Unit 1 Homework Solutions

Unit One Homework Solutions, September 2, 2010 1 Complete the following table for H2O Row 1 2 3 4 T(oC) 50 P(kPa) 200 400 600 250 110 v(m3/kg) 4.16 Phase description Saturated vapor Use the Tables A-4 through A-8 for water starting on page 914. For the first row data, we first try to see if the given specific volume of 4.16 m3/kg lies in the liquid, gas, or mixed region. We do this by looking at the values of vf and vg at the given temperature of 50oC. On page 914, we find that vf(50oC) = 0.001012 m3/kg and vg(50oC) = 12.026 m3/kg. Since the given volume of 4.16 m3/kg lies between the saturated liquid volume, vf, and the saturated vapor volume, vg, at the given temperature of 50oC, we conclude that the state is in the mixed region. Since we are in the mixed region, the pressure is the saturation pressure at the given temperature of 50oC. From the same location in Table A-4 we find Psat(50oC) = 12.352 kPa. The quality at this point can be found as follows. x= v−vf vg − v f 4.16 = m3 m3 − 0.001012 kg kg m3 m3 − 0.001012 12.026 kg kg = 0.346 The point in row two is defined as saturated vapor. This means that the temperature is the saturation pressure at 200 kPa and the volume is the volume of the saturated vapor at this pressure. From Table A-5 on page 916 we find Tsat(200 kPa) = 120.21oC and vg(200 kPa) = 0.88578 m3/kg. The data in row three are found in the superheat table, A-6, on page 918; the volume at this point is 0.59520 m3/kg. Note that we the pressures in this table are in MPa so we have to recognize that 400 kPa = 0.4 MPa. In searching for the state in row four, we find that the temperature of 110oC is less than the saturation temperature at 600 kPa, which is found in Table A-5 on page 916 to be 158.83oC. This can also be found if we guessed that this was a gas and tired to locate this pressure and temperature in the superheat region: Table A-6 on page 918. We see that that the minimum temperature in the section of this table for 600 kPa (stated as 0.6 MPa) is 200oC. To see if this is a liquid we can find the saturation temperature at the top of the table section, next to the 0.6 MPa value. Because the given temperature of 110oC is less than the saturation temperature of 158.83oC, the state is a compressed liquid. The simplest approach is to approximate the specific volume at this state as the volume of the saturated liquid at the given temperature of 110oC. This volume is found to be 0.001052 m3/kg (Table A-4, page 914). Unit one homework solutions Page 2 The alternative approach would be a double interpolation in the compressed liquid table on page 924. Because this table does not have a temperature of 110oC, we have to interpolate data between the table temperatures of 100oC and 120oC. Because the minimum pressure in this table is 5 MPa, we have to interpolate between the saturation pressure and the pressure of 5 MPa to get the values at the desired pressure of 600 kPa. First interpolate to find the value at 100oC and 600 kPa. ( ) o ( o ) ( v 100 C ,600 kPa = v f 100 C + ) ( )[600 kPa − P (100 C )] (100 C ) v 100 o C ,5000 kPa − v f 100 o C sat o o 5000 kPa − Psat 0.0010410 m 0.001052 m 3 − 3 0.001052 m 3 kg kg (600 kPa − 101.42 kPa ) = 0.0010509 m = + kg kg 5000 kPa − 101.42 kPa 3 Next interpolate to find the value at 120oC and 600 kPa. ( ) o ( o ) ( v 120 C ,600 kPa = v f 120 C + ) ( ) [600 kPa − P (120 C )] (120 C ) v 120 o C ,5000 kPa − v f 120 o C sat o o 5000 kPa − Psat 0.0010576 m 0.001060 m 3 − 3 0.001060 m 3 kg kg (600 kPa − 198.67 kPa ) = 0.0010598 m = + kg kg 5000 kPa − 198.67 kPa 3 Finally we can interpolate between the 100oC and 120oC at the pressure of 600 kPa. ( ) v(120 ) ( v 110 o C ,600 kPa = v 100 o C ,600 kPa + o ) ( C ,600 kPa − v 100 o C ,600 kPa o o ) (110 o C − 100 o C 120 C − 100 C 0.0010598 m 3 0.0010509 m 3 − 0.0010509 m 3 0.0010553 m 3 kg kg o o = + − = C C 110 100 kg kg 120 o C − 100 o C ( ) We see that this answer is close to the value of 0.001052 m3/kg for the saturated liquid volume at the given temperature of 1100C. The answers for this row and the previous rows are shown in the table below. Row 1 2 3 4 T(oC) 50 120.23 250 110 P(kPa) 12.352 200 400 600 v(m3/kg) 4.16 0.88578 0.59520 0.001052 Phase description Mixed region with x = 0.346 Saturated vapor Superheated vapor Compressed liquid ) Unit one homework solutions 2 Page 3 Complete the following table for refrigerant 134a: Row 1 2 3 4 5 T(oF) 15 10 P(psia) 80 h(Btu/lbm) 78 70 180 129.467 110 x Phase description 0.6 1.0 Use the Tables A-11E through A-13E for refrigerant 134a starting on page 976. For the first row data, we find that the specific enthalpy of 78 Btu/lbm lies between the saturated liquid and saturated vapor enthalpies at the given pressure of 80 psia. (On page 9777, Table A12E , we find that hf(80 psia) = 33.394 Btu/lbm and hg(80 psia) = 112.20 Btu/lbm; the difference between these two, hfg(80 psia) = 78.804 Btu/lbm is given in the table.) Thus, the state is in the mixed region and the temperature is the saturation temperature at the given pressure of 80 psia. From the same location in Table A12E we find Tsat(80 psia) = 65.89oF. The quality at this point can be found as follows. x= h − hf h fg 78 = Btu Btu − 33.394 lbm lbm = 0.566 Btu 78.804 lbm The point in row two has a quality of 0.6. Thus, it must be in the mixed region. We use Table A11E on page 976 to get the saturation data. The pressure is the saturation pressure at the given temperature, Psat(15oF) = 29.759 psia, and we compute the enthalpy from the given quality and the tabulated values of enthalpy of the saturated liquid, hf(15oF) = 16.889 Btu/lbm, and the enthalpy change between saturated liquid and saturated vapor, hfg(15oF) = 88.377 Btu/lbm. h = h f + xh fg = 16.889 Btu lbm  = 69.915 Btu + 0.6 88.377 Btu  lb lbm m  If we look for the data in row three (T = 10oF, P = 70 psia) in the superheat table on page 978, we find that the saturation temperature at 70 psia is 58.30oF. Since the given temperature is less than this we know that we have a compressed liquid. We can approximate the compressed liquid properties as those of the saturated liquid state at the given temperature of 10oF. From Table A-11E on page 976, we find that hf(10oF) = 15.318 Btu/lbm. The data in row four are found in the superheat table on page 979. We look for an enthalpy of 129.46 Btu/lbm in the small table for a pressure of 180 psia and find that this occurs at a temperature of 160oF. (Normally we would not be so lucky and would have to interpolate.) The data in the last row specify a quality of one. This means that we have a saturated vapor. We use the saturation table on page 976 to find P = Psat(110oF) = 161.16 kPa and hg(110oF) = 117.23 Btu/lbm. The results for all rows are summarized in the table below. Row 1 2 3 4 5 T(oF) 65.89 15 10 160 110 P(psia) 80 29.759 70 180 161.16 h(Btu/lbm) 78 69.915 15.318 129.46 117.23 x 0.566 0.6 N/A N/A 1.0 Phase description Mixed region Mixed region Compressed liquid Superheated vapor Saturated vapor Unit one homework solutions A piston-cylinder device initially contains 50L of liquid water at 25oC and 300 kPa. Heat is added to the water at constant pressure until the entire liquid is vaporized. (a) What is the mass of the water? (b) What is the final temperature? (c) Determine the total enthalpy change. (d) Show the process on a T-v diagram with respect to saturation lines. We find that the initial state of 25oC and 300 kPa is a compressed liquid, because the temperature of 25oC is less than the saturation temperature at 300 kPa. (See Table A-5 on page 916 that gives Tsat(300 300 kPa = 133.52oC. Alternatively we could find that the pressure of 300 kPa is greater than the saturation pressure at 25oC, which is found as 3.1598 kPa in Table A-4 on page 915.) We approximate this compressed liquid state as the saturated liquid at the given temperature of 25oC, from Table A-4. This gives an initial volume, v1 ≈ vf(25oC) = 0.001003 m3/kg and an initial enthalpy, h1 ≈ hf(25oC) = 104.89 kJ/kg. Since we know the total volume, V = 50L = 0.05 m3 and the initial specific volume, v1 = 0.001003 m3/kg, we can find the mass in the system as the ratio of these two quantities. V m= 1 = v1 T-v Diagram for Water 1000 Temperature (degrees Celsius) 3 Page 4 final point 100 10 initial point 1 0.1 0.0001 0.01 1 100 10000 Specific volume (m3/kg) 0.05 m 3 m3 0.001003 kg m = 49.85 kg We are told that the entire liquid is vaporized at the final state. If we assume that there is no superheat, the final state must be that of saturated vapor. Since we are also told that the process occurs at constant pressure, the final pressure, P2 = P1 = 300 kPa. The final temperature is the saturation temperature at 300 kPa from Table A-5; i.e., T2 = Tsat(300 kPa) = 133.55C . The total enthalpy change is the difference between the initial and final enthalpy, multiplied by the system mass. We found the initial enthalpy above; we know that the final state is a saturated vapor at 300 kPa; thus h2 = hg(300 kPa) = 2724.9 kJ/kg, and the total enthalpy change is given by the following equation. ∆H = m(h2 – h1) = (49.85 kg)(2724.9 kJ/kg – 104.89 kJ/kg = 1.306x105 kJ/kg The T-v diagram with the process indicated is shown above. On this diagram the process is a solid line (blue if you view this in color) and the saturation curve is the dotted line (purple in color). The temperature of the compressed liquid increases, at constant pressure, until the saturation temperature at the constant pressure of 300 kPa (133.55oC) is reached. This initial part of the process is so close to the saturation line that we cannot distinguish the two on the plot. Once the saturation temperature is reached, the additional heat vaporizes the liquid water. This part of the Unit one homework solutions Page 5 process takes place at constant temperature. (In the mixed region a constant pressure process is a constant temperature pressure process and vice versa.) 4 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25oC, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the temperature in the tire rises to 50oC. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa. For this problem we can use the ideal gas equation. The gas constant for air is found from Table A-1 on page 908 to be 0.2870 kJ/kg·. At the initial state with an atmospheric pressure of 100 kPa and a gage pressure of 210 kPa, the absolute pressure is 210 kPa + 100 kPa = 310 kPa. The absolute temperature initially is 25 + 273.15 = 298.15 K. The absolute temperature at the final state is 50 + 273.15 = 323.15 K. In the first part of this problem, the mass is constant. We are not told anything about how the tire expands as the temperature increases. Let’s make the simple assumption that the volume remains constant. According to the ideal gas equation, PV = mRT. If we compare two states for which mass and volume are the same we have the following result. T T V = const = 1 = 2 mR P1 P2 ⇒ P2 = P1 T2 T1 Substituting the given data into the final equation gives the pressure at the temperature of 50oC. P2 = P1 T2 323.15 K = (310 kPa ) = 336 kPa T1 298.15 K The pressure rise is simply the difference P2 – P1 = (336 – 310) kPa = 26 kPa . We contemplate a third state, after we bleed off the air. In this final state the pressure equals the initial pressure (P3 = P1 = 310 kPa), while the final temperature remains at 50oC. (T3 = T2 = 50oC.) The mass difference between these two states is computed from the ideal gas equation as follows.  P1V P3V P1V  1 1  (310 kPa )(0.025 m 3 )  1 1 ∆m = m1 − m3 = − = − = −    0.2870 kJ RT1 RT3 R  T1 T3   298.15 K 323.15 K  kg K ∆m = 0.0070 kg Unit one homework solutions 5 Page 6 A rigid tank contains 20 lbm of air at 20 psia and 70oF. More air is added to the tank until the pressure and temperature rise to 35 psia and 90 oF, respectively. Determine the amount of air added to the tank. We can solve this problem by assuming that air is an ideal gas. The gas constant for air is given as 0.06855 Btu/(lbm•R) = 0.3704 psia•ft3/(lbm•R) in Table A-1E on page 958 of the text. The latter value and units are used here because they are consistent with the units specified for other variables. If we denote the initial state as state 1 and the final state as state 2, we can apply the ideal-gas equation of state to both the initial and final state as shown below. m1 = P1 V1 RT1 m2 = P2 V2 RT2 In this problem we are given m1 = 20 lbm, P1 = 20 psia, T1 = 70oF = 529.67 R, P2 = 35 psia, T2 = 90oF = 549.67 R. We are also told that the contained is rigid so that we can assume that V2 = V1. Use the symbol, V = V1 = V2, to denote this constant volume. If we solve the ideal-gas equation at state 1 for this constant volume, V, and substitute the result into the equation at state 2 we get the following equation for the mass added. ∆m = m 2 − m1 = P2 V2 − m1 = RT2 P2 m1 RT1 P T  P1 PT − m1 = m1 2 1 − m1 = m1  2 1 − 1 RT2 P1T2  P1T2  Substituting the known values of P1, T1, P2, T2, and m1 into this equation gives the final result. P T   (35 psia )(529.67 R )  − 1 = 13.73 lbm ∆m = m1  2 1 − 1 = (20 lb m )   (20 psia )(549.67 R )   P1T2  Unit one homework solutions 6 Page 7 Determine the specific volume of superheated water vapor at 10 MPa and 400oC using (a) the ideal gas equations, (b) the generalized compressibility charts and (c) the steam tables. Also, determine the error involved in the first two cases. We can solve part (c) first. The steam-table specific volume is found from Table A-6 on page 908 for the given pressure and temperature. vsteam tables = 0.026436 m3/kg To use the ideal-gas equation of state, we first obtain the gas constant for H2O from table A-1 on page 884: R = 0.4615 kJ/(kg•K) for water. Applying this to the pressure of 10 MPa = 10,000 kPa and a temperature of 400oC = 673.15 K gives the following specific volume. v ideal gas 0.4615 kJ (673.15 K ) RT kg ⋅ K = 0.03106 m3/kg = = 1 kJ P (10,000 kPa ) kPa ⋅ m 3 The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables. Error = 0.03106 m 3 − 0.026416 m kg 0.026416 m 3 3 kg = 17.6% kg In order to use the generalized compressibility chart, we first have to compute the reduced pressure and temperature. These are simply the ratio of the given pressure and temperature to the critical pressure and temperature, respectively. We find the critical values for water from Table A-1 on page 908: Tc = 647.1 K and Pc = 22.06 MPa. The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables. Pr = 10 MPa P = = 0.453 Pc 22.06 MPa Tr = T 673.15 K = = 1.04 Tc 647.1 K With these values of reduced pressure and temperature, we can find the compressibility factor from the chart in Figure A-15 on page 932. This gives Z = 0.84. With this value of Z, we can find the desired specific volume as follows. v Z chart = 3 ZRT  = 0.026 m3/kg = Zvideal gas = (0.84) 0.03106 m kg  P  As with the ideal-gas calculation, the error is found by comparing the result to the specific volume obtained from the steam tables. Error = 0.026 m 3 kg − 0.026416 m 0.026416 m 3 kg 3 kg = 1.2%