ME 322 - Homework 3A Study & Problem Mastery Suite

📌 Official Reference Constraints:
All governing equations, state lookups, and unit conversion constants are derived strictly from:

ME 322 Equation Sheet (Jan 14, 2025 Revision): Heat engines ($\eta_{\text{th}} = \frac{\dot{W}_{\text{net}}}{\dot{Q}_H}$, $\eta_{\text{th,c}} = 1 - \frac{T_L}{T_H}$), Refrigerators & Heat Pumps ($\text{COP}_{\text{R}} = \frac{\dot{Q}_L}{\dot{W}_{\text{in}}}$, $\text{COP}_{\text{HP}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}}$, $\text{COP}_{\text{HP}} = \text{COP}_{\text{R}} + 1$), and Carnot limits.
Thermodynamic Property Tables (6th Edition): Page 1 & 2 Conversion Factors ($1\text{ kW} = 3,412.14\text{ Btu/h}$, $1\text{ hp} = 42.408\text{ Btu/min} = 0.7068\text{ Btu/s}$, $1\text{ hp} = 0.7457\text{ kW}$), Table A-4 ($h_f$ for saturated liquid water), and Table A-11 (saturated R-134a at $12^\circ\text{C}$).

📐 Quick Reference: Governing Equations & Conversions

System Type	Actual Performance	Carnot (Reversible) Limit	Key Relations & Unit Factors
Heat Engine (HE)	$\eta_{\text{th}} = \frac{\dot{W}_{\text{net,out}}}{\dot{Q}_H} = 1 - \frac{\dot{Q}_L}{\dot{Q}_H}$	$\eta_{\text{th,c}} = 1 - \frac{T_L}{T_H}$	$\dot{W}_{\text{net,out}} = \dot{Q}_H - \dot{Q}_L$
Heat Pump (HP)	$\text{COP}_{\text{HP}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}} = \frac{\dot{Q}_H}{\dot{Q}_H - \dot{Q}_L}$	$\text{COP}_{\text{HP,c}} = \frac{1}{1 - T_L/T_H} = \frac{T_H}{T_H - T_L}$	$\text{COP}_{\text{HP}} = \text{COP}_{\text{R}} + 1$
Refrigerator (R) / AC	$\text{COP}_{\text{R}} = \frac{\dot{Q}_L}{\dot{W}_{\text{in}}} = \frac{\dot{Q}_L}{\dot{Q}_H - \dot{Q}_L}$	$\text{COP}_{\text{R,c}} = \frac{1}{T_H/T_L - 1} = \frac{T_L}{T_H - T_L}$	$\dot{Q}_H = \dot{Q}_L + \dot{W}_{\text{in}}$
Units & Absolute Temps	$T(\text{K}) = T(^\circ\text{C}) + 273.15 \quad \| \quad T(\text{R}) = T(^\circ\text{F}) + 460$		$1\text{ kW} = 3,412.14\text{ Btu/h}$ $1\text{ hp} = 42.408\text{ Btu/min} = 0.7457\text{ kW}$

Problem 1: Diesel Engine Add-On Steam Generator

Evaluation of secondary cycle heat recovery claim

10 Points Total

An after-market manufacturer of diesel engine add-on parts claims that they can add a steam generator to replace the radiator on a large power generation diesel engine. This generator will keep the engine block at $250^\circ\text{F}$ (it will generate steam at that temperature), but the steam will drive a turbine. The manufacturer claims the overall engine efficiency will increase from $35\%$ to $55\%$. Evaluate the claim. Does the claim violate the laws of thermodynamics? Assume an outdoor heat rejection temperature of $68^\circ\text{F}$.

Figure 1.1: Thermodynamic block diagram for the original diesel engine and the combined secondary steam turbine add-on.

Part a: Maximum Possible Combined Efficiency 6 Points

Calculate the theoretical maximum possible overall thermal efficiency ($\eta_{\text{combined,max}}$) of the combined diesel engine + steam turbine system (in percent, $\%$):

💡 Hint: Formulas & Approach

Step 1: For the original engine with $\eta_1 = 0.35$, express the work $W_{\text{DE}}$ and heat rejection $Q_{L1}$ in terms of fuel heat input $Q_{H1}$.

Step 2: The rejected heat $Q_{L1}$ enters the secondary steam generator at $T_{H2} = 250^\circ\text{F}$. Convert temperatures to absolute Rankine: $T(\text{R}) = T(^\circ\text{F}) + 460$.

Step 3: Calculate the Carnot limit of the secondary heat engine: $\eta_{\text{th,c2}} = 1 - \frac{T_{L2}}{T_{H2}}$.

Step 4: Find total work $W_{\text{net,max}} = W_{\text{DE}} + \eta_{\text{th,c2}} Q_{L1}$, then divide by $Q_{H1}$.

📝 Show Full Worked Solution

1. Original Diesel Engine Analysis:

$$\eta_1 = \frac{W_{\text{DE}}}{Q_{H1}} = 0.35 \implies W_{\text{DE}} = 0.35 Q_{H1}$$ $$Q_{L1} = Q_{H1} - W_{\text{DE}} = (1 - 0.35) Q_{H1} = 0.65 Q_{H1}$$

2. Secondary Steam Engine (Add-On) Carnot Limit:

$$T_{H2} = 250^\circ\text{F} + 460 = 710\text{ R}$$ $$T_{L2} = 68^\circ\text{F} + 460 = 528\text{ R}$$ $$\eta_{\text{th,c2}} = 1 - \frac{T_{L2}}{T_{H2}} = 1 - \frac{528}{710} = 1 - 0.74366 = 0.25634 \quad (25.63\%)$$

3. Maximum Secondary Engine Work:

$$W_{\text{SE,max}} = \eta_{\text{th,c2}} \cdot Q_{L1} = 0.25634 \times (0.65 Q_{H1}) = 0.16662 Q_{H1}$$

4. Combined System Efficiency:

$$W_{\text{net,max}} = W_{\text{DE}} + W_{\text{SE,max}} = 0.35 Q_{H1} + 0.16662 Q_{H1} = 0.51662 Q_{H1}$$ $$\eta_{\text{combined,max}} = \frac{W_{\text{net,max}}}{Q_{H1}} = 0.51662 = \mathbf{51.6\%} \; (\text{or } 51.7\%)$$

Answer: $\eta_{\text{combined,max}} = 51.6\%$ (or $51.7\%$)

Part b: Claim Evaluation & Thermodynamic Law 4 Points

Does the manufacturer's claim of 55% overall efficiency violate the laws of thermodynamics?

A. Yes, it violates the Second Law of Thermodynamics because the claimed efficiency of 55% exceeds the Carnot theoretical maximum of 51.6%. B. Yes, it violates the First Law of Thermodynamics because energy is not conserved when combining cycles. C. No, it does not violate any thermodynamic laws because 55% is strictly less than 100%. D. No, because steam turbines operate with water which has an infinite thermal reservoir capacity.

📝 Show Full Worked Solution

Thermodynamic Conclusion:

Under the Carnot principles (Second Law), no heat engine can be more efficient than a completely reversible heat engine operating between the same two thermal reservoirs.

Even if the secondary steam cycle were 100% reversible (Carnot cycle), the maximum attainable overall efficiency is only 51.6%.

Because the claimed overall efficiency is 55% > 51.6%, the claim represents an impossible machine and violates the Second Law of Thermodynamics.

Correct Choice: A (Violates the 2nd Law)

Problem 2: Geothermal Heat Pump System with R-134a

Energy balance, property table lookups (Table A-4 & A-11), and Carnot limits

15 Points Total

A heat pump operating with R-134a as the working fluid is used to keep a space at $25^\circ\text{C}$ by absorbing heat from geothermal water that enters the evaporator at $60^\circ\text{C}$ at a flow rate of $0.065\text{ kg/s}$ and leaves at $40^\circ\text{C}$. Refrigerant enters the evaporator at $12^\circ\text{C}$ with quality $x = 15\%$ ($0.15$) and leaves as a saturated vapor at the same pressure. The compressor consumes $1.6\text{ kW}$ of power.

Figure 2.1: Schematic diagram of the vapor-compression heat pump coupled to geothermal water at the evaporator.

Part a: Rate of Heat Absorption in Evaporator ($\dot{Q}_L$) 3 Points

Determine the rate of heat absorption from the geothermal water in the evaporator (in $\text{kW}$):

💡 Hint: Formulas & Approach

From Table A-4 (Saturated water temperature table), look up saturated liquid enthalpy $h_f$ at $60^\circ\text{C}$ and $40^\circ\text{C}$ (or use $c_p = 4.18\text{ kJ/kg}\cdot\text{K}$ for liquid water from Table A-3).

Apply the steady-flow thermal energy balance on the geothermal water: $\dot{Q}_L = \dot{m}_w (h_{w,\text{in}} - h_{w,\text{out}})$.

📝 Show Full Worked Solution

Property Lookups from Table A-4:

At $T_{w,\text{in}} = 60^\circ\text{C}$: $h_{w,\text{in}} \approx h_f(60^\circ\text{C}) = 251.18\text{ kJ/kg}$
At $T_{w,\text{out}} = 40^\circ\text{C}$: $h_{w,\text{out}} \approx h_f(40^\circ\text{C}) = 167.53\text{ kJ/kg}$

Energy Balance:

$$\dot{Q}_L = \dot{m}_w \left(h_{w,\text{in}} - h_{w,\text{out}}\right) = 0.065\text{ kg/s} \times (251.18 - 167.53)\text{ kJ/kg}$$ $$\dot{Q}_L = 0.065 \times 83.65 = \mathbf{5.44\text{ kW}} \quad (5.437\text{ kW})$$

Answer: $\dot{Q}_L = 5.44\text{ kW}$

Part b: Mass Flow Rate of R-134a ($\dot{m}_R$) 3 Points

Determine the mass flow rate of R-134a through the heat pump system (in $\text{kg/s}$):

kg/s

💡 Hint: Formulas & Approach

Look up saturated R-134a properties from Table A-11 at $T = 12^\circ\text{C}$:

Find $h_f$, $h_{fg}$, and $h_g$. Then find inlet enthalpy $h_1 = h_f + x_1 h_{fg}$ and exit enthalpy $h_2 = h_g$.

Equate heat absorbed by refrigerant to geothermal heat transfer: $\dot{Q}_L = \dot{m}_R (h_2 - h_1)$.

📝 Show Full Worked Solution

Property Lookups from Table A-11 (R-134a at $12^\circ\text{C}$):

$h_f = 68.18\text{ kJ/kg}$
$h_{fg} = 189.09\text{ kJ/kg}$
$h_g = 257.27\text{ kJ/kg}$

Enthalpy at Evaporator Inlet ($x_1 = 0.15$):

$$h_1 = h_f + x_1 h_{fg} = 68.18 + 0.15(189.09) = 68.18 + 28.36 = 96.54\text{ kJ/kg}$$

Enthalpy at Evaporator Exit (Saturated vapor):

$$h_2 = h_g = 257.27\text{ kJ/kg}$$ $$\Delta h_{\text{evap}} = h_2 - h_1 = 257.27 - 96.54 = 160.73\text{ kJ/kg} = (1 - x_1)h_{fg}$$

Mass Flow Rate:

$$\dot{m}_R = \frac{\dot{Q}_L}{\Delta h_{\text{evap}}} = \frac{5.437\text{ kW}}{160.73\text{ kJ/kg}} = \mathbf{0.0338\text{ kg/s}}$$

Answer: $\dot{m}_R = 0.0338\text{ kg/s}$

Part c: Rate of Heat Supply to Heated Space ($\dot{Q}_H$) 3 Points

Determine the rate of heat supplied to the heated room by the condenser (in $\text{kW}$):

💡 Hint: Formulas & Approach

From the First Law energy balance on the overall heat pump cycle: $\dot{Q}_H = \dot{Q}_L + \dot{W}_{\text{in}}$.

📝 Show Full Worked Solution

From the First Law of Thermodynamics for a cyclical heat pump:

$$\dot{Q}_H = \dot{Q}_L + \dot{W}_{\text{in}} = 5.437\text{ kW} + 1.60\text{ kW} = \mathbf{7.04\text{ kW}}$$

Answer: $\dot{Q}_H = 7.04\text{ kW}$

Part d: Actual Coefficient of Performance ($\text{COP}_{\text{HP}}$) 2 Points

Calculate the actual coefficient of performance for this heat pump:

(dimensionless)

💡 Hint: Formulas & Approach

From equation sheet: $\text{COP}_{\text{HP}} = \frac{\text{Desired Output}}{\text{Required Input}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}}$.

📝 Show Full Worked Solution

$$\text{COP}_{\text{HP}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}} = \frac{7.037\text{ kW}}{1.60\text{ kW}} = \mathbf{4.40}$$

Answer: $\text{COP}_{\text{HP}} = 4.40$

Part e: Ideal Coefficient of Performance 2 Points

Determine the ideal (Carnot) coefficient of performance for these two reservoir temperatures (using $60^\circ\text{C}$ and $25^\circ\text{C}$):

(dimensionless)

💡 Hint: Formulas & Approach

From the equation sheet: $\text{COP}_{\text{HP,c}} = \frac{1}{1 - T_L/T_H} = \frac{T_H}{T_H - T_L}$.

Use $T_H = 60^\circ\text{C} + 273.15 = 333.15\text{ K}$ and $T_L = 25^\circ\text{C} + 273.15 = 298.15\text{ K}$.

📝 Show Full Worked Solution

Converting temperatures to Kelvin:

$$T_H = 60^\circ\text{C} + 273.15 = 333.15\text{ K} \quad (\text{or } 333\text{ K})$$ $$T_L = 25^\circ\text{C} + 273.15 = 298.15\text{ K} \quad (\text{or } 298\text{ K})$$ $$\text{COP}_{\text{HP,ideal}} = \frac{T_H}{T_H - T_L} = \frac{333.15}{60 - 25} = \frac{333.15}{35} = \mathbf{9.51}$$

Answer: $\text{COP}_{\text{HP,ideal}} = 9.51$

Part f: Minimum Power Input to Compressor 2 Points

Determine the minimum power input to the compressor (in $\text{kW}$) for the same heating supply rate ($\dot{Q}_H = 7.04\text{ kW}$):

💡 Hint: Formulas & Approach

Minimum power occurs when the heat pump operates at ideal Carnot performance: $\dot{W}_{\text{in,min}} = \frac{\dot{Q}_H}{\text{COP}_{\text{HP,ideal}}}$.

📝 Show Full Worked Solution

$$\dot{W}_{\text{in,min}} = \frac{\dot{Q}_H}{\text{COP}_{\text{HP,ideal}}} = \frac{7.037\text{ kW}}{9.51} = \mathbf{0.74\text{ kW}}$$

Answer: $\dot{W}_{\text{in,min}} = 0.74\text{ kW}$

Problem 3: Carnot Heat Engine Waste Heat Rejection

Carnot cycle power conversions (hp) and absolute Rankine/Fahrenheit source temperature

10 Points Total

A heat engine is operating on a Carnot cycle and has a thermal efficiency of $75\%$. The waste heat from this engine is rejected to a nearby lake at $60^\circ\text{F}$ at a rate of $800\text{ Btu/min}$.

Figure 3.1: Energy flow diagram of the Carnot heat engine rejecting heat to a sink at $60^\circ ext{F}$.

Part a.1: Net Power Output in Btu/min 3 Points

Determine the net power output of the Carnot heat engine in $\text{Btu/min}$:

Btu/min

💡 Hint: Formulas & Approach

From the equation sheet: $\eta_{\text{th}} = 1 - \frac{\dot{Q}_L}{\dot{Q}_H}$. Solve for $\dot{Q}_H$: $\dot{Q}_H = \frac{\dot{Q}_L}{1 - \eta_{\text{th}}}$.

Then compute net work: $\dot{W}_{\text{net,out}} = \dot{Q}_H - \dot{Q}_L$.

📝 Show Full Worked Solution

Given $\eta_{\text{th}} = 0.75$ and $\dot{Q}_L = 800\text{ Btu/min}$:

$$\dot{Q}_H = \frac{\dot{Q}_L}{1 - \eta_{\text{th}}} = \frac{800\text{ Btu/min}}{1 - 0.75} = \frac{800}{0.25} = 3,200\text{ Btu/min}$$ $$\dot{W}_{\text{net,out}} = \dot{Q}_H - \dot{Q}_L = 3,200 - 800 = \mathbf{2,400\text{ Btu/min}}$$

Answer: $\dot{W}_{\text{net,out}} = 2,400\text{ Btu/min}$

Part a.2: Net Power Output in Horsepower (hp) 2 Points

Convert the net power output into horsepower ($\text{hp}$):

💡 Hint: Formulas & Approach

From the conversion table on page 1 of Property Tables: $1\text{ hp} = 0.7068\text{ Btu/s} = 42.408\text{ Btu/min}$ (or $2,544.5\text{ Btu/h}$).

📝 Show Full Worked Solution

Using unit factor $1\text{ hp} = 0.7068\text{ Btu/s} \times 60\text{ s/min} = 42.408\text{ Btu/min}$:

$$\dot{W}_{\text{net,out}} = \frac{2,400\text{ Btu/min}}{42.408\text{ Btu/min}\cdot\text{hp}^{-1}} = \mathbf{56.6\text{ hp}}$$

Answer: $\dot{W}_{\text{net,out}} = 56.6\text{ hp}$

Part b.1: Source Temperature in Rankine (R) 3 Points

Determine the absolute temperature of the high-temperature heat source ($T_H$) in Rankine ($\text{R}$):

💡 Hint: Formulas & Approach

From the equation sheet: $T_R = T_{^\circ\text{F}} + 460$. Calculate $T_L$ in Rankine.

Apply Carnot efficiency: $\eta_{\text{th,c}} = 1 - \frac{T_L}{T_H} \implies T_H = \frac{T_L}{1 - \eta_{\text{th,c}}}$.

📝 Show Full Worked Solution

$$T_L = 60^\circ\text{F} + 460 = 520\text{ R}$$ $$\eta_{\text{th,c}} = 1 - \frac{T_L}{T_H} \implies 0.75 = 1 - \frac{520\text{ R}}{T_H}$$ $$\frac{520\text{ R}}{T_H} = 0.25 \implies T_H = \frac{520}{0.25} = \mathbf{2,080\text{ R}}$$

Answer: $T_H = 2,080\text{ R}$

Part b.2: Source Temperature in Fahrenheit (°F) 2 Points

Convert the source temperature to Fahrenheit ($^\circ\text{F}$):

°F

💡 Hint: Formulas & Approach

$T(^\circ\text{F}) = T(\text{R}) - 460$.

📝 Show Full Worked Solution

$$T_H(^\circ\text{F}) = T_H(\text{R}) - 460 = 2,080 - 460 = \mathbf{1,620^\circ\text{F}}$$

Answer: $T_H = 1,620^\circ\text{F}$

Problem 4: Inventor Refrigerator Claim Evaluation

Carnot maximum coefficient of performance and engineering feasibility

5 Points Total

An inventor claims to have developed a refrigeration system that removes heat from a closed cold space maintained at $-12^\circ\text{C}$ and rejects it to the surrounding environment at $25^\circ\text{C}$ while maintaining a coefficient of performance of $6.5$. Is this claim reasonable? Why?

Part a: Maximum Theoretical Carnot COP 3 Points

Calculate the maximum possible theoretical coefficient of performance ($\text{COP}_{\text{R,Carnot}}$) operating between these two temperatures:

(dimensionless)

💡 Hint: Formulas & Approach

From the equation sheet: $\text{COP}_{\text{R,c}} = \frac{1}{T_H/T_L - 1} = \frac{T_L}{T_H - T_L}$.

Convert both temperatures to Kelvin: $T_L = -12 + 273.15\text{ K}$ and $T_H = 25 + 273.15\text{ K}$.

📝 Show Full Worked Solution

Convert to absolute temperature:

$$T_L = -12^\circ\text{C} + 273.15 = 261.15\text{ K} \quad (\text{or } 261\text{ K})$$ $$T_H = 25^\circ\text{C} + 273.15 = 298.15\text{ K} \quad (\text{or } 298\text{ K})$$ $$\Delta T = T_H - T_L = 25 - (-12) = 37\text{ K}$$ $$\text{COP}_{\text{R,Carnot}} = \frac{T_L}{T_H - T_L} = \frac{261.15\text{ K}}{37\text{ K}} = \mathbf{7.05}$$

Answer: $\text{COP}_{\text{R,Carnot}} = 7.05$

Part b: Claim Evaluation & Thermodynamic Assessment 2 Points

Select the best thermodynamic evaluation of the inventor's claim:

A. The claim is theoretically possible because $\text{COP}_{\text{claimed}} = 6.5 \le \text{COP}_{\text{Carnot}} = 7.05$ (does not violate the Second Law), but achieving $92.2\%$ of the Carnot limit is practically unrealistic for a real mechanical refrigeration system due to friction, motor losses, and heat transfer across finite $\Delta T$. B. The claim strictly violates the Second Law of Thermodynamics because any COP greater than 1.0 is impossible. C. The claim violates the First Law of Thermodynamics because energy is destroyed in the evaporator. D. The claim is standard and unremarkable because ordinary domestic refrigerators operate at a COP of 12 or higher.

📝 Show Full Worked Solution

Evaluation:

1. Second Law Feasibility: Because $\text{COP}_{\text{claimed}} = 6.5 < \text{COP}_{\text{Carnot}} = 7.05$, the machine does not violate the Second Law.

2. Practical Engineering Reality: The second-law efficiency is $\eta_{II} = \frac{6.5}{7.05} = 92.2\%$. Real-world vapor compression cycles operate at only $30\%$ to $60\%$ of the Carnot COP (typical COP would be $2.1$ to $4.2$) due to throttling valve irreversibility, compressor friction, and necessary temperature gradients across coils. Thus, claiming 6.5 is practically unrealistic.

Correct Choice: A

Problem 5: Split Air Conditioner System Analysis

Energy balances, unit conversion (Btu/h to kW), Heat Pump vs AC COP, and Carnot benchmark

10 Points Total

A $18,000\text{ Btu/h}$ split air conditioner running at full load is used to keep a room at $22^\circ\text{C}$ in an ambient environment at $33^\circ\text{C}$. The electrical power input to the compressor is $3.4\text{ kW}$.

Part a: Rate of Heat Rejected in Condenser ($\dot{Q}_H$) 2 Points

Calculate the rate of heat rejected in the condenser to the outdoors (in $\text{kW}$):

💡 Hint: Formulas & Approach

Convert cooling capacity from $\text{Btu/h}$ to $\text{kW}$ using the Property Tables conversion factor: $1\text{ kW} = 3,412.14\text{ Btu/h}$.

Apply the First Law for a refrigeration cycle: $\dot{Q}_H = \dot{Q}_L + \dot{W}_{\text{in}}$.

📝 Show Full Worked Solution

1. Cooling Rate Conversion:

$$\dot{Q}_L = \frac{18,000\text{ Btu/h}}{3,412.14\text{ Btu/h}\cdot\text{kW}^{-1}} = 5.275\text{ kW}$$

2. Condenser Heat Rejection:

$$\dot{Q}_H = \dot{Q}_L + \dot{W}_{\text{in}} = 5.275\text{ kW} + 3.40\text{ kW} = \mathbf{8.68\text{ kW}} \quad (8.675\text{ kW})$$

Answer: $\dot{Q}_H = 8.68\text{ kW}$

Part b: COP of the Air Conditioner ($\text{COP}_{\text{R}}$) 1.5 Points

Determine the coefficient of performance of the air conditioner (cooling mode):

(dimensionless)

💡 Hint: Formulas & Approach

From equation sheet: $\text{COP}_{\text{R}} = \frac{\text{Desired Cooling}}{\text{Required Work Input}} = \frac{\dot{Q}_L}{\dot{W}_{\text{in}}}$.

📝 Show Full Worked Solution

$$\text{COP}_{\text{R}} = \frac{\dot{Q}_L}{\dot{W}_{\text{in}}} = \frac{5.275\text{ kW}}{3.40\text{ kW}} = \mathbf{1.55}$$

Answer: $\text{COP}_{\text{R}} = 1.55$

Part c: COP If Operating as a Heat Pump ($\text{COP}_{\text{HP}}$) 1.5 Points

Calculate the coefficient of performance if the same unit operates as a heat pump:

(dimensionless)

💡 Hint: Formulas & Approach

Use the governing relationship between heat pump and refrigerator COPs: $\text{COP}_{\text{HP}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}} = \text{COP}_{\text{R}} + 1$.

📝 Show Full Worked Solution

$$\text{COP}_{\text{HP}} = \frac{\dot{Q}_H}{\dot{W}_{\text{in}}} = \frac{8.675\text{ kW}}{3.40\text{ kW}} = \mathbf{2.55}$$ $$\text{Or: } \text{COP}_{\text{HP}} = \text{COP}_{\text{R}} + 1 = 1.55 + 1 = 2.55$$

Answer: $\text{COP}_{\text{HP}} = 2.55$

Part d: Carnot COP for Given Temperatures 1.5 Points

Calculate the theoretical Carnot COP for a refrigerator operating between $22^\circ\text{C}$ and $33^\circ\text{C}$:

(dimensionless)

💡 Hint: Formulas & Approach

From the equation sheet: $\text{COP}_{\text{R,Carnot}} = \frac{T_L}{T_H - T_L}$.

Convert to absolute Kelvin: $T_L = 22 + 273.15\text{ K}$ and $T_H = 33 + 273.15\text{ K}$.

📝 Show Full Worked Solution

$$T_L = 22^\circ\text{C} + 273.15 = 295.15\text{ K}$$ $$T_H = 33^\circ\text{C} + 273.15 = 306.15\text{ K}$$ $$\Delta T = T_H - T_L = 33 - 22 = 11\text{ K}$$ $$\text{COP}_{\text{R,Carnot}} = \frac{T_L}{T_H - T_L} = \frac{295.15\text{ K}}{11\text{ K}} = \mathbf{26.8}$$

Answer: $\text{COP}_{\text{R,Carnot}} = 26.8$

Part e: Rate of Cooling for Carnot Refrigerator 1.5 Points

Determine the rate of cooling (in $\text{kW}$) if the unit operated as a Carnot refrigerator for the same $3.4\text{ kW}$ power input:

💡 Hint: Formulas & Approach

Multiply the Carnot COP by the supplied power: $\dot{Q}_{L,\text{Carnot}} = \text{COP}_{\text{R,Carnot}} \times \dot{W}_{\text{in}}$.

📝 Show Full Worked Solution

$$\dot{Q}_{L,\text{Carnot}} = \text{COP}_{\text{R,Carnot}} \times \dot{W}_{\text{in}} = 26.832 \times 3.40\text{ kW}} = \mathbf{91.2\text{ kW}}$$

Answer: $\dot{Q}_{L,\text{Carnot}} = 91.2\text{ kW}$

Part f: Physical Explanation for Performance Discrepancy 2 Points

Why is the cooling rate so vastly different between the ideal Carnot cycle ($91.2\text{ kW}$) and the actual machine ($5.28\text{ kW}$)?

A. Real refrigeration systems suffer from fluid friction, compressor isentropic inefficiencies, irreversible throttling through the expansion valve, and require significant temperature differences ($\Delta T$) across the evaporator and condenser coils to transfer heat, resulting in a much wider internal temperature lift than $11\text{ K}$. B. The Carnot cycle assumes that electricity contains twice as much energy as thermal energy. C. Real air conditioners violate the First Law of Thermodynamics by losing mass to the atmosphere. D. Split air conditioners use alternating current (AC) rather than direct current (DC).

📝 Show Full Worked Solution

Physical Engineering Reasons:

Finite $\Delta T$ for Heat Transfer: To transfer heat from a $22^\circ\text{C}$ room into the evaporator, the refrigerant must boil at a much colder temperature (e.g. $5^\circ\text{C}$). Similarly, to reject heat to $33^\circ\text{C}$ outdoor air, the condenser must condense at e.g. $50^\circ\text{C}$. The actual cycle operates across $50 - 5 = 45\text{ K}$ instead of the external $11\text{ K}$ reservoir gap!
Throttling Valve Irreversibility: Real systems use an isenthalpic expansion valve (Joule-Thomson throttling) which is highly irreversible, unlike an isentropic turbine.
Compressor Losses: Real compressors have isentropic efficiencies around $70-80\%$, electric motor losses, and mechanical friction.

Correct Choice: A