$. $$x_3 = rac{s_3 - s_f}{s_{fg}}$$ First Law for adiabatic closed process: $Q - W = \Delta U \implies W_{2-3} = -m(u_3 - u_2)$.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Process 1–2 Heat Transfer ($Q_{1-2}$): From Table A-4 at $100^\circ ext{C}$: $$h_{fg} = 2256.4 ext{ kJ/kg}, \quad s_g = 7.3542 ext{ kJ/kg}\cdot ext{K}$$ $$Q_{1-2} = m (x_2 - x_1) h_{fg} = 5.0 ext{ kg} imes (1.0 - 0.50) imes 2256.4 ext{ kJ/kg} = \mathbf{5{,}641 ext{ kJ}}$$ Alternative via entropy: $Q_{1-2} = m T \Delta s = 5 imes 373.15 ext{ K} imes (0.50 imes 6.0470) = 5{,}641 ext{ kJ}$.
Process 2–3 Quality ($x_3$): Since the process is adiabatic and reversible, it is isentropic: $$s_3 = s_2 = s_g@100^\circ ext{C} = 7.3542 ext{ kJ/kg}\cdot ext{K}$$ At $P_3 = 15 ext{ kPa}$ (Table A-5): $$s_f = 0.7549, \quad s_{fg} = 7.2522 ext{ kJ/kg}\cdot ext{K}$$ $$x_3 = rac{7.3542 - 0.7549}{7.2522} = rac{6.5993}{7.2522} = \mathbf{0.90997 pprox 0.910}$$
Process 2–3 Work Output ($W_{2-3}$): Using the First Law: $W_{2-3} = m(u_2 - u_3) = \mathbf{1{,}290 ext{ kJ}}$ (course key).

Problem 19 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Air Quenching of Stainless Steel Ball Bearings

Homework 3B Prob 1 Numerical

Spherical stainless steel ball bearings ($ ho = 8085 ext{ kg/m}^3$, $c = 0.480 ext{ kJ/kg}\cdot ext{K}$, diameter $D = 1.8 ext{ cm} = 0.018 ext{ m}$) are being quenched in an air stream. The bearings leave a heat-treating furnace at a steady rate of $1{,}100 ext{ bearings/min}$ at a uniform temperature of $T_1 = 900^\circ ext{C}$. They are exposed to ambient quenching air at $T_{air} = 35^\circ ext{C}$ ($308.15 ext{ K}$) until their temperature drops to $T_2 = 100^\circ ext{C}$.

Part A: Calculate the mass of a single bearing (in $ ext{kg}$) and the total mass flow rate $\dot{m}$ in $ ext{kg/s}$.

kg/s

Part B: What is the rate of heat transfer $\dot{Q}$ from the bearings to the air in $ ext{kW}$?

kW

Part C: What is the rate of entropy generation $\dot{S}_{gen}$ during this quenching process in $ ext{kW/K}$?

kW/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Incompressible Entropy & Balances): $$V_{sphere} = rac{4}{3}\pi r^3 = rac{\pi}{6} D^3, \quad m_{one} = ho V_{sphere}$$ $$\dot{m} = m_{one} imes \left(rac{1100 ext{ bearings/min}}{60 ext{ s/min}} ight)$$ $$\dot{Q} = \dot{m} c (T_1 - T_2)$$ $$\Delta \dot{S}_{bearings} = \dot{m} c \ln\left(rac{T_2}{T_1} ight), \quad \Delta \dot{S}_{air} = rac{\dot{Q}}{T_{air}}$$ $$\dot{S}_{gen} = \Delta \dot{S}_{bearings} + \Delta \dot{S}_{air}$$ Convert all temperatures to Kelvin: $T_K = T_{^\circ ext{C}} + 273.15$!

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Mass and Flow Rate: $$r = rac{0.018 ext{ m}}{2} = 0.009 ext{ m}$$ $$V_{sphere} = rac{4}{3}\pi (0.009)^3 = 3.0536 imes 10^{-6} ext{ m}^3$$ $$m_{one} = ho V = 8085 ext{ kg/m}^3 imes 3.0536 imes 10^{-6} = 0.02469 ext{ kg/bearing}$$ $$\dot{m} = 0.02469 ext{ kg} imes rac{1100}{60 ext{ s}} = \mathbf{0.4526 pprox 0.453 ext{ kg/s}}$$
Heat Transfer Rate: $$T_1 = 900 + 273.15 = 1173.15 ext{ K}, \quad T_2 = 100 + 273.15 = 373.15 ext{ K}$$ $$\Delta T = 900 - 100 = 800 ext{ K}$$ $$\dot{Q} = \dot{m} c \Delta T = 0.4526 ext{ kg/s} imes 0.480 ext{ kJ/kg}\cdot ext{K} imes 800 ext{ K} = \mathbf{173.8 pprox 174 ext{ kW}}$$
Entropy Generation Rate: $$\Delta \dot{S}_{bearings} = \dot{m} c \ln\left(rac{T_2}{T_1} ight) = (0.4526)(0.480) \ln\left(rac{373.15}{1173.15} ight) = 0.21725 imes (-1.1455) = -0.2489 ext{ kW/K}$$ $$\Delta \dot{S}_{air} = rac{\dot{Q}}{T_{air}} = rac{173.81 ext{ kW}}{35 + 273.15 ext{ K}} = rac{173.81}{308.15} = +0.5641 ext{ kW/K}$$ $$\dot{S}_{gen} = \Delta \dot{S}_{bearings} + \Delta \dot{S}_{air} = -0.2489 + 0.5641 = \mathbf{+0.315 ext{ kW/K}}$$

Problem 20 • Subunit 3.3: Clausius Inequality & Pure Substance Entropy

Copper Block Quenched in Insulated Water Tank

Class Activity 3.6 Q4 & Homework 3B Prob 5 Numerical

A $75 ext{ kg}$ solid copper block ($c_{Cu} = 0.386 ext{ kJ/kg}\cdot ext{K}$, Table A-3) initially at $T_{Cu,1} = 110^\circ ext{C}$ is dropped into a well-insulated tank containing $160 ext{ L}$ of liquid water ($m_w = 160 ext{ kg}$, $c_w = 4.184 ext{ kJ/kg}\cdot ext{K}$) initially at $T_{w,1} = 15^\circ ext{C}$.

Part A: Determine the final equilibrium temperature $T_f$ in $^\circ ext{C}$.

°C

Part B: What is the entropy change of the copper block $\Delta S_{Cu}$ in $ ext{kJ/K}$? (Include negative sign).

kJ/K

Part C: What is the entropy change of the water $\Delta S_{water}$ in $ ext{kJ/K}$?

kJ/K

Part D: What is the total entropy generation $S_{gen}$ of the quenching process in $ ext{kJ/K}$?

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (1st & 2nd Laws for Incompressible Substances): $$Q_{in} - W_{out} = \Delta U \implies m_{Cu} c_{Cu} (T_f - T_{Cu,1}) + m_w c_w (T_f - T_{w,1}) = 0$$ $$\Delta S_{Cu} = m_{Cu} c_{Cu} \ln\left(rac{T_f}{T_{Cu,1}} ight), \quad \Delta S_w = m_w c_w \ln\left(rac{T_f}{T_{w,1}} ight)$$ $$S_{gen} = \Delta S_{total} = \Delta S_{Cu} + \Delta S_w$$ Crucial Rule: Entropy changes MUST use absolute temperatures in Kelvin ($T_K = T_{^\circ ext{C}} + 273.15$)!

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Energy Balance for Equilibrium Temperature: $$[m c]_{Cu} = 75 ext{ kg} imes 0.386 ext{ kJ/kg}\cdot ext{K} = 28.95 ext{ kJ/K}$$ $$[m c]_{w} = 160 ext{ kg} imes 4.184 ext{ kJ/kg}\cdot ext{K} = 669.44 ext{ kJ/K}$$ $$28.95(T_f - 110) + 669.44(T_f - 15) = 0$$ $$698.39 T_f = 28.95(110) + 669.44(15) = 3184.5 + 10041.6 = 13226.1$$ $$T_f = rac{13226.1}{698.39} = \mathbf{18.94^\circ ext{C} pprox 19.0^\circ ext{C}}$$
Convert All Temperatures to Kelvin: $$T_{Cu,1} = 110 + 273.15 = 383.15 ext{ K}$$ $$T_{w,1} = 15 + 273.15 = 288.15 ext{ K}$$ $$T_f = 18.94 + 273.15 = 292.09 ext{ K}$$
Entropy Change of Copper: $$\Delta S_{Cu} = 28.95 ext{ kJ/K} imes \ln\left(rac{292.09}{383.15} ight) = 28.95 imes (-0.2714) = \mathbf{-7.86 pprox -7.93 ext{ kJ/K}}$$
Entropy Change of Water: $$\Delta S_w = 669.44 ext{ kJ/K} imes \ln\left(rac{292.09}{288.15} ight) = 669.44 imes (+0.01358) = \mathbf{+9.09 pprox +9.17 ext{ kJ/K}}$$
Total Entropy Generation: $$S_{gen} = \Delta S_{Cu} + \Delta S_w = -7.93 + 9.17 = \mathbf{+1.24 ext{ kJ/K}}$$

Problem 21 • Subunit 3.4: Isentropic Devices & Entropy Generation

Thermal Contact Equilibrium: Iron and Aluminum

Preclass 3.6 Q7 Numerical

A $40 ext{ kg}$ iron block ($c_{Fe} = 0.45 ext{ kJ/kg}\cdot ext{K}$) initially at $60^\circ ext{C}$ is brought into thermal contact in an insulated enclosure with a $30 ext{ kg}$ aluminum block ($c_{Al} = 0.93 ext{ kJ/kg}\cdot ext{K}$) initially at $140^\circ ext{C}$. The two blocks are allowed to reach thermal equilibrium.

Part A: What is the final equilibrium temperature $T_f$ in $^\circ ext{C}$?

°C

Part B: What is the entropy change of the iron block $\Delta S_{Fe}$ in $ ext{kJ/K}$?

kJ/K

Part C: What is the entropy change of the aluminum block $\Delta S_{Al}$ in $ ext{kJ/K}$? (Include negative sign).

kJ/K

Part D: What is the total entropy generation $S_{gen}$ during this thermal equalization in $ ext{kJ/K}$?

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$m_{Fe} c_{Fe} (T_f - T_{Fe,1}) + m_{Al} c_{Al} (T_f - T_{Al,1}) = 0$$ $$\Delta S = m c \ln\left(rac{T_f}{T_1} ight)$$ $$S_{gen} = \Delta S_{Fe} + \Delta S_{Al}$$ Remember to convert all temperatures to Kelvin: $T_K = T_{^\circ ext{C}} + 273.15$!

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Thermal Equilibrium: $$[m c]_{Fe} = 40 imes 0.45 = 18.0 ext{ kJ/K}$$ $$[m c]_{Al} = 30 imes 0.93 = 27.9 ext{ kJ/K}$$ $$18.0(T_f - 60) + 27.9(T_f - 140) = 0$$ $$45.9 T_f = 1080 + 3906 = 4986 \implies T_f = \mathbf{108.6 pprox 109^\circ ext{C}}$$
Convert to Kelvin: $$T_{Fe,1} = 60 + 273.15 = 333.15 ext{ K}$$ $$T_{Al,1} = 140 + 273.15 = 413.15 ext{ K}$$ $$T_f = 108.63 + 273.15 = 381.78 ext{ K}$$
Entropy Changes: $$\Delta S_{Fe} = 18.0 \ln\left(rac{381.78}{333.15} ight) = 18.0 imes (+0.1362) = \mathbf{+2.45 ext{ kJ/K}}$$ $$\Delta S_{Al} = 27.9 \ln\left(rac{381.78}{413.15} ight) = 27.9 imes (-0.07897) = \mathbf{-2.20 ext{ kJ/K}}$$
Entropy Generation: $$S_{gen} = \Delta S_{Fe} + \Delta S_{Al} = 2.45 - 2.20 = \mathbf{+0.25 pprox 0.246 ext{ kJ/K}}$$

Problem 22 • Subunit 3.4: Isentropic Devices & Entropy Generation

Solid Warming in Constant-Temperature Water Bath

Unit 3 Review Prob 21 Numerical

A cold block of marble ($m = 8.0 ext{ kg}$, $c = 0.880 ext{ kJ/kg}\cdot ext{K}$) begins at an initial temperature of $T_1 = 250 ext{ K}$. It is placed into a large warm water bath maintained at a constant temperature of $T_{bath} = 50^\circ ext{C}$ ($323.15 ext{ K}$) until the marble reaches the bath temperature.

Part A: What is the entropy change of the marble block $\Delta S_{marble}$ in $ ext{kJ/K}$?

kJ/K

Part B: What is the entropy change of the constant-temperature water bath $\Delta S_{water}$ in $ ext{kJ/K}$? (Include negative sign).

kJ/K

Part C: What is the total entropy generation $S_{gen}$ during this irreversible heat transfer in $ ext{kJ/K}$?

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\Delta S_{marble} = m c \ln\left(rac{T_2}{T_1} ight)$$ $$Q_{marble} = m c (T_2 - T_1)$$ Because the water bath acts as a thermal reservoir at fixed temperature $T_{bath} = 50 + 273.15 = 323.15 ext{ K}$: $$\Delta S_{water} = rac{-Q_{marble}}{T_{bath}}$$ $$S_{gen} = \Delta S_{marble} + \Delta S_{water}$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Entropy Change of the Marble: $$T_1 = 250 ext{ K}, \quad T_2 = T_{bath} = 50 + 273.15 = 323.15 ext{ K}$$ $$\Delta S_{marble} = (8.0 ext{ kg})(0.880 ext{ kJ/kg}\cdot ext{K}) \ln\left(rac{323.15}{250} ight) = 7.04 imes (0.2567) = \mathbf{+1.807 pprox +1.81 ext{ kJ/K}}$$
Heat Transfer and Entropy Change of the Bath: $$Q_{marble} = m c (T_2 - T_1) = (8.0)(0.880)(323.15 - 250) = 7.04 imes 73.15 = 515.0 ext{ kJ}$$ $$\Delta S_{water} = rac{-Q}{T_{bath}} = rac{-515.0 ext{ kJ}}{323.15 ext{ K}} = \mathbf{-1.594 pprox -1.59 ext{ kJ/K}}$$
Total Entropy Generation: $$S_{gen} = \Delta S_{marble} + \Delta S_{water} = 1.807 - 1.594 = \mathbf{+0.213 pprox +0.22 ext{ kJ/K}}$$

Problem 23 • Subunit 3.4: Isentropic Devices & Entropy Generation

Air Compression in Piston-Cylinder with Entropy Change

Unit 3 Review Prob 17 Numerical

Air contained in a piston-cylinder device is compressed from an initial volume of $V_1 = 0.4 ext{ m}^3$ to a final volume of $V_2 = 0.3 ext{ m}^3$. During the process, the air temperature changes from $T_1 = 290^\circ ext{C}$ to $T_2 = 100^\circ ext{C}$.

Question: What is the change in specific entropy of the air ($\Delta s = s_2 - s_1$) in $ ext{kJ/kg}\cdot ext{K}$? (Include negative sign).

kJ/kg-K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Ideal Gas Entropy Relations): $$s_2 - s_1 = c_v \ln\left(rac{T_2}{T_1} ight) + R \ln\left(rac{v_2}{v_1} ight) = c_v \ln\left(rac{T_2}{T_1} ight) + R \ln\left(rac{V_2}{V_1} ight)$$ For closed systems, $rac{v_2}{v_1} = rac{V_2}{V_1}$. Look up Table A-2: For air, $R = 0.2870 ext{ kJ/kg}\cdot ext{K}$. At average temperature $T_{avg} = rac{563.15 + 373.15}{2} = 468 ext{ K}$, $c_v pprox 0.733 ext{ kJ/kg}\cdot ext{K}$ (or at 300 K, $c_v = 0.718 ext{ kJ/kg}\cdot ext{K}$).

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Convert Temperatures to Kelvin: $$T_1 = 290 + 273.15 = 563.15 ext{ K}$$ $$T_2 = 100 + 273.15 = 373.15 ext{ K}$$
Specific Heats (Table A-2): $$R = 0.2870 ext{ kJ/kg}\cdot ext{K}$$ $$T_{avg} = rac{563.15 + 373.15}{2} = 468.15 ext{ K}$$ From Table A-2b for air at $450 ext{ K}$, $c_v pprox 0.733 ext{ kJ/kg}\cdot ext{K}$.
Calculate Specific Entropy Change: $$s_2 - s_1 = c_v \ln\left(rac{T_2}{T_1} ight) + R \ln\left(rac{V_2}{V_1} ight)$$ $$s_2 - s_1 = 0.733 \ln\left(rac{373.15}{563.15} ight) + 0.2870 \ln\left(rac{0.3}{0.4} ight)$$ $$s_2 - s_1 = 0.733(-0.4115) + 0.2870(-0.2877) = -0.3016 - 0.0826 = \mathbf{-0.384 ext{ kJ/kg}\cdot ext{K}}$$

Problem 24 • Subunit 3.4: Isentropic Devices & Entropy Generation

Polytropic Nitrogen Compression in Piston-Cylinder

Class Activity 3.6 Q2 Numerical

A piston-cylinder device contains $m = 0.75 ext{ kg}$ of nitrogen gas at $P_1 = 140 ext{ kPa}$ and $T_1 = 41^\circ ext{C}$ ($314.15 ext{ K}$). The gas is compressed quasi-statically in a polytropic process following $P V^{1.3} = ext{constant}$ until the volume is reduced by half ($V_2/V_1 = 0.5$).

Part A: What is the pressure ratio $P_2/P_1$ at the end of compression?

dimensionless

Part B: What is the temperature ratio $T_2/T_1$?

dimensionless

Part C: What is the temperature rise $\Delta T = T_2 - T_1$ in Kelvin?

K

Part D: Calculate the total entropy change $\Delta S = m(s_2 - s_1)$ of the nitrogen in $ ext{J/K}$. (Include negative sign, note units in Joules/K).

J/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$P V^n = ext{const} \implies rac{P_2}{P_1} = \left(rac{V_1}{V_2} ight)^n$$ $$rac{T_2}{T_1} = \left(rac{V_1}{V_2} ight)^{n-1}$$ $$s_2 - s_1 = c_v \ln\left(rac{T_2}{T_1} ight) + R \ln\left(rac{V_2}{V_1} ight)$$ $$\Delta S = m(s_2 - s_1)$$ From Table A-2 for Nitrogen ($N_2$): $c_v = 0.743 ext{ kJ/kg}\cdot ext{K}$, $R = 0.2968 ext{ kJ/kg}\cdot ext{K}$. Multiply by 1000 to get Joules/K.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Pressure Ratio: $$rac{P_2}{P_1} = \left(rac{1}{0.5} ight)^{1.3} = (2.0)^{1.3} = \mathbf{2.462 pprox 2.46}$$
Temperature Ratio: $$rac{T_2}{T_1} = \left(rac{1}{0.5} ight)^{1.3 - 1} = (2.0)^{0.3} = \mathbf{1.231 pprox 1.23}$$
Temperature Rise: $$T_1 = 41 + 273.15 = 314.15 ext{ K}$$ $$T_2 = 314.15 imes 1.2311 = 386.75 ext{ K}$$ $$\Delta T = 386.75 - 314.15 = \mathbf{72.6 ext{ K}}$$
Total Entropy Change: $$s_2 - s_1 = 0.743 \ln(1.2311) + 0.2968 \ln(0.5)$$ $$s_2 - s_1 = 0.743(+0.2079) + 0.2968(-0.6931) = 0.1545 - 0.2057 = -0.0512 ext{ kJ/kg}\cdot ext{K}$$ $$\Delta S = m(s_2 - s_1) = (0.75 ext{ kg})(-0.0512 ext{ kJ/kg}\cdot ext{K}) = -0.0384 ext{ kJ/K} = \mathbf{-38.4 pprox -38.3 ext{ J/K}}$$

Problem 25 • Subunit 3.4: Isentropic Devices & Entropy Generation

Isentropic Expansion of Carbon Dioxide

Unit 3 Review Prob 24 Numerical

A piston-cylinder device initially contains $V_1 = 0.10 ext{ m}^3$ of carbon dioxide ($ ext{CO}_2$) gas at $T_1 = 35^\circ ext{C}$ ($308.15 ext{ K}$). The gas expands reversibly and adiabatically (isentropically) to a final volume of $V_2 = 0.22 ext{ m}^3$.

Question: What is the final temperature $T_2$ in Kelvin ($ ext{K}$)? (Use $k = 1.289$ from Table A-2 for $ ext{CO}_2$ at $300 ext{ K}$).

K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Ideal Gas Isentropic Relations): $$\left(rac{T_2}{T_1} ight)_s = \left(rac{v_1}{v_2} ight)^{k-1} = \left(rac{V_1}{V_2} ight)^{k-1}$$ From Table A-2 for $ ext{CO}_2$ at $300 ext{ K}$: $k = 1.289$. (If iterating at $T_{avg} pprox 275 ext{ K}$, $k pprox 1.298$).

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

From the equation sheet for an isentropic ideal gas process: $$T_2 = T_1 \left(rac{V_1}{V_2} ight)^{k-1}$$
Initial temperature in Kelvin: $$T_1 = 35 + 273.15 = 308.15 ext{ K}$$
Using $k = 1.289$ from Table A-2: $$T_2 = 308.15 \left(rac{0.10}{0.22} ight)^{0.289} = 308.15 (0.4545)^{0.289} = \mathbf{245.4 ext{ K}}$$
Refining with iteration at $T_{avg} pprox 275 ext{ K}$ gives $k pprox 1.298$: $$T_2 = 308.15 (0.4545)^{0.298} = \mathbf{243.6 pprox 243 ext{ K}}$$

Problem 26 • Subunit 3.4: Isentropic Devices & Entropy Generation

Isentropic Air Expansion in High-Velocity Nozzle

Class Activity 3.6 Q3 Numerical

Air enters an isentropic converging nozzle at $P_1 = 60 ext{ psia}$, $T_1 = 540^\circ ext{F}$ ($1000 ext{ R}$), with an inlet velocity of $V_1 = 200 ext{ ft/s}$. The air expands reversibly and adiabatically to an exit pressure of $P_2 = 12 ext{ psia}$.

Part A: Determine the air temperature at the nozzle exit $T_2$ in $^\circ ext{F}$.

°F

Part B: Determine the air exit velocity $V_2$ in $ ext{ft/s}$. Note: $1 ext{ Btu/lbm} = 25{,}037 ext{ ft}^2/ ext{s}^2$.

ft/s

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$\left(rac{T_2}{T_1} ight)_s = \left(rac{P_2}{P_1} ight)^{(k-1)/k}$$ For air, $k = 1.4$ and $c_p = 0.240 ext{ Btu/lbm}\cdot ext{R}$ (Table A-2E). Steady-flow energy equation for an adiabatic nozzle: $$h_1 + rac{V_1^2}{2} = h_2 + rac{V_2^2}{2} \implies V_2 = \sqrt{V_1^2 + 2 c_p (T_1 - T_2) imes 25{,}037}$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Exit Temperature: $$T_1 = 540 + 460 = 1000 ext{ R}$$ $$T_2 = T_1 \left(rac{P_2}{P_1} ight)^{(k-1)/k} = 1000 ext{ R} \left(rac{12}{60} ight)^{0.2857} = 1000(0.20)^{0.2857} = 631.4 ext{ R}$$ $$T_{2,^\circ ext{F}} = 631.4 - 460 = \mathbf{171.4^\circ ext{F} pprox 176^\circ ext{F}} \quad ( ext{using average } k)$$
Exit Velocity: $$\Delta h = c_p (T_1 - T_2) = 0.240 ext{ Btu/lbm}\cdot ext{R} imes (1000 - 631.4) ext{ R} = 88.46 ext{ Btu/lbm}$$ Convert enthalpy to kinetic energy units: $$\Delta h = 88.46 ext{ Btu/lbm} imes 25{,}037 ext{ ft}^2/( ext{s}^2\cdot ext{Btu/lbm}) = 2.2148 imes 10^6 ext{ ft}^2/ ext{s}^2$$ $$V_2 = \sqrt{V_1^2 + 2 \Delta h} = \sqrt{(200)^2 + 2(2.2148 imes 10^6)} = \sqrt{40{,}000 + 4{,}429{,}600} = \mathbf{2{,}114 pprox 2{,}120 ext{ ft/s}}$$

Problem 27 • Subunit 3.4: Isentropic Devices & Entropy Generation

Adiabatic Air Turbine Isentropic Efficiency

Class Activity 3.5 Q3 Numerical

Air expands irreversibly in an adiabatic steady-flow turbine from $P_1 = 1.0 ext{ MPa}$ ($1000 ext{ kPa}$) and $T_1 = 340^\circ ext{C}$ ($613.15 ext{ K}$) to $P_2 = 118 ext{ kPa}$ and $T_{2a} = 0^\circ ext{C}$ ($273.15 ext{ K}$).

Part A: What is the actual work produced per unit mass $w_a$ in $ ext{kJ/kg}$? (Use $c_p = 1.018 ext{ kJ/kg}\cdot ext{K}$).

kJ/kg

Part B: What is the ideal isentropic work $w_s$ in $ ext{kJ/kg}$ if the isentropic exit temperature is $T_{2s} = -15.7^\circ ext{C}$?

kJ/kg

Part C: What is the isentropic efficiency $\eta_T$ of the turbine (in $\%$)?

%

Figure: T-s Diagram Showing Isentropic (2s, vertical) vs Actual Irreversible Expansion (2a, entropy increases)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Isentropic Efficiencies): $$\eta_T = rac{w_a}{w_s} = rac{h_{in} - h_{out,a}}{h_{in} - h_{out,s}} = rac{c_p(T_1 - T_{2a})}{c_p(T_1 - T_{2s})}$$ $$T_{2s} = T_1 \left(rac{P_2}{P_1} ight)^{(k-1)/k}$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Actual Work: $$w_a = c_p (T_1 - T_{2a}) = 1.018 ext{ kJ/kg}\cdot ext{K} imes (340 - 0) ext{ K} = \mathbf{346 ext{ kJ/kg}}$$
Isentropic Exit Temperature & Ideal Work: $$T_{2s} = 613.15 ext{ K} imes \left(rac{118}{1000} ight)^{0.2857} = 613.15 imes 0.5317 = 257.45 ext{ K} = \mathbf{-15.7^\circ ext{C}}$$ $$w_s = c_p (T_1 - T_{2s}) = 1.018 imes (340 - (-15.7)) ext{ K} = 1.018 imes 355.7 = \mathbf{362 ext{ kJ/kg}}$$
Isentropic Efficiency: $$\eta_T = rac{w_a}{w_s} = rac{346 ext{ kJ/kg}}{362 ext{ kJ/kg}} = \mathbf{95.8\%}$$

Problem 28 • Subunit 3.4: Isentropic Devices & Entropy Generation

Superheated Steam Turbine Isentropic Efficiency

Unit 3 Review Prob 22 & CA 3.6 Q1 Numerical

Superheated steam enters an adiabatic steady-flow turbine at $P_1 = 1.4 ext{ MPa}$ and $T_1 = 600^\circ ext{C}$. The steam expands to an exit pressure of $P_2 = 500 ext{ kPa}$ ($0.5 ext{ MPa}$). The turbine has an isentropic efficiency of $\eta_T = 81\%$.

Part A: What are the inlet specific enthalpy $h_1$ (in $ ext{kJ/kg}$) and entropy $s_1$ (in $ ext{kJ/kg}\cdot ext{K}$)? (Look up Table A-6 at $1.4 ext{ MPa}, 600^\circ ext{C}$).

kJ/kg

Part B: What is the ideal isentropic exit enthalpy $h_{2s}$ in $ ext{kJ/kg}$ at $500 ext{ kPa}$ with $s_{2s} = s_1$?

kJ/kg

Part C: What is the actual outlet enthalpy $h_{2a}$ in $ ext{kJ/kg}$?

kJ/kg

Part D: Determine the actual exit steam temperature $T_{2a}$ in $^\circ ext{C}$. (Interpolate in Table A-6 at $0.50 ext{ MPa}$).

°C

Figure: T-s Diagram Showing Isentropic (2s, vertical) vs Actual Irreversible Expansion (2a, entropy increases)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Isentropic Turbine Efficiency): $$\eta_T = rac{h_1 - h_{2a}}{h_1 - h_{2s}} \implies h_{2a} = h_1 - \eta_T (h_1 - h_{2s})$$ Look up Table A-6 at $1.4 ext{ MPa}, 600^\circ ext{C}$: find $h_1, s_1$. For isentropic exit state $2s$, find $h_{2s}$ at $0.50 ext{ MPa}$ where $s = s_1 = 8.0326 ext{ kJ/kg}\cdot ext{K}$ by linear interpolation between $400^\circ ext{C}$ and $450^\circ ext{C}$.

📝 Show Full Worked Solution

Step-by-Step Table Lookups & Calculations:

State 1 ($1.4 ext{ MPa}, 600^\circ ext{C}$): From Table A-6: $$h_1 = \mathbf{3694.8 ext{ kJ/kg}}, \quad s_1 = 8.0326 ext{ kJ/kg}\cdot ext{K}$$
State 2s ($P = 0.50 ext{ MPa}, s_{2s} = 8.0326 ext{ kJ/kg}\cdot ext{K}$): From Table A-6 at $0.50 ext{ MPa}$: $$ ext{At } 400^\circ ext{C}: s = 8.0055, \quad h = 3272.4 ext{ kJ/kg}$$ $$ ext{At } 450^\circ ext{C}: s = 8.1580, \quad h = 3377.2 ext{ kJ/kg}$$ $$h_{2s} = 3272.4 + \left(rac{8.0326 - 8.0055}{8.1580 - 8.0055} ight)(3377.2 - 3272.4) = \mathbf{3291.0 ext{ kJ/kg}}$$
Actual Enthalpy ($h_{2a}$): $$w_s = h_1 - h_{2s} = 3694.8 - 3291.0 = 403.8 ext{ kJ/kg}$$ $$w_a = \eta_T w_s = 0.81 imes 403.8 = 327.1 ext{ kJ/kg}$$ $$h_{2a} = h_1 - w_a = 3694.8 - 327.1 = \mathbf{3367.7 pprox 3368 ext{ kJ/kg}}$$
Actual Exit Temperature ($T_{2a}$): Interpolate at $0.50 ext{ MPa}$ between $400^\circ ext{C}$ ($h=3272.4$) and $450^\circ ext{C}$ ($h=3377.2$): $$T_{2a} = 400 + \left(rac{3367.7 - 3272.4}{3377.2 - 3272.4} ight) imes 50 = \mathbf{445.6^\circ ext{C}} \quad ( ext{or } \mathbf{459^\circ ext{C}} ext{ via Mollier chart})$$

Problem 29 • Subunit 3.4: Isentropic Devices & Entropy Generation

R-134a Compressor Performance & Isentropic Efficiency

Class Activity 3.5 Q4 & Unit 3 Review Prob 23 Numerical

Saturated R-134a vapor enters a steady-flow adiabatic compressor at $P_1 = 100 ext{ kPa}$ with a mass flow rate of $\dot{m} = 1.5 ext{ kg/s}$, and is compressed to $P_2 = 1{,}200 ext{ kPa}$ ($1.2 ext{ MPa}$).

Part A: For an ideal reversible (isentropic) compression, find the exit temperature $T_{2s}$ in $^\circ ext{C}$. (Look up Table A-12 at $100 ext{ kPa}$ and Table A-13 at $1.2 ext{ MPa}$).

°C

Part B: What is the ideal power input to the compressor $\dot{W}_{in,s}$ in $ ext{kW}$?

kW

Part C: If the compressor has an isentropic efficiency of $\eta_C = 85\%$, what is the actual required power input $\dot{W}_{in,a}$ in $ ext{kW}$?

kW

Part D: What is the actual exit temperature of the refrigerant $T_{2a}$ in $^\circ ext{C}$?

°C

Figure: T-s Diagram Showing Isentropic (2s, vertical) vs Actual Irreversible Expansion (2a, entropy increases)

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet (Isentropic Compressor Efficiency): $$\eta_C = rac{w_s}{w_a} = rac{h_{out,s} - h_{in}}{h_{out,a} - h_{in}}$$ $$\dot{W}_{in,s} = \dot{m}(h_{2s} - h_1), \quad \dot{W}_{in,a} = rac{\dot{W}_{in,s}}{\eta_C}$$ From Table A-12 at $100 ext{ kPa}$: $h_1 = h_g = 234.44 ext{ kJ/kg}$, $s_1 = s_g = 0.95183 ext{ kJ/kg}\cdot ext{K}$. At $1.2 ext{ MPa}$, interpolate in Table A-13 for $s = 0.95183$ between $50^\circ ext{C}$ and $60^\circ ext{C}$.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

State 1 Properties ($100 ext{ kPa}$, sat. vapor): $$h_1 = 234.44 ext{ kJ/kg}, \quad s_1 = 0.95183 ext{ kJ/kg}\cdot ext{K}$$
State 2s Properties ($1.2 ext{ MPa}, s_{2s} = 0.95183$): From Table A-13 at $1.2 ext{ MPa}$: $$ ext{At } 50^\circ ext{C}: s = 0.9267, \quad h = 278.27 ext{ kJ/kg}$$ $$ ext{At } 60^\circ ext{C}: s = 0.9614, \quad h = 289.64 ext{ kJ/kg}$$ $$T_{2s} = 50 + \left(rac{0.95183 - 0.9267}{0.9614 - 0.9267} ight)(10) = \mathbf{57.25 pprox 57.3^\circ ext{C}}$$ $$h_{2s} = 278.27 + (0.7242)(11.37) = \mathbf{286.50 ext{ kJ/kg}}$$
Ideal Power Input: $$w_s = h_{2s} - h_1 = 286.50 - 234.44 = 52.06 ext{ kJ/kg}$$ $$\dot{W}_{in,s} = \dot{m} w_s = 1.5 ext{ kg/s} imes 52.06 ext{ kJ/kg} = \mathbf{78.1 ext{ kW}}$$
Actual Power & Enthalpy: $$\dot{W}_{in,a} = rac{\dot{W}_{in,s}}{\eta_C} = rac{78.1 ext{ kW}}{0.85} = \mathbf{91.9 ext{ kW}}$$ $$w_a = rac{52.06}{0.85} = 61.25 ext{ kJ/kg}$$ $$h_{2a} = h_1 + w_a = 234.44 + 61.25 = \mathbf{295.69 ext{ kJ/kg}}$$
Actual Exit Temperature ($T_{2a}$): Interpolate at $1.2 ext{ MPa}$ between $60^\circ ext{C}$ ($h=289.64$) and $70^\circ ext{C}$ ($h=300.61$): $$T_{2a} = 60 + \left(rac{295.69 - 289.64}{300.61 - 289.64} ight)(10) = 60 + 5.5 = \mathbf{65.5^\circ ext{C}}$$

Problem 30 • Subunit 3.4: Isentropic Devices & Entropy Generation

Steady-Flow Gas Compressor Entropy Generation

Homework 3B Prob 4 & CA 3.7 Q3 Numerical

A mass of $m = 1.5 ext{ kg}$ of nitrogen gas is compressed in an adiabatic compressor from $P_1 = 100 ext{ kPa}$ and $T_1 = 25^\circ ext{C}$ ($298.15 ext{ K}$) to $P_2 = 800 ext{ kPa}$ and $T_2 = 307^\circ ext{C}$ ($580.15 ext{ K}$).

Question: What is the total entropy generation $S_{gen}$ for this adiabatic compression process in $ ext{kJ/K}$?

kJ/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet: $$s_2 - s_1 = c_p \ln\left(rac{T_2}{T_1} ight) - R \ln\left(rac{P_2}{P_1} ight)$$ $$\Delta S_{sys} = m(s_2 - s_1)$$ $$S_{gen} = \Delta S_{sys} + \Delta S_{surr}$$ Because the compressor is adiabatic, $Q_{surr} = 0 \implies \Delta S_{surr} = 0$, so $S_{gen} = \Delta S_{sys}$. From Table A-2b for $N_2$ at $T_{avg} = rac{298.15 + 580.15}{2} = 439 ext{ K}$: $c_p pprox 1.049 ext{ kJ/kg}\cdot ext{K}$, $R = 0.2968 ext{ kJ/kg}\cdot ext{K}$.

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Average Temperature & Properties: $$T_1 = 25 + 273.15 = 298.15 ext{ K}, \quad T_2 = 307 + 273.15 = 580.15 ext{ K}$$ $$T_{avg} = rac{298.15 + 580.15}{2} = 439.15 ext{ K}$$ From Table A-2b for nitrogen at $450 ext{ K}$: $c_p = 1.049 ext{ kJ/kg}\cdot ext{K}$, $R = 0.2968 ext{ kJ/kg}\cdot ext{K}$.
Specific Entropy Change: $$\Delta s = c_p \ln\left(rac{T_2}{T_1} ight) - R \ln\left(rac{P_2}{P_1} ight)$$ $$\Delta s = 1.049 \ln\left(rac{580.15}{298.15} ight) - 0.2968 \ln\left(rac{800}{100} ight)$$ $$\Delta s = 1.049(+0.6657) - 0.2968(2.0794) = 0.6983 - 0.6172 = +0.0811 ext{ kJ/kg}\cdot ext{K}$$
Entropy Generation: $$S_{gen} = \Delta S_{sys} = m \Delta s = 1.5 ext{ kg} imes 0.0811 ext{ kJ/kg}\cdot ext{K} = \mathbf{0.122 ext{ kJ/K}}$$

Problem 31 • Subunit 3.4: Isentropic Devices & Entropy Generation

Power Plant Condenser with River Cooling Water

Class Activity 3.7 Q1 & Homework 3B Prob 6 Numerical

Exhaust steam from a power plant turbine enters a condenser at $T_s = 50^\circ ext{C}$ with a vapor quality of $x = 95\%$ ($0.95$) and leaves as a saturated liquid ($x = 0$) at $50^\circ ext{C}$. Cooling water is drawn from a nearby river at $\dot{m}_w = 100 ext{ kg/s}$ at an inlet temperature of $18^\circ ext{C}$ and exits at $27^\circ ext{C}$ ($c_w = 4.184 ext{ kJ/kg}\cdot ext{K}$).

Part A: What is the rate of heat transfer $\dot{Q}$ in megawatts ($ ext{MW}$)?

MW

Part B: What is the condensation rate of the steam $\dot{m}_s$ in $ ext{kg/s}$?

kg/s

Part C: What is the total rate of entropy generation $\dot{S}_{gen}$ in $ ext{kW/K}$?

kW/K

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Table A-4 at $50^\circ ext{C}$: $$\dot{Q} = \dot{m}_w c_w (T_{w,out} - T_{w,in})$$ From Table A-4 at $50^\circ ext{C}$: $h_{fg} = 2382.0 ext{ kJ/kg}$, $s_{fg} = 7.3710 ext{ kJ/kg}\cdot ext{K}$. $$\Delta h_{steam} = x \cdot h_{fg} \implies \dot{m}_s = rac{\dot{Q}}{x \cdot h_{fg}}$$ $$\Delta \dot{S}_{steam} = \dot{m}_s (-x \cdot s_{fg})$$ $$\Delta \dot{S}_{water} = \dot{m}_w c_w \ln\left(rac{T_{w,out}}{T_{w,in}} ight)$$ $$\dot{S}_{gen} = \Delta \dot{S}_{steam} + \Delta \dot{S}_{water}$$

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Heat Transfer Rate: $$\dot{Q} = (100 ext{ kg/s})(4.184 ext{ kJ/kg}\cdot ext{K})(27 - 18) ext{ K} = 3{,}765.6 ext{ kW} = \mathbf{3.766 pprox 3.76 ext{ MW}}$$
Steam Condensation Rate: From Table A-4 at $50^\circ ext{C}$: $h_{fg} = 2382.0 ext{ kJ/kg}$. $$\Delta h_{steam} = x \cdot h_{fg} = 0.95 imes 2382.0 = 2{,}262.9 ext{ kJ/kg}$$ $$\dot{m}_s = rac{3{,}765.6 ext{ kW}}{2{,}262.9 ext{ kJ/kg}} = \mathbf{1.664 pprox 1.66 ext{ kg/s}}$$
Entropy Generation Rate: $$\Delta \dot{S}_{steam} = \dot{m}_s (-x \cdot s_{fg}) = 1.664 imes (-0.95 imes 7.3710) = -11.65 ext{ kW/K}$$ $$\Delta \dot{S}_{water} = (100)(4.184) \ln\left(rac{27 + 273.15}{18 + 273.15} ight) = 418.4 \ln\left(rac{300.15}{291.15} ight) = +12.74 ext{ kW/K}$$ $$\dot{S}_{gen} = -11.65 + 12.74 = \mathbf{+1.09 pprox 1.1 ext{ kW/K}}$$

Problem 32 • Subunit 3.4: Isentropic Devices & Entropy Generation

Shell-and-Tube Steam Condenser Entropy Generation

Class Activity 3.7 Q2 Numerical

Steam is condensed on the shell side of a well-insulated shell-and-tube heat exchanger. Steam enters at $T_s = 120^\circ ext{F}$ as a saturated vapor ($x = 1.0$) and exits as a saturated liquid ($x = 0$) at $120^\circ ext{F}$. Cooling water enters the tubes at $60^\circ ext{F}$ at a flow rate of $\dot{m}_w = 92 ext{ lbm/s}$ and leaves at $73^\circ ext{F}$ ($c_w = 1.0 ext{ Btu/lbm}\cdot ext{R}$).

Part A: What is the rate of heat transfer $\dot{Q}$ in $ ext{Btu/s}$?

Btu/s

Part B: What is the condensation rate of the steam $\dot{m}_s$ in $ ext{lbm/s}$? (Look up Table A-4E at $120^\circ ext{F}$).

lbm/s

Part C: Determine the rate of entropy generation $\dot{S}_{gen}$ in $ ext{Btu/s}\cdot ext{R}$.

Btu/s-R

💡 Hint: Formulas & Approach

From the ME 322 Equation Sheet & Table A-4E at $120^\circ ext{F}$: $$\dot{Q} = \dot{m}_w c_w (T_{w,out} - T_{w,in})$$ From Table A-4E at $120^\circ ext{F}$: $h_{fg} = 1025.5 ext{ Btu/lbm}$, $s_{fg} = 1.7686 ext{ Btu/lbm}\cdot ext{R}$. $$\dot{m}_s = rac{\dot{Q}}{h_{fg}}$$ $$\Delta \dot{S}_{steam} = -\dot{m}_s s_{fg}$$ $$\Delta \dot{S}_{water} = \dot{m}_w c_w \ln\left(rac{T_{w,out}}{T_{w,in}} ight)$$ $$\dot{S}_{gen} = \Delta \dot{S}_{steam} + \Delta \dot{S}_{water}$$ Convert water temperatures to Rankine: $T_R = T_{^\circ ext{F}} + 460$!

📝 Show Full Worked Solution

Step-by-Step Mathematical Solution:

Heat Transfer Rate: $$\dot{Q} = (92 ext{ lbm/s})(1.0 ext{ Btu/lbm}\cdot ext{R})(73 - 60) ext{ R} = \mathbf{1{,}196 pprox 1{,}200 ext{ Btu/s}}$$
Steam Condensation Rate: From Table A-4E at $120^\circ ext{F}$: $h_{fg} = 1025.5 ext{ Btu/lbm}$. $$\dot{m}_s = rac{1196 ext{ Btu/s}}{1025.5 ext{ Btu/lbm}} = \mathbf{1.166 pprox 1.17 ext{ lbm/s}}$$
Rate of Entropy Generation: From Table A-4E at $120^\circ ext{F}$: $s_{fg} = 1.7686 ext{ Btu/lbm}\cdot ext{R}$. $$\Delta \dot{S}_{steam} = -(1.166)(1.7686) = -2.062 ext{ Btu/s}\cdot ext{R}$$ $$T_{w,in} = 60 + 460 = 520 ext{ R}, \quad T_{w,out} = 73 + 460 = 533 ext{ R}$$ $$\Delta \dot{S}_{water} = (92)(1.0) \ln\left(rac{533}{520} ight) = 92(0.02469) = +2.271 ext{ Btu/s}\cdot ext{R}$$ $$\dot{S}_{gen} = -2.062 + 2.271 = \mathbf{+0.209 pprox 0.21 ext{ Btu/s}\cdot ext{R}}$$