Thermodynamics Quiz: Steady Flow Energy Equation Sfee
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Steady Flow Energy Equation SfeeQuestion 1 of 7

In a steam ejector, high-pressure steam at 2 MPa and 300°C with velocity 800 m/s entrains low-pressure steam at 20 kPa and 60°C with velocity 50 m/s. The mass flow rate ratio is 1:3 (high pressure to low pressure). If the mixed stream exits at 150 kPa and 200 m/s, what is the exit temperature?

96.4°C
101.8°C
107.3°C
112.9°C
118.6°C
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Thermodynamics Quiz

Thermodynamics Quiz: Steady Flow Energy Equation Sfee

Practice Steady Flow Energy Equation Sfee in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Steady Flow Energy Equation Sfee, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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Question 1

In a steam ejector, high-pressure steam at 2 MPa and 300°C with velocity 800 m/s entrains low-pressure steam at 20 kPa and 60°C with velocity 50 m/s. The mass flow rate ratio is 1:3 (high pressure to low pressure). If the mixed stream exits at 150 kPa and 200 m/s, what is the exit temperature?

  1. 96.4°C
  2. 101.8°C
  3. 107.3°C (correct answer)
  4. 112.9°C
  5. 118.6°C
Explanation: When you encounter a steam ejector problem, you're dealing with a mixing process where two fluid streams combine. The key principle is applying conservation of mass and energy while accounting for kinetic energy effects due to the high velocities involved. For this mixing process, you need to apply the steady flow energy equation. Since the ejector operates adiabatically (no heat transfer) and has no work interactions, energy conservation requires that the total enthalpy of the incoming streams equals the total enthalpy of the exit stream, including kinetic energy terms. First, establish the mass flow rates: if the ratio is 1:3 (high to low pressure), let m˙1=1\dot{m}_1 = 1 kg/s and m˙2=3\dot{m}_2 = 3 kg/s. From steam tables, the specific enthalpies are: high-pressure steam at 2 MPa, 300°C has h1=3024h_1 = 3024 kJ/kg, and low-pressure steam at 20 kPa, 60°C has h2=2610h_2 = 2610 kJ/kg. Applying energy conservation with kinetic energy terms: m˙1(h1+V122)+m˙2(h2+V222)=(m˙1+m˙2)(h3+V322)\dot{m}_1(h_1 + \frac{V_1^2}{2}) + \dot{m}_2(h_2 + \frac{V_2^2}{2}) = (\dot{m}_1 + \dot{m}_2)(h_3 + \frac{V_3^2}{2}) Solving gives h3=2730h_3 = 2730 kJ/kg. At 150 kPa with this enthalpy, the steam tables show the exit temperature is 107.3°C. Answer A (96.4°C) likely neglects kinetic energy effects. Answer B (101.8°C) probably uses incorrect property values or approximations. Answer D (112.9°C) may result from errors in the energy balance setup. Remember: steam ejector problems always require careful attention to kinetic energy terms due to high velocities, and accurate steam property data is essential for correct solutions.

Question 2

In a mixing chamber, 3 kg/s of steam at 300 kPa and 150°C mixes adiabatically with 1 kg/s of steam at 300 kPa and 200°C. Both inlet streams have negligible velocities, while the exit stream has a velocity of 60 m/s. What is the temperature of the mixed steam at the exit?

  1. 161.2°C (correct answer)
  2. 163.4°C
  3. 165.8°C
  4. 168.1°C
  5. 170.5°C
Explanation: When you encounter adiabatic mixing problems, you're applying conservation of mass and energy to find exit conditions. Since the process is adiabatic (no heat transfer), all energy changes come from the mixing itself and kinetic energy effects. Start with the energy balance for steady flow: m˙1h1+m˙2h2=m˙3h3+m˙3V322\dot{m}_1 h_1 + \dot{m}_2 h_2 = \dot{m}_3 h_3 + \dot{m}_3 \frac{V_3^2}{2} From steam tables at 300 kPa: Stream 1 (150°C) has h1=2768.8h_1 = 2768.8 kJ/kg, and Stream 2 (200°C) has h2=2865.6h_2 = 2865.6 kJ/kg. The kinetic energy term is V322=6022000=1.8\frac{V_3^2}{2} = \frac{60^2}{2000} = 1.8 kJ/kg (converting from J/kg). Solving for exit enthalpy: h3=3(2768.8)+1(2865.6)4−1.8=111724−1.8=2791.2h_3 = \frac{3(2768.8) + 1(2865.6)}{4} - 1.8 = \frac{11172}{4} - 1.8 = 2791.2 kJ/kg At 300 kPa with h3=2791.2h_3 = 2791.2 kJ/kg, interpolating steam tables gives 161.2°C, confirming answer A. The wrong answers represent common calculation errors: B (163.4°C) likely results from neglecting the kinetic energy correction entirely. C (165.8°C) might come from using incorrect enthalpy values or mass flow rates. D (168.1°C) could result from sign errors in the energy balance or incorrect unit conversions. Remember that in mixing problems, always account for exit kinetic energy when velocities are significant (>10 m/s). The kinetic energy term reduces the available thermal enthalpy, lowering the final temperature. Practice identifying when kinetic effects matter—they're often the key to getting the right answer.

Question 3

Steam flows through a throttling valve from 2 MPa and 280°C to 500 kPa. The inlet velocity is 30 m/s and the exit velocity is 90 m/s. If the process loses 2 kJ/kg of heat to the surroundings, what is the exit temperature?

  1. 147.8°C
  2. 149.2°C (correct answer)
  3. 150.6°C
  4. 152.1°C
  5. 153.5°C
Explanation: When you encounter a throttling valve problem, you're dealing with a steady-flow process where enthalpy plus kinetic energy remains constant, but heat loss complicates the energy balance. The key is applying the steady-flow energy equation properly. For this throttling process with heat loss, use: h1+V122=h2+V222+qouth_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} + q_{out} First, find the inlet enthalpy. At 2 MPa and 280°C, steam is superheated with h1=2957.2 kJ/kgh_1 = 2957.2 \text{ kJ/kg} (from steam tables). Next, calculate the kinetic energy terms: V122=3022000=0.45 kJ/kg\frac{V_1^2}{2} = \frac{30^2}{2000} = 0.45 \text{ kJ/kg} and V222=9022000=4.05 kJ/kg\frac{V_2^2}{2} = \frac{90^2}{2000} = 4.05 \text{ kJ/kg} Solving for exit enthalpy: h2=h1+V122−V222−qout=2957.2+0.45−4.05−2=2951.6 kJ/kgh_2 = h_1 + \frac{V_1^2}{2} - \frac{V_2^2}{2} - q_{out} = 2957.2 + 0.45 - 4.05 - 2 = 2951.6 \text{ kJ/kg} At 500 kPa with h2=2951.6 kJ/kgh_2 = 2951.6 \text{ kJ/kg}, interpolating steam tables gives T2=149.2°CT_2 = 149.2°C, which is answer B. Answer A (147.8°C) likely neglects the kinetic energy difference. Answer C (150.6°C) probably ignores the heat loss entirely. Answer D (152.1°C) might incorrectly add the heat loss instead of subtracting it. Remember: throttling problems aren't truly isenthalpic when heat loss or significant velocity changes occur. Always account for all energy terms in your steady-flow energy equation, and pay careful attention to signs—heat loss reduces the exit enthalpy.

Question 4

A compressor operates with air entering at 100 kPa, 27°C, and 120 m/s, and exiting at 800 kPa, 180°C, and 80 m/s. The mass flow rate is 2.5 kg/s. If the cooling water removes 45 kW of heat from the compressor, what is the required power input? Assume cp=1.005c_p = 1.005 kJ/kg·K.

  1. 428 kW accounting for both kinetic energy changes and heat removal effects (correct answer)
  2. 438 kW with heat removal reducing required input by approximately 10%
  3. 423 kW neglecting kinetic energy effects which contribute less than 2%
  4. 433 kW with kinetic energy effects reducing required input by 1.2%
Explanation: Using SFEE for compressor: W˙in=m˙[(h2−h1)+v22−v122]−Q˙\dot{W}_{in} = \dot{m}[(h_2 - h_1) + \frac{v_2^2 - v_1^2}{2}] - \dot{Q}. Enthalpy change: Δh=cpΔT=1.005×153=153.8\Delta h = c_p \Delta T = 1.005 × 153 = 153.8 kJ/kg. Kinetic energy change: v22−v122=802−12022000=−4.0\frac{v_2^2 - v_1^2}{2} = \frac{80^2 - 120^2}{2000} = -4.0 kJ/kg. Power: W˙in=2.5×(153.8−4.0)+45=419.5+45=428\dot{W}_{in} = 2.5 × (153.8 - 4.0) + 45 = 419.5 + 45 = 428 kW. B incorrectly treats heat as positive work. C neglects kinetic energy. D uses wrong sign for heat transfer.

Question 5

A diffuser slows air from 300 m/s to 50 m/s while pressure increases from 80 kPa to 95 kPa. The inlet temperature is 250 K. The process is adiabatic with a mass flow rate of 5.2 kg/s. What is the outlet temperature and the percentage of kinetic energy converted to internal energy?

  1. T₂ = 287 K with 97.2% of kinetic energy converted to internal energy increase (correct answer)
  2. T₂ = 291 K with 89.6% of kinetic energy contributing to temperature rise
  3. T₂ = 284 K with 91.8% kinetic energy conversion efficiency to thermal energy
  4. T₂ = 295 K with 85.4% of initial kinetic energy recovered as enthalpy increase
Explanation: For adiabatic diffuser: h1+v122=h2+v222h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}. Kinetic energy change: Δke=v12−v222=3002−5022000=43.75\Delta ke = \frac{v_1^2 - v_2^2}{2} = \frac{300^2 - 50^2}{2000} = 43.75 kJ/kg. This equals enthalpy increase: Δh=cp(T2−T1)=43.75\Delta h = c_p(T_2 - T_1) = 43.75, so T2=250+43.751.005=287T_2 = 250 + \frac{43.75}{1.005} = 287 K. Initial kinetic energy = 45.0 kJ/kg, final = 1.25 kJ/kg. Energy converted = 43.75 kJ/kg. Percentage = 43.7545.0×100%=97.2%\frac{43.75}{45.0} × 100\% = 97.2\%.

Question 6

A throttling valve reduces steam pressure from 2 MPa, 300°C to 500 kPa. The inlet velocity is 50 m/s and outlet velocity is 180 m/s. The valve is well-insulated. What is the outlet temperature, and what assumption is most critical for this analysis?

  1. 151.2°C assuming constant enthalpy process with kinetic energy effects negligible
  2. 148.6°C with kinetic energy increase causing slight enthalpy reduction during expansion (correct answer)
  3. 153.8°C assuming adiabatic process but accounting for kinetic energy change effects
  4. 149.4°C with isenthalpic assumption valid only when kinetic energy changes are small
Explanation: For throttling, SFEE gives: h1+v122=h2+v222h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2} (adiabatic, no work). From steam tables: h₁ = 3023.5 kJ/kg at 2 MPa, 300°C. Kinetic energy change: v22−v122=1802−5022000=14.95\frac{v_2^2 - v_1^2}{2} = \frac{180^2 - 50^2}{2000} = 14.95 kJ/kg. Therefore: h₂ = 3023.5 - 14.95 = 3008.6 kJ/kg. At 500 kPa with h = 3008.6 kJ/kg, T₂ = 148.6°C. A neglects kinetic energy effects. C incorrectly adds kinetic energy. D misunderstands the isenthalpic assumption validity.

Question 7

An ejector uses high-pressure steam to entrain and compress low-pressure steam. Primary steam: 1.5 kg/s at 2 MPa, 300°C, velocity 600 m/s. Secondary steam: 0.8 kg/s at 15 kPa, 90% quality, velocity 100 m/s. The mixed stream exits at 150 kPa and 200 m/s. Assuming adiabatic mixing, what is the exit temperature?

  1. 142.6°C with momentum conservation requiring velocity adjustment for accurate temperature
  2. 138.4°C accounting for kinetic energy effects in high-velocity steam mixing (correct answer)
  3. 145.8°C with secondary steam contribution limited by low initial enthalpy
  4. 140.2°C assuming ideal mixing with momentum effects negligible compared to energy balance
Explanation: Primary steam: h₁ = 3023.5 kJ/kg, ke₁ = 600²/2000 = 180 kJ/kg. Secondary steam: h₂ = 0.9 × 2599.1 + 0.1 × 225.9 = 2361.8 kJ/kg, ke₂ = 100²/2000 = 5 kJ/kg. Exit kinetic energy: ke₃ = 200²/2000 = 20 kJ/kg. Total mass flow: 2.3 kg/s. Energy balance: 1.5(3023.5+180)+0.8(2361.8+5)=2.3(h3+20)1.5(3023.5 + 180) + 0.8(2361.8 + 5) = 2.3(h₃ + 20). Solving: 4805.3+1893.4=2.3h3+464805.3 + 1893.4 = 2.3h₃ + 46, so h3=6652.72.3=2892.5h₃ = \frac{6652.7}{2.3} = 2892.5 kJ/kg. At 150 kPa with h = 2892.5 kJ/kg, T₃ ≈ 138.4°C. A neglects proper kinetic energy accounting. C overestimates temperature. D underestimates the effect of high kinetic energy in primary stream.