Mechanical Engineering - Notes

March 11, 2003 Homework Solutions 4-79 Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450oC, and 80 m/s. The exit conditions are 10 kPa, 92% quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the change in kinetic energy, (b) the power output, and (c) the turbine inlet area. Here we have a steady problem with one inlet and one outlet. We are also told that the turbine is adiabatic so that the heat transfer is zero. The general first law and mass balance equations are shown below. dE system dt 2 2     V V i = Q − W u − ∑ m i  hi + + gzi  + ∑ m i  hi + i + gzi  2 2 outlet   inlet   dmsystem dt For the steady problem we know that = ∑ m i − inlet dE system dt = ∑ m outlet dmsystem dt i = 0 . We are given no elevation data to compute the potential energy changes, but these are usually small so we will make the assumption that they are negligible. Since we have only one inlet and one outlet we have only  in = m out = m . Finally, the stipulation that the turbine is adiabatic one unique mass flow rate: m means that follows: Q = 0. With these assumptions, the first law energy balance can be written as 2 2     V V in  −  hout + out  W u = m  hin + 2   2   The first question asks us to find the change in kinetic energy. If we interpret this as the change in kinetic energy per unit mass, we can find this change as shown below.  2 2 2 Vin2 Vout 1  80 m   50 m   1 kJ ⋅ s 2 = 1.95 kJ/kg ∆KE = − =  −     2 2 2  s   s   1000 kg ⋅ m 2 We find the power output from the first law equation. Here we need the inlet and outlet enthalpy values which are found from the property tables for water. hin = h(10 MPa, 450oC) = 3230.9 kJ/kg from the superheat tables hout = hf(10 kPa) + xout hfg(10 kPa) = 2393.2 kJ/kg with xout = 0.92 and data from the saturation table with pressure as the look-up variable. We then find the power output as follows  2   12 kg  3240.9 kJ 2393.2 kJ 1.95 kJ Vin2 Vout  = − − + Wu = m  hin − hout +  s  kg kg kg 2 2   10.2 MW  1 MW ⋅ s =   1000 kJ The relationship between mass flow rate and cross sectional area of the flow channel is given by  VA  = the following relationship. m . To apply this equation to the turbine inlet, we have to find the v specific volume at that point. From the superheat tables, we find that vin = v(10 MPa, 450oC) = 0.02975 m3/kg. Thus, the area of the turbine inlet is found as follows. vm 0.02975 m 3 12 kg s = 0.00466 m2 . A=  = kg s 80 m V 4-84 Argon gas enters an adiabatic turbine steadily at 900 kPa and 450oC with a velocity of 80 m/s and leaves at 150 kPa with a velocity of 150 m/s. The inlet area of the turbine is is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon. The application of the first law and the mass-balance equation to this problem proceeds in exactly the same way as in the previous problem problem (4-79). The resulting equation for the work will therefore be the same.  2  Vin2   Vout   −  hout + Wu = m  hin + 2 2        This problem is different from the previous one in that we will use ideal gas properties for agron. In addition, we are not given the mass flow rate, however we can find it from the data given on the inlet velocity and area. We have to find the specific volume, using the ideal gas equation at the inlet, to apply this formula. Using v = RT/P for the specific volume, with R = 0.2081 kJ/kg∙K for argon gives the mass flow rate as follows.   VA VAP 80 m 10 −4 m 3 kg ⋅ K 1 1 kJ kg ( = = = 2.874 m = 60 cm 2 ) (900 kPa ) 2 3 v RT s cm 0.2081 kJ 723.15 K kPa ⋅ m s For noble gases like argon, the heat capacity is constant over a wide temperature range. (See figure 2-68 on page 102.) Thus we can use the equation that hout – hin = cp(Tout – Tin). Where the value of cp for argon is 0.5203 kJ/kg∙K. We can combine this ideal gas relationship for the enthalpy change with the first law equation to get the following result.  2   Vin2 Vout    Wu = m c p (Tin − Tout ) +  − 2 2    ⇒ Tout  2  W u  Vin2 Vout  −  − m  2 2  = Tin − cp We can solve the final equation for Tout. Note that we could solve for the temperature in either kelvins or degrees Celsius. Tout  2  W u  Vin2 Vout  −  − 2  m  2 = Tin − = cp 2 2 250 kW 1 kJ 1  80 m   1500 m   kJ ⋅ s 2 −   −  2.874 kg kW ⋅ s 2  s   s   1000 kg ⋅ m 2   s 723.15 K − 0.5203 kJ kg ⋅ K Tout = 540 K 4-86 Refrigerant-134a enters an adiabatic compressor as saturated vapor at -20oC and leaves at 0.7 MPa and 70oC. The mass flow rate of the refrigerant is 1.2 kg/s. Determine (a) the power input to the compressor and (b) the volume flow rate of the refrigerant at the compressor inlet. The application of the first law and the mass-balance equation to this problem proceeds in almost exactly the same way as in the first problem (4-79). The only difference is that we are given no data on the inlet and outlet velocities with which to compute the kinetic energy terms. Based on our experience, we expect that these terms will make only a small contribution to the overall energy balance so we will assume that they are zero. With this assumption, plus all the assumptions that were used in problem 4-79, the equation for the work can be written as follows. W u = m (hin − hout ) We can find the inlet and outlet enthalpies from the property data given, using R-134a tables. hin = hg(-20oC) = 253.31 kJ/kg, from the saturation tables with temperature as the look-up variable hout =h(0.7 MPa, 70oC) = 307.01 kJ/kg from the superheat tables. We can now find the power input to the compressor. 1.2 kg  253.31 kJ 307.01 kJ − W u = m (hin − hout ) =  s  kg kg  kW ⋅ s = -86.04 kW   1 kJ The negative value for indicates that this is a power input to the compressor. The volume flow rate is the product of the mass flow rate (kg/s) and the specific volume (m3/kg). The specific volume at the inlet is found from the saturation tables; vin = vg(-20oC) = 0.1464 m3/kg. We can then find the volume flow rate at the inlet by multiplying this by the mass flow rate of 1.2 kg/s. 1.2 kg 0.1464 m 3 0.176 m 3 = V = m v = s kg s 4-86 Refrigerant-134a is throttled from the saturated liquid state at 800 kPa to a pressure of 140 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant. After this introduction to steady processes, you can simply approach this problem by starting with the statement that the enthalpy is constant in a throttling process. To see how we reach this conclusion, we start with our usual equations for the first law and the mass balance. dE system dt 2 2     V V = Q − W u − ∑ m i  hi + i + gzi  + ∑ m i  hi + i + gzi  2 2 outlet   inlet   dmsystem = ∑ m i − dt inlet ∑ m outlet i Here our system is a valve. We assume that this is a steady flow, so dE system dt = dmsystem dt = 0. We are given no elevation data to compute the potential energy changes or velocity data to compute kinetic energy changes, but these are usually small so we will make the assumption that they are negligible. Since we have only one inlet and one outlet, we have only one unique mass  in = m out = m . The valve has no work output device such as a shaft that rotates or flow rate: m electrical leads producing power, so we say that the work is zero. Finally, we assume that the definition of a throttling process is one in which the valve is well-insulated so that the heat transfer is zero. With these assumptions, the first law energy balance can be written as follows: W u = Q + m (hin − hout ) 0 = 0 + m (hin − hout ) so that or hin = hout We can find the inlet enthalpy from the R-134a tables: hin = hf(800 kPa) = 93.42 kJ/kg from the saturation tables with temperature as the look-up variable. Since hout = hin and we are given Pout = 140 kPa, we know that the final state is one where P = 140 kPa and h = 93.42 kJ/kg. From the saturation tables at the outlet pressure of 140 kPa (0.14 MPa), we find that hf(140 kPa) = 25.77 kJ/kg and hg(140 kPa) = 236.04 kJ/kg. Thus the outlet enthalpy of 93.42 kJ/kg is in the mixed region at this pressure. Since the inlet is a saturated liquid, we know that Tin = Tsat(Pin = 800 kPa) = 31.33oC. Similarly, since the outlet is in the mixed region, we know that Tout = Tsat(Pout = 140 kPa) = -18.80oC. We can then find the temperature drop. Tin − Tout = 31.33o C − 18.80o C = −50.13o C We can find the final specific volume from the final quality, where we use the final enthalpy to find the quality. To compute the specific volume, we need the specific volumes of the saturated liquid and saturated vapor at the final pressure of 140 kPa. From the saturation tables we find vf = 0.0007381 m3/kg and vg = 0.1395 kJ/kg. Thus the final specific volume can be found as follows. h − hf m3 (v g − v f ) = 0.0007381 + v = v f + x (v g − v f ) = v f + hg − h f kg 93.42 kJ 25.77 kJ − kg kg 210.27 kJ 25.77 kJ − kg kg 3  0.13876 m 0.0007381 m  − kg kg  3    v = 0.0454 m3/kg 4-110 Refrigerant-134a at 1 MPa and 80oC is to be cooled to 1 MPa and 30oC in a condenser by air. The air enters at 100 kPa and 27oC with a volume flow rate of 800 m3/min and leaves at 95 kPa and 60oC. Determine the mass flow rate of the refrigerant. One approach that we will not take here is to define two systems: the flowing air and the flowing refrigerant. In this approach we would compute the heat transfer from the air and set that equal to the heat transfer to the refrigerant. We will not take this approach. Instead we will define one system as shown in the diagram. In this one system, we have two inlet flows and two outlet flows. However, the air and R-134a flows do not mix. Thus we conclude, for a steady system, in which dmsystem dt = 0, that we have only two unique mass flow rates: Air in R-134a in ---system boundary R-134a out Air out m air ,in = m air ,out =m air , and m R ,in = m R ,out =m R . (In this problem we will use the subscript, “R” to indicate R-134a. We can apply this result to our analysis of the first law equation. dE system dt For this steady flow, = Q − W u − dE system dt       Vi 2 Vi 2 m i  hi + + gzi  + ∑ m i  hi + + gzi  ∑ 2 2 outlet   inlet   = 0 . We are given no elevation data to compute the potential energy changes or velocity data to compute kinetic energy changes, but these are usually small so we will make the assumption that they are negligible. The valve has no cooler has no work output device such as a shaft that rotates or electrical leads producing power, so we say that the work is zero. Finally, we assume that the cooler is designed so that the heat transfer from the cooler, defined as a single system, is negligible, compared to the internal heat transfer between the air and the refrigerant. So we will take the heat transfer to be zero. With these assumptions, the first law energy balance can be written as follows: 0 = −m air ,out hair ,out − m R ,out hR ,out + m air ,in hair ,in + m R ,in hR ,in Using the results that m air ,in = m air ,out = m air , and m R ,in = m R ,out =m R from our analysis of the mass balance equation (and noting that the streams do not mix) our first law energy balance becomes. m air (hair ,out − hair ,in ) = m R (hR ,in − hR ,out ) We can find the enthalpies of R-134a from the property tables: hR,in = hR(1 MPa, 80oC) = 313.20 kJ/kg from superheat tables on page 845. The value of hR,out = h(1 MPa, 30oC) is the enthalpy of a compressed liquid. We can see this since the temperature of a saturated liquid at 1 MPa is 39.39oC, which is higher than the outlet temperature. We can approximate the outlet refrigerant enthalpy as the enthalpy of a saturated liquid at the given temperature of 300C: hR,out = h(1 MPa, 30oC) ≈ hf(30oC) = 91.49 kJ/kg from Table A-11 on page 842. We will assume that air behaves as an ideal gas at the temperatures and pressures of the air in this problem. Since the temperature difference of the air is small and air is at low temperature, we will use a constant heat capacity for air of 1.005 kJ/kg∙K, from Table A-2 on page 825. From the same table we can find the gas constant for air: R = 0.2870 kJ/kg∙K. We do not know the mass flow rate of the air, but we can compute it from the inlet volume flow rate, Vin , and the inlet specific volume, vin = RTin/Pin. m air = Vair ,in Vair ,in Pair ,in 800 m 3 kg ⋅ K 100 kPa 1 kJ = = = 929.8 kg 3 min min 0.287 kJ 300.15 K kPa ⋅ m vair ,in RTair ,in We can how substitute ∆h = cp∆T for air into our first law and solve for the refrigerant flow rate. m air (hair ,out − hair ,in ) = m air c p ,air (Tair ,out − Tair ,in ) = m R (hR ,in − hR ,out ) In using this equation, we can compute the temperature difference in degrees Celsius or in kelvins and get the same result. 929.8 kg 1.005 kJ (60 − 27)K m air c p ,air (Tair ,out − Tair ,in ) min kg ⋅ K = 139.0 kg/min . = m R = 313.20 kJ 91.49 kJ (hR,in − hR,out ) − kg kg 4-111E Air enters the evaporator section of a window air conditioner at 14.7 psia and 90 oF with a volume flow rate of 200 ft3/min. Refrigerant-134a at 20 psia with a quality of 30% enters the evaporator at a rate of 4 lbm/min and leaves as a saturated vapor at the same pressure. Determine (a) the exit temperature of the air and (b) the rate of heat transfer from the air. This problem is similar to the previous one. For the first part of the problem we will analyze the system as a combined system. We will make the same assumptions that we did in problem 4110 – steady state, air and refrigerant streams do not mix, negligible kinetic and potential energies, heat and work both equal zero – so that we will arrive at the same equation that we had for the previous problem. m air (hair ,out − hair ,in ) = m R (hR ,in − hR ,out ) We can find the enthalpies of R-134a from the property tables: hR,out = hg(20 psia) = 101.39 Btu/lbm from the saturation table, A-12E. At the same pressure, the enthalpies of the saturated liquid and vapor are found from Table A-12E to compute the inlet enthalpy from the given quality, xin = 0.3. hin = h f (90 psia ) + xin h fg (90 psia ) = 10.89 Btu 90.50 Btu 38.04 Btu + (0.3) = lbm lbm lbm We will assume that air behaves as an ideal gas at the temperatures and pressures of the air in this problem. Since the temperature difference of the air is small and air is at low temperature, we will use a constant heat capacity of 0.240 Btu/lbm∙R, from Table A-2E. From the Table A-1E, we can find the gas constant for air: R = 0.3704 psia∙ft3/lbm∙R. We do not know the mass flow rate of the air, but we can compute it from the inlet volume flow rate, Vin , and the inlet specific volume, vin = RTin/Pin. (Add 459.67 to the 90oF inlet temperature to get the inlet temperature as 549.67 R.) m air = Vair ,in Vair ,in Pair ,in 200 ft 3 lbm ⋅ R 14.7 psia lb = = = 14.43 m 3 min vair ,in RTair ,in min 0.3704 psia ⋅ ft 549.67 R We can how substitute ∆h = cp∆T for air into our first law and solve for the outlet air temperature. m air (hair ,out − hair ,in ) = m air c p ,air (Tair ,out − Tair ,in ) = m R (hR ,in − hR ,out ) Tair ,out = Tair ,in + m R (hR ,in − hR ,out ) = 90 0 F +  mair c p ,air 4 lbm min  38.04 Btu 101.39 Btu    − lbm lbm  = 16.8oF .  14.43 lbm 0.24 Btu lbm ⋅ R min To get the heat transfer from the air to the refrigerant, we have to analyze either the air only or the refrigerant only as the system. Here we choose to analyze the refrigerant only. With the assumptions discussed previously – steady flow, negligible kinetic and potential energy changes, no work – plus the fact that for the refrigerant only we have one inlet and one outlet, the first law reduces to the following expression. 4 lbm Q from air to refrigerant = Q = m R (hR ,out − hR ,in ) = min  101.39 Btu 38.04 Btu   = 253.5 Btu/min .  − lbm lbm   Here the positive sign indicates that the direction of heat flow is from the air to the refrigerant.