ME Thermodynamics I

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Kimberly Pitts
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1 HW-22 (25 points) Given: 1 A gas power cycle with initial properties as listed on the EFD. The compressor pressure ratio is 25:1 Find: 1 Sketch all the processes on a p-h diagram and calculate the enthalpy, entropy change and entropy generation for all processes. EFD: 1 Assumptions: 2 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow T boundary for the combustor is the outlet temperature Outlet pressure is the same as inlet pressure Basic Equations: 2 dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o dt = j s = s o 2 s o 1 R ln p 2 p 1 : a) Q j T j + i ṁ i (s) i o ṁo(s) o + σ 1

4 b) From the ideal gas tables the following information is retrieved: s o 1 = 1.559 kj/kg K h 1 = 260 kj/kg s o 3 = 3.

287 ln(0.7/17.5) s o 2 = 2.4828 kj/kgk 0 = s 3 s 4 = s o 3 s o 4 Rln(17.5/0.7) 0 = 3.597 s o 4 0.287 ln(17.5/0.7) s o 4 = 2.

2 4 b) From the ideal gas tables the following information is retrieved: s o 1 = kj/kg K h 1 = 260 kj/kg s o 3 = kj/kg K h 3 = 1880 kj/kg For the compressor and the turbine 0 = s 1 s 2 = s o 1 s o 2 Rln(0.7/17.5) = s o ln(0.7/17.5) s o 2 = kj/kgk 0 = s 3 s 4 = s o 3 s o 4 Rln(17.5/0.7) 0 = s o ln(17.5/0.7) s o 4 = kj/kgk Using the previously found values and interpolating we find T 2 = K h 2 = kj/kg T 4 = kj/kg h 4 = kj/kg 2

3 Calculating the enthalpies changes: Compressor h 2 h 1 = kj/kg 1 Combustor h 3 h 2 = kj/kg 1 h 4 h 3 = kj/kg 2 c) For the compressor from the second law and using the given information Combustor d) Compressor 0 = ṁ(s 1 s 2 ) + Q T + σ s = s 1 s 2 = 0 1 s = s 2 s 3 = s o 2 s o 3 R ln(p 3 /p 2 ) s = kj/kgk 3 s = s 3 s 4 = 0 1 σ compressor = 0 1 σ turbine = 0 1 For the combustor, from the first law it can be found that q = h 3 h 2 = kj/kg. Also, assuming T bound = 1700K K dt = Q T + ṁ(s 2 s 3 ) + σ 0 = q T + (s 2 s 3 ) + σ σ = kj/kgk 3 3

EFD: 3 Assumptions: 2 Neglect PE and KE changes Steady State and Steady Flow The mixer is at standard pressure Basic

4 HW-23 (25 points) Given: 1 An un-insulated mixer with two inlets and one outlet with properties as shown in the EFD Find: 1 Calculate the heat transfer to the surroundings and the entropy generation for the process. EFD: 3 Assumptions: 2 Neglect PE and KE changes Steady State and Steady Flow The mixer is at standard pressure Basic Equations: 2 dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o dt = j : a) Q j T j + i ṁ i (s) i o ṁo(s) o + σ 4

5 0 = ṁ 1 + ṁ 2 ṁ 3 m 3 = 1.3 kg/s Ẇ elec = W 1 Ẇ rot = = 500 W 1 For the enthalpies, we use the compressed liquid approximation h 1 = h f + ν f (P P (T sat )) = ( ) = kj/kg 2 h 2 = h f + ν f (P P (T sat )) = ( ) = kj/kg 2 h 3 = h f + ν f (P P (T sat )) = ( ) = kj/kg 2 Applying the first law to the control volume and remembering work is negative since it is going into the system 0 = Q Ẇ + ṁ 1h 1 + ṁ 2 h 2 ṁ 3 h 3 Q = W 3 For the saturated liquid tables, we obtain the entropy values s 1 = kj/kgk s 2 = kj/kgk s 3 = kj/kgk Using the second law and using the boundary temperature to be 25 o C 0 = Q T + ṁ 1s 1 + ṁ 2 s 2 ṁ 3 s 3 + σ σ = W/K 3 5

Decide which component is critical to improve EFD: 1 Assumptions: 2 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow The compressor and turbine are

6 HW-24 (25 points) Given: 1 A gas power cycle with initial properties as listed on the EFD. The compressor pressure ratio is 25:1 Find: 1 Sketch all the processes on a p-h diagram and calculate the enthalpy, entropy change and entropy generation for all processes. Decide which component is critical to improve EFD: 1 Assumptions: 2 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow The compressor and turbine are adiabatic T boundary for the combustor is the outlet temperature Outlet pressure is the same as inlet pressure Basic Equations: 2 dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o dt = j s = s o 2 s o 1 R ln p 2 p 1 : a) Q j T j + i ṁ i (s) i o ṁo(s) o + σ 6

7 4 b) From the ideal gas tables the following information is retrieved: Calculating the enthalpies changes: Compressor Combustor c) s o 1 = kj/kg K h 1 = 260 kj/kg s o 2 = kj/kg K h 2 = kj/kg s o 3 = kj/kg K h 3 = 1880 kj/kg s o 4 = kj/kg K h 4 = kj/kg h 2 h 1 = kj/kg 2 h 3 h 2 = kj/kg 2 h 4 h 3 = kj/kg 1 7

8 Compressor Combustor d) Compressor s = s 1 s 2 = s o 1 s o 2 R ln(p 1 /p 2 ) = ln(0.7/17.5) s = kj/kgk 2 s = s 2 s 3 = s o 2 s o 3 R ln(p 2 /p 3 ) = ln(17.5/17.5) s = kj/kgk 2 s = s 3 s 4 = s o 3 s o 4 R ln(p 3 /p 4 ) = ln(17.5/0.7) s = kj/kgk 1 σ compressor = kj/kgk 1 σ turbine = kj/kgk 1 For the combustor, from the first law it can be found that q = h 3 h 2 = kj/kg. Also, assuming T bound = 1700K K dt = Q T + ṁ(s 2 s 3 ) + σ 0 = q T + (s 2 s 3 ) + σ σ = kj/kgk 2 8

ME Thermodynamics I

ME Thermodynamics I Homework - Week 01 HW-01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.