Isentropic Efficiency

Solutions to Group Exercise for Unit Ten – Isentropic Efficiency A steam turbine has an inlet temperature and pressure of 900oC and 3.0 MPa. Its outlet pressure is 600 kPa. The turbine is an adiabatic, steady steady-flow system. Its isentropic efficiency is 80%. The turbine exhaust is fed into a constant pressure heater that increases the temperature of the steam to its initial temperature of 900oC. Find (a) the work output of the turbine and (b) the heat transfer required to heat the steam to 900oC in the heater. Repeat the calculation three times using the following assumptions for the property calculations: 1. Property data for water from the steam tables. For a steady-state system with one inlet and one outlet and negligible changes in kinetic and potential energy, the first law reduces to q = w + hout – hin. For an adiabatic process, q = 0, and w = hin – hout. The isentropic efficiency, ηs, is used to find the actual work, w, from the maximum or isentropic work, ws. ηs = w/ws. We find the following properties from the Table A-6 on page 921, for Pin = 3 MPa and Tin = 900oC: hin = 4387.5 kJ/kg and sin = 8.2028 kJ/kg∙K. The isentropic work is found when sout = sin. At the ideal outlet state of sout = sin = 8.2028 kJ/kg∙K, and Pout = 600 kPa, the final temperature is between 500oC and 600oC. We can interpolate to find the enthalpy at this point as follows: hout ,s 3701.7 kJ 3483.4 kJ − 3483.4 kJ kg kg = + 8.2695 kJ 8.0041 kJ kg − kg ⋅ K kg ⋅ K  8.2028 kJ 8.0041 kJ  3646.8 kJ  = − kg ⋅ K  kg  kg ⋅ K This enthalpy is used to compute the ideal work: ws = hin – hout,s = 4387.5 kJ/kg – 3646.8 kJ/kg = 740.7 kJ/kg. Multiplying this by the isentropic efficiency of 80% gives the actual work: w = ηsws.=(80%)(740.7 kJ/kg) or w = 592.5 kJ/kg . We can find the actual enthalpy outlet form the turbine using the first law with this actual work: hout = hin – w = 4387.5 kJ/kg – 592.5 kJ/kg = 3795.0 kJ/kg. The subsequent heater will have no useful work, so the first law for the heater is q = hout – hin. The inlet enthalpy to the heater is the actual outlet enthalpy from the turbine, 3795.0 kJ/kg. Since the pressure is constant, the outlet conditions for the heater are Pout = Pin = 600 kPa and the given value of Tout = 900 oC. At this point we find hout = 4396.2 kJ/kg from Table A-6 on page 920. Thus the heat transfer in the heater is computed as q = q = hout – hin. = 4396.2 kJ/kg – 3795.0 kJ/kg or q = 601.2 kJ/kg . 2. Constant-heat-capacity ideal gas behavior for water. Here we apply the same first law analyses that we used in answering problem one. The only difference is that we use the constant heat capacity ideal gas relations. With these equations we find enthalpy changes, ∆h, as cp∆T, and we find the end states of an isentropic process are related by the equation T2,s = T1(P2/P1)(k-1)/k. We could use the power series equations for cp in Table A-2(c) on page 913, to get a value of cp at an average temperature for this process. These equations actually give a molar heat capacity. We would have to divide by the molecular weight of water to get a heat capacity per unit mass. Finally we would have to compute k = cp/cv = cp/(cp – R). To simplify this example let’s use the low temperature value of cp from Table A-1(a), page 911: cp = 1.8723 kJ/kg∙K and k = 1.327. With these properties we compute the ideal outlet temperature as follows: Tout,s = (1173.15 K) (600 kPa / 3000 kPa)(1.327-1)/1.327 = 789.1 K. Thus the ideal work, ws = cp (Tin – Tout,s) = (1.8723 kJ/kg∙K)(1173.15 K – 789.1 K) = 719.1 kJ/kg. Multiplying this by the isentropic efficiency of 80% gives the actual work: w = ηsws.=(80%)(719.1 kJ/kg) or w = 575.3 kJ/kg . We can find the actual outlet temperature from the turbine using the first law with this actual work. Since w = cp(Tin – Tout), the actual outlet temperature is found as follows Tout 575.3 kJ w kg = Tin − = 1173.15 K − = 865.9 K kJ 1 . 8723 cp kg ⋅ K As in problem one, the first law for the heater is q = hout – hin. The inlet temperature to the heater is the actual outlet temperature from the turbine, 865.9 K. Thus the heat transfer in the heater is computed as q = q = hout – hin = cp(Tout – Tin) = (1.8723 kJ/kg∙K)(1173.15 K – 865.9 K) or q = 575.3 kJ/kg . Since the enthalpy for an ideal gas is a function of temperature only, and the turbine and heater have the same temperature difference (although in opposite directions), the turbine work is the same as the subsequent heat addition. This will also be true for the case of variable heat capacity done next. 3. Variable-heat-capacity ideal gas behavior using the ideal gas tables. Here we again apply the same first law analyses that we used in answering problem one. The only difference is that we use the ideal gas tables to get the enthalpy changes using the h(T) values in those tables. In addition, we use the tabulated values of so(T) find the end states of an isentropic process by the equation so(T2) = so(T1) + R ln(P2/P1). (We cannot use the usual relationships for Pr(T) because table A-23 (page 947) does not contain this quantity.) At the initial temperature of 1173.15 K, we interpolate between 1160 and 1180 K in Table A-23 on page 861 to obtain the following (molar) values: hin sino = 239.346 kJ kmol ⋅ K = 43,212 kJ kmol and . We can compute the outlet value of so(T), using the universal gas constant for the use of molar units, as follows.  P  239.346 kJ 8.314 kJ  600 kPa  225.965 kJ o = = sino + R ln 2  = ln + sout kmol ⋅ K kmol ⋅ K  3000 kPa  kmol ⋅ K  P1  This value lies between 840 K and 850 K. We can interpolate between the enthalpy values at these two temperatures to get the following result for the ideal outlet enthalpy. hout ,s 29846 kJ 29454 kJ − 29454 kJ  225.969 kJ 225.592 kJ  29769 kJ kmol kmol = + −  = 226.057 kJ 225.592 kJ  kmol ⋅ K ⋅ kmol kmol K kmol  − kmol ⋅ K kmol ⋅ K With this result we can compute the ideal work per mole and multiply it by the isentropic efficiency to compute the actual work per mole. 13443 kJ 10755 kJ  43212 kJ 29769 kJ  − = w = η s ws = η s (hin − hout ,s ) = 80%  = 80% kmol  kmol kmol  kmol To get the work on a per unit mass basis, we divide by the molar mass of water from Table A-1 on page 910. 10755 kJ w 597.0 kJ w= = kmol = kg M 18.015 kg kmol The actual molar enthalpy from the turbine is the inlet enthalpy minus the work done. This value is 43212 kJ/kmol – 10755 kJ/kmol = 32,457 kJ/kmol. As in previous problems the heat transfer in the heater is given by the equation q = hout – hin, and the enthalpy into the heater is the outlet enthalpy from the turbine = 32457 kJ/kmol. The outlet enthalpy from the heater is the ideal gas enthalpy at 1173.15 K. This is the same as the enthalpy into the turbine which was found above to be 43,212 kJ/kmol. The heat transfer in the heater is then found by dividing the molar enthalpy difference by the molar mass of water. 43212 kJ 32457 kJ 10755 kJ − 597.0 kJ h − hin q kmol = kmol or q = = out = kmol q= kg 18.015 kg 18.015 kg M M kmol kmol 4. Find the work required to pump 20 lbm/s of liquid refrigerant 134a from an initial temperature and pressure of 0oF and 30 psia to pressure of 300 psia if the pump has an isentropic efficiency of 80%. Assume the heat transfer is negligible. We see that the initial temperature of 0oF is less than the saturation temperature of 15.37oF at 30 psia confirming that the initial state is a compressed liquid. We will assume that the properties of this compressed liquid are those of a saturated liquid at the given temperature of 0oF. This gives vin = 0.01185 ft3/lbm and hin = 12.199 Btu/lbm. To compute the isentropic work, we use the equation that ∆h = v∆P for an isentropic process with a liquid. This gives the ideal work, ws = hin – hout,s = vin(Pin – Pout) = (0.01185 ft3/lbm)(30 psia – 300 psia)(1 Btu/5.40395 psia⋅ft3) = –0.0592 Btu/lbm. Since this is a work input device we divide by the isentropic efficiency to get the actual work as w = ws/ηs = (–0.0592 Btu/lbm) / (0.8) = –0.740 Btu/lbm. Multiplying by the mass flow rate of 20 lbm/sec gives the actual power input as –14.8 Btu/s = 20.94 hp . What is the actual outlet enthalpy? From the first law we have hout = hin – w = 12.199.Btu/lbm – z(–0.740 Btu/lbm) = 12.94 Btu/lbm . 5. The inlet to a steady, adiabatic, R-134a compressor is 3000 lbm/h of saturated vapor at -40oF. If the power input to the compressor is 50 hp, what is the outlet pressure if the compressor has an isentropic efficiency of 80%? In order to solve this problem we have to determine the outlet pressure for the hypothetical isentropic process that is used to define the isentropic efficiency. This outlet pressure is the same as the actual outlet pressure. In the actual process the work input is 50 hp; in the isentropic process, the ideal work input would be only 80% of the actual work input, which is (80%)(50 hp) = 40 hp. Thus the ideal (isentropic) work is -40 hp. For a steady flow device with one inlet and one outlet, zero heat transfer, and negligible changes in kinetic and potential energy, the first law is W u = m (hin − hout ) . For the inlet saturated vapor at Tin = -40oF, hin = hg(-40oF) = 97.10 Btu/lbm, and sin = sg(-40oF) = 0.23135 Btu/lbm·R. The ideal outlet enthalpy is found from the following equation for the ideal work input. hout , s = hin − W u , s 97.10 Btu − 40 hp 2544.4 Btu 131.03 Btu = − = 3000 lbm m lbm hp·h lbm h So we now know the ideal outlet state is defined by hout,s = 131.03 Btu/lbm and sout,s = sin = 0.23135 Btu/lbm·R. Location of the final pressure requires a double interpolation after finding the place in the tables where the values of enthalpy and entropy bracket the required values of hout,s and sout,s. It appears that the first place to start interpolation is at pressures of 300 and 400 psia. At 300 psia, the enthalpy that corresponds to the value of sout,s = sin = 0.23135 Btu/lbm·R is found by the following interpolation. h300 psia ,s = sOut 134.60 Btu 128.60 Btu −  0.23135 Btu 0.22802 Btu  130.76 Btu lbm lbm 128.65 Btu  = − = + 0.23733 Btu 0.22802 Btu  lbm ⋅ R lbm lbm ⋅ R  lbm − lbm ⋅ R lbm ⋅ R At 400 psia, the enthalpy that corresponds to the value of sout,s = sin = 0.23135 Btu/lbm·R is found by the following interpolation. h400 psia , s = sOut 135.88 Btu 128.99 Btu −  0.23135 Btu 0.22471 Btu  133.44 Btu 128.99 Btu lbm lbm   = − = + 0.23500 Btu 0.22471 Btu  lbm ⋅ R lbm ⋅ lb R lbm m  − lbm ⋅ R lbm ⋅ R Now that we have the enthalpy values at two pressures for sout = sin, we can interpolate to find the pressure at the given value of hout,s = 131.03 Btu/lbm. Pout = 300 psia + 400 psia − 300 psia  131.03 Btu 130.76 Btu    = 309.8 psia − 133.44 Btu 130.76 Btu  lbm lbm  − lbm lbm The outlet pressure is 309.8 psia .