ME 322 • Unit 2 Control Volumes
Homework 2B Interactive Review
Targeted practice across all 5 engineering devices from Homework 2B: Steam Nozzle, Argon/CO Turbine, R-134a Throttling Valve, Non-Insulated Heat Exchanger, and Water Pump. Includes separate formula hints, full step-by-step mathematical workthroughs, and the extracted diagram.
Problem 1 • Steam Nozzle
Nozzle
Steam at \(4\text{ MPa}\) and \(400^\circ\text{C}\) enters a nozzle with a velocity of \(60\text{ m/s}\). The steam leaves the nozzle at \(2\text{ MPa}\) and \(300^\circ\text{C}\). The inlet area of the nozzle is \(50\text{ cm}^2\). Heat is being lost at a rate of \(75\text{ kJ/s}\).
a. Determine the mass flow rate of the steam:
kg/s
💡 Hint: Formulas & Approach
Recall the mass flow rate relation for steady flow: \(\dot{m} = \frac{A_1 V_1}{v_1}\).
State 1 is given as \(P_1 = 4\text{ MPa}\) and \(T_1 = 400^\circ\text{C}\). Look up the specific volume \(v_1\) in the Superheated Water vapor tables (Table A-6). Don't forget to convert the area from \(\text{cm}^2\) to \(\text{m}^2\) (\(1\text{ m}^2 = 10^4\text{ cm}^2\)).
State 1 is given as \(P_1 = 4\text{ MPa}\) and \(T_1 = 400^\circ\text{C}\). Look up the specific volume \(v_1\) in the Superheated Water vapor tables (Table A-6). Don't forget to convert the area from \(\text{cm}^2\) to \(\text{m}^2\) (\(1\text{ m}^2 = 10^4\text{ cm}^2\)).
📝 Show Full Worked Solution
1. State 1 Property Lookup (Table A-6):
At \(P_1 = 4.0\text{ MPa}\) and \(T_1 = 400^\circ\text{C}\):
\(v_1 = 0.07343\text{ m}^3/\text{kg}\), \(\quad h_1 = 3214.5\text{ kJ/kg}\).
At \(P_1 = 4.0\text{ MPa}\) and \(T_1 = 400^\circ\text{C}\):
\(v_1 = 0.07343\text{ m}^3/\text{kg}\), \(\quad h_1 = 3214.5\text{ kJ/kg}\).
2. Area Conversion:
\(A_1 = 50\text{ cm}^2 \times \frac{1\text{ m}^2}{10{,}000\text{ cm}^2} = 0.0050\text{ m}^2\).
\(A_1 = 50\text{ cm}^2 \times \frac{1\text{ m}^2}{10{,}000\text{ cm}^2} = 0.0050\text{ m}^2\).
3. Mass Flow Rate Calculation:
\(\dot{m} = \frac{A_1 V_1}{v_1} = \frac{(0.0050\text{ m}^2)(60\text{ m/s})}{0.07343\text{ m}^3/\text{kg}} = 4.0855\text{ kg/s}\).
\(\dot{m} = \frac{A_1 V_1}{v_1} = \frac{(0.0050\text{ m}^2)(60\text{ m/s})}{0.07343\text{ m}^3/\text{kg}} = 4.0855\text{ kg/s}\).
Answer: \(\dot{m} = 4.086\text{ kg/s}\)
b. Determine the exit velocity of the steam:
m/s
💡 Hint: Formulas & Approach
Apply the First Law (steady-state energy balance) for a single-stream control volume:
\(\dot{Q} - \dot{W} = \dot{m}\left[(h_2 - h_1) + \frac{V_2^2 - V_1^2}{2000}\right]\).
Nozzles produce no work (\(\dot{W} = 0\)). Since heat is lost, \(\dot{Q} = -75\text{ kW}\). Lookup \(h_2\) at \(P_2 = 2\text{ MPa}\) and \(T_2 = 300^\circ\text{C}\).
\(\dot{Q} - \dot{W} = \dot{m}\left[(h_2 - h_1) + \frac{V_2^2 - V_1^2}{2000}\right]\).
Nozzles produce no work (\(\dot{W} = 0\)). Since heat is lost, \(\dot{Q} = -75\text{ kW}\). Lookup \(h_2\) at \(P_2 = 2\text{ MPa}\) and \(T_2 = 300^\circ\text{C}\).
📝 Show Full Worked Solution
1. State 2 Property Lookup (Table A-6):
At \(P_2 = 2.0\text{ MPa}\) and \(T_2 = 300^\circ\text{C}\):
\(v_2 = 0.12551\text{ m}^3/\text{kg}\), \(\quad h_2 = 3024.2\text{ kJ/kg}\).
At \(P_2 = 2.0\text{ MPa}\) and \(T_2 = 300^\circ\text{C}\):
\(v_2 = 0.12551\text{ m}^3/\text{kg}\), \(\quad h_2 = 3024.2\text{ kJ/kg}\).
2. Energy Balance Formulation:
\(\frac{\dot{Q}}{\dot{m}} = (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2000}\).
\(\frac{-75\text{ kJ/s}}{4.0855\text{ kg/s}} = (3024.2 - 3214.5)\text{ kJ/kg} + \frac{V_2^2 - (60)^2}{2000}\).
\(\frac{\dot{Q}}{\dot{m}} = (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2000}\).
\(\frac{-75\text{ kJ/s}}{4.0855\text{ kg/s}} = (3024.2 - 3214.5)\text{ kJ/kg} + \frac{V_2^2 - (60)^2}{2000}\).
3. Algebraic Solution for \(V_2\):
\(-18.358 = -190.3 + \frac{V_2^2 - 3600}{2000}\).
\(171.942 = \frac{V_2^2 - 3600}{2000} \implies V_2^2 - 3600 = 343{,}884\).
\(V_2 = \sqrt{347{,}484} \approx 589.5\text{ m/s}\).
\(-18.358 = -190.3 + \frac{V_2^2 - 3600}{2000}\).
\(171.942 = \frac{V_2^2 - 3600}{2000} \implies V_2^2 - 3600 = 343{,}884\).
\(V_2 = \sqrt{347{,}484} \approx 589.5\text{ m/s}\).
Answer: \(V_2 \approx 590\text{ m/s}\)
c. Determine the exit area of the nozzle:
cm²
💡 Hint: Formulas & Approach
By conservation of mass, \(\dot{m}_1 = \dot{m}_2 = \dot{m}\).
Rearrange the mass flow formula for exit area: \(A_2 = \frac{\dot{m} v_2}{V_2}\).
Then convert \(\text{m}^2\) to \(\text{cm}^2\) by multiplying by \(10^4\).
Rearrange the mass flow formula for exit area: \(A_2 = \frac{\dot{m} v_2}{V_2}\).
Then convert \(\text{m}^2\) to \(\text{cm}^2\) by multiplying by \(10^4\).
📝 Show Full Worked Solution
1. Calculate Exit Area in \(\text{m}^2\):
\(A_2 = \frac{\dot{m} v_2}{V_2} = \frac{(4.0855\text{ kg/s})(0.12551\text{ m}^3/\text{kg})}{589.5\text{ m/s}} = 0.00086985\text{ m}^2\).
\(A_2 = \frac{\dot{m} v_2}{V_2} = \frac{(4.0855\text{ kg/s})(0.12551\text{ m}^3/\text{kg})}{589.5\text{ m/s}} = 0.00086985\text{ m}^2\).
2. Convert to \(\text{cm}^2\):
\(A_2 = 0.00086985\text{ m}^2 \times 10{,}000\frac{\text{cm}^2}{\text{m}^2} = 8.6985\text{ cm}^2\).
\(A_2 = 0.00086985\text{ m}^2 \times 10{,}000\frac{\text{cm}^2}{\text{m}^2} = 8.6985\text{ cm}^2\).
Answer: \(A_2 = 8.699\text{ cm}^2\)
Problem 2 • Gas Turbine
Turbine
Argon enters an adiabatic turbine at \(1600\text{ kPa}\) and \(450^\circ\text{C}\) with a velocity of \(55\text{ m/s}\). The gas leaves the turbine at \(150\text{ kPa}\) with a velocity of \(150\text{ m/s}\). The inlet area is \(60\text{ cm}^2\). The power output is \(190\text{ kW}\).
a. For a turbine power output of 190 kW, what is the exit temperature of the argon?
°C
💡 Hint: Formulas & Approach
1. Treat Argon as a noble / ideal gas: \(R = 0.2081\text{ kJ/kg}\cdot\text{K}\), \(c_p = 0.5203\text{ kJ/kg}\cdot\text{K}\).
2. Compute inlet specific volume using ideal gas law: \(v_1 = \frac{R T_1}{P_1}\) with \(T_1\) in Kelvin.
3. Find mass flow rate: \(\dot{m} = \frac{A_1 V_1}{v_1}\).
4. Apply adiabatic energy balance: \(-\dot{W}_{\text{out}} = \dot{m}\left[c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2000}\right]\). Solve for \(T_2\).
2. Compute inlet specific volume using ideal gas law: \(v_1 = \frac{R T_1}{P_1}\) with \(T_1\) in Kelvin.
3. Find mass flow rate: \(\dot{m} = \frac{A_1 V_1}{v_1}\).
4. Apply adiabatic energy balance: \(-\dot{W}_{\text{out}} = \dot{m}\left[c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2000}\right]\). Solve for \(T_2\).
📝 Show Full Worked Solution
1. Gas Properties (Argon, Table A-2a):
\(R = 0.2081\text{ kPa}\cdot\text{m}^3/\text{kg}\cdot\text{K}\), \(\quad c_p = 0.5203\text{ kJ/kg}\cdot\text{K}\).
\(T_1 = 450 + 273.15 = 723.15\text{ K}\).
\(R = 0.2081\text{ kPa}\cdot\text{m}^3/\text{kg}\cdot\text{K}\), \(\quad c_p = 0.5203\text{ kJ/kg}\cdot\text{K}\).
\(T_1 = 450 + 273.15 = 723.15\text{ K}\).
2. Specific Volume & Mass Flow Rate:
\(v_1 = \frac{R T_1}{P_1} = \frac{(0.2081)(723.15)}{1600} = 0.09406\text{ m}^3/\text{kg}\).
\(A_1 = 60\text{ cm}^2 = 0.0060\text{ m}^2\).
\(\dot{m} = \frac{(0.0060)(55)}{0.09406} = 3.5084\text{ kg/s}\).
\(v_1 = \frac{R T_1}{P_1} = \frac{(0.2081)(723.15)}{1600} = 0.09406\text{ m}^3/\text{kg}\).
\(A_1 = 60\text{ cm}^2 = 0.0060\text{ m}^2\).
\(\dot{m} = \frac{(0.0060)(55)}{0.09406} = 3.5084\text{ kg/s}\).
3. First Law Energy Balance:
\(-\dot{W}_{\text{out}} = \dot{m}\left[c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2000}\right]\).
\(-190\text{ kW} = (3.5084)\left[0.5203(T_2 - 450) + \frac{150^2 - 55^2}{2000}\right]\).
\(\Delta\text{ke} = \frac{22500 - 3025}{2000} = 9.7375\text{ kJ/kg}\).
\(\frac{-190}{3.5084} = -54.156\text{ kJ/kg} = 0.5203(T_2 - 450) + 9.7375\).
\(-63.893 = 0.5203(T_2 - 450) \implies T_2 - 450 = -122.8^\circ\text{C}\).
\(T_2 = 450 - 122.8 = 327.2^\circ\text{C}\).
\(-\dot{W}_{\text{out}} = \dot{m}\left[c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2000}\right]\).
\(-190\text{ kW} = (3.5084)\left[0.5203(T_2 - 450) + \frac{150^2 - 55^2}{2000}\right]\).
\(\Delta\text{ke} = \frac{22500 - 3025}{2000} = 9.7375\text{ kJ/kg}\).
\(\frac{-190}{3.5084} = -54.156\text{ kJ/kg} = 0.5203(T_2 - 450) + 9.7375\).
\(-63.893 = 0.5203(T_2 - 450) \implies T_2 - 450 = -122.8^\circ\text{C}\).
\(T_2 = 450 - 122.8 = 327.2^\circ\text{C}\).
Answer: \(T_2 \approx 327^\circ\text{C}\)
b. What is the exit temperature if Carbon Monoxide (CO) is the working fluid?
°C
💡 Hint: Formulas & Approach
For Carbon Monoxide, specific heats vary significantly over this temperature change.
1. Use ideal gas properties of CO from Table A-23.
2. Compute \(v_1 = \frac{R_{\text{CO}} T_1}{P_1}\) with \(R_{\text{CO}} = \frac{8.314}{28.011} = 0.2968\text{ kJ/kg}\cdot\text{K}\) to obtain \(\dot{m}\).
3. Find \(h_1\) at \(723.15\text{ K}\).
4. Solve for \(h_2\) from energy balance, then interpolate in Table A-23 to find \(T_2\).
1. Use ideal gas properties of CO from Table A-23.
2. Compute \(v_1 = \frac{R_{\text{CO}} T_1}{P_1}\) with \(R_{\text{CO}} = \frac{8.314}{28.011} = 0.2968\text{ kJ/kg}\cdot\text{K}\) to obtain \(\dot{m}\).
3. Find \(h_1\) at \(723.15\text{ K}\).
4. Solve for \(h_2\) from energy balance, then interpolate in Table A-23 to find \(T_2\).
📝 Show Full Worked Solution
1. Carbon Monoxide Inflow State:
\(R_{\text{CO}} = 0.2968\text{ kJ/kg}\cdot\text{K}\).
\(v_1 = \frac{(0.2968)(723.15)}{1600} = 0.13414\text{ m}^3/\text{kg}\).
\(\dot{m} = \frac{(0.0060)(55)}{0.13414} = 2.4599\text{ kg/s}\).
\(R_{\text{CO}} = 0.2968\text{ kJ/kg}\cdot\text{K}\).
\(v_1 = \frac{(0.2968)(723.15)}{1600} = 0.13414\text{ m}^3/\text{kg}\).
\(\dot{m} = \frac{(0.0060)(55)}{0.13414} = 2.4599\text{ kg/s}\).
2. Enthalpy at State 1 (Table A-23):
At \(T_1 = 723.15\text{ K}\), interpolate in Table A-23: \(\bar{h}_1 \approx 21{,}385\text{ kJ/kmol}\), so \(h_1 = \frac{21{,}385}{28.011} = 763.45\text{ kJ/kg}\).
At \(T_1 = 723.15\text{ K}\), interpolate in Table A-23: \(\bar{h}_1 \approx 21{,}385\text{ kJ/kmol}\), so \(h_1 = \frac{21{,}385}{28.011} = 763.45\text{ kJ/kg}\).
3. Energy Balance for \(h_2\):
\(h_2 = h_1 - \frac{\dot{W}_{\text{out}}}{\dot{m}} - \Delta\text{ke} = 763.45 - \frac{190}{2.4599} - 9.74 = 763.45 - 77.24 - 9.74 = 676.47\text{ kJ/kg}\).
\(\bar{h}_2 = (676.47)(28.011) = 18{,}948.6\text{ kJ/kmol}\).
\(h_2 = h_1 - \frac{\dot{W}_{\text{out}}}{\dot{m}} - \Delta\text{ke} = 763.45 - \frac{190}{2.4599} - 9.74 = 763.45 - 77.24 - 9.74 = 676.47\text{ kJ/kg}\).
\(\bar{h}_2 = (676.47)(28.011) = 18{,}948.6\text{ kJ/kmol}\).
4. Exit Temperature from Table A-23:
Interpolating in Table A-23 between \(640\text{ K}\) and \(680\text{ K}\) yields \(T_2 \approx 671.3\text{ K} = 398.2^\circ\text{C}\).
Interpolating in Table A-23 between \(640\text{ K}\) and \(680\text{ K}\) yields \(T_2 \approx 671.3\text{ K} = 398.2^\circ\text{C}\).
Answer: \(T_2 \approx 398^\circ\text{C}\)
Problem 3 • Throttling Valve
Throttling Valve
R-134a at \(800\text{ kPa}\) and \(25^\circ\text{C}\) is throttled to a temperature of \(-20^\circ\text{C}\).
a. What is the pressure at the exit of the throttle?
kPa
💡 Hint: Formulas & Approach
Throttling is an isenthalpic expansion: \(h_1 = h_2\).
Check whether State 1 is compressed liquid or superheated by comparing \(T_1 = 25^\circ\text{C}\) to \(T_{\text{sat}}\) at \(800\text{ kPa}\). Approximate \(h_1 \approx h_f(T_1)\).
Then compare \(h_2\) to \(h_f\) and \(h_g\) at \(-20^\circ\text{C}\) in the saturation tables (Table A-11) to determine the phase and corresponding saturation pressure.
Check whether State 1 is compressed liquid or superheated by comparing \(T_1 = 25^\circ\text{C}\) to \(T_{\text{sat}}\) at \(800\text{ kPa}\). Approximate \(h_1 \approx h_f(T_1)\).
Then compare \(h_2\) to \(h_f\) and \(h_g\) at \(-20^\circ\text{C}\) in the saturation tables (Table A-11) to determine the phase and corresponding saturation pressure.
📝 Show Full Worked Solution
1. State 1 Analysis (Table A-12):
At \(P_1 = 800\text{ kPa}\), the saturation temperature is \(T_{\text{sat}} = 31.31^\circ\text{C}\).
Since \(T_1 = 25^\circ\text{C} < T_{\text{sat}}\), State 1 is a compressed liquid.
Approximating as saturated liquid at \(25^\circ\text{C}\) (Table A-11):
\(h_1 \approx h_f(25^\circ\text{C}) = 86.41\text{ kJ/kg}\).
At \(P_1 = 800\text{ kPa}\), the saturation temperature is \(T_{\text{sat}} = 31.31^\circ\text{C}\).
Since \(T_1 = 25^\circ\text{C} < T_{\text{sat}}\), State 1 is a compressed liquid.
Approximating as saturated liquid at \(25^\circ\text{C}\) (Table A-11):
\(h_1 \approx h_f(25^\circ\text{C}) = 86.41\text{ kJ/kg}\).
2. State 2 Analysis (Table A-11 at \(T_2 = -20^\circ\text{C}\)):
Because throttling is isenthalpic, \(h_2 = h_1 = 86.41\text{ kJ/kg}\).
At \(-20^\circ\text{C}\): \(h_f = 25.49\text{ kJ/kg}\) and \(h_g = 238.41\text{ kJ/kg}\).
Since \(h_f < h_2 < h_g\), the refrigerant exits as a saturated liquid-vapor mixture.
Because throttling is isenthalpic, \(h_2 = h_1 = 86.41\text{ kJ/kg}\).
At \(-20^\circ\text{C}\): \(h_f = 25.49\text{ kJ/kg}\) and \(h_g = 238.41\text{ kJ/kg}\).
Since \(h_f < h_2 < h_g\), the refrigerant exits as a saturated liquid-vapor mixture.
3. Exit Pressure:
For any saturated mixture, the pressure equals the saturation pressure at that temperature:
\(P_2 = P_{\text{sat}}(-20^\circ\text{C}) = 132.82\text{ kPa}\).
For any saturated mixture, the pressure equals the saturation pressure at that temperature:
\(P_2 = P_{\text{sat}}(-20^\circ\text{C}) = 132.82\text{ kPa}\).
Answer: \(P_2 = 132.8\text{ kPa}\)
b. What is the specific internal energy at the exit?
kJ/kg
💡 Hint: Formulas & Approach
First, find the vapor quality at the exit: \(x_2 = \frac{h_2 - h_f}{h_{fg}}\) using values at \(-20^\circ\text{C}\).
Then calculate internal energy using the mixture equation: \(u_2 = u_f + x_2 u_{fg}\).
Then calculate internal energy using the mixture equation: \(u_2 = u_f + x_2 u_{fg}\).
📝 Show Full Worked Solution
1. Vapor Quality Calculation (Table A-11 at \(-20^\circ\text{C}\)):
\(h_f = 25.49\text{ kJ/kg}\), \(\quad h_{fg} = 212.91\text{ kJ/kg}\).
\(x_2 = \frac{h_2 - h_f}{h_{fg}} = \frac{86.41 - 25.49}{212.91} = \frac{60.92}{212.91} = 0.28613\).
\(h_f = 25.49\text{ kJ/kg}\), \(\quad h_{fg} = 212.91\text{ kJ/kg}\).
\(x_2 = \frac{h_2 - h_f}{h_{fg}} = \frac{86.41 - 25.49}{212.91} = \frac{60.92}{212.91} = 0.28613\).
2. Specific Internal Energy Lookup:
At \(-20^\circ\text{C}\): \(u_f = 25.29\text{ kJ/kg}\), \(\quad u_{fg} = 193.81\text{ kJ/kg}\).
\(u_2 = u_f + x_2 u_{fg} = 25.29 + (0.28613)(193.81) = 25.29 + 55.45 = 80.74\text{ kJ/kg}\).
At \(-20^\circ\text{C}\): \(u_f = 25.29\text{ kJ/kg}\), \(\quad u_{fg} = 193.81\text{ kJ/kg}\).
\(u_2 = u_f + x_2 u_{fg} = 25.29 + (0.28613)(193.81) = 25.29 + 55.45 = 80.74\text{ kJ/kg}\).
Answer: \(u_2 = 80.74\text{ kJ/kg}\)
Problem 4 • Heat Exchanger
Heat Exchanger
Hot exhaust gases of an internal combustion engine are used to produce saturated water vapor at \(2\text{ MPa}\). The exhaust gases enter the heat exchanger at \(400^\circ\text{C}\) at a rate of \(32\text{ kg/min}\), while water enters at \(15^\circ\text{C}\). The heat exchanger is not well insulated, and \(10\%\) of the heat given up by the exhaust gases is lost to surroundings. The mass flow rate of the exhaust gases is 15 times that of the water. (Approximate exhaust gas as air, \(c_p = 1.045\text{ kJ/kg}\cdot\text{K}\)).
a. Heat transfer rate to the water (\(\dot{Q}_{\text{water}}\)):
kW
💡 Hint: Formulas & Approach
1. Find the water mass flow rate: \(\dot{m}_{\text{water}} = \frac{\dot{m}_{\text{gas}}}{15}\) (remember to convert \(\text{kg/min}\) to \(\text{kg/s}\)).
2. Inlet water: \(h_{\text{w,in}} \approx h_f(15^\circ\text{C})\) (compressed liquid).
3. Exit water: saturated vapor at \(2\text{ MPa} \implies h_{\text{w,out}} = h_g(2\text{ MPa})\).
4. \(\dot{Q}_{\text{water}} = \dot{m}_{\text{water}}(h_{\text{w,out}} - h_{\text{w,in}})\).
2. Inlet water: \(h_{\text{w,in}} \approx h_f(15^\circ\text{C})\) (compressed liquid).
3. Exit water: saturated vapor at \(2\text{ MPa} \implies h_{\text{w,out}} = h_g(2\text{ MPa})\).
4. \(\dot{Q}_{\text{water}} = \dot{m}_{\text{water}}(h_{\text{w,out}} - h_{\text{w,in}})\).
📝 Show Full Worked Solution
1. Mass Flow Rates:
\(\dot{m}_{\text{gas}} = \frac{32\text{ kg/min}}{60\text{ s/min}} = 0.5333\text{ kg/s}\).
\(\dot{m}_{\text{water}} = \frac{\dot{m}_{\text{gas}}}{15} = \frac{0.5333}{15} = 0.035556\text{ kg/s}\).
\(\dot{m}_{\text{gas}} = \frac{32\text{ kg/min}}{60\text{ s/min}} = 0.5333\text{ kg/s}\).
\(\dot{m}_{\text{water}} = \frac{\dot{m}_{\text{gas}}}{15} = \frac{0.5333}{15} = 0.035556\text{ kg/s}\).
2. Water Property Lookups (Steam Tables):
Inlet (\(15^\circ\text{C}\), Table A-4): \(h_{\text{w,in}} \approx h_f(15^\circ\text{C}) = 62.98\text{ kJ/kg}\).
Outlet (\(2.0\text{ MPa}\) sat. vapor, Table A-5): \(h_{\text{w,out}} = h_g(2\text{ MPa}) = 2798.3\text{ kJ/kg}\).
Inlet (\(15^\circ\text{C}\), Table A-4): \(h_{\text{w,in}} \approx h_f(15^\circ\text{C}) = 62.98\text{ kJ/kg}\).
Outlet (\(2.0\text{ MPa}\) sat. vapor, Table A-5): \(h_{\text{w,out}} = h_g(2\text{ MPa}) = 2798.3\text{ kJ/kg}\).
3. Heat Transfer Rate to Water:
\(\dot{Q}_{\text{water}} = \dot{m}_{\text{water}}(h_{\text{w,out}} - h_{\text{w,in}}) = (0.035556\text{ kg/s})(2798.3 - 62.98)\text{ kJ/kg}\).
\(\dot{Q}_{\text{water}} = (0.035556)(2735.32) = 97.26\text{ kW}\).
\(\dot{Q}_{\text{water}} = \dot{m}_{\text{water}}(h_{\text{w,out}} - h_{\text{w,in}}) = (0.035556\text{ kg/s})(2798.3 - 62.98)\text{ kJ/kg}\).
\(\dot{Q}_{\text{water}} = (0.035556)(2735.32) = 97.26\text{ kW}\).
Answer: \(\dot{Q}_{\text{water}} = 97.3\text{ kW}\)
b. Heat transfer rate from the exhaust gas (\(\dot{Q}_{\text{gas}}\)):
kW
💡 Hint: Formulas & Approach
Set up the heat transfer balance between the streams and surroundings:
\(\dot{Q}_{\text{gas}} = \dot{Q}_{\text{water}} + \dot{Q}_{\text{loss}}\).
Since \(10\%\) of the gas heat is lost to surroundings, \(90\%\) is received by water: \(\dot{Q}_{\text{water}} = 0.90 \times \dot{Q}_{\text{gas}}\).
\(\dot{Q}_{\text{gas}} = \dot{Q}_{\text{water}} + \dot{Q}_{\text{loss}}\).
Since \(10\%\) of the gas heat is lost to surroundings, \(90\%\) is received by water: \(\dot{Q}_{\text{water}} = 0.90 \times \dot{Q}_{\text{gas}}\).
📝 Show Full Worked Solution
1. Heat Balance Relation:
\(\dot{Q}_{\text{water}} = 0.90 \dot{Q}_{\text{gas}}\).
\(\dot{Q}_{\text{gas}} = \frac{\dot{Q}_{\text{water}}}{0.90} = \frac{97.26\text{ kW}}{0.90} = 108.07\text{ kW}\).
\(\dot{Q}_{\text{water}} = 0.90 \dot{Q}_{\text{gas}}\).
\(\dot{Q}_{\text{gas}} = \frac{\dot{Q}_{\text{water}}}{0.90} = \frac{97.26\text{ kW}}{0.90} = 108.07\text{ kW}\).
Answer: \(\dot{Q}_{\text{gas}} = 108.1\text{ kW}\)
c. Temperature of the exhaust gas at the exit:
°C
💡 Hint: Formulas & Approach
Use the heat loss from the exhaust gas using constant specific heat: \(\dot{Q}_{\text{gas}} = \dot{m}_{\text{gas}} c_p (T_{\text{in}} - T_{\text{out}})\).
Solve directly for \(T_{\text{out}}\).
Solve directly for \(T_{\text{out}}\).
📝 Show Full Worked Solution
1. Exhaust Gas Energy Balance:
\(\dot{Q}_{\text{gas}} = \dot{m}_{\text{gas}} c_p (T_{\text{g,in}} - T_{\text{g,out}})\).
\(108.07\text{ kW} = (0.5333\text{ kg/s})(1.045\text{ kJ/kg}\cdot\text{K})(400^\circ\text{C} - T_{\text{g,out}})\).
\(\dot{Q}_{\text{gas}} = \dot{m}_{\text{gas}} c_p (T_{\text{g,in}} - T_{\text{g,out}})\).
\(108.07\text{ kW} = (0.5333\text{ kg/s})(1.045\text{ kJ/kg}\cdot\text{K})(400^\circ\text{C} - T_{\text{g,out}})\).
2. Solve for Temperature:
\(400 - T_{\text{g,out}} = \frac{108.07}{(0.5333)(1.045)} = \frac{108.07}{0.5573} = 193.9^\circ\text{C}\).
\(T_{\text{g,out}} = 400 - 193.9 = 206.1^\circ\text{C}\).
\(400 - T_{\text{g,out}} = \frac{108.07}{(0.5333)(1.045)} = \frac{108.07}{0.5573} = 193.9^\circ\text{C}\).
\(T_{\text{g,out}} = 400 - 193.9 = 206.1^\circ\text{C}\).
Answer: \(T_{\text{g,out}} = 206^\circ\text{C}\)
Problem 5 • Water Pump
Liquid Pump
A pump increases water pressure from \(70\text{ kPa}\) at the inlet to \(700\text{ kPa}\) at the outlet. Water enters at \(15^\circ\text{C}\) through a \(1\text{ cm}\) diameter opening and exits through a \(1.5\text{ cm}\) diameter opening. Mass flow rate is \(0.5\text{ kg/s}\).
a. Inlet and outlet velocities (\(V_1\) and \(V_2\)):
m/s (Inlet)
m/s (Outlet)
💡 Hint: Formulas & Approach
Liquid water is incompressible: \(v \approx v_f(15^\circ\text{C}) \approx 0.001001\text{ m}^3/\text{kg}\).
Calculate circular opening cross-sectional areas: \(A = \frac{\pi D^2}{4}\).
Then use the volumetric and mass flow relation: \(V = \frac{\dot{m} v}{A}\).
Calculate circular opening cross-sectional areas: \(A = \frac{\pi D^2}{4}\).
Then use the volumetric and mass flow relation: \(V = \frac{\dot{m} v}{A}\).
📝 Show Full Worked Solution
1. Cross-Sectional Areas:
\(D_1 = 0.01\text{ m} \implies A_1 = \frac{\pi (0.01)^2}{4} = 7.854 \times 10^{-5}\text{ m}^2\).
\(D_2 = 0.015\text{ m} \implies A_2 = \frac{\pi (0.015)^2}{4} = 1.767 \times 10^{-4}\text{ m}^2\).
\(D_1 = 0.01\text{ m} \implies A_1 = \frac{\pi (0.01)^2}{4} = 7.854 \times 10^{-5}\text{ m}^2\).
\(D_2 = 0.015\text{ m} \implies A_2 = \frac{\pi (0.015)^2}{4} = 1.767 \times 10^{-4}\text{ m}^2\).
2. Velocity Calculations:
Specific volume \(v \approx v_f(15^\circ\text{C}) = 0.001001\text{ m}^3/\text{kg}\).
\(V_1 = \frac{\dot{m} v}{A_1} = \frac{(0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})}{7.854 \times 10^{-5}\text{ m}^2} = 6.372\text{ m/s}\).
\(V_2 = \frac{\dot{m} v}{A_2} = \frac{(0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})}{1.767 \times 10^{-4}\text{ m}^2} = 2.832\text{ m/s}\).
Specific volume \(v \approx v_f(15^\circ\text{C}) = 0.001001\text{ m}^3/\text{kg}\).
\(V_1 = \frac{\dot{m} v}{A_1} = \frac{(0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})}{7.854 \times 10^{-5}\text{ m}^2} = 6.372\text{ m/s}\).
\(V_2 = \frac{\dot{m} v}{A_2} = \frac{(0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})}{1.767 \times 10^{-4}\text{ m}^2} = 2.832\text{ m/s}\).
Answers: \(V_1 = 6.37\text{ m/s}\), \(\quad V_2 = 2.83\text{ m/s}\)
b. Pump work input:
Watts
💡 Hint: Formulas & Approach
Incompressible pump work equation: \(\dot{W}_{\text{in}} = \dot{m}\left[v(P_2 - P_1) + \frac{V_2^2 - V_1^2}{2000}\right]\).
Note that standard textbook convention often reports pump shaft work primarily from pressure rise (\(\dot{W} \approx \dot{m} v \Delta P\)), converted to Watts (\(\text{kW} \times 1000\)).
Note that standard textbook convention often reports pump shaft work primarily from pressure rise (\(\dot{W} \approx \dot{m} v \Delta P\)), converted to Watts (\(\text{kW} \times 1000\)).
📝 Show Full Worked Solution
1. Pressure Rise Work:
\(\dot{W}_{\Delta P} = \dot{m} v (P_2 - P_1) = (0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})(700 - 70)\text{ kPa}\).
\(\dot{W}_{\Delta P} = (0.5)(0.001001)(630) = 0.3153\text{ kW} = 315.3\text{ W}\).
\(\dot{W}_{\Delta P} = \dot{m} v (P_2 - P_1) = (0.5\text{ kg/s})(0.001001\text{ m}^3/\text{kg})(700 - 70)\text{ kPa}\).
\(\dot{W}_{\Delta P} = (0.5)(0.001001)(630) = 0.3153\text{ kW} = 315.3\text{ W}\).
2. Kinetic Energy Difference:
\(\Delta\text{KE} = \frac{V_2^2 - V_1^2}{2000} = \frac{2.832^2 - 6.372^2}{2000} = -0.0163\text{ kJ/kg}\).
\(\dot{m}\Delta\text{KE} = -0.0081\text{ kW} = -8.1\text{ W}\).
Total work: \(\approx 307.2\text{ W}\) (or \(315\text{ W}\) neglecting \(\Delta\text{KE}\)).
\(\Delta\text{KE} = \frac{V_2^2 - V_1^2}{2000} = \frac{2.832^2 - 6.372^2}{2000} = -0.0163\text{ kJ/kg}\).
\(\dot{m}\Delta\text{KE} = -0.0081\text{ kW} = -8.1\text{ W}\).
Total work: \(\approx 307.2\text{ W}\) (or \(315\text{ W}\) neglecting \(\Delta\text{KE}\)).
Answer: \(\dot{W}_{\text{in}} \approx 315\text{ W}\)