Thermodynamics - Homework Solutions

Homework Solutions 6-42E Steam enters an adiabatic turbine at 800 psia and 900oF and leaves at a pressure of 40 psia. Determine the maximum amount of work that can be delivered by this turbine. Assume that the turbine is a steady flow device with negligible changes in kinetic and potential energies. For this adiabatic turbine, with one inlet and one outlet, the first law gives the work as w = hin – hout. The maximum work in an adiabatic process occurs when the outlet entropy is the same as the inlet entropy. Use the steam tables to find the properties. In this case, hin = h(800 psia 900oF) = 1455.6 Btu/lbm and sin = s(800 psia 900oF) = 1.6408 Btu/lbm∙R. At the outlet pressure of 40 psia, the entropy value of 1.6408 Btu/lbm∙R is seen to be in the mixed region, so we have to compute the quality to find the outlet enthalpy. xout = sout − s f ( 40 psia ) s fg ( 40 psia ) = 1.6408 Btu hout = h f ( 40 psia ) + xout h fg ( 40 psia ) = lbm ⋅ R − 0.39214 kJ 1.2845 kJ kg ⋅ K = 0.972 kg ⋅ K  933.8 Btu  1143.8 Btu 236.16 Btu  = + (0.972) lbm lbm   lbm We can now compute the maximum work. w = hin − hout = 6-68 1455.6 Btu 1143.8 Btu − lbm lbm w= 311.8 Btu lbm Air is compressed steadily by a 5-kW compressor form 100 kPa and 17oC to 600 kPa and 167oC at a rate of 1.6 kg/min. During this process some heat transfer takes place between the compressor and the surrounding medium at 17oC. Determine the rate of entropy change of air during this process. It appears that we have a large amount of extraneous information in this problem. Since we are only asked to find the entropy change of the air and the process is steady, the desired entropy change is given by the following equation if we use the air tables to find the entropy change, accounting for the variable heat capacity.  o  P  o Sair = m air (sout − sin ) = m air  sair (T2 ) − sair (T1 ) − R ln 2   P1   Using the values of so from the air tables and R = 0.287 kJ/kg∙K for air gives the answer as follows. Page 2  o  P  1.6 kg min  2.0877 kJ o Sair = m air  sair ( 440.15 K ) − sair ( 290.15 K ) − R ln 2  = min 60 s  kg ⋅ K  P1   − 6-76 1.66802 kJ 0.287 kJ  600 kPa  kW ⋅ s  − ln kg ⋅ K kg ⋅ K  100 kPa  1 kJ 0.00250 kW Sair = − K Air enters a nozzle steadily at 280 kPa and 77oC with a velocity of 50 m/s and exits at 85 kPa and 320m/s. The heat losses from the nozzle to the surrounding medium at at 20oC are estimated to be 3.2 kJ/kg. Determine (a) the exit temperature and (b) the total entropy change for this process. Here we have a steady flow system with one inlet and one outlet. There is no useful work in the nozzle. We neglect changes in potential energy and write the first law as follows   V2  V2   −  h +  q =  h + 2  out  2  in  We use the air tables to account for the temperature variation of the heat capacity and we note that q = -3.2 kJ/kg since this is a heat loss. With the inlet enthalpy, hin = h(350 K) = 350.49 kJ/kg from the air tables, we can solve our first law for the outlet enthalpy as follows. 2 hout = q + hin + 2 Vin2 − Vout 2 2  50 m   320 m    −  3.2 kJ 350.49 kJ  s   s  kJ ⋅ s 2 =− + + kg kg 2 1000 kg ⋅ m 2 This gives hout = 297.34 kJ/kg. Interpolating in the air tables to find this enthalpy, we find that it occurs at Tout = 297.2 K . The total entropy change, ∆stotal is the sum of the entropy change of the air, ∆sair, plus that of the surroundings, ∆ssurround. We find the entropy change of the air from the usual equation for the air tables. P   85 kPa  o o o o  = ∆sair = sair (T2 ) − sair (T1 ) − R ln 2  = sair ( 297.2 K ) − sair (350 K ) − R ln  280 kPa   P1  1.6924 kJ 0.287 kJ  85 kPa  0.1775 kJ  = ln − kg ⋅ K kg ⋅ K kg ⋅ K  280 kPa  The entropy change of the surroundings (per kilogram of air) is the heat transfer divided by the constant temperature of 293.15 K for the surroundings. Note that the heat transfer to the surroundings is the negative of the heat transfer to the nozzle: qsurround = 3.2 kJ/kg. ∆s surr = qsurr Tsurr 3.2 kJ 0.0109 kJ kg = = 293.15 K kg ⋅ K Adding the entropy change of the air and the surroundings gives the total entropy change. Page 3 ∆stotal = ∆sair + ∆s surr = 6-95 0.1775 kJ 0.0109 kJ + kg ⋅ K kg ⋅ K ∆stotal = 0.1884 kJ kg ⋅ K Steam enters an adiabatic turbine at 8 MPa and 500oC with a mass flow rate of 3 kg/s and leaves at 30 kPa. The isentropic efficiency of the turbine is 90%. Neglecting the kinetic energy change of the steam, determine (a) the temperature at the turbine exit and (b) the power output of the turbine. We assume that we have a steady flow process in which the kinetic and potential energies are negligible. We are given that the turbine is adiabatic and we note that is has only one inlet and one outlet. We thus have the simple form of the first law, w = hin = hout. This equation applies to both the actual and the ideal process. We first compute the ideal work, ws = hin – hout,s. From the steam tables we find that hin = h(8 MPa, 500oC) = 3398.3 kJ/kg and sin = s(8 MPa, 500oC) = 6.7240 kJ/kg∙K. The ideal outlet state of 30 kPa and sout,s = sin is seen to be in the mixed region, so we have to compute the quality to find the outlet enthalpy. xout = sout − s f (30 kPa ) s fg (30 kPa ) = 6.7240 kJ − 0.9439 kJ kg ⋅ K kg ⋅ K = 0.847 6.8247 kJ kg ⋅ K hout = h f (30 kPa ) + xout h fg (30 kPa ) =  2336.1 kJ 289.23 kJ + (0.847) kg kg   2267.9 kJ  = kg  We can now compute the maximum work. ws = hin − hout ,s = 3398.3 kJ 2267.9 kJ 1130.4 kJ − = kg kg kg Multiplying this by the isentropic efficiency gives the actual work and multiplying that result by the mass flow rate gives the power output of the turbine. 3 kg (90% ) 1130.4 kJ MW ⋅ s W = 3.052 MW W = m w = m η s ws = s kg 1000 kJ Applying the first law to the actual process gives hout,a as follows. hout ,a = hin − w = hin − 2380.9 kJ 3.052 MW W 3398.3 kJ = − = 3 kg MW ⋅ s kg m kg s 1000 kJ At the outlet pressure of 30 kPa, the value of hout = 2380.9 kJ/kg is seen to be in the mixed region so the final temperature, Tout = Tsat(30 kPa) = 69.10oC . 6-100 Refrigerant 134a enters an adiabatic compressor as a saturated vapor at 120 kPa at a rate of 0.3 m3/min and exits at 1 MPa pressure. If the isentropic efficiency of the compressor is 80%, determine (a) the temperature of the refrigerant at the exit of the compressor and (b) the power input in kW. Also, show the process on a T-s diagram with respect to the saturation lines. Page 4 We assume a steady process with negligible changes in kinetic and potential energies. We are given that the compressor is adiabatic and we see that it has one inlet and one outlet. Thus the first law for this compressor reduces to W = m ( hin − hout ) . We use the property tables for refrigerant 134a. On a per unit mass basis we can compute the ideal work, ws = hin – hout,s and the actual work wa = hin – hout. At the inlet state, a saturated vapor at 120 kPa, we find the following properties: hin = hg(120 kPa) = 233.86 kJ/kg; sin = sg(120 kPa) = 0.9354 kJ/kg∙K; vin = vg(120 kPa) = 0.1614 m3/kg. We can use the inlet volume flow rate and the inlet specific volume to compute the mass flow rate. 0.3 m 3 min V min 60 s 0.031 kg = m = in = 0.1614 m 3 s vin kg At the ideal outlet state of Pout = 1 MPa and sout,s = sin = 0.9354 kJ/kg∙K, we find the ideal outlet enthalpy to be 277.84 kJ/kg. This allows us to compute the ideal work per unit mass. ws = hin − hout ,s = 233.86 kJ 277.84 kJ 43.98 kJ − =− kg kg kg The negative work confirms out understanding that this is a work input device so that we have to divide the ideal work by the isentropic efficiency to find the actual work. Multiplying the actual work by the mass flow rate gives the power. 0.031 kg  43.98 kJ  −  s kg m w  s = W = m w = ηs 80%  kW ⋅ s   1 kJ W = −1.70 kW The negative sign on the power confirms that this is a power input. Applying the first law to the actual process gives hout,a as follows. hout ,a = hin − w = hin − (− 1.70 kW ) = 288.84 kJ W 233.86 kJ = − 0.031 kg kW ⋅ s kg m kg 1 kJ s At the outlet pressure of 1 MPa, the value of hout = 288.84 kJ/kg gives an outlet temperature which can be found from interpolation in the refrigerant-134a tables: Tout = 57.7oC .