SOLUTION MANUAL ENGLISH UNIT ROBLEMS CHATER 5 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition
CHATER 5 SUBSECTION ROB NO. Concept-Study Guide roblems 39-44 Kinetic and otential Energy 45-47 roperties from General Tables 48-50 Simple rocesses 5-57 Multistep rocesses and Reiew roblems 58-6, and 8 Solids and Liquids 64-67 Ideal Gas 68-7, and 63 olytropic rocesses 73-78 Energy equation in Rate Form 79-8 New 5th SI New 5th SI New 5th SI 39 new 54 new 44 69 95 40 new - 55 50 70 new 89 a 4 new 7 56 5 48 7 5 0 4 new 8 57 5 7 30 93 43 new 58 0 6 73 9 44 new 7 59 09 64 74 3 07 45 new 60 3 9 75 7 04 46 0 0 6 4 6 76 new 06 47 03 6 8 8 77 3 4mod 48 04 mod 3 63 4 85 78 3 5 49 05 mod 30 64 9 78 79 35 50 04 mod - 65 new 77 80 new 5 5 07 37 66 0 76 8 36 7 5 08 38 67 new 8 8 34 38 53 06 39 68 97
Concept roblems 5.39E What is cal in english units, what is Btu in ft-lbf? Look in Table A. for the conersion factors under energy Btu = 778.693 lbf-ft cal = 4.868 J = 4.868 055 Btu = 0.00397 Btu = 3.088 lbf-ft 5.40E Work as F x has units of lbf-ft, what is that in Btu? Look in Table A. for the conersion factors under energy lbf-ft =.8507 0-3 Btu 5.4E A 500 lbm car is accelerated from 5 mi/h to 40 mi/h. How much work is that? The work input is the increase in kinetic energy. E E = (/)m[v - V ] = W = 0.5 500 lbm [40 5 ] mi h = 50 [ 600 65 ] lbm 609.3 3.8084 ft 3600 s lbf 3.74 lbm ft/s = 6 53 lbf-ft = 3369 Btu
5.4E A crane use 7000 Btu/h to raise a 00 lbm box 60 ft. How much time does it take? ower = Ẇ = FV = mgv = mg L t F = mg = 00 3.74 3.74 lbf = 00 lbf t = FL 00 lbf 60 ft = Ẇ 7000 Btu/h = 7.9 s = 00 60 3600 7000 778.7 s Recall Eq. on page 0: lbf = 3.74 lbm ft/s, Btu = 778.7 lbf-ft (A.)
5.43E I hae 4 lbm of liquid water at 70 F, 5 psia. I now add 0 Btu of energy at a constant pressure. How hot does it get if it is heated? How fast does it moe if it is pushed by a constant horizontal force? How high does it go if it is raised straight up? a) Heat at 00 ka. Energy equation: E E = Q W = Q (V V ) = H H = m(h h ) h = h + Q /m = 38.09 + 0/4 = 43.09 Btu/lbm Back interpolate in Table F.7.: T = 75 F (We could also hae used T = Q /mc = 0 / (4.00) = 5 F) b) ush at constant. It gains kinetic energy. 0.5 m V = W V = W /m = 0 778.7 lbf-ft/4 lbm = 0 778.7 3.74 lbm-(ft/s) /4 lbm = 500 ft/s c) Raised in graitational field m g Z = W 0 778.7 lbf-ft lbm-ft/s Z = W /m g = 4 lbm 3.74 ft/s 3.74 lbf = 389 ft
5.44E Air is heated from 540 R to 640 R at V = C. Find q? What if from 400 to 500 R? rocess: V = C W = Ø Energy Eq.: u u = q 0 q = u u Read the u-alues from Table F.5 a) q = u u = 09.34 9.6 = 7.8 Btu/lbm b) q = u u = 474.33 45.64 =.7 Btu/lbm case a) C 7.8/00 = 0.7 Btu/lbm R, see F.4 case b) C.7/00 = 0.7 Btu/lbm R (6 % higher)
Kinetic and otential Energy 5.45E Airplane takeoff from an aircraft carrier is assisted by a steam drien piston/cylin-der with an aerage pressure of 00 psia. A 38 500 lbm airplane should be accelerated from zero to a speed of 00 ft/s with 30% of the energy coming from the steam piston. Find the needed piston displacement olume. Solution: C.V. Airplane. No change in internal or potential energy; only kinetic energy is changed. E E = m (/) (V - 0) = 38 500 lbm (/) 00 (ft/s) = 9 500 000 lbm-(ft/s) = 5 983 09 lbf-ft The work supplied by the piston is 30% of the energy increase. W = dv = ag V = 0.30 (E E ) = 0.30 5 983 09 lbf-ft = 794 98 lbf-ft V = W 794 98 lbf-ft = ag 00 44 lbf/ft = 6.3 ft3
5.46E A hydraulic hoist raises a 3650 lbm car 6 ft in an auto repair shop. The hydraulic pump has a constant pressure of 00 lbf/in. on its piston. What is the increase in potential energy of the car and how much olume should the pump displace to delier that amount of work? Solution: C.V. Car. No change in kinetic or internal energy of the car, neglect hoist mass. 3650 3.74 6 E E = E - E = mg (Z Z ) = = 900 lbf-ft 3.74 The increase in potential energy is work into car from pump at constant. W = E E = dv = V V = E E = 900 00 44 =.5 ft3
5.47E A piston motion moes a 50 lbm hammerhead ertically down 3 ft from rest to a elocity of 50 ft/s in a stamping machine. What is the change in total energy of the hammerhead? Solution: C.V. Hammerhead The hammerhead does not change internal energy i.e. same,t E E = m(u u ) + m( V - 0) + mg (h - 0) = 0 + [ 50 (/) 50 + 50 3.74 (-3)] / 3.74 = [56500 486]/3.74 = 7 333 lbf-ft 7 333 = ( 778 ) Btu =.8 Btu
roperties General Tables 5.48E Find the missing properties and gie the phase of the substance. a. H O u = 000 Btu/lbm, T = 70 F h =? =? x =? b. H O u = 450 Btu/lbm, = 500 lbf/in. T =? x =? =? c. R- T = 30 F, = 75 lbf/in. h =? x =? Solution: a) Table F.7.: u f < u < u g => -phase mixture of liquid and apor x = (u u f )/ u fg = (000 38.8)/854.4 = 0.89 = f + x fg = 0.077 + 0.89 0.0483 = 8.97 ft 3 /lbm h = h f + x h fg = 38.95 + 0.89 93.95 = 069.5 Btu/lbm ( = 000 + 4.848 8.97 44/778) b) Table F.7.: u < u f so compressed liquid B..3, x = undefined T = 47.8 F, = 0.09689 ft 3 /lbm c) Table F.9.: > sat => x = undef, compr. liquid Approximate as saturated liquid at same T, h h f = 8.6 Btu/lbm States shown are placed relatie to the two-phase region, not to each other. C.. b c a T T c b C.. a = const.
5.49E Find the missing properties among (, T,, u, h) together with x, if applicable, and gie the phase of the substance. a. R- T = 50 F, u = 85 Btu/lbm b. H O T = 600 F, h = 3 Btu/lbm c. R- = 50 lbf/in., h = 5.5 Btu/lbm Solution: a) Table F.9.: u < u g => L+V mixture, = 98.77 lbf/in x = (85-4.04)/ 74.75 = 0.855 = 0.08 + 0.855 0.543 = 0.4558 ft 3 /lbm h = 4.7 + 0.855 84.68 = 93.33 Btu/lbm b) Table F.7.: h > h g => superheated apor follow 600 F in F.7. 00 lbf/in ; = 3.058 ft 3 /lbm ; u = 08.9 Btu/lbm c) Table F.9.: h > h g => superheated apor so in F.9. T 00 F ; = 0.3953 ft 3 /lbm 44 u = h - = 5.5 50 0.3953 = 04.5 Btu/lbm 778 States shown are placed relatie to the two-phase region, not to each other. C.. a T C.. = const. b, c T a b, c
5.50E Find the missing properties and gie the phase of the substance. a. R-34a T = 40 F, h = 85 Btu/lbm =? x =? b. NH 3 T = 70 F, = 60 lbf/in. u =? =? x =? c. R-34a T = 00 F, u = 75 Btu/lbm Solution: a) Table F.0.: h > hg => x = undef, superheated apor F.0., find it at gien T between saturated 43.9 psi and 00 psi to match h: 0.836 + (0.459-0.836) 43.93 + (00-43.93) 85-83.63 86.8-83.63 = 0.04 ft3 /lbm 85-83.63 86.8-83.63 = 5 lbf/in b) Table F.8.: < sat x = undef. superheated apor F.8., = (6.3456 + 6.5694)/ = 6.457 ft 3 /lbm u = h- = (/)(694.59 + 705.64) 60 6.4575 (44/778) = 700.5 7.7 = 68.405 Btu/lbm c) Table F.0.:: u > u g => sup. apor, calculate u at some to end with 55 lbf/in ; 0.999 ft 3 /lbm; h = 85. Btu/lbm This is a double linear interpolation States shown are placed relatie to the two-phase region, not to each other. C.. a b T c T C.. a = const. b c
Simple rocesses 5.5E A cylinder fitted with a frictionless piston contains 4 lbm of superheated refrigerant R-34a apor at 400 lbf/in., 00 F. The cylinder is now cooled so the R-34a remains at constant pressure until it reaches a quality of 75%. Calculate the heat transfer in the process. Solution: C.V.: R-34a m = m = m; Energy Eq.5. m(u - u ) = Q - W rocess: = const. W = dv = V = (V - V ) = m( - ) T V V State : Table F.0. h = 9.9 Btu/lbm State : Table F.0. h = 40.6 + 0.75 43.74 = 73.45 Btu/lbm Q = m(u - u ) + W = m(u - u ) + m( - ) = m(h - h ) = 4 (73.45 9.9) = -77.98 Btu
5.5E Ammonia at 30 F, quality 60% is contained in a rigid 8-ft 3 tank. The tank and ammonia are now heated to a final pressure of 50 lbf/in.. Determine the heat transfer for the process. Solution: C.V.: NH 3 V Continuity Eq.: m = m = m ; Energy Eq.5.: m(u - u ) = Q - W rocess: Constant olume = & W = 0 State : Table F.8. two-phase state. = 0.050 + 0.6 4.7978 =.904 ft 3 /lbm u = 75.06 + 0.6 49.7 = 369.75 Btu/lbm m = V/ = 8/.904 =.755 lbm State :, = T 58 F u = h - = 74.03-50.904 44/778 = 66.4 Btu/lbm Q =.755 (66.4-369.75) = 803.6 Btu
5.53E Water in a 6-ft 3 closed, rigid tank is at 00 F, 90% quality. The tank is then cooled to 0 F. Calculate the heat transfer during the process. Solution: C.V.: Water in tank. m = m ; m(u - u ) = Q - W rocess: V = constant, =, W = 0 State : = 0.0663 + 0.9 33.646 = 30.7 ft 3 /lbm u = 68.03 + 0.9 906.5 = 983.6 Btu/lbm State : T, = mix of sat. solid + ap. Table C.8.4 = 30.7 = 0.0744 + x 5655 => x = 0.00535 u = -49.3 + 0.00535 66.5 = -43.07 Btu/lbm m = V/ = 6 / 30.7 = 0.98 lbm Q = m(u - u ) = 0.98 (-43.07-983.6) = -3 Btu C.. T C.. = const. T T S L L + V S + V C.. V
5.54E A constant pressure piston/cylinder has lbm water at 00 F and.6 ft 3. It is now cooled to occupy /0 of the original olume. Find the heat transfer in the process. C.V.: Water m = m = m; Energy Eq.5. m(u - u ) = Q - W rocess: = const. W = dv = V = (V - V ) = m( - ) State : Table F.7. (T, = V/m =.6/ =.3 ft 3 /lbm) = 800 psia, h = 567.8 Btu/lbm State : Table F.7. (, = /0 = 0.3 ft 3 /lbm) two-phase state x = ( f )/ fg = (0.3 0.0087)/0.5488 = 0.679 h = h f + x h fg = 509.63 + x 689.6 = 65.4 Btu/lbm Q = m(u - u ) + W = m(h - h ) = (65.4 567.8) = -884.8 Btu T V V
5.55E A piston/cylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 0 lbf/in., shown in Fig 5.50. It contains water at 5 F, which is then heated until the water becomes saturated apor. Find the final temperature and specific work and heat transfer for the process. Solution: C.V. Water in the piston cylinder. Continuity: m = m, Energy: u - u = q - w rocess: = const. =, => w = d = ( - ) State : T, => Table F.7.4 compressed solid, take as saturated solid. = 0.0746 ft 3 /lbm, u = -46.84 Btu/lbm State : x =, = = 0 psia due to process => Table F.7. = g ( ) = 0.09 ft 3 /lbm, T = 8 F ; u = 08 Btu/lbm w = ( - ) = 0(0.09-0.0746) 44/778 = 74.3 Btu/lbm q = u - u + w = 08 - (-46.84) + 74.3 = 303 Btu/lbm L C.. C.. T C.. T S L+V V = C S+V
5.56E A water-filled reactor with olume of 50 ft 3 is at 000 lbf/in., 560 F and placed inside a containment room, as shown in Fig. 5.48. The room is well insulated and initially eacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room olume so the final pressure does not exceed 30 lbf/in.. C.V.: Containment room and reactor. Mass: m = m = V reactor / = 50/0.07 = 95.7 lbm Energy m(u - u ) = Q - W = 0/ u = u = 55.5 Btu/lbm State : 30 lbf/in., u < ug phase Table F.7. u = 55.5 = 8.48 + x 869.4 x = 0.384 = 0.07 + 0.384 3.808 = 5.3 ft 3 /lbm V = m = 95.7 5.3 = 8 ft 3 T 000 30 30 psia u = const L C.. T 00 ka
5.57E A piston/cylinder contains lbm of liquid water at 70 F, and 30 lbf/in.. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 300 lbf/in. with a olume of 4 ft 3. Find the final temperature and plot the - diagram for the process. Calculate the work and the heat transfer for the process. Solution: Take CV as the water. m = m = m ; m(u u ) = Q - W State : Compressed liquid, take saturated liquid at same temperature. = f (0) = 0.0605 ft 3 /lbm, State : = V /m = 4/ = ft 3 /lbm and = 300 psia u = u f = 38.09 Btu/lbm => Superheated apor T = 600 F ; u = 03. Btu/lbm Work is done while piston moes at linearly arying pressure, so we get W = dv = area = ag (V V ) 44 = 0.5 (30+3000)(4-0.03) =.8 Btu 778 Heat transfer is found from the energy equation Q = m(u u ) + W = (03. 38.09) +.8 = 45.4 Btu cb
Multistep and Reiew roblems 5.58E A twenty pound-mass of water in a piston/cylinder with constant pressure is at 00 F and a olume of.6 ft 3. It is now cooled to 00 F. Show the diagram and find the work and heat transfer for the process. Solution: C.V. Water Energy Eq.: Q = m(u u ) + W = m(h - h ) rocess Eq.: Constant pressure W = m( - ) roperties from Table F.7. and F.7.3 State : T, =.6/0 =.3 ft 3 /lbm, = 800 lbf/in, h = 567.8 State : 800 lbf/in, 00 F = 0.0609 ft 3 /lbm, h = 70.5 Btu/lbm T 800 psia W = 0 800 (0.0609 -.3) 44/778 = -399 Btu Q = 0 (70.5-567.8) = -9 953 Btu
5.59E A ertical cylinder fitted with a piston contains 0 lbm of R- at 50 F, shown in Fig. 5.64. Heat is transferred to the system causing the piston to rise until it reaches a set of stops at which point the olume has doubled. Additional heat is transferred until the temperature inside reaches 0 F, at which point the pressure inside the cylinder is 00 lbf/in.. a. What is the quality at the initial state? b. Calculate the heat transfer for the oerall process. Solution: C.V. R-. Control mass goes through process: -> -> 3 As piston floats pressure is constant ( -> ) and the olume is constant for the second part ( -> 3). So we hae: 3 = = State 3: Table F.9. (,T) 3 = 0.959 ft 3 /kg, u 3 = h - = 7.0-00 0.959 44/778 = 06. Btu/lbm o 3 cb R- V So we can determine state and Table F.9.: = 0.4795 = 0.08 + x (0.543) => x = 0.49 u = 4.04 + 0.49 74.75 = 4.6 Btu/lbm State : = 0.959 ft 3 /lbm, = = 98.7 psia, this is still -phase. W 3 = W = dv = (V - V ) = 98.7 0(0.95948-0.47974) 44/778 = 7.0 Btu Q 3 = m(u 3 - u ) + W 3 = 0(06. - 4.6) + 7.0 = 66 Btu
5.60E A piston/cylinder contains lbm of water at 70 F with a olume of 0. ft 3, shown in Fig. 5.9. Initially the piston rests on some stops with the top surface open to the atmosphere, o, so a pressure of 40 lbf/in. is required to lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated apor find the final temperature, olume, and the heat transfer. Solution: C.V. Water. This is a control mass. m = m = m ; m(u - u ) = Q - W State : 0 C, = V/m = 0./ = 0.05 ft 3 /lbm x = (0.05-0.0605)/867.579 = 0.000393 a u = 38.09 + 0.000393 995.64 = 38.3 Btu/lbm To find state check on state a: = 40 psia, = = 0.05 ft 3 /lbm Table F.7.: f < < g = 0.50 State is saturated apor at 40 psia as state a is two-phase. T = 67.3 F V = g = 0.50 ft 3 /lbm, V = m =.0 ft 3, u = ug= 09.7 Btu/lbm ressure is constant as olume increase beyond initial olume. W = dv = lift (V -V ) = 40 (.0 0.) 44 / 778 = 54.75 Btu Q = m(u - u ) + W = (09.7 38.3) + 54.75 = 63 Btu
5.6E Two tanks are connected by a ale and line as shown in Fig. 5.6. The olumes are both 35 ft 3 with R-34a at 70 F, quality 5% in A and tank B is eacuated. The ale is opened and saturated apor flows from A into B until the pressures become equal. The process occurs slowly enough that all temperatures stay at 70 F during the process. Find the total heat transfer to the R-34a during the process. C.V.: A + B State A: Table F.0., u A = 98.7 + 0.5 69.3 = 5.6 Btu/lbm A = 0.033 + 0.5 0.545 = 0.494 ft 3 /lbm => m A = V A / A = 34.3 lbm rocess: Constant T and total olume. m = m A ; V = V A + V B = 70 ft 3 State : T, = V /m = 70/34.3 = 0.988 ft 3 /lbm => x = (0.988-0.033)/0.545 = 0.54 ; u = 98.7 + 0.54*69.3 = 34.6 Btu/lbm The energy equation gies the heat transfer Q = m u - m A u A - m B u B + W = m (u - u A ) = 34.3 (34.6-5.6) = 445 Btu A B
5.6E Ammonia, NH 3, is contained in a sealed rigid tank at 30 F, x = 50% and is then heated to 00 F. Find the final state, u and the specific work and heat transfer. Solution: Continuity Eq.: m = m ; Energy Eq.5.: E - E = Q ; ( W = 0/ ) rocess: V = V = = 0.050 + 0.5 4.7945 =.4 ft 3 /lbm State : Table F.8., Table F.8.: u = 75.06 + 0.5 49.7 = 30.65 Btu/lbm & T between 50 psia and 75 psia = 63 lbf/in, h = 706.6 Btu/lbm linear interpolation u = h - = 706.6 63.4 44/778 = 633.5 Btu/lbm V rocess equation gies no displacement: w = 0 ; The energy equation then gies the heat transfer as q = u - u = 633.5-30.65 = 3.85 Btu/lbm
5.63E Water at 70 F, 5 lbf/in., is brought to 30 lbf/in., 700 F. Find the change in the specific internal energy, using the water table and the ideal gas water table in combination. State : Table F.7. u u f = 38.09 Btu/lbm State : Highest T in Table F.7. is 400 F Using a u from the ideal gas table F.6, we get - h 700 - h - 000 = 600-769 = 433 Btu/lbmol= 790 Btu/lbm u 700 - u 400 = h- R(700-400) = 790-53.34 300 778 = 700.9 Since ideal gas change is at low we use 400 F, lowest aailable lbf/in from steam tables, F.7., u x = 543. Btu/lbm as the reference. u - u = (u - u x ) ID.G. + (u x - u ) = 700.9 + 543. - 38.09 = 06 Btu/lbm
Solids and Liquids 5.64E A car with mass 350 lbm dries with 60 mi/h when the brakes are applied to quickly decrease its speed to 0 mi/h. Assume the brake pads are lbm mass with heat capacity of 0. Btu/lbm R and the brake discs/drums are 8 lbm steel where both masses are heated uniformly. Find the temperature increase in the brake assembly. C.V. Car. Car looses kinetic energy and brake system gains internal u. No heat transfer (short time) and no work term. m = constant; E - E = 0-0 = m car (V V ) + m brake(u u ) The brake system mass is two different kinds so split it, also use C since we do not hae a u table for steel or brake pad material. m steel C T + m pad C T = m car (V V ) (8 0. + 0.) T = 350 0.5 300.46667 /(3.74 778) = 446.9 Btu => T = 44 F
5.65E A lbm steel pot contains lbm liquid water at 60 F. It is now put on the stoe where it is heated to the boiling point of the water. Neglect any air being heated and find the total amount of energy needed. Solution: Energy Eq.: U U = Q W The steel does not change olume and the change for the liquid is minimal, so W 0. State : T = T sat (atm) = F Tbl F.7. : u = 8. Btu/lbm, u = 80. Btu/lbm Tbl F. : C st = 0. Btu/lbm R Sole for the heat transfer from the energy equation Q = U U = m st (u u ) st + m HO (u u ) HO = m st C st (T T ) + m HO (u u ) HO Q = lbm 0. Btu Btu lbm R ( 60) R + lbm (80. 8.) lbm = 33.4 + 304 = 337.4 Btu
5.66E A copper block of olume 60 in. 3 is heat treated at 900 F and now cooled in a 3- ft 3 oil bath initially at 70 F. Assuming no heat transfer with the surroundings, what is the final temperature? C.V. Copper block and the oil bath. m met (u - u ) met + m oil (u - u ) oil = Q - W = 0/ solid and liquid u C V T m met C Vmet (T - T,met ) + m oil C Voil (T - T,oil ) = 0/ m met = Vρ = 60-3 555 = 9.7 lbm m oil = Vρ = 3.5 57 = 99.5 lbm Energy equation becomes 9.7 0.09(T -900) + 99.5 0.43(T -70) = 0/ T = 86.8 F
5.67E An engine consists of a 00 lbm cast iron block with a 40 lbm aluminum head, 40 lbm steel parts, 0 lbm engine oil and lbm glycerine (antifreeze). Eerything begins at 40 F and as the engine starts, it absorbs a net of 7000 Btu before it reaches a steady uniform temperature. We want to know how hot it becomes. Energy Eq.: U U = Q W rocess: The steel does not change olume and the change for the liquid is minimal, so W 0. So sum oer the arious parts of the left hand side in the energy equation m Fe (u u ) + m Al (u u ) Al + m st (u u ) st + m oil (u u ) oil + m gly (u u ) gly = Q Tbl F. : C Fe = 0., C Al = 0.5, C st = 0. all units of Btu/lbm R Tbl F.3 : C oil = 0.46, C gly = 0.58 all units of Btu/lbm R So now we factor out T T as u u = C(T T ) for each term [ m Fe C Fe + m Al C Al + m st C st + m oil C oil + m gly C gly ] (T T ) = Q T T = Q / Σm i C i 7000 = 00 0. + 40 0.5 + 40 0. + 0 0.46 + 0.58 = 7000 44.56 = 57 R T = T + 57 = 40 + 57 = 97 F Shaft power Air intake filter Fan Radiator Atm. air Exhaust flow Coolant flow
Ideal Gas 5.68E A cylinder with a piston restrained by a linear spring contains 4 lbm of carbon dioxide at 70 lbf/in., 750 F. It is cooled to 75 F, at which point the pressure is 45 lbf/in.. Calculate the heat transfer for the process. Solution: C.V. The carbon dioxide, which is a control mass. Continuity Eq.: m m = 0 Energy Eq.: m (u u ) = Q - W rocess Eq.: = A + BV (linear spring) W = dv = ( + )(V - V ) Equation of state: State : V = mrt / = State : V = mrt / = V = mrt (ideal gas) 4 35. (750 + 460) 70 44 4 35. (75 + 460) 45 44 = 6.85 ft 3 =.59 ft 3 W = (70 + 45)(.59 6.85) 44/778 = -55.98 Btu To ealuate u - u we will use the specific heat at the aerage temperature. From Table F.6: C po (T ag ) = h T = 697-0 M 00-537 = 0.45 44.0 = 0.347 Btu/lbm R C V = C p R = 0.375 35.0/778 = 0.94 Btu/lbm R For comparison the alue from Table F.4 at 77 F is C o = 0.56 Btu/lbm R Q = m(u - u ) + W = mc o (T - T ) + W = 4 0.94(75-750) - 55.98 = -575.46 Btu CO
5.69E An insulated cylinder is diided into two parts of 0 ft 3 each by an initially locked piston. Side A has air at atm, 600 R and side B has air at 0 atm, 000 R as shown in Fig. 5.95. The piston is now unlocked so it is free to moe, and it conducts heat so the air comes to a uniform temperature T A = T B. Find the mass in both A and B and also the final T and. C.V. A + B. Then Q = 0/, W = 0/. Force balance on piston: A A = B A, so final state in A and B is the same. State A: u A = 0.457 ; m A = V RT = 9.4 0 44 =.33 lbm 53.34 600 State B: u B = 367.64 ; m A (u - u ) A + m B (u - u ) B = 0/ (m A + m B )u = m A u A + m B u B m B = V RT = 47 0 44 53. 34 000 =.984 lbm =.33 0.457 +.984 367.64 = 864.95 Btu u = 864.95/3.307 = 6.55 T = 475 R = m tot RT /V tot = 3.307 53.34 475 0 44 = 90.34 lbf/in A B
5.70E A 65 gallons rigid tank contains methane gas at 900 R, 00 psia. It is now cooled down to 540 R. Assume ideal gas and find the needed heat transfer. Solution: Ideal gas and recall from Table A. that gal = 3 in 3, m = V/RT = 00 65 3 96.35 900 =.886 lbm rocess: V = constant = V => W = 0 Use specific heat from Table F.4 Energy Equation u - u = C (T T ) = 0.45 (900 540) = 49.4 Btu/lbm Q = m(u - u ) =.886 (-49.4) = 43. Btu
5.7E Air in a piston/cylinder at 30 lbf/in., 080 R, is shown in Fig. 5.69. It is expanded in a constant-pressure process to twice the initial olume (state ). The piston is then locked with a pin, and heat is transferred to a final temperature of 080 R. Find, T, and h for states and 3, and find the work and heat transfer in both processes. C.V. Air. Control mass m = m 3 = m : u -u = q - w ; w = d = ( - ) = R(T -T ) Ideal gas = RT T = T / = T = 60 R = = 30 lbf/in, h = 549.357 w = RT = 74.05 Btu/lbm Table F.5 h = 549.357 Btu/lbm, h 3 = h = 6.099 Btu/lbm q = u - u + w = h - h = 549.357-6.099 = 88.6 Btu/lbm 3: 3 = = w 3 = 0, 3 = T 3 /T = / = 5 lbf/in q 3 = u 3 - u = 87.058-40.76 = -4. Btu/lbm o 30 60 T Air cb 5 3 080 3
5.7E A 30-ft high cylinder, cross-sectional area ft, has a massless piston at the bottom with water at 70 F on top of it, as shown in Fig. 5.93. Air at 540 R, olume 0 ft 3 under the piston is heated so that the piston moes up, spilling the water out oer the side. Find the total heat transfer to the air when all the water has been pushed out. Solution o HO cb air 0 V Vmax V The water on top is compressed liquid and has mass V HO = V tot - V air = 30-0 = 0 ft 3 m HO = V HO / f = 0/0.0605 = 46 lbm g Initial air pressure is: = 0 + m HO g/a = 4.7 + = 3.353 psia 44 and then m air = V 3.353 0 44 RT = =.675 lbm 53.34 540 State : = 0 = 4.7 lbf/in, V = 30 = 30 ft 3 W = dv = ( + )(V - V ) = (3.353 + 4.7)(30-0) 44 / 778 = 70.43 Btu State :, V T = T V V = 540 4.7 30 3.353 0 = 09.7 R Q = m(u - u ) + W =.675 0.7 (09.7-540) + 70.43 = 66. Btu
olytropic rocess 5.73E An air pistol contains compressed air in a small cylinder, as shown in Fig. 5.. Assume that the olume is in. 3, pressure is 0 atm, and the temperature is 80 F when armed. A bullet, m = 0.04 lbm, acts as a piston initially held by a pin (trigger); when released, the air expands in an isothermal process (T = constant). If the air pressure is atm in the cylinder as the bullet leaes the gun, find a. The final olume and the mass of air. b. The work done by the air and work done on the atmosphere. c. The work to the bullet and the bullet exit elocity. C.V. Air. Air ideal gas: m air = V /RT = 0 4.7 53.34 539.67 = 4.6 0-5 lbm rocess: V = const = V = V V = V / = 0 in 3 W = dv = V (/V) dv = V ln( V /V ) = 0.036 Btu W,ATM = 0 (V - V ) = 0.04 Btu W bullet = W - W,ATM = 0.0 Btu = m bullet (V ex ) V ex =(W bullet /m B ) / = ( 0.0 778 3.74 / 0.04) / = 65.9 ft/s
5.74E A piston/cylinder in a car contains in. 3 of air at 3 lbf/in., 68 F, shown in Fig. 5.66. The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent n =.5 to a final olume six times smaller. Determine the final pressure, temperature, and the heat transfer for the process. C.V. Air. This is a control mass going through a polytropic process. Cont.: m = m ; Energy: E - E = m(u - u ) = Q - W rocess: n = const. ; Ideal gas: = RT n = n = n = 3 (6).5 =.08 lbf/in T = T ( / ) = 57.67(.08/3 6) = 85.9 R = C -.5 T T = C -0.5 m = V 3 - RT = 53.34 57.67 = 4.69 0-4 lbm w = d = - n ( R - ) = - n (T - T ) = 53.34 85.9-57.67 ( -.5) 778 = -8.79 Btu/lbm q = u - u + w = 4.64-90.05-8.79 = -30. Btu/lbm Q = m q = 4.69 0-4 (-30.)= -0.039 Btu
5.75E Oxygen at 50 lbf/in., 00 F is in a piston/cylinder arrangement with a olume of 4 ft 3. It is now compressed in a polytropic process with exponent, n =., to a final temperature of 400 F. Calculate the heat transfer for the process. Continuity: m = m ; Energy: E - E = m(u - u ) = Q - W State : T, and ideal gas, small change in T, so use Table C.4 rocess: V n = constant m = V 50 4 44 RT = = 0.9043 lbm 48.8 659.67 W = -n ( V - V ) = mr -n (T - T 0.9043 48.8 400 00 ) =. 778 = - 56. Btu Q = m(u - u ) + W mc (T - T ) + W = 0.9043 0.58 (400-00) 56. = - 7.54 Btu = C -. T T T T = C -0.
5.76E Helium gas expands from 0 psia, 600 R and 9 ft 3 to 5 psia in a polytropic process with n =.667. How much heat transfer is inoled? Solution: C.V. Helium gas, this is a control mass. Energy equation: m(u u ) = Q W rocess equation: V n = constant = V n = V n 0 9 44 Ideal gas (F.4): m = V/RT = = 0. lbm 386 600 Sole for the olume at state V = V ( / ) /n = 9 0 5 0.6 = 0.696 ft 3 5 0.696 T = T V /( V ) = 600 0 9 = 534.8 R Work from Eq.4.4 W = V - V 5 0.696-0 9 -n = -.667 psia ft 3 = 9.33 psia ft 3 = 43 lbf-ft = 5.43 Btu Use specific heat from Table F.4 to ealuate u u, C = 0.744 Btu/lbm R Q = m(u u ) + W = m C (T T ) + W = 0. 0.744 (534.8 600) + 5.43 = -0.003 Btu
5.77E A cylinder fitted with a frictionless piston contains R-34a at 00 F, 80% quality, at which point the olume is 3 Gal. The external force on the piston is now aried in such a manner that the R-34a slowly expands in a polytropic process to 50 lbf/in., 80 F. Calculate the work and the heat transfer for this process. Solution: C.V. The mass of R-34a. roperties in Table F.0. = f + x fg = 0.0387 + 0.8 0.378 = 0.76 ft 3 /lbm u = 08.5 + 0.8 6.77 = 58.73 Btu/lbm; = 38.96 psia m = V/ = 3 3-3 / 0.76 = 0.40/ 0.76 =.455 lbm State : =.0563 ft 3 /lbm (sup.ap.); u = 8. 50.0563 44/778 = 7.3 Btu/lbm rocess: n = ln / ln V V = ln 38.96 50 / ln.0563 9.76 = 0.766 W = dv = V - V - n = 50.0563 38.96 0.76 0.766.455 44 778 = 6.3 Btu Q = m(u u ) + W =.455 (7.3 58.73) + 6.3 = 34.6 Btu
5.78E A piston cylinder contains argon at 0 lbf/in., 60 F, and the olume is 4 ft 3. The gas is compressed in a polytropic process to 00 lbf/in., 550 F. Calculate the heat transfer during the process. Find the final olume, then knowing, V,, V the polytropic exponent can be determined. Argon is an ideal monatomic gas (C is constant). V = V = ( / )/(T /T ) = 4 0 00 009.67 59.67 =.554 ft3 rocess: V n = const. => n = ln / ln V V = ln 00 0 / ln 4.554 =.70 W = -n ( V V 00.554-0 4 ) = -.70 44 778 = -9.9 Btu m = V/RT = 0 4 44 / (38.68 59.67) = 0.573 lbm Q = m(u u ) + W = m C (T T ) + W = 0.573 0.0745 (550 60) 9.9 =.0 Btu
Rates of Energy 5.79E A small eleator is being designed for a construction site. It is expected to carry four 50 lbm workers to the top of a 300-ft-tall building in less than min. The eleator cage will hae a counterweight to balance its mass. What is the smallest size (power) electric motor that can drie this unit? m = 4 50 = 600 lbm ; Z = 300 ft ; t = minutes -Ẇ = Ė = m g Z t = 600 3.74 300 3.74 60 550 =.73 hp
5.80E Water is in a piston cylinder maintaining constant at 330 F, quality 90% with a olume of 4 ft 3. A heater is turned on heating the water with 0 000 Btu/h. What is the elapsed time to aporize all the liquid? Solution: Control olume water. Continuity Eq.: m tot = constant = m apor + m liq on a rate form: ṁ tot = 0 = ṁ apor + ṁ liq ṁ liq = -ṁ apor Energy equation: U. = Q. - Ẇ = ṁ apor u fg = Q. - ṁ apor fg Rearrange to sole for ṁ apor ṁ apor (u fg + fg ) = ṁ apor h fg = Q. From table F.7. h fg = 887.5 Bt/lbm, = 0.0776 + 0.9 4.938 = 3.88 ft 3 /lbm m = V / = 4/3.88 =.0303 lbm, m liq = (-x )m = 0.0303 lbm ṁ apor = Q. /h fg = 0 000 887.5 Btu/h Btu/lbm =.676 lbm/h = 0.0033 lbm/s t = m liq / ṁ apor = 0.0303 / 0.0033 = 3.9 s
5.8E A computer in a closed room of olume 5000 ft 3 dissipates energy at a rate of 0 hp. The room has 00 lbm of wood, 50 lbm of steel and air, with all material at 540 R, atm. Assuming all the mass heats up uniformly how long time will it take to increase the temperature 0 F? C.V. Air, wood and steel. m = m ; U - U = Q = Q. t The total olume is nearly all air, but we can find olume of the solids. V wood = m/ρ = 00/44.9 =.3 ft 3 ; V steel = 50/488 = 0.0 ft 3 V air = 5000 -.3-0.0 = 4997.7 ft 3 m air = V/RT = 4.7 4997.7 44/(53.34 540) = 367.3 lbm We do not hae a u table for steel or wood so use heat capacity. U = [m air C + m wood C + m steel C ] T = (367.3 0.7 + 00 0.3 + 50 0.) 0 = 56. + 600 +0 = 966 Btu = Q. t = 0 (550/778) t => t = [966/0] 778 550 = 78 sec = 4.6 minutes
5.8E A closed cylinder is diided into two rooms by a frictionless piston held in place by a pin, as shown in Fig. 5.38. Room A has 0.3 ft 3 air at 4.7 lbf/in., 90 F, and room B has 0 ft 3 saturated water apor at 90 F. The pin is pulled, releasing the piston and both rooms come to equilibrium at 90 F. Considering a control mass of the air and water, determine the work done by the system and the heat transfer to the cylinder. Solution: C.V. A + B, control mass of constant total olume. Energy equation: m A (u u ) A + m B (u B u B ) = Q W rocess equation: V = C W = 0 T = C (u u ) A = 0 (ideal gas) The pressure on both sides of the piston must be the same at state. Since two-phase: = g H O at 90 F = A = B = 4.46 ka Air, I.G.: A V A = m A R A T = A V A = g H O at 90 F V A V A = 4.7 0.3 0.6988 = 6.3 ft3 Now the water olume is the rest of the total olume V B = V A + V B - V A = 0.30 + 0-6.3 = 3.99 ft 3 m B = V B 0 = B 467.7 = 0.038 lbm B = 86.6 ft3 /lbm 86.6 = 0.06099 + x B (467.7-0.06) => x B = 0.39895 u B = 58.07 + 0.39895 98. = 449.9 Btu/lbm; u B = 040. Q = m B (u B u B ) = 0.038 (449.9-040.) = -.6 Btu A B