ME Thermodynamics I

Transcription

1 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. EFD: 7 Assumptions: 5 Steady State, de dt =0 Basic Equations: 3 None : See EFD. 1

2 Homework - Week 01 HW-02 (25 points) Given: 2 Schematic of a manual automobile transmission with its description. Find: 3 Identify the system and the interactions with the surroundings. EFD: 7 Assumptions: 5 Steady State, de =0 dt Basic Equations: 3 None : 5 See EFD. The 5% power loss goes to the surroundings ( Q loss ). 2

3 Homework - Week 02 HW-03 (25 points) Given: 1 Schematic of a hand pump used to raise water. Find: 1 Free Body Diagram, the pressure on the piston surface (kpa) and the work done to lift the water. EFD: 5 Assumptions: 3 The piston is frictionless The mass of the piston is negligible Steady State Basic Equations: 4 W = pdv : a) Force Body Diagram for the piston 3 1

4 Homework - Week 02 b) From the definition of pressure F = m g F = ρ V g F = ρ A h g c) W = P = F A P = ρ A h g A P = ρ h g P = P = kp a 4 V 2 V 1 P dv W = P (V 2 V 1 ) W = P (A h) W = (π (0.02) ) W = J 4 2

5 Homework - Week 02 HW-04 (25 points) Given: 2 Schematic of a powder roller compactor. Thickness of the material going in and out v in =0.75 m 2 /s, ρ=700 kg/m 3 Find: 3 Velocity of the material at the exit (v out ) and mass flow rate of the material (ṁ) EFD: 5 Assumptions: 2 Steady State Steady Flow (SSSF). Basic Equations: 3 dm dt = Σm in Σm out : a) The mass flow rate can be expressed as dm dt = Σm in Σm out 0 = Σm in Σ m in = m out m out 3

6 Homework - Week 02 ṁ = ρ V A Then, replacing the previous expression b) ρv in A in = ρv out A out A in = 1.2cm 10cm = 12cm 2 = m 2 A out = 0.3cm 10cm = 3cm 2 = m 2 v out = 3 m/s 5 ṁ = m in = m out = ṁ = 0.63 kg/s 5 4

7 Homework - Week 02 HW-05 (25 points) Given: 1 Schematic and descriptions of different devices such as a solar panel, an automobile transmission, a roller compactor and a manual pump Find: Identify the system and the heat/work interactions associated with the different devices. Show the direction of the interactions. EFD: 6 6 5

8 Homework - Week Assumptions: None Basic Equations: None : See EFDs 6

9 Homework - Week 02 HW-06 (25 points) Given: 1 Schematic of a rigid, insulated and impermeable tank containing a changing partition. Find: 1 Identify whether the system is at equilibrium or not for every case. EFD: 6 Assumptions: 1 None Basic Equations: 1 None : (a) No equilibrium: Since the partition is not impermeable to mass diffusion, there will be a induced flow of mass between the N 2 and O 2 5 (b) No equilibrium: Since the partition is not insulated, there will be net heat flux between the two substances. 5 (c) No equilibrium: In this case there are two possible potentials, thermal and mechanical. Since the partition is only insulated, there will be net movement. 5 1

10 Homework - Week 02 HW-07 (25 points) Given: 1 Three different processes: (a) Constant specific volume process from 1 bar to 5 bar at a specific volume of 0.01 m 3 /kg. (b) Constant pressure process from a specific volume of 0.01 m 3 /kg to one of m 3 /kg at 3 bar (c) Constant temperature process from STP to a pressure of 2.5 bar. Find: 1 Draw the corresponding processes in a p-v diagram EFD: 4 Assumptions: 1 None Basic Equations: 1 None : (a) 6 2

11 Homework - Week 02 (b) 6 w = v2 v1 pdv = 0 w = v2 v1 pdv = p v2 v1 = 300 ( ) = 2.25 kj/kg dv = (c) 5 First, we calculate the specific volumes at the initial state, i.e, 25 o C and 1 bar: pv = RT 100 v 1 = v 1 = m 3 /kg 3

12 Homework - Week 02 For the second state, the temperature is 25 o C and 2.5 bar: pv = RT 250 v 2 = v 2 = m 3 /kg w = v2 v1 pdv = R T ln v 2 v 1 = = 0.287(298.15)ln = kj/kg 4

13 Homework - Week 03 HW-08 (25 points) Given: 1 Three different states for water: (a) p = 1 bar and ρ = 1000 kg/m 3 (b) p = 1 bar and ρ = 500 kg/m 3 (c) p = 1 bar and ν = 1.67 m 3 /kg Find: 1 Identify the state of water at the specified conditions. EFD: N/A Assumptions: None Basic Equations: None : (a) Given that the saturated vapor point at 1 bar corresponds to m 3 /kg, it can be concluded that the state of water at m 3 /kg is compressed liquid. 4 From the chart provided, the following temperature and enthalpy can be estimated: T 100 o C 2 h 420 kj/kg 1 (b) Given that the saturated vapor point at 1 bar corresponds to m 3 /kg, it can be concluded that the state of water at m 3 /kg is a Saturated Liquid Vapor Mixture. 4 1

14 Homework - Week 03 From the chart provided we can estimate the temperature: T 100 o C 2 Also, from the chart the estimation of the quality can be obtained, estimating v f and v g 1.7: Quality at state b, v b = (1 x b )v fb + x b v gb x b = v b v fb v gb v fb x b = h = h f + x b (h g h f ) = ( ) = kj/kg 1 (c) Given that the saturated vapor point at 1 bar corresponds to m 3 /kg, it can be concluded that the state of water at 1.67 m 3 /kg is a saturated liquid vapor mixture. 4 2

15 Homework - Week 03 From the charts the temperature is estimated, T 100 o C. 2 Quality at state 1, v 1 = (1 x 1 )v f1 + x 1 v g1 x 1 = v 1 v f1 v g1 v f1 x 1 = h ( ) kj/kg 1 3

16 Homework - Week 03 HW-09 (25 points) Given: 1 Water as working fluid at the following conditions: (a) p = 1 bar and ν = 1000 m 3 /kg (b) p = 1 bar and ν = 500 m 3 /kg (c) p = 1 bar and ν = 1.67 m 3 /kg (d) p = 1 bar and ν = 2.00 m 3 /kg Find: 1 Temperature and enthalpy of water at the corresponding states EFD: N/A Assumptions: None Basic Equations: None : (a) p = 1 bar and ν = m 3 /kg T = o C 2 The enthalpy is approximated as the enthalpy of the saturated liquid. (b) p = 1 bar and ν = m 3 /kg h h f kj/kg 3 Quality at state, T = o C 2 v = (1 x)v f + xv g x = v v f v g v f x = h = h f + x b (h g h f ) = ( ) = kj/kg 3 4

17 Homework - Week 03 (c) p = 1 bar and ν = 1.67 m 3 /kg The state of water is SLVM (ν < ν g ), hence, the temperature is the same, T sat =99.61 o C 3 Interpolation for h: ν h h h (d) p = 1 bar and ν = 2.00 m 3 /kg = h = ( ) h = kj/kg 3 ( ) Interpolation for T: ν T 160 T 200 Interpolation for h: T = T = ( ) T = C 3 ( ) ν h h h = h = ( ) h = kj/kg 3 ( )

18 Homework - Week 03 HW-10 (25 points) Given: 1 Water changes from 1 bar and 25 o C to 100 bar and 35 o C Find: 1 Calculate the change in internal energy and specific enthalpy of the water per unit mass EFD: 3 Assumptions: 3 (a) Quasi-equilibrium process (b) Steady state Basic Equations: None : For the first state, since T sat > T the water is in compressed liquid (CL). Using the approximation as saturated liquid we get u 1 = kj/kg 2 Using the compressed liquid approximation for the enthalpy, we get h 1 = h liq + ν liq (P P sat ) = ( ) = kj/kg 3 For the second state, from the compressed liquid tables: Interpolation for u 2 : 6

19 Homework - Week 03 T u u Interpolation for h 2 : u = u 2 = ( ) u 2 = kj/kg 3 ( ) T h h Then, h = h 2 = ( ) h 2 = kj/kg 3 ( ) h = h 2 h 1 = kj/kg 3 u = u 2 u 1 = kj/kg 3 7

20 Homework - Week 03 HW-11 (25 points) Given: 1 Two different inflows, one of cold water at 15 o C and the other of hot water at 60 o C and one outflow at 40 o C. Heat loss of q=5 kj/kg of outflow water. Find: 1 Ratio of cold water to hot water. EFD: 3 Assumptions: 2 Steady State Pressure = 1 bar Neglect PE and KE changes. 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 : First, the enthalpy at each point is calculated: 1

21 Homework - Week 03 h = h f + ν f (P P (T sat )) h 1 = ( ) = kj/kg 3 h 2 = ( ) = kj/kg 3 h 3 = ( ) = kj/kg 3 We apply the Conservation of mass and the First Law to our Control Volume 0 = ṁ 1 + ṁ 2 ṁ = ṁ 3 q + ṁ 1 h 1 + ṁ 2 h 2 ṁ 3 h 3 0 = (ṁ 1 + ṁ 2 )q + ṁ 1 h 1 + ṁ 2 h 2 (ṁ 1 + ṁ 2 )h 3 0 = (ṁ1 ṁ 2 + 1)q + ṁ1 ṁ 2 h 1 + h 2 (ṁ1 ṁ 2 + 1)h 3 Solving for the ratio of cold water to hot water, remember that q= -5 kj/kg ṁ 1 ṁ 2 =

22 Homework - Week 03 HW-12 (25 points) Given: 1 Inlet gases of a gas turbine at 10 bar and 1500 K and exhaust gases at 2 bar and 900 K. Heat loss of 15 kw. Find: 1 Calculate the output turbine shaft work. EFD: 3 Assumptions: 2 Steady state Air is an ideal gas Neglect KE and PE changes Basic Equations: 2 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o : First, the enthalpy of air at state 1 and state 2 is retrieved from the ideal gas tables: h 1 = 1636 kj/kg 4 h 2 = kj/kg 4 3

23 Homework - Week 03 Using the first law de dt = Q Ẇ + ṁ 1h 1 ṁ 2 h = Q Ẇ + ṁ 1h 1 ṁ 2 h 2 Ẇ = Q + ṁ 1 h 1 ṁ 2 h 2 Ẇ = ( ) Ẇ = kw 4 4

24 Homework - Week 04 HW-13 (25 points) Given: 1 The inlet and outlet states of water and refrigerant R-134a flowing through a condenser Find: 1 Mass flow of cooling water EFD: 3 Assumptions: 2 Steady State The condenser is adiabatic. Neglect PE and KE changes. Inlet water is at 25 o C and 1 bar 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 : First, the enthalpy at each point is calculated. For state 3, since T> T sat at 5 bar it is in Super Heated Vapor. As given, State 4 is Saturated liquid h 3 = kj/kg 3 h 4 = kj/kg 3 1

25 Homework - Week 04 For the states 1 and 2 the Compressed liquid approximaiton is used: h = h f + ν f (P P (T sat )) h 2 = ( ) = kj/kg 3 h 1 = ( ) = kj/kg 3 Applying the First Law to our Control Volume 0 = ṁ w (h 1 h 2 ) + ṁ 134a (h 3 h 4 ) ṁ w = ṁ134a(h 3 h 4 ) (h 1 h 2 ) ṁ w = 1.18 kg/s 4 2

26 Homework - Week 04 HW-14 (25 points) Given: 1 Initial and final states of air to be stored within a cavern of m o 3 Find: 1 Calculate the work required to compress air from atmospheric pressure EFD: 3 Assumptions: 2 Air is an ideal gas Neglect KE and PE changes Basic Equations: 2 de dt = Q Ẇ + i PV=nRT : ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o R = R u = MW air 29.8 = kj/kg K 3

27 Homework - Week 04 Using the ideal gas law P 2 V = mrt 2 m 2 = P V RT = = kg 2 P 1 V = mrt 1 m 1 = P V RT = = kg 2 Retrieve the specific internal energy values from the ideal gas tables u 1 (300K) = kj/kg 2 u 2 (750K) = 552 kj/kg 2 h 1 (300K) = kj/kg 2 Applying the first law with the assumptions made t2 t 1 de dt = Ẇ + ṁ i(h) de t2 dt = Ẇ + ṁ i(h) t 1 m 2 u 2 m 1 u 1 = W + (m 2 m 1 )h 1 3 Notice that h 1 is constant through the integration given the selection of system boundary. W = (m 2 m 1 )h 1 m 2 u 2 + m 1 u 1 W = 1.22x10 8 kj W = MJ 3 4

28 Homework - Week 05 HW-16 (25 points) Given: 1 Air inside an IC engine undergoing 5 processes: (a) An insulated compression from p 1 = 1 bar and T 1 = 300 K to p 2 = 10 bar along the path pν 1.4 = constant. (b) Constant volume heat addition until T 3 = 1500 K. (c) Constant pressure heating until ν 4 = 2 ν 3. (d) Insulated expansion from ν 4 to ν 5 = ν 1 the path pν 1.4 = constant. (e) Constant volume cooling until p = p 1 Find: 1 Heat, work and internal energy changes for the different processes that the air undergoes. EFD: 1 Assumptions: 2 Quasi-equilibrium No mass flows in or out of the system Friction-less piston Air behaves as an ideal gas 1

29 Homework - Week 05 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 PV=nRT w = pdv : a) First we calculate v 1 and v 2 p 1 v 1 = RT 1 v 1 = RT 1 p 1 v 1 = m 3 /kg p 2 v = p 1 v v 2 = m 3 /kg Using the calculated values to find the specific work Using the first law for a closed system Then we find u 2 for later calculations w = 2 1 pdv = c dv ν1.4 = p 2v 2 p 1 v = kj/kg 1 u = q w u = w u = kj/kg 2 u 1 (300K) = kj/kg u 2 = u 2 = kj/kg b) Retrieving from ideal gas tables u 3 and calculating u u 3 (1500K) = 1205 kj/kg u = u 3 u 2 u = kj/kg 1 2

30 Homework - Week 05 From the first law analysis u = q w u = q q = kj/kg 2 c) Calculate v 4 = 2v 3 = 2v 2 = = m 3 /kg, then calculate p 3 through the ideal gas law Evaluate the specific work integral p 3 v 3 = RT 3 p 3 = kp a w = Then we use the ideal gas law to calculate T pdv = p 3 (v 4 v 3 ) = kj/kg 1 p 4 v 4 = RT T 4 = T 4 = 3000 K Since u(3000k) is not on the ideal gas tables we extrapolate to get the required value. for u 4 : Interpolation T u 1873 u Then u = u 4 = ( ) u 4 = kj/kg ( ) u = u 4 u 3 u = u = 1436 kj/kg 1 3

31 Homework - Week 05 Applying the first law, we obtain: q = u + w q = q = kj/kg 1 d) Calculate p 5, knowing that v 5 = v 1 = m 3 /kg p 5 v5 1.4 = p 4 v4 1.4 p 5 = kp a Calculate the specific work using the integral over the respective path Applying first law e) w = p 5v 5 p 4 v w = kj/kg 2 u = q w u = kj/kg 1 Applying the first law u 5 = u + u 4 u 5 = u 5 = kj/kg f) q = u 5 u 1 q = q = kj/kg 1 u = kj/kg 2 4

32 Homework - Week

33 Homework - Week 05 HW-17 (25 points) Given: 1 Schematic of a turbojet engine with different properties across its different sections. Find: 1 (a) Temperature at the end of the intake section (b) The compressor exit pressure and temperature (c) Turbine exit temperature (d) Nozzle exit velocity EFD: 3 Assumptions: 2 Air is an ideal gas Neglect PE changes All sections except the combustor are adiabatic Basic Equations: 2 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o : a) Applying the first law to the diffuser, we obtain: 0 = ṁ(h a + V a h 1 V ) 2 h 1 = h a + V a h 1 = kj/kg 6

34 Homework - Week 05 From the ideal gas tables T 1 = 285 K 2 b) Assuming the inlet pressure is 0.3 bar we can find that P outlet = 6 bar and from the given information we know that T exit =600 K 4 c) First we need to calculate the specific work required in the compressor Now, we apply the first law to the turbine Ẇ ṁ = h 1 h 2 (600K) Ẇ = ṁ Ẇ = kj/kg 2 ṁ 0 = Ẇ ṁ + h 3 h 4 h 4 = Ẇ ṁ + h 3(1750K) h 4 = h 4 = kj/kg 2 Using h 4 we interpolate to find T 4 = K d) Applying the first law to the nozzle we obtain: V 2 5 h 4 = h 5 (850K) + V = V 5 = m/s 2 7

35 Homework - Week 05 HW-18 (25 points) Given: 1 A 5 kg block of iron at 95 o C comes into equilibrium with a water vat at 25 o C. Water should be treated as a thermal reservoir. Find: 1 The entropy generation for water and iron as the system. EFD: 3 Assumptions: 2 The water vat is adiabatic. Neglect PE and KE changes. Water and iron are incompressible substances Basic Equations: 2 1

36 Homework - Week 05 dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o ds dt = Q j j T j + i ṁ i (s) i o ṁo(s) o + σ : We start with the closed system entropy balance S = j σ = S We evaluate the entropy change for the iron Q T + σ = S water + S F e S F e = m c F e ln T F inal T initial = ln = kj/kgk 4 For the water entropy change we apply the second law to the EFD II, taking the water as the system and assuming the process is reversible (σ = 0) for the water given that its temperature changes slightly: S water = Q T To find the heat going into the water we apply the following equation to the iron block: Q = m c T = ( 70) = kj 4 Evaluating the entropy generation of the water, notice that the heat is negative for the block but positive for the water. S water = = kj/k 4 σ = S = = kj/k 4 2

37 Homework - Week 05 HW-20 (25 points) Given: 1 Initial temperature and mass for a water vat and for an iron block block. T i,water = 25C T i,f e = 95C m w = 100kg m F e = 5kg Find: 1 Final temperature of the water and entropy changes for the iron and the water EFD: 1 Assumptions: 2 No mass flows in or out of the system Friction-less piston Water is an incompressible fluid 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 ds dt = Q j j T j + i ṁ i (s) i o ṁo(s) o + σ S = c ln T f T i : 1

38 Homework - Week 05 Applying the first law to our system U = 0 U w + U F e = 0 m w C w (T f T i ) + m F e C F e (T f T i ) = (T f ) 25) (T f 95) = 0 T f = 25.38C 6 For the entropy changes S F e = C F e ln T f T i S F e = 0.46 ln = kj/kgk 6 S W ater = C W ater ln T f T i S W ater = 4.18 ln = kj/kgk 6 2

39 Homework - Week 05 HW-21 (25 points) Given: 1 Schematic of a turbojet engine with different properties across its different sections. Find: 1 The entropy change of all the devices that are part of the turbojet. EFD: 3 Assumptions: 2 Air is an ideal gas Neglect PE changes All sections except the combustor are adiabatic There is no pressure drop in the combustor The turbine is isentropic Basic Equations: 2 ds dt = Q j j T j + i ṁ i (s) i o ṁo(s) o + σ S = s o 2 s o 1 R ln p 2 p 1 : a) For the diffuser s = s in s out = ln = kj/kgk 3 3

40 Homework - Week 05 b) For the compressor c) For the combustor s = s in s out = ln = kj/kgk 3 s = s in s out d) For the turbine = ln 6 6 = kj/kgk 3 s = 0 2 e) For the nozzle, the outlet pressure of the turbine is required. Interpolation for s o : T s o s o Solving for the pressure s o = s o = ( ) s o = kj/kg ( ) s = 0 = ln 6 p oulet p outlet = 3 bar 2 Now, using the turbine outlet pressure we find the entropy change for the nozzle = s in s out = ln = kj/kgk 3 4

41 Homework - Week 06 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 ds 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

42 Homework - Week 06 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

43 Homework - Week 06 Calculating the enthalpies changes: Compressor h 2 h 1 = kj/kg 1 Combustor h 3 h 2 = kj/kg 1 Turbine h 4 h 3 = kj/kg 2 c) For the compressor from the second law and using the given information Combustor Turbine d) Compressor Turbine 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 ds dt = Q T + ṁ(s 2 s 3 ) + σ 0 = q T + (s 2 s 3 ) + σ σ = kj/kgk 3 3

44 Homework - Week 06 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 ds dt = j : a) Q j T j + i ṁ i (s) i o ṁo(s) o + σ 4

45 Homework - Week 06 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

46 Homework - Week 06 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 ds 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

47 Homework - Week 06 4 b) From the ideal gas tables the following information is retrieved: Calculating the enthalpies changes: Compressor Combustor Turbine 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

48 Homework - Week 06 Compressor Combustor Turbine d) Compressor Turbine 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 ds dt = Q T + ṁ(s 2 s 3 ) + σ 0 = q T + (s 2 s 3 ) + σ σ = kj/kgk 2 8

49 Homework - Week 07 HW-25 (25 points) Given: 1 A turbojet nozzle with know inlet and outlet parameters Find: 1 Calculate isentropic efficiency and entropy generation. EFD: 2 Assumptions: 1 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow Nozzle is adiabatic and does not have any boundary work 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 η nozzle = h h s = s o 2 s o 1 R ln p 2 p 1 : a) From HW 17 we retrieve that s o 1 = kj/kg Interpolation for h 1 : 1

50 Homework - Week 07 T h 1612 h h = h 1 = ( ) h 1 = K ( ) s = s o 2 s o 1 R ln p 2 p 1 0 = s o ln s o 2s = kj/kgk Retrieving from the ideal tables we obtain T 2 = 840 K and h 2s = 866 kj/kg 2, also from the ideal gas table we retrieve the enthalpy value at the oulet of the nozzle h 2 = kj/kg 2 η nozzle = h h s η nozzle = h 1 h 2 h 1 h 2 s = = 98.52% 6 b) Applying the second law, and using the assumptions of SSSF and adiabatic nozzle we obtain σ = s 2 s 1 = ln = kj/kgk 6 2

51 Homework - Week 07 HW-27 (25 points) Given: 1 Schematic of a turbojet engine with inlet and outlet properties Find: 1 Calculate the isentropic efficiency and the entropy generation of the turbine and compressor EFD: 2 Assumptions: 1 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow The compressor and turbine are adiabatic 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 ds dt = j s = s o 2 s o 1 R ln p 2 p 1 : Q j T j + i ṁ i (s) i o ṁo(s) o + σ 3

52 Homework - Week 07 a) From previous homework and using the given information we obtain for the first state T 1 = 285 K h 1 = kj/kg p 1 = 0.3 bar s o 1 = kj/kgk For the second state T 2 = 700 K h 2 = kj/kg p 2 = 6 bar s = s o 2s s o 1 R ln p 2 p 1 0 = s o 2s ln s o 2s = From the ideal gas tables we obtain h 2s = kj/kg 3, calculating the isentropic efficiency for the compressor η compressor = h s h η compressor = h 1 h 2 s h 1 h 2 = = 89.99% 4 For the turbine, from HW 21 we assume the turbine to be isentropic, hence: η turbine = 100% 3 For the specific entropy generation calculations, using the assumption that the compressor and the turbine are adiabatic we obtain σ comp = s 2 s 1 = s o 2 s o 1 R ln p 2 p 1 = ln = kj/kgk 4 σ turb = s 4 s 3 = 0 kj/kgk 4 4

53 Homework - Week 07 HW-28 (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 Calculate the isentropic effciency of the turbine and the compressor EFD: 2 Assumptions: 1 Neglect PE and KE Air behaves as an ideal gas Steady state and steady flow The compressor and turbine are adiabatic 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 ds dt = j s = s o 2 s o 1 R ln p 2 p 1 : Q j T j + i ṁ i (s) i o ṁo(s) o + σ 5

54 Homework - Week 07 From the ideal gas tables the following information is retrieved: 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 Compressor s = s 1 s 2s = s o 1 s o 2s R ln(p 1 /p 2 ) 0 = s o 2s ln(0.7/17.5) s o 2s = kj/kgk 3 Turbine s = s 3 s 4s = s o 3 s o 4s R ln(p 3 /p 4 ) = s o 4s ln(17.5/0.7) s o 4s = kj/kgk 3 Interpolation for h 2s : s o h h 2s Interpolation for h 4s : h 2s = h 2s = ( ) h 2s = kj/kg ( )

55 Homework - Week 07 s o h h 4s h 4s = h 4s = ( ) h 4s = kj/kg ( ) η compressor = h s h η compressor = h 1 h 2 s h 1 h 2 = = 82.64% 3 η turbine = h h s η turbine = h 3 h 4 h 3 h 4 s = = 95.52% 3 7

56 Homework - Week 07 HW-29 (25 points) Given: 1 Air flowing through a compressor at steady-state with inlet and outlet properties as stated. Find: 1 The amount of work required for different processes and its entropy generation.. EFD: 2 Assumptions: 1 Neglect PE and KE changes Steady State and Steady Flow 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 ds dt = j s = s o 2 s o 1 R ln p 2 p 1 Q j T j + i ṁ i (s) i o ṁo(s) o + σ : Since the prompt is asking for the minimum required work, σ = 0. i) From the ideal gas tables we retrieve h 1 = kj/kg s o 1 = kj/kgk 8

57 Homework - Week 07 Applying the isentropic assumption Interpolation for h 2 s: s = s o 2s s o 1 R ln p 2 p 1 0 = s o 2s ln s o 2s = kj/kgk s o h h 2 s h 2 s = h 2 s = ( ) h 2 s = kj/kgk ( ) w = h 1 h 2s = = kj/kgk 3 For the entropy change from the definition of isentropic we know, s = 0 3 ii) From the definition of flow work for an internally reversible process, w = = vdp RT p dp = RT ln p 1 p 2 = ln = kj/kg 3 This result may also be obtained by applying the second law 0 = q T + s 1 s 2 and the first law 0 = q w For the entropy change s = s o 2 s o 1 R ln p 2 p 1 = ln = kj/kgk 3 9

58 Homework - Week 07 iii) For a polytropic process pv n = c and for an internally reversible process w = vdp = n n 1 (p 2v 2 p 1 v 1 ) = n n 1 R(T 2 T 1 ) Using the ideal gas law, the definition of a polytropic process and assuming n = 1.2 T 2 T 1 = ( p 2 p 1 ) 1 (1/n) T 2 = 433 K Interpolation for s o 2: w = kj/kg 3 T s o s o s o Calculating the entropy change: = s o 2 = ( ) s o 2 = kj/kgk s = ln = kj/kgk 3 ( ) The process that requires the less amount of work is the isothermal process. 10

59 Homework - Week 05 HW-30 (25 points) Given: 1 A Rankine cycle with inlet and outlet values as shown on the EFD Find: 1 Calculate the net total work and the efficiency of the cycle EFD: 1 Assumptions: 2 Steady State Steady Flow State 3 is in Saturated Liquid 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 ds dt = Q j j T j + i ṁ i (s) i o ṁo(s) o + σ η isen = h hs : From the Super Heated Vapor tables h 1 = 3323 kj/kg s 1 = kj/kgk 1

60 Homework - Week 05 s 1 = s 2s = s f + x(s g s f ) = x 2s ( ) x 2s = Calculating h 2s h 2s = ( ) = kj/kg From the definition of isentropic efficiency for a turbine η turb = h 1 h 2 h 1 h 2s h 2 = kj/kg From the Saturated Liquid Vapor Mixture Tables h 3 = kj/kg s 3 = kj/kgk From the Compressed tables, using interpolation and the fact that s 3 = s 4s h 4s = kj/kg η pump = h 3 h 4s h 3 h 4 h 4 = kj/kg Ẇ net = ṁ(h 1 h 2 h 4 + h 3 ) = kw From the lecture book the efficiency may be calculated as η cycle = (h 1 h 2 ) + (h 3 h 4 ) (h 1 h 4 ) = = 35.57% 2

61 Homework - Week 05 HW-31 (25 points) Given: 1 A Rankine cycle with reheating, with given values as shown in the EFD Find: 1 Calculate the efficiency of the cycle and compare it to the Carnot cycle EFD: 3 Assumptions: 2 Neglect PE changes Pump and turbines are isentropic Assume Cold reservoir is at 25 C 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 ds dt = Q j j T j + i ṁ i (s) i o ṁo(s) o + σ η isen = h hs : For calculating the efficiency of the cycle we need to calculate all the enthalpies, from the SHV 3

62 Homework - Week 05 tables we obtain, h 1 = kj/kg s 1 = kj/kgk From the SLVM tables we obtain the values for the state 4 and 5 h 4 = ( ) h 4 = 2325 kj/kg s 4 = ( ) s 4 = kj/kgk h 5 = kj/kg s 5 = kj/kgk From the prompt we know that s 1 = s 2 s 3 = s 4 s 5 = s 6 From the SHV tables and using interpolation we obtain the values for state 3 Interpolation for h 3 : P h h h = h 3 = ( ) h 3 = kj/kg ( ) s 3 = s 4 = kj/kgk For State 2 p 2 = p 3 and s 2 = s 1, for simplicity we may assume p 2 15 bar. Interpolation for h 2 : 4

63 Homework - Week 05 s h h h = h 2 = ( ) ( ) h 2 = kj/kg For state 6 we use the fact that p 6 = p 1 and that s 5 = s 6. Interpolation from the Compressed liquid tables gives us Interpolation for h 2 : s h h h = h 2 = ( ) ( ) h 2 = kj/kg Finally, using all the calculated enthalpies we obtain the cycle efficiency For the Carnot efficiency η = (h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 ) (h 1 h 6 ) + (h 3 h 2 ) = 44.3% η carnot = 1 T cold T hot = = 65.8% 5