Chapter 2 Solutions. 2.1 (D) Pressure and temperature are dependent during phase change and independent when in a single phase.

Transcription

1 Chater Solutions.1 (D) Pressure and temerature are deendent during hase change and indeendent when in a single hase.. (B) Sublimation is the direct conversion of a solid to a gas. To observe this rocess, set a iece of dry ice out on a table to.. (B) From Table B-1 at 9000 m, the atmosheric ressure is 0% of atmosheric ressure at sea level which is 0 kpa. From Table C- the saturation temerature for 0 kpa or 0.0 MPa is 69.1ºC..4 (A) The water is subcooled. If it were saturated it would be at 100ºC (1ºF) and cook you..5 (A) From Table C-1 at 00 C and v = 0.00 m /kg, the water is a saturated mixture. At 00ºC and v = 0. m /kg the water is a suerheated vaor. Grah (A) describes this rocess. Constant ressure goes u to the right and constant temerature goes down to the right..6 (D) The mass is kg and the volume is 0.08 m. The secific volume is 0.08 m v = = 0.04 m /kg kg From Table C-1 we get v f = m /kg and v g = m /kg. Calculating the quality we get v vf x = = = 0.1 v v g f.7 (A) From Table C- for P = 10 MPa and T = 00 C, the water is subcooled since saturation temerature at 10 MPa is 11 C. This means that the secific volume is v f at 00 C. From Table C-1 we get vf = m / kg. V = 0 kg m /kg = 0.08 m.8 (C) The secific volume at 400 C and 4 MPa is v 1 = m /kg = v. The volume doesn t change if it is rigid. Either Table C-1 or C- will contain the desired ressure. Table C-1 is referred since the v-entries are closer together. The entry at 0 C is v 1 = 0.07 m /kg, quite close to 0.07 m /kg. Interolation need not be accurate in order to select the correct answer: P.8 MPa..9 (A) (1000 liters = 1 m ) V = L = 0.00 m. Remember that quality is calculated on a mass basis, not a volume basis. From Table C-1 for a temerature of 00 C we get v f = m /kg and v g = m /kg. The liquid mass = V liquid /v f = m / m /kg = kg. The vaor mass = V vaor /v g = 0.00 m /0.174 m /kg = kg. Now we can calculate the quality: mvaor kg x = = = = 1.8% m kg kg total 9

2 .10 (B) Average the h-values between 6 MPa and 7 MPa for 600 C and 700 C in Table C-. There is only a small change in enthaly with ressure so a calculator is not needed: At 600 C: h= 654 At 700 C: h= Now, interolate between 654 and 891: h = ( ) = 8 kj/kg 4.11 (A) The tire is a torus with a minor radius r of 0.01 m and a major radius R of 1 m. The volume of the tire is the circumference times the area of the tire: 4 V = πd A= π 1 ( π 0.01 ) = m Treat air as an ideal gas (to check the units, use kpa = kn/m and kj = kn m): 4 ( 500 kpa)( m ) PV m = = = kg RT kj 0.87 ( 9 K) kg K.1 (C) The ressure in the cylinder can be found using the ideal gas law: mrt P = = = 18 kpa or 8 kpa gage V π The ressure balances the weight of the iston but we must use the gage ressure or add a force on the to of the iston due to the atmosheric ressure: W = PA = π 0.15 = 4070 N.1 (A) Because the temerature is so low and the ressure quite high, we anticiate non-ideal gas behavior. So, let s aroximate the Z-factor using Aendix G: P T 10 PR = 0.80, TR Z 0.59 P =.77 = = T = 1 = = cr PV m = = =.7 kg ZRT P (B) For an ideal gas ρ = =. ρ = = kg/m outside v RT ρ = = kg/m inside Difference = = 0.4 kg/m cr 10

3 Δh (B) C = = =.8 kj/kg C using a central-difference ΔT aroximation. Comare with the value from Table B- where steam is treated as an ideal gas (near atmosheric temerature and low ressure)..16 (D) From Table B-: C = 1.87 kj/kg K. Δ h= C Δ T = = kj/kg P From Table C- h 00 = kj/kg and h 600 = 704 kj/kg Δ h = = 8.5 kj/kg %difference = 100% = 10% (A) The enthaly change to warm u the ice is found by aroximating C to be.05 kj/kg C between 0 C and 0 C (see Table B-4): Δ Hice = mδ h= = 410 kj The ice is now at 0 C at which state it melts, requiring the heat of fusion, 0 kj/kg: Δ Hmelt = mδ hfusion = 10 0 = 00 kj If the temerature of the water is now raised from 0 C to 100 C, the enthaly change is found using C = 4.18 kj/kg C to be Δ Hmelt = mδ hwater = mcδ T = = 4180 kj Finally, the steam is suerheated to 00 C. From Table C- at P = 0.1 MPa, we find Δ H = mδ h= 10 ( ) = 1998 kj The total enthaly change to warm the ice, melt the ice, vaorize the water, and suerheat the steam is Δ H = = 990 kj.18 a) From Table C-1 for T = 140 C, P sat = 0.61 MPa = 61 kpa b) For T = 00 C, P sat = MPa = 1554 kpa c) For T = 0 C, P sat = 11.7 MPa = kpa.19 a) From Table C-E for P = 1 sia, T sat is calculated by interolating between 10 sia and 14.7 sia: T = 1 sia 10 sia sat T sat = 01. F sia 10 sia b) For P = 8 sia, interolate between 5 sia and 0 sia: o T sat 8 sia 5 sia = 0 sia 5 sia o o T sat = 46. F 11

4 c) For P = 50 sia, T sat = 81.0 F.0 For a temerature of 90 C, we use Table C-1 to find P sat = MPa = 70.1 kpa. From Table B-1 we see that P/P 0 = between 000 m and 000 m of altitude. We interolate to find the exact altitude h: h 000 m = 000 m 000 m h = 880 m.1 We use Table C-1E to find the ressure and Table C-E to find the corresonding saturation temerature. a) For h = 1,000 ft we get P/P 0 by interolating between 10,000 ft and 15,000 ft. P P ,000 ft 10,000 ft =. P/P 0 = ,000 ft 10,000 ft P = sia = 9.4 sia For this ressure we get the saturation temerature by interolation between 8 sia and 10 sia: T sat 9.4 sia 8 sia =. T sat = 190 o F sia 8 sia b) For h = 0,000 ft we get P/P 0 = 0.46 so P = sia = 6.8 sia. We have to interolate between 6 and 8 sia to find the saturation temerature: T sat 6.8 sia 6 sia =. T sat = 175. o F sia 6 sia c) For h =,000 ft we have to interolate between 0,000 ft and 5,000 ft to get P/P 0. P P ,000 ft 0,000 ft =. P/P o = 0.7 and P = sia = 4 sia , 000 ft 0, 000 ft For a ressure of 4 sia the saturation temerature T sat = 15.9 o F. mvaor 7 kg x = = = 0.41 m 10 kg + 7 kg total. Quality is determined on a mass basis so we have to use Table C-1E to get the secific volume of each hase in order to calculate the mass of each hase: 1

5 m V liquid 1 ft = = = 60.1 lbm v ft /lbm liquid f m V vaor 5 ft = = = 0.15 lbm v.6 ft /lbm vaor g mvaor 0.15 lbm x = = = = 0.5% m 60.1 lbm lbm total.4 At state 1 the water is a saturated liquid (x = 0). The secific volume v f at 00 kpa (0. MPa) can be found in Table C-: v f = m /kg. Then V 1 = mv f = 4 kg ( m /kg) = m At state the water is a saturated vaor at 00 kpa. The secific volume v g at 00 kpa (0. MPa) can be found in Table C-: v g = m /kg. Then V = mv g = 4 kg ( m /kg) =.54 m..5 At state 1 the water is a saturated liquid (x = 0). The secific volume v f at 400 kpa (0.4 MPa) can be found in Table C-: v f = m /kg. So V m m = = = 0.7 kg. (This mass stays constant for the rocess.) v m /kg f At state the water is a saturated vaor at 400 kpa. The secific volume v g at 400 kpa (0.4 MPa) can be found in Table C-: v g = m /kg. So V = mv g = 0.7 kg m /kg = m.6 We have a fixed mass of 10 kg and a quality x = 0.85 (V vaor = 8.5 kg) so the system is a saturated mixture. Table C- lists the secific volumes for saturated mixtures. a) For a ressure of 140 kpa: v = v f + x(v g v f ) v = ( ) = 1.05 m /kg V = 10 kg (1.05 m /kg) = 10.5 m V vaor = 8.5 kg (1.7 m /kg) = m and T = T sat = 109. C b) For a ressure of 00 kpa: v = v f + x(v g v f ) v = ( ) = 0.75 m /kg. V = 10 kg (0.75 m /kg) = 7.5 m V vaor = 8.5 kg ( m /kg) = 7.58 m and T = T sat = 10. C c) For a ressure of 000 kpa: v = v f + x(v g v f ) v = ( ) = m /kg. V = 10 kg ( m /kg) = m 1

6 V vaor = 8.5 kg ( m /kg) = m and T = T sat = 1.4 C.7 The water is 10 kg of a saturated mixture with a quality of 0.6 = m vaor /m total. The roerties can be obtained from Table C-1 for T = 40 C: v = v f + x(v g v f ) = ( ) = m /kg V = mv = 10 kg (11.71 m /kg) = m. V liquid = = m V vaor = 6 kg (19.5 m /kg) = m (The volume is essentially all vaor.) P = P sat = MPa = 7.8 kpa.8 Table C-E gives us English unit saturation information if we know the ressure. a) For P = 70 sia, T sat = 0.0 F b) For P = 150 sia, T sat = 58.5 F c) For P = 400 sia, T sat = F.9 P = atm 100 kpa/atm = 0.0 kpa or MPa which is below the lowest ressure in Table C-. The saturation temerature would be very near 0 C..0 To resond to this roblem, lease Google ressure cooker..1 We know that the water is a saturated mixture with a volume of 5 m and a quality of 70%. Table C-1 will give the roerties. a) For T = 10 C, P = P sat = MPa = kpa. V V 5 m = = = = kg v v + x ( v v ) ( ) f g f b) For T = 40 C, P = P sat =.44 MPa = 44 kpa. V V 5 m = = = = kg v v + x ( v v ) ( ) f g f c) For T = 70 C, P = P sat = 1.0 MPa. V V 5 m = = = = 115 kg v v + x ( v v ) ( ) f g f. Table C-1 lists the roerties for saturated water in SI units. For T = 50 C we get v f = m /kg and v g = 1.0 m /kg Initial volume: V 1 = 0.5 kg ( m /kg) = m Final volume: V = 0.5 kg (1.0 m /kg) = m A big change!. Since the container is rigid, the volume and secific volume stay constant. Table C-1 or C- gives the roerties at the critical oint: v = m /kg. a) Table C- gives roerties for saturated water. At MPa 14

7 v vf x = = = =.01% v v g f b) Table C- gives roerties for saturated water. At 1. MPa v vf x = = = = 1.4% v v g f c) Table C- gives roerties for saturated water. At 0.4 MPa v vf x = = = = 0.449% v v g f.4 The water is initially at 00 kpa and x = 0.1. We can calculate the constant secific volume of the rigid container using the information from Table C-: ( ) ( ) v= v + x v v = = m /kg f g f a) At 140 o C from Table C-1: P = 61. kpa v vf x = = = or 17.4% v v g f b) At 180 o C from Table C-1: P = 100 kpa v vf x = = = or 45.8% v v c) At 10 o C from Table C-1: P = 1906 kpa g f v vf x = = = or 85.6% v v g f.5 At state 1 the water is a subcooled liquid at 5⁰C, so an accurate estimate of the secific volume is to use vf at 5 C. From Table C-1, v 1 = m /kg. The secific volume at 10 C from Table C-1 is v f = m /kg. Hence, ΔV = m(v v 1) = 4 ( ) = m. Very little change..6 State 1: m = kg, T 1 = 0 C, and P 1 = 5 MPa. This is a subcooled liquid. State : m = kg, T = 00 C, and P = P 1 = 5 MPa. At this temerature the water remains a subcooled liquid since P < P sat = 1.55 MPa. i) Using Table C-4 for the secific volume at state 1, v 1 = m /kg. For state, v = m /kg. 15

8 ΔV = ( ) = m ii) Use Table C-1 to get the secific volumes using the v f values: v 1 = m /kg, v = m /kg ΔV = ( ) = m (essentially the same answer) Conclusion: Table C-1 can be used to find v for comressed liquids. Simly ignore the ressure..7 At state 1 the quality is zero and the ressure is 00 kpa or 0. MPa. Use Table C- to find v 1 = v f = m /kg. At state the ressure is MPa and the temerature is 400 C. Here the water is a suerheated vaor. Use Table C- to get the secific volume: v = m /kg. Then ΔV = ( ) = 0.00 m.8 At state 1 the water is a saturated mixture with a temerature of 10 C and a quality of 0.4. The ressure is the saturation ressure at 10 C which, from Table C-1, is MPa or 70.1 kpa. From this table we get the secific volumes for a saturated liquid v f and a saturated vaor v g to calculate v = v f + x(v g v f ) = ( ) = 0.68 m /kg V total = =.68 m The volume of liquid is V f = m /kg = m.9 The secific volume of the water is v = V/m = 70 ft /10 lbm = 7 ft /lbm. i) For v = 7 ft /lbm and P = 50 sia, use Table C-E to find that v f < 7 < v g. The water is a saturated mixture. ii) For v = 7 ft /lbm and P = 6 sia, use Table C-E to find that 7 > v g. The water is a suerheated vaor. At 61 sia, it would be a saturated mixture. iii) For v = 7 ft /lbm and P = 100 sia, use Table C-E to find that v > v g. The water is a suerheated vaor..40 The roblem states that the water is initially a saturated water vaor so x = 1. Knowing that T = 00 C we use table C-1 to get P 1 = P sat = 1554 kpa. The secific volume is v = v g = m /kg. a) If the temerature stays constant and the volume increases by 50%, v 1 = 1.5v = m /kg = m /kg which is in the suerheat range being greater than v g. We can use Table C- to get the roerties but this would require interolating between ressures of 1 MPa and 1. MPa. Use the IRC Calculator: P = 1070 kpa. b) If the temerature stays constant and we increase the volume by 100% v 1 = v 1 = m /kg = m /kg which is in the suerheat range being greater than ν g. We can use Table C- to get the roerties but this would require interolating between ressures of 0.8 MPa and 1 MPa. Use the IRC Calculator: P = 818 kpa. 16

9 c) If the temerature stays constant and we increase the volume by 00% v 1 = v = m /kg = 0.8 m /kg which is in the suerheat range being greater than ν g. We can use Table C- to get the roerties but this would require interolating between ressures of 0.4 MPa and 0.6 MPa. Use the IRC Calculator: P = 554 kpa..41 At 40 sia and 400 F the water is suerheated (P < P sat at 400 ). From Table C-E we get v = 1.6 ft /lbm: m = V/ v = 10 ft /1.6 ft /lbm = 0.79 lbm.4 a) For P = 5 MPa and T = 5 C the water is solid ice. Table C-5 gives the secific volume for ice to be m /kg: V = mv = 8 kg m /kg = m b) For P = 5 MPa and T = 0 C the water is a subcooled liquid. Table C-4 gives the secific volume for the water to be m /kg: V = mv = 8 kg m /kg = m c) For P = 5 MPa and T = 400 C the water is a suerheated vaor. Table C- gives the secific volume for the vaor to be m /kg: V = mv = 8 kg m /kg = m d) For P = 5 MPa and T = 800 C the water is a suerheated vaor. Table C- gives the secific volume for the vaor to be m /kg: V = mv = 8 kg m /kg = m.4 The weight of the iston is W = mg = 160 kg (9.81 m/s ) = N. The ressure caused by the weight of the iston: N P iston = W/A iston = π 0.1 m = Pa or 50 kpa The initial ressure is P 1 = P iston + P atm = 50 kpa kpa = 150 kpa. At the initial state, the ressure is 150 kpa and x = 1. The temerature is the saturation temerature at 150 kpa, which from Table C- is 111 o C, by interolation. Also by interolation, v 1 = v g = m /kg. Then V 1 = 0.0 kg m /kg = 0.0 m. a) To comress the sring 6 cm, the additional ressure is ( N/m)( 0.06 m) Kx P = A = π = N/m = 115 kpa 0.1 m So, P = 150 kpa kpa = 65 kpa. The volume is V = V + ΔV = 0.0 m m π(0.1 m) = 0.04 m The secific volume is then v = 0.04 m /0.0 kg = 1. m /kg. Using the IRC calculator with P = 65 kpa, the temerature is T = 418 C. 17

10 b) To comress the sring 10 cm, the additional ressure is ( N/m)( 0.1 m) Kx P = A = π = N/m = 191 kpa 0.1 m So, P = 150 kpa kpa = 41 kpa. The volume is V = V + ΔV = 0.0 m m π(0.1 m) = 0.05 m. The secific volume is then v = 0.05 m /0.0 kg = 1.5 m /kg. Using the IRC calculator with P = 41 kpa, the temerature is T = 65 C. c) To comress the sring 15 cm, the additional ressure is ( N/m)( 0.15 m) Kx P = A = π = N/m = 86 kpa 0.1 m So, P = 150 kpa + 86 kpa = 46 kpa. The volume is V = V + ΔV = 0.0 m m π(0.1 m) = 0.07 m The secific volume is then v = 0.07 m /0.0 kg = 1.5 m /kg. Using the IRC Calculator with P = 46 kpa, the temerature is T = 1010 C..44 The weight of the iston is W = mg = 160 kg 9.81 m/s = 1570 N. The ressure caused by the weight of the iston is W P = A = 1570 N π = N/m = 50 kpa 0.1 m The initial ressure P 1 = P iston + P atm = 50 kpa kpa = 150 kpa. For a ressure of 150 kpa and x = 0.8, we can calculate the secific volume using the data in Table C- between 0.14 MPa and 0.16 MPa. By interolation: v 1 = ( ) = 0.91 m /kg Also by interolation the saturation temerature is 111 o C. For the final state, the temerature is 800 o C which means the final state is robably a suerheated vaor. The only thing that stays constant in this roblem is the mass. We have to balance the osition of the iston, the ressure, and the temerature to find the final state. This must be done by trial and error. The initial volume in the cylinder can be calculated from state 1: For the final state T = 800 o C and P = 150 kpa + V 1 = 0.91 m /kg 0.0 kg = m ( 60 kn / m)x π 0.1 m, V m + ( π 0.1 x) m v = = m 0.0 kg 18

11 We must choose a value of the iston dislacement x so that all three roerties match in the steam tables. This occurs for x = 15 cm and P = 44 kpa. The quickest and easiest way to solve this is to use the IRC Calculator..45 From Problem.4, P initial = 150 kpa, x = 1, T 1 = 111. C, v 1 = m /kg and V 1 = 0.0 m. The iston must rise cm to reach the iston. This will be at constant ressure: P = 150 kpa, V = 0.0 m + [0.0 π (0.1) ] m = 0.06 m 0.06 m v = = 0.0 kg 1.1 m /kg Using the IRC Calculator with P = 150 kpa and v = 1.1 m /kg we get T = 111 o C and x = 0.975, a saturated mixture. At the final state, the sring is comressed 6 cm. Then V m = 0.06 m π 0.1 = m, v = = 1.5 m /kg 0.0 kg P = 150 kpa kpa = 65 kpa (from Problem.4) For P = 65 kpa and v = 1.5 m /kg we use the IRC Calculator to get T = 4 C, a suerheated vaor..46 Draw the diagrams to scale. Not many oints are needed to observe the general shae. Each diagram will resemble this sketch:.47 Use Table C-E to find the roerties for this saturated mixture at 50 sia by interolation between 00 sia and 00 sia:.48 The steam is at 1.86 MPa and 40 C. T = T sat = ( )/ = 400 F u f = ( )/ = 74 Btu/lbm u g = ( )/ = Btu/lbm h f = ( )/ = 74.9 Btu/lbm h g = ( )/ = Btu/lbm u = ( ) = 671 Btu/lbm h = ( ) = Btu/lbm i) Using Table C- we must double interolate for both ressure and temerature. P = 1.8 MPa P = MPa T = 400 C h = 50.9 h = 47.6 T = 500 C h = h = h 1.86MPa, 40 C = 94 kj/kg ii) For P = 1860 kpa and T = 40 C, use the IRC calculator: h = 90 kj/kg, u = 980 kj/kg, and v = m /k 19

12 iii) Using values from Part (ii) h = = 90 kj/kg (The numbers are to significant digits, so the answer is exressed with significant digits. All the answers are accetable. The fourth digit in Part (i) undoubtedly is meaningless. Significant digits are always lost when using straight-line interolations..49 At state 1 the water is saturated vaor (x = 1) at 00 F. At state the water is saturated liquid (x = 0). a) Use Table C-1E for the roerties. The saturation ressure at 00 F is sia and: v f = ft /lbm, u f = 69.5 Btu/lbm, h f = 69.7 Btu/lbm v g = 6.47 ft /lbm, u g = 1100 Btu/lbm, h g = Btu/lbm ΔV = 10 ( ) = ft ΔU = 10 ( ) = 805 Btu ΔH = 10 ( ) = 9105 Btu ΔH = ΔU + PΔV = (66.98 lbf/in )(144 in /ft ) (64.55 ft ) /778 ft-lbf/btu) = 9105 Btu b) Use the IRC Calculator for the roerties: P sat = 67 sia and: v f = ft /lbm, u f = 70 Btu/lbm, h f = 70 Btu/lbm v g = 6.47 ft /lbm, u g = 1100 Btu/lbm, h g = 1180 Btu/lbm ΔV = 10 ( ) = 64.5 ft ΔU = 10 ( ) = 800 Btu ΔH = 10 ( ) = 9100 Btu ΔH = ΔU + PΔV = (67 lbf/in )(144 in /ft ) (64.5 ft ) (1 Btu/778 ft-lbf) = 9100 Btu.50 State 1 is a saturated vaor (x = 1) at 00 o C. State is a suerheated vaor at MPa and 600 o C. a) The roerties for state 1 come from Table C-1: v 1 = v g = m /kg, h 1 = h g = 79. kj/kg The roerties for state come from Table C-: v = m /kg, h = kj/kg Δv = = 0.07 m /kg Δh = = kj/kg b) Use the IRC Calculator: v 1 = 0.17 m /kg, h 1 = 790 kj/kg v = 0. m /kg, h = 690 kj/kg Δv = = 0.07 m /kg Δh = = 900 kj/kg 0

13 .51 The ressure stays constant at 00 kpa: State 1: T 1 = 0 C, P 1 = 00 kpa (treat this as a saturated liquid at 0 C) State : T = 400 C, P = P 1 = 00 kpa or 0. MPa (suerheated vaor) a) The roerties for state 1 are from Table C-1 at 0 C: v 1 = v f = m /kg, u 1 = u f = 8.9 kj/kg, h 1 = h f = 8.9 kj/kg The roerties for state are from Table C- at 400 C: v = m /kg, u = kj/kg, Δρ = = 997 kg/m ΔU = 4 ( ) = kj ΔH = 4 ( ) = kj h = 76.6 kj/kg b) Use the IRC Calculator: At 0 C: v 1 = v f = m /kg, u 1 = u f = 8.9 kj/kg, h 1 = h f = 84.1 kj/kg At 400 C: v = 1.55 m /kg, u = 970 kj/kg, h = 80 kj/kg Δρ = = 999 kg/m ΔU = 4 ( ) = kj (the Calculator uses only significant digits) ΔH = 4 ( ) = kj.5 State 1: m = 0 kg, T 1 = 00 C, P 1 = 0.6 MPa suerheated State : T = 400 C, P = P 1 = 0.6 MPa suerheated a) Use Table C- for the enthalies: h 1 = kj/kg, h = 70. kj/kg. ΔH = 0 ( ) = 4174 kj b) Use the IRC Calculator for the enthalies: h 1 = 060 kj/kg, h = 70 kj/kg. ΔH = 0 (70 060) = 400 kj.5 You must determine the secific roerties in this roblem since you do not know the mass of the system. State 1: T 1 = 100 F, P 1 = 500 sia subcooled liquid State : T = 500 F P = P 1 = 500 sia suerheated vaor a) Use Table C-4E for the roerties at state 1: u 1 = Btu/lbm, h 1 = 69.6 Btu/lbm Use Table C-E for the roerties at state : u = Btu/lbm h = 11.5 Btu/lbm Δu = = 107 Btu/lbm Δh = = 116 Btu/lbm b) Use the IRC Calculator: 1

14 u 1 = 67.9 Btu/lbm, h 1 = 69.4 Btu/lbm, u = 1140 Btu/lbm, h = 10 Btu/lbm Δu = = 1070 Btu/lbm, Δh = = 1160 Btu/lbm.54 At state 1 the water is solid ice at 100 kpa and 0 C. The secific internal energy can be obtained from Table C-5: u 1 = 74 kj/kg At state the water is a subcooled liquid at 100 kpa and 0 C. The secific internal energy can be obtained from Table C-1 for T = 0 C: u = u f = 8.9 kj/kg ΔU = m(u g u i ) = 10 [8.9 ( 74)] = 4580 kj.55 The water is undergoing sublimation from solid ice at 0 C to a saturated vaor at 0 C. Table C-5 gives the roerties for these states:.56 We are given V = 0.05 m and m = 50 kg so ΔU = m(u g u i ) = 4 [48 ( 74)] = kj 0.05 m v = = m /kg 50 kg a) For P =.5 MPa or 500 kpa, from Table D-, we find the secific volume for a saturated liquid and a saturated vaor. v f = m /kg, v g = m /kg Since v is less than v f, the R14a is a subcooled liquid. b) For P = 40 kpa we use Table D- to obtain the saturation secific volumes. v f = m /kg, v g = m /kg Again, since v is greater than v f but less than v g, the R14a is a saturated mixture with x = = or 0.88% The secific volume of the refrigerant is 10 ft v = = 100 lbm 0.1 ft /lbm Use Table D-1E for T = 0 F: v f = ft /lbm, v g =.417 ft /lbm Since v f < v < v g the R14a is a saturated mixture with x = = or.60%

15 .58 At 00 kpa the saturation temerature of R14a is 1 C so the refrigerant is a subcooled liquid. Using Table D-1, the saturated liquid roerties at 1 C are v 1 = v f = m /kg, u 1 = u f = 4.5 kj/kg, h 1 = h f = 4.9 kj/kg At state the ressure remains constant at 00 kpa and T = 60 C. The refrigerant is now a suerheated vaor and we use Table D-: v = 0.1 m /kg, u = 78.1 kj/kg, h = 04.5 kj/kg. i) ΔV = 100 kg ( ) = 1.1 m ii) ΔU = 100 kg ( ) = 4 80 kj iii) ΔH = 100 kg ( ) = kj.59 State 1: T 1 = 80 F and x 1 = 0. See Table D-1E: P 1 = P sat = sia, h 1 = h f = 7.7 Btu/lbm State : P = 10 sia and h = h 1 = 7.7 Btu/lbm. We can get the enthaly data from Table D-E at 10 sia: h f =.91 Btu/lbm, h g = 97.7 Btu/lbm. Since h is between these two values, the state is a saturated mixture. The temerature at state is the saturation temerature of 9.7 F (very cold) and h= h + xh. 7.7 =.91 + x(94.45). x= 0.64 or 6.4% f fg.60 The working fluid is ammonia. State 1: T 1 = 40 C and x = From Table E-1: h f = 0.0 kj/kg, h fg = 189 kj/kg h 1 = = 10 kj/kg State : P = 400 kpa, T = 40 C. Use Table E- for suerheat: h = kj/kg. Δh = = 5 kj/kg.61 We know the mass, the ressure and the temerature of the air. Use the ideal gas law to determine the secific volume. Use Table B- to get the gas constant for air: v 1 kj K RT1 kg K = = = 0.7 m /kg. V 1 = mv 1 = = 0.07 m P 10 kpa 1 At the final state T = 5 C or 08 K kj K RT kg K v = = = 0.77 m /kg. V = mv = = m P 10 kpa In the above, the units convert as kj K kn m K m = =. kg K kpa kg K kn/m kg

16 .6 V = 4/ πr = (4/) π (1) = 4.19 ft, T = 7 F = 5 R, P = 14.7 sia ft lbf 5. 5 R RT lbm- R ft v = = = 1.40, P 14.7 lbf/in 144 in /ft lbm V m = = 4.19 ft = 0.1 lbm v 1.40 ft / lbm At 5000 ft the ressure is 0.8P 0 = sia) = 1. sia from Table B-1E. We assume that the temerature does not change: v RT 5. 5 R = = = 16.1 ft /lbm, V = mv= = 5.05 ft P Since V = πr = 5.05 ft, r = 1.06 ft and d=.1 ft..6 For carbon dioxide, Table B- gives R = kj/kg K. For P = 1 kpa and T = 60 C CO or 1 K. Then RT K v = = = 40.4 m /kg (See Problem.61 for units) P 1.64 The container has a constant volume of 5 m and an initial temerature of 00 K. Use the ideal gas law to find the final mass. The final ressure is 500 kpa: m PV = = = 9.04 kg (See Problem.61 for units.) RT Δm = = 8.8 kg.65 Initially, m = 10 lbm, P 1 = 14.7 sia, and T 1 = 80 F or 540 R: ft-lbf 10 lbm R mrt V = = lbm- R = 16 ft 1 1 P1 ( ) lbf/ft The final volume V = V 1 /4 = 16/4 = 4 ft. With P = 40 sia, lbf (40 144) 4 ft PV = = ft o T = 67 R mr ft lbf 10 lbm 5. lbm R.66 The gas is unknown so the gas constant is unknown. 4 v= = m /kg, From Table B- we see that the gas is helium. Pv 400 R = = =.077 kj/kg K T 85 4

17 .67 Convert the temerature to 9 K: 6 PV 100 ( 10 ) 6 m = = =.8 10 kg RT ( ) 6 7 W = mg=.8 10 kg 9.81 m/s =. 10 N or. MN.68 Given: V = 10 m i) ii) RT v = = = m /kg, ρ = = = 1.47 kg / m P 00 v W = mg = ρvg = = 145 N RT P = = = 410 kpa, ρ = = = 1.5 kg / m v 0.08 v 0.08 W = mg = ρvg = = 10 N,, iii) 40 = mg = ρvg = ρ ρ = kg/m, 1 1 v = = = ρ m /kg, Pv T = = = 510 K,or 4850 C R 0.87 iv) 1 1 RT 0.87 v= = = 0.05 m /kg, P = = = 17 kpa, ρ 0 v 0.05 W = mg = ρvg = = 1960 N,.69 Given: V = 0 ft i) ii) RT v = = = 15.9 ft /lbm, ρ = = = lbm/ft P v 15.9 W = mg = ρvg = = 60.7 lbf RT P 144 = = = 67 sia, ρ = = = 0.8 lbm/ft v 1. v 1. W = mg = ρvg = = 805 lbf,, iii) W = mg= ρvg= ρ 0. = 80. ρ = lbm/ft, 1 1 Pv ( ) 1.07 v= = = 1.07 ft /lbm, T = = = 61 R, or 800 F ρ R 5. iv) 1 1 P 144 = ρrt = = 7 sia, v= = = 0.65 lbm/ft ρ 1.6 W = mg = ρvg = = 1546 lbf, 5

18 .70 Inside the house T = 7 F or 5 R. If we assume that the ressure is 14.7 sia, ρ inside P = = = lbm/ft RT 5. 5 Outside the house T = 0 F or 440 R. We again assume that the ressure is atmosheric. ρ outside P = = = lbm/ft RT Δ ρ = = lbm/ft The change in density will cause the chimney effect: lighter articles rise and heavier articles fall. Hence, there exists a migration of air articles from the high density region to the low density region, i.e., from the inside to the outside near the to of the house. The heavier articles then enter near the bottom of the walls to take the lace of the vacating articles..71 P RT v T 1 60 = = = = 1.8 P v RT T with v = v. Then, the ressure in Arizona is 1 P = P 1.8 = ( ) 1.8 = 68.6 sia or 5.9 si gage 1 Don t multily 5 si by 1.8. You must use absolute ressures..7 Since the drag force is directly roortional to the density of the air, the higher the density the greater the drag force, which will cause the ball to not travel as far. The question is: Is humid air more dense or less dense than dry air? Consider the ideal gas law ρ = PRT / : P P P P ρ = = and ρ = =. ρ = 1.61ρ R T 0.87T R T 0.46T air humid dry air humid dry air steam The density of the dry air is greater than the density of humid air so the drag is greater on a dry day, contrary to what all those crazy golfers think! It is more uncomfortable on a humid day but we re talking about distance! If it weren t for the ideal gas law, the ball would travel further on that dry day. Was it the liberals or the conservatives who made u that law?.7 Find the gas constant R for nitrogen in Table B-: i) Using the ideal gas law, RT v = = = m/kg P 4000 ii) From Fig.. for P = 4 MPa and T = 10 K, we get Z = 0.4. Then ZRT v = = = m/kg P 4000 The ideal gas law does not give accetable results at high ressures and low temeratures. That combination is uncommon in our introductory course so it is not encountered very often. 6

19 .74 Given: m = 10 kg, V = 1 m, and T = 40 C or K. 1 v = = m/kg. a) i) Use Table B- for the gas constant of air. The ressure, using the ideal gas law, is RT 0.87 P = = = 669 kpa v 0.1 ii) The critical temerature and ressure for air are found in Table B-: 1 K,.77 MPa. The coefficients for van der Waals equation (.17) are found in Table B-10: RT a P = = = 661 kpa v b v b) Table. in Section.5 gives the coefficients of the Beattie-Bridgeman equation (.19) for air. A 0 = 11.84, B 0 = , a = , b = , c = The molar secific volume for air is m kg v = vm = =.90 m /kmol kg kmol RT c b A a = v vt v 1 v v = u 0 P v B 0 = 66 kpa c) Use the estimate of the ressure from Part (i) to find P R : P R = P T T 175 R P = 669 = 0.177, = T = = cr From Aendix Fig. H-1 for T R = 1.75 and P R = 0.177, the comressibility factor is about This rovides a ressure of P = = 66 kpa. d) The IRC calculator for dry air at 40 C and a density of 10 kg/m rovides P = 664 kpa..75 The working fluid is nitrogen at 80 F or 80 R with a secific volume of 0.6 ft /lbm. i) The ideal gas law is used to find RT P = = = 4,900 lbf/ft or 4 sia v 0.6 ii) Find the van der Waals coefficients in Table B-10. Equation (.17) rovides cr 7

20 RT a P = = =, 670 sfa or 4 sia v b v iii) Use the estimate of the ressure from Part (i) to find P R :.76 For air at 60 kpa and 80 C or 19K. P R = P T T 1 R P = 4 = 0.49, = T = 80 = cr From Fig. H-1, Z 0.96 so P = = sia. i) The ideal gas law is used to give RT v = = = 0.9 m/kg P 60 ii) The critical constants for air from Table B-are T cr = 1 K and P cr =.77 MPa. Use Table B-10 for the van der Waals coefficients and Eq..17 for v: RT a P = or 60 = v b v v v This equation is solved for v by trial and error. Use the value from Part (i) as a start: cr For v = 0.9, 60 =? For v = 0.9, 60 =? v = 0.91 m /kg. P T iii) P = T 1 R R P = 60 = 0.016, = T = 19 = cr Using Table H- we get Z v = 0.9 m /kg cr iv) The IRC Calculator for dry air at 80 C and a ressure of 60 kpa gives v = 0.9 m /kg..77 The secific heats are listed in Table B-. a) For air C P = 1.00 kj/kg C, Δh = 1.00 (450 C 0 C) = 40 kj/kg b) For nitrogen C P = 1.04 kj/kg C, Δh = 1.04 (450 0) = 448 kj/kg c) For hydrogen C P = kj/kg C, Δh = (450 0) = 6110 kj/kg d) For roane C P = kj/kg C, Δh = (450 0) = 7 kj/kg e) For steam C P = 1.87 kj/kg C, Δh = 1.87 (450 0) = 805 kj/kg.78 State 1: T 1 = 0 C and x 1 = 1 (Steam is very near an ideal gas at this state: check v = RT/ P.) State : T = 150 C and P = 00 kpa i) For steam, we use Table B- to get C P = 1.87 kj/kg C Δh = C P ΔT = 1.87(150 0) = 5 kj/kg g 8

21 ii) From Table C-1, h 1 = h g = 556. kj/kg. From Table C-, h = kj/kg Δh = = 1 kj/kg iii) Using the IRC calculator, we obtain h 1 = 560 kj/kg and h = 770 kj/kg Δh = = 10 kj/kg.79 a) i) Table B- gives the constant secific heats for nitrogen: Δu = kj/kg C(500 0 ) = 57.6 kj/kg Δh = 1.04 kj/kg C(500 0 ) = 479 kj/kg ii) Use Table F- for the molar internal energies and enthalies: kj / kmol 6190 kj / kmol Δ u = = 71 kj/kg 8 kg / kmol 085 kj / kmol 8669 kj / kmol Δ h = = 514 kj/kg 8 kg / kmol b) The average temerature is T av = ( )/ = 60 C or 5K 5 9 C = C = 9.9 kj/kmol K 9. 9 kj / kmol K = = kj/kg K 8 kg / kmol C = C R= = kj/kg K v Δ u = ( 500 0) = 70 kj/kg, Δ h = ( 500 0) = 51 kj/kg c) = = kj/kg K C C = C R= = kj / kg K v P Δ u= ( 500 0) = 70 kj/kg, Δ h= ( 50 ) = 51 kj/kg.80 m = 0 lbm of air and ΔT = 975 R a) i) From Table B- for air Cv = Btu/lbm- R and C = 0.4 Btu/lbm- R: Δ U = mc Δ T = = 0 Btu v Δ H = mc Δ T = = 4680 Btu ii) From Aendix F-1E: for air at 485 o R we get u 1 = 8.6 Btu/lbm and h 1 = Btu/lbm by interolation. For air at 1460 o R, u = 58.6 Btu/lbm and h = 58.6 Btu/lbm. ΔU = 0 ( ) = 50 Btu 9

22 ΔH = 0 ( ) = 4854 Btu b) Table B-6 is in SI units only. The roblem will have to be converted to SI, solved in SI and then converted back to English units. T 1 = 485 R or 69 K, T = 1460 R or 811 K, T av = 540 K For air: Table B-6 rovides Cv = kj/kg K and C = 1.04 kj/kg K (it is not necessary to be exact in the relatively easy interolation), so Δ U = mc Δ T = ( ) (811 69) = 69 kj or 500 Btu v Δ H = mc Δ T = ( ) 1.04 (811 69) = 5114 kj or 4850 Btu c) Again, we solve the roblem in SI units: C = T T T At the average temerature of 540 K, C = kj/kgk. Then C = C R= = kj/kg K v Δ U = mc Δ T = ( ) (811 69) = 80 kj or 60 Btu v Δ H = mc Δ T = ( ) (811 69) = 54 kj or 4670 Btu d) The IRC Calculator rovides the internal energies and enthalies for dry air at both temeratures. Using English units, ΔU = 0 (59 8.6) = 50 Btu ΔH = 0 (59 116) = 4860 Btu.81 For steam at 1600 kpa and 400 C, we use Table C- for the following enthalies: T ( C) h (kj/kg) i) Using backward differencing at 50 C and 400 C, there results C Δh = = =.18 kj/kg C Δ T 50 C ii) Use forward differencing at 50 C and 400 C: C Δh = = =.18 kj/kg C ΔT 100 iii) Use central differencing at 00 C and 500 C (the intervals must be the same): C Δh = = =.18 kj/kg C ΔT 150 0

23 iv) For steam at 00 K from Table B-, C = 1.87 kj/kg C The curve C (T) does not have sufficient curvature for a difference to aear in the three difference methods; however, central differencing should give the most accurate value of C at 400 C..8 Use the enthaly data for steam from Table C-. a) P = 0. MPa and T = 400 C. We will use central differencing between 00 and 500 to estimate the secific heat at 400 : C Δh = = =.08 kj/kg C ΔT 00 b) P = 1.8 MPa and T = 400 C. We will use central differencing between 00 and 500 to estimate the secific heat at 400 : C Δh = = =.0 kj/kg C ΔT 00 c) P = 7 MPa and T = 400 C. We will use central differencing between 50 and 450 to estimate the secific heat at 400 : C Δh = = =.71 kj/kg C ΔT Using enthaly data for steam from Table C-E. a) P = 40 sia and T = 600 F. We will use central differencing between 500 F and 700 F to get the secific heat at 600 F. C Δh = = = Btu/lbm- F ΔT 00 b) P = 140 sia and T = 600 F. We will use central differencing between 500 F and 700 F to get the secific heat at 600 F. C Δh = = = Btu/lbm- F ΔT 00 c) P = 600 sia and T = 600 F. We will use central differencing between 500 F and 700 F to get the secific heat at 600 F. C Δh = = = 0.67 Btu/lbm- F Δ T lbm of water is heated at 14.7 sia from 60 to 00. The water will exist as a subcooled liquid. The comressed liquid roerties will be aroximated by the saturated liquid roerties. Use Table C.1E to get the internal energies and enthalies: At 60 F, u 1 = u f = 8.08 Btu/lbm and h 1 = h f = 8.08 Btu/lbm. At 00 F, u = u f = Btu/lbm and h = h f = Btu/lbm 1

24 ΔU = 100 ( ) = 14,000 Btu ΔH = 100 ( ) = 14,000 Btu The actual difference is Δh Δ u=δ ( Pv) = PΔ v = ( ) ( ) = 1.7 ft-lbf/lbm or Btu/lbm.85 From Table B-4 we get for mercury at 10 C and 1 atm a value of C = 0.18 kj/kg C. ΔH = mc ΔT = ΔT = 00 kj ΔT = 90 C T final = 5 C + 90 C = 15 C. (Mercury boils at 57 C.).86 In this roblem, 10 kg of ice at 0 C is heated to become liquid water at 50 C. The hase change occurs at 0 C since the ressure is resumably atmosheric. a) If C =.1 kj/kg C, the enthaly change of the ice is ΔH ice = mc ΔT = 10.1 [0 ( 0)] = 40 kj As stated on Section..4, the heat of fusion for ice is 0 kj/kg, so that ΔH melt = 10 kg 0 kj/kg = 00 kj To heat the liquid water from 0 C to 50 C, use C from Table B-4: ΔH liquid = mc ΔT = = 090 kj ΔH total = 40 kj + 00 kj kj = 5810 kj b) Use the equation C, ice = T: Δ H = m C dt ice = 10 ( T) dt = 10.1(0 + 0) = 44 kj 0 As stated on Section..4, the heat of fusion for ice is 0 kj/kg, so that ΔH melt = 10 kg 0 kj/kg = 00 kj To heat the liquid water from 0 C to 50 C, use C from Table B-4: ΔH liquid = mc ΔT = = 090 kj ΔH total = 44 kj + 00 kj kj = 580 kj