Solutions to Midterm Examination

Solutions to Midterm Examination 1. Problems a through e below apply to an ideal gas for which the gas constant, R = 0.25 kJ/kg; the heat capacity of this gas is given by the following equation: cp = a + b/T with a = 1 kJ/kg-K and b = -50 kJ/kg. a. Compute the heat transfer in a closed system expansion from 1 m3/kg to 10 m3/kg at a constant temperature of 500 K. For a closed system, Q = ∆U + W = m(u2 – u1) + ∫PdV. Dividing these equations by the mass gives q = u2 – u1 + w = u2 – u1 + ∫Pdv. For an ideal gas, the internal energy is a function of temperature only. So for the isothermal process, ∆U = 0 and q = w. For an ideal gas, Pv = RT so that the path of an isothermal process in an ideal gas is P = RT/v where RT is a constant. Integrating this equation gives the work. w= v2 ∫ v1  10 m 3   v 2  0.25 kJ dv RT kg (500 K ) ln dv = RT ∫ = RT ln  =  1 m3 v v  v 1  kg ⋅ K v1   kg v2    = 287.8 kJ  kg   Since q = w in this problem, we have q = 287.8 kJ/kg . b. Compute the heat transfer in a closed-system compression from 1 MPa to 10 MPa at a constant temperature of 400 K. This is the same problem, with different data, as given in part a. Thus we can use the same equations to find the work and the heat transfer. In this case we are not given the initial and final volumes that are required to find the work. However, we can get them from the ideal gas equation of state: Pv = RT. Substituting v = RT/P in the equation for the work from part a gives us the work for this problem as follows.  RT   v2   P  0.25 kJ  1 MPa   P2  230.3 kJ  = − = RT ln 1  = w = RT ln  = RT ln ( 400 K ) ln  kg  RT P   10 MPa   v1   P2  kg ⋅ K 1   Since q = w in this problem, we have q = -230.3 kJ/kg . c. Find the entropy changes for the processes in questions a and b. The entropy change in an ideal gas is given by either of the following equations. T2 T  v  2 dT P  dT + R ln 2  = ∫ c p − R ln 2  s2 − s1 = ∫ cv T T  v 1  T1  P1  T1 Midterm exam solutions Page 2 Since the temperature is constant in questions a and b, we can find the entropy change for each process by considering only the pressure or specific volume terms. This gives the following results for the entropy changes in a and b.  10 m 3   v 2  0.25 kJ  kg ln = R ln  = 3  v 1  kg ⋅ K  1 m   kg ∆sa = (s2 − s1 )a       ∆sa = P  0.25 kJ  10 MPa   ∆sb = (s2 − s1 )b = − R ln 2  = − ln P kg K MPa ⋅ 1    1 0.5756 kJ kg ⋅ K ∆sb = − 0.5756 kJ kg ⋅ K d. The gas undergoes an isentropic compression starting from an initial state of 300 K and 1 MPa to a final temperature of 400 K. What is the final pressure? We can use the equation for the ideal gas entropy from part c and set s2 = s1 to get the results for an isentropic process. This gives the following equation to solve for the final pressure, P2. T2 T P  dT − R ln 2  = 0 s2 − s1 = ∫ c p T  P1  T1 ⇒  P  1 2 dT ln 2  = ∫ c p T  P1  R T1 Since cp = a + b/T, we can evaluate the integral, ∫(cpdT/T) as follows: T T T  P2  1 2 dT 1 2  1  b  dT 1  b  2 1   T2   1 = ∫ a +  = a ln(T ) −  = a ln  − b 2 − 2  ln  = ∫ c p T R T1  T T R T  T1 R   T1   T2 T1   P1  R T1 Plugging in the data given for R, a, and b gives the log of the pressure ratio.  P  1  T   1 1  ln 2  = a ln 2  − b 2 − 2  =  P1  R   T1   T2 T1  kg ⋅ K 0.25 kJ  1 kJ  400 K   − 50 kJ  −  ln   kg ⋅ K  300 K   kg  1 1   = 0.98406 −   400 K 300 K  This gives P2 = P1 e0.98406 = (1 MPa)(2.6753) or P2 = 2.657 MPa . e. Find the work and the heat transfer, if the isentropic compression in part d takes place in a closed system. We can use the closed system first law from parts a and b: q = u2 – u1 + w. Because this is an isentropic process, q = 0 so we have w = u1 – u2. Since this is an ideal gas, du = cvdT = (cp – R)dT. We can find the value of u1 – u2 by integrating (cp – R)dT from T2 = 400 K to T1 = 300 K. T1 w = u1 − u2 = ∫ cv dT = T2 T1 ∫ (c T2 T1 p T  b   − R )dT = ∫  a + − R dT = (a − R )(T1 − T2 ) + b ln 1  T   T2  T2  We find the work by substituting the given values for a, b, and R.  T   1 kJ 0.25 kJ w = (a − R )(T1 − T2 ) + b ln 1  =  − kg ⋅ K  T2   kg ⋅ K   50 kJ (300 K − 400 K ) +  − kg     300 K    ln   400 K  Midterm exam solutions Page 3 w=− 60.62 kJ kg ⋅ K As expected, the work is negative indicating the work input required to compress the gas. 2. This problem deals with a turbine in which 12 kg/s of steam enters the turbine at a temperature and pressure of 600oC and 10 MPa. There are two exit streams: the first is at a pressure of 1 MPa and the second is at a pressure of 100 kPa. These data apply to all three parts of the problem given below. a. The first (1 MPa) exit stream has a flow rate of 2 kg/s at a temperature of 300oC and the second is in the mixed region with a quality of 95%. The turbine has a heat loss of 500 kW. What is the power output of the turbine? We assume that the turbine is a steady flow device with negligible changes in kinetic and potential energy. With this assumption, the first law for open systems can be written as follows for the turbine. In this equation the subscript 0 denotes the entrance and the subscripts 1 and 2 denote the first and second exit streams respectively. Q = W u + m 1h1 + m 2 h2 − m 0 h0 The mass balance equation for this system can be written as follows: m 1 + m 2 = m 0 ⇒ m 2 = m 0 − m 1 We see that the flow rate of the second exit stream is just the difference between the inlet flow  2 = 12 kg/s – 2 kg/s = 10 kg/s. and the first outlet stream so that m The required enthalpy values for the first law are found from the steam tables: h0 = h(600oC, 10 MPa) = 3625.3 kJ/kg; h1 = h(300oC, 1 MPa) = 3051.2 kJ/kg; h2 = h(100 kPa, x = 0.96) = hf(100 kPa) + xhfg(100 kPa) = 417.46 kJ/kg + 0.95(2258.0 kJ/kg) = 2562.6 kJ/kg. We can solve the first law for the turbine work and note that the heat transfer term is –500 kW since it is a heat loss (i.e. heat transferred from the turbine.) This gives the turbine power output as follows. 12 kg 3625.3 kJ W u = Q + m 0 h0 − m 1h1 − m 2 h2 = −500 kW +  kg  s − 2 kg 3051.2 kJ 10 kg 2562.6 kJ  kW ⋅ s −  kJ s kg s kg  W u = 11275 kW = 11.28 MW b. The first (1 MPa) exit stream has a temperature of 300oC and the second leaves is in the mixed region with a quality of 95%. The turbine has a heat loss of 500 kW and a power output of 12 MW. What are the flow rates of the exit streams? In this problem, we have the same inlet and exit states as in the previous problem. The only difference is that we are given the work here, but we do not know either of the exit mass-flow rates. If we substitute all the given data, and the enthalpy values that we found in part a into our first law and our mass balance, we obtain the following results. 12 kg 3625.3 kJ W u = 12 MW = 12000 kW = Q + m 0 h0 − m 1h1 − m 2 h2 = −500 kW +  kg  s − m 2 kg 3051.2 kJ m 2 kg 2562.6 kJ  kW ⋅ s −  kJ s kg s kg  m 0 = 12 kg = m 1 + m 2 s Midterm exam solutions Page 4 This gives two simultaneous equations that we can solve for the mass flow rates. We note that the equations are dimensionally consistent once we convert the power output to kW, so we can drop the explicit consideration of units as we solve the equations. To solve this pair of equations  2 = 12 − m 1 , and substitute this result into the first law equation to solve for m 1 . we first write m 12000 = −500 + 43503.6 − 3051.2m 1 − 2562.5(12 − m 1 ) We then find m 2 = 12 − m 1 = 12 kg s − 0.5166 kg s so that m 1 = 0.5166 kg m 2 = 11.4834 kg s s c. Find the maximum work output of the turbine under the following conditions: it is adiabatic and the outflow of the first exit stream is 2 kg/s. We know that the maximum work in an adiabatic process occurs when the entropy is constant. In this problem, with one inlet and two exit streams, the entropy of both exit streams has to be the same as that of the inlet stream for an isentropic process. We start by finding the inlet entropy, s0 = s(1 MPa, 600oC) = 6.9029 kJ/kg∙K. We then have to find the enthalpy values for the outlet streams when they have their given pressure and an entropy value of 6.9029 kJ/kg∙K. For the first outlet stream, with P = 1 MPa, we have to interpolate between 200oC and 250oC to get h1 = h(1 MPa, s1 = s0 = 6.9029 kJ/kg∙K) = 2931.8 kJ/kg. At the second outlet, the entropy value of 6.9029 kJ/kg∙K is in the mixed region so we have to compute the quality to get the enthalpy, h2. x2 = s2 − s f (100 kPa ) s fg (100 kPa ) = 6.9029 kJ h2 = h f (100 kPa ) + x 2 h fg (100 kPa ) = − 1.3026 kJ kg ⋅ K kg ⋅ K = 0.9246 6.0568 kJ kg ⋅ K  2258.0 kJ 417.46 kJ + (0.9246) kg kg   2505.3 kJ  = kg  We can still use the first law analysis used for the previous parts, except that the heat transfer is now zero. 12 kg 3625.3 kJ W u = Q + m 0 h0 − m 1h1 − m 2 h2 = 0 kW +  kg  s − 2 kg 2931.8 kJ 10 kg 2505.3 kJ  MW ⋅ s −  1000 kJ s kg s kg  W u = 12.59 MW