Preclass 3.2 & Class Activity 3.2 Practice Suite
Interactive mastery suite covering heat engines, refrigerators, heat pumps, power plant logistics, food sensible refrigeration enthalpy (Table A-7), and HVAC life-cycle economic optimization.
📋 Key Concept & Formula Matrix (Copy into Notes)
Governing relations strictly from the ME 322 Equation Sheet and Property Tables (6th Edition).
1. Thermal Cycles & Governing Performance Relations
| Thermal Device | Energy Balance (1st Law) | Performance Metric | Equation Sheet Formula | Reversible (Carnot) Limit |
|---|---|---|---|---|
| Heat Engine (HE) | $\dot{W}_{net} = \dot{Q}_H - \dot{Q}_L$ | Thermal Efficiency $\eta_{th}$ | $$\eta_{th} = \frac{\dot{W}_{net}}{\dot{Q}_H} = 1 - \frac{\dot{Q}_L}{\dot{Q}_H}$$ | $$\eta_{th,C} = 1 - \frac{T_L}{T_H}$$ |
| Refrigerator (R) | $\dot{Q}_H = \dot{Q}_L + \dot{W}_{net,in}$ | Cooling COP $\text{COP}_R$ | $$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{net,in}} = \frac{\dot{Q}_L}{\dot{Q}_H - \dot{Q}_L}$$ | $$\text{COP}_{R,C} = \frac{T_L}{T_H - T_L}$$ |
| Heat Pump (HP) | $\dot{Q}_H = \dot{Q}_L + \dot{W}_{net,in}$ | Heating COP $\text{COP}_{HP}$ | $$\text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{net,in}} = \text{COP}_R + 1$$ | $$\text{COP}_{HP,C} = \frac{T_H}{T_H - T_L}$$ |
2. Specialized Engineering Applications & Unit Conversions
| Application Topic | Governing Equation | Property Lookup / Key Constant | Units & Equation Sheet Conversion |
|---|---|---|---|
| Power Unit Conversion | $$\dot{W}_{hp} = \frac{\dot{W}_{Btu/hr}}{2544.5}$$ | ME 322 Equation Sheet Conversion Factors | $$1\text{ hp} = 2544.5\text{ Btu/hr} = 0.7457\text{ kW}$$ |
| Coal Furnace Combustion Logistics | $$\dot{Q}_H = \dot{m}_{coal} \cdot HV$$ $$\dot{m}_{air} = AF \cdot \dot{m}_{coal}$$ |
Heating Value: $HV$ in $\text{kJ/kg}$ Gravimetric Air-Fuel Ratio: $AF$ in $\text{kg air/kg coal}$ |
$$1\text{ MW} = 1000\text{ kW} = 1000\text{ kJ/s}$$ $$1\text{ day} = 86,400\text{ s}$$ |
| Food Sensible Refrigeration | $$\Delta \dot{H} = \dot{m} \cdot c_p \cdot \Delta T$$ $$\dot{Q}_L = \Delta \dot{H}$$ |
Table A-7 (Food Properties): Bananas above freezing ($-0.8^\circ\text{C}$): $$c_p = 3.35\text{ kJ/kg}\cdot^\circ\text{C}$$ |
$$\dot{m}\text{ [kg/s]} = \frac{\text{kg/hr}}{3600}$$ $$\text{kW} = \text{kJ/s}$$ |
| HVAC Life-Cycle Economics | $$W_{in} = \frac{Q_{cooling}}{\text{COP}}$$ $$\text{Annual Cost} = W_{in} \times \text{Rate}$$ $$\text{Total}_{10\text{yr}} = \text{Capital} + 10(\text{Annual})$$ |
Electricity Rate (e.g. $\$0.10/\text{kWh}$) Capital Equipment Cost |
Payback Period:$$\text{Payback} = \frac{\Delta \text{Capital}}{\Delta \text{Annual Savings}}$$ |
Part A: Preclass 3.2 Quiz Problems (Questions 1 – 6)
Problems 2 through 7 from the official Canvas Preclass 3.2 (ignoring the video study reflection prompt).
If $15\text{ Btu/min}$ of heat are transferred to a heat engine, and $5\text{ Btu/min}$ are rejected to the environment, what is the net power output of the engine?
💡 Hint: Heat Engine First Law Balance
From the First Law energy balance for a closed cycle heat engine:
$$\dot{W}_{net,out} = \dot{Q}_H - \dot{Q}_L$$
where $\dot{Q}_H$ is the heat input from the high-temperature source and $\dot{Q}_L$ is the heat rejected to the low-temperature sink.
📝 Show Full Worked Solution
Step 1: Identify Given Values:
$$\dot{Q}_H = 15\text{ Btu/min}$$ $$\dot{Q}_L = 5\text{ Btu/min}$$Step 2: Apply Energy Balance:
$$\dot{W}_{net,out} = \dot{Q}_H - \dot{Q}_L = 15\text{ Btu/min} - 5\text{ Btu/min} = 10\text{ Btu/min}$$What is the thermal efficiency of the heat engine in the previous problem?
💡 Hint: Thermal Efficiency Definition
Thermal efficiency is defined as the net work output divided by the total heat input:
$$\eta_{th} = \frac{\dot{W}_{net}}{\dot{Q}_H} = 1 - \frac{\dot{Q}_L}{\dot{Q}_H}$$
📝 Show Full Worked Solution
Step 1: Apply Thermal Efficiency Formula:
$$\eta_{th} = \frac{\dot{W}_{net}}{\dot{Q}_H} = \frac{10\text{ Btu/min}}{15\text{ Btu/min}} = \frac{2}{3} \approx 0.6667$$Rounding to two decimal places yields $0.66$ (or $66.7\%$).
A refrigerator that provides $5\text{ kW}$ of cooling is powered by a $2\text{ kW}$ compressor. What is the heat rejection rate from the refrigerator?
💡 Hint: Refrigerator Energy Accounting
For a refrigeration cycle, the heat rejected to the warm surroundings ($\dot{Q}_H$) must balance all energy inputs into the refrigerant:
$$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}$$
📝 Show Full Worked Solution
Step 1: Identify Energy Streams:
$$\dot{Q}_L = 5\text{ kW}\text{ (cooling load extracted from refrigerated space)}$$ $$\dot{W}_{in} = 2\text{ kW}\text{ (electric power input to compressor)}$$Step 2: Solve for Total Heat Rejected:
$$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in} = 5\text{ kW} + 2\text{ kW} = 7\text{ kW}$$What is the Coefficient of Performance ($\text{COP}_R$) of the refrigerator in the previous problem?
💡 Hint: Definition of COP for Refrigeration
The COP of a refrigerator is defined as the desired cooling effect divided by the required work input:
$$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{net,in}}$$
📝 Show Full Worked Solution
Step 1: Compute COP:
$$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}} = \frac{5\text{ kW}}{2\text{ kW}} = 2.5$$A heat engine has an efficiency of $30\%$ and a rated power output of $5\text{ MW}$. The rate heat is being rejected from the engine is closest to?
💡 Hint: Two-Step Heat Engine Accounting
1. Solve for total heat input: $\dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}}$
2. Solve for heat rejected: $\dot{Q}_L = \dot{Q}_H - \dot{W}_{net}$
📝 Show Full Worked Solution
Step 1: Calculate Heat Input Rate $\dot{Q}_H$:
$$\dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}} = \frac{5\text{ MW}}{0.30} = 16.667\text{ MW}$$Step 2: Calculate Heat Rejection Rate $\dot{Q}_L$:
$$\dot{Q}_L = \dot{Q}_H - \dot{W}_{net} = 16.667\text{ MW} - 5.0\text{ MW} = 11.667\text{ MW}$$The closest listed option is $12\text{ MW}$.
Heat pumps are often designed as reversible systems: they can provide heating in cold weather and cooling (air conditioning) in warm weather. If the Coefficient of Performance for a $10,000\text{ Btu/hr}$ heat pump when acting as a heat pump is $4.0$, what is its COP when acting as an air conditioner?
💡 Hint: Relationship Between COP_HP and COP_R
From the ME 322 Equation Sheet:
$$\text{COP}_{HP} = \text{COP}_R + 1$$
Note: An air conditioner is thermodynamically identical to a refrigerator (both cool the low-temperature space $\dot{Q}_L$). Thus $\text{COP}_{AC} = \text{COP}_R$.
📝 Show Full Worked Solution
Step 1: Apply the HP-Refrigeration Identity:
$$\text{COP}_{HP} = \frac{\dot{Q}_H}{\dot{W}_{in}} = \frac{\dot{Q}_L + \dot{W}_{in}}{\dot{W}_{in}} = \frac{\dot{Q}_L}{\dot{W}_{in}} + 1 = \text{COP}_R + 1$$Step 2: Solve for Cooling COP:
$$\text{COP}_{AC} = \text{COP}_R = \text{COP}_{HP} - 1 = 4.0 - 1.0 = 3.0$$Part B: Class Activity 3.2 Short Answer Problems (Problems 7 – 10)
Multi-step engineering calculations directly from the in-class problem set with interactive numerical validation and full step-by-step solutions.
A heat engine has a heat input of $30,000\text{ Btu/hr}$ and a thermal efficiency of $40\%$. How much net power will this engine produce?
💡 Hint: Efficiency & Horsepower Conversion Factor
1. Power in Btu/hr: $\dot{W}_{net} = \eta_{th} \times \dot{Q}_H$
2. Horsepower conversion from ME 322 Equation Sheet:
$$1\text{ hp} = 2544.5\text{ Btu/hr}$$
$$\dot{W}_{hp} = \frac{\dot{W}_{Btu/hr}}{2544.5\text{ Btu/hp}\cdot\text{hr}}$$
📝 Show Full Worked Solution
Part (a): Calculate Power Output in Btu/hr:
$$\dot{W}_{net} = \eta_{th} \cdot \dot{Q}_H = 0.40 \times 30,000\text{ Btu/hr} = 12,000\text{ Btu/hr}$$Part (b): Convert to Horsepower:
$$\dot{W}_{net} = \frac{12,000\text{ Btu/hr}}{2544.5\text{ Btu/hp}\cdot\text{hr}} = 4.716\text{ hp} \approx 4.72\text{ hp}$$A coal-burning steam power plant produces a net power of $\dot{W}_{net} = 300\text{ MW}$ with an overall thermal efficiency of $\eta_{th} = 32\%$. The heating value ($HV$) of the coal is $28,000\text{ kJ/kg}$, a standard train car carries approximately $105,000\text{ kg}$ of coal, and the gravimetric air-fuel ratio in the furnace is $12\text{ kg air / kg coal}$.
💡 Hint: Step-by-Step Logistics & Rate Accounting
1. Heating Load: $\dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}}$ (with $\text{MW} = 1000\text{ kJ/s}$)
2. Daily Energy & Coal Mass: $Q_{H,24hr} = \dot{Q}_H \times (24 \times 3600\text{ s})$. Then $m_{coal} = \frac{Q_{H,24hr}}{HV}$.
3. Train Cars: $\text{Cars/day} = \frac{m_{coal}}{105,000\text{ kg/car}}$.
4. Air Mass Flow: Coal flow rate is $\dot{m}_{coal} = \frac{\dot{Q}_H}{HV}$. Then $\dot{m}_{air} = AF \times \dot{m}_{coal}$.
📝 Show Full Worked Solution
Part (a): Thermal Heating Load $\dot{Q}_H$:
$$\dot{Q}_H = \frac{\dot{W}_{net}}{\eta_{th}} = \frac{300\text{ MW}}{0.32} = 937.5\text{ MW} = 937,500\text{ kJ/s}$$Part (b): Mass of Coal Consumed in 24 Hours:
$$\Delta t = 24\text{ hr} \times 3600\text{ s/hr} = 86,400\text{ s}$$ $$Q_H = \dot{Q}_H \cdot \Delta t = 937,500\text{ kJ/s} \times 86,400\text{ s} = 8.10 \times 10^{10}\text{ kJ}$$ $$m_{coal} = \frac{Q_H}{HV} = \frac{8.10 \times 10^{10}\text{ kJ}}{28,000\text{ kJ/kg}} = 2,892,857\text{ kg/day} \approx 2.89 \times 10^6\text{ kg/day}$$Part (c): Daily Train Cars Required:
$$\text{Number of train cars} = \frac{2,892,857\text{ kg/day}}{105,000\text{ kg/car}} = 27.55 \approx 28\text{ cars/day}$$Part (d): Air Mass Flow Rate $\dot{m}_{air}$:
$$\dot{m}_{coal} = \frac{\dot{Q}_H}{HV} = \frac{937,500\text{ kJ/s}}{28,000\text{ kJ/kg}} = 33.482\text{ kg/s}$$ $$\dot{m}_{air} = AF \times \dot{m}_{coal} = 12\text{ kg air/kg coal} \times 33.482\text{ kg coal/s} = 401.785\text{ kg/s} \approx 402\text{ kg/s}$$Bananas are to be cooled from $24^\circ\text{C}$ to $13^\circ\text{C}$ at a mass rate of $215\text{ kg/hr}$ by a refrigeration system. The electric power input to the refrigerator compressor is $1.4\text{ kW}$.
💡 Hint: Property Table Lookup (Table A-7) & Sensible Cooling
1. In Table A-7, locate Bananas:
Freezing point is $-0.8^\circ\text{C}$. Since $24^\circ\text{C} \to 13^\circ\text{C}$ is above freezing, read:
$$c_p = 3.35\text{ kJ/kg}\cdot^\circ\text{C}$$
2. Convert mass flow rate to $\text{kg/s}$: $\dot{m} = \frac{215\text{ kg/hr}}{3600\text{ s/hr}}$.
3. $\Delta \dot{H} = \dot{m} \cdot c_p \cdot \Delta T$.
4. Cooling rate $\dot{Q}_L = \Delta \dot{H}$, $\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}}$, and $\dot{Q}_H = \dot{Q}_L + \dot{W}_{in}$.
📝 Show Full Worked Solution
Property Table Lookup:
From Property Tables (6th Edition) Table A-7: Properties of foods:
For bananas above freezing: $c_p = 3.35\text{ kJ/kg}\cdot\text{K}$.
Part (a): Enthalpy Change Rate $\Delta \dot{H}$:
$$\dot{m} = \frac{215\text{ kg/hr}}{3600\text{ s/hr}} = 0.05972\text{ kg/s}$$ $$\Delta T = 24^\circ\text{C} - 13^\circ\text{C} = 11^\circ\text{C} = 11\text{ K}$$ $$\Delta \dot{H} = \dot{m} c_p \Delta T = (0.05972\text{ kg/s})(3.35\text{ kJ/kg}\cdot\text{K})(11\text{ K}) = 2.2007\text{ kJ/s} = 2.20\text{ kW}$$Part (b): Cooling Rate $\dot{Q}_L$:
$$\dot{Q}_L = \Delta \dot{H} = 2.20\text{ kJ/s}$$Part (c): Refrigerator COP:
$$\text{COP}_R = \frac{\dot{Q}_L}{\dot{W}_{in}} = \frac{2.2007\text{ kW}}{1.4\text{ kW}} = 1.5719 \approx 1.57$$Part (d): Heat Rejected to Environment $\dot{Q}_H$:
$$\dot{Q}_H = \dot{Q}_L + \dot{W}_{in} = 2.2007\text{ kW} + 1.4\text{ kW} = 3.6007\text{ kW} \approx 3.60\text{ kW}$$Consider a commercial building whose annual air conditioning cooling load is estimated to be $120,000\text{ kWh/year}$ in a region where the unit cost of electricity is $\$0.10/\text{kWh}$. Two air conditioning units are being evaluated for purchase over a $10\text{-year}$ expected operating life:
- Air Conditioner A: Seasonal average $\text{COP} = 3.2$, Initial capital cost = $\$5,000$.
- Air Conditioner B: Seasonal average $\text{COP} = 5.0$, Initial capital cost = $\$8,000$.
💡 Hint: Life-Cycle Energy & Cost Equations
1. Annual Electricity Needed: $W_{electric} = \frac{Q_{load}}{\text{COP}}$ in $\text{kWh/year}$.
2. Annual Cost: $\text{Cost/yr} = W_{electric} \times \$0.10/\text{kWh}$.
3. 10-Year Total: $\text{Total} = \text{Initial Cost} + 10 \times (\text{Annual Cost})$.
4. Savings: $\text{Savings} = \text{Total}_A - \text{Total}_B$.
📝 Show Full Worked Solution
Part (a): Annual Operating Cost:
For Unit A ($\text{COP} = 3.2$):
$$W_{in,A} = \frac{120,000\text{ kWh/yr}}{3.2} = 37,500\text{ kWh/year}$$ $$\text{Operating Cost}_A = 37,500\text{ kWh/yr} \times \$0.10/\text{kWh} = \$3,750/\text{year}$$For Unit B ($\text{COP} = 5.0$):
$$W_{in,B} = \frac{120,000\text{ kWh/yr}}{5.0} = 24,000\text{ kWh/year}$$ $$\text{Operating Cost}_B = 24,000\text{ kWh/yr} \times \$0.10/\text{kWh} = \$2,400/\text{year}$$Part (b): 10-Year Life-Cycle Total Cost:
$$\text{Total Cost}_A = \$5,000 + 10(\text{\$3,750}) = \$5,000 + \$37,500 = \$42,500$$ $$\text{Total Cost}_B = \$8,000 + 10(\text{\$2,400}) = \$8,000 + \$24,000 = \$32,000$$Part (c): 10-Year Savings & Recommendation:
$$\text{Total 10-Year Savings} = \$42,500 - \$32,000 = \$10,500$$ $$\text{Annual Savings} = \$3,750 - \$2,400 = \$1,350/\text{year}$$ $$\text{Payback Period} = \frac{\Delta \text{Initial Cost}}{\text{Annual Savings}} = \frac{\$8,000 - \$5,000}{\$1,350/\text{year}} = \frac{\$3,000}{\$1,350} = 2.22\text{ years}$$Conclusion: Recommend purchasing Unit B. The $\$3,000$ higher initial investment is fully recouped in just $2.2$ years, yielding a net savings of $\$10,500$ over the equipment life.