Solution to extra problem in chapter 8 Noember 9, 000 Fall 000 J. Murthy ME 4- HERMODYNAMICS I 8.5 Water i ued a the working fluid in a Carnot cycle heat engine, where it change from aturated liquid to aturated apor at 00 C a heat i added. Heat i rejected in a contant preure proce (alo contant ) at 0 ka. he heat engine power a Carnot cycle refrigerator that operate between -5 C and +0 C. Find the heat added to the water per kg water. How much heat hould be added to the water in the heat engine o the refrigerator can remoe kj from the cold pace? Solution: Carnot cycle: q H = H ( ) = h fg = 473.5 (4.04) = 940 kj/kg L = at (0 ka) = 60.06 o C β ref = Q L / W = L / ( H - L ) = (73-5) / (0 - (-5)) = 58 / 35 = 7.37 W = Q L / β = / 7.37 = 0.36 kj W = η HE Q H HO η HE = - 333/473 = 0.9 Q H HO = 0.36 / 0.96 = 0.46 kj 8. A cylinder fitted with a piton contain ammonia at 50 C, 0% quality with a olume of L. he ammonia expand lowly, and during thi proce heat i tranferred to maintain a contant temperature. he proce continue until all the liquid i gone. Determine the work and heat tranfer for thi proce. C.V. Ammonia in the cylinder. NH3 able B..: = 50 C, x = 0.0, V = L = 0.00777 + 0. 0.0659 = 0.04095 50 C =.5 + 0. 3.493 =.60 m = V / = 0.00/0.04095 = 0.07 kg S = G = 0.06336, = G = 4.763 roce: = contant to x =.0, = contant =.033 Ma W = dv = m( - ) = 033 0.07 (0.06336-0.04095) = 7. kj Q = ds = m( - ) = 33. 0.07(4.763 -.60) = 59.65 kj or Q = m(u - u ) + W = m(h - h ) h = 4.48 + 0. 050.0 = 63.48, h = 47.49 Q = 0.07(47.49-63.48) = 59.65 kj
8.4 A cylinder fitted with a frictionle piton contain water. A contant hydraulic preure on the back face of the piton maintain a cylinder preure of 0 Ma. Initially, the water i at 700 C, and the olume i 00 L. he water i now cooled and condened to aturated liquid. he heat releaed during thi proce i the Q upply to a cyclic heat engine that in turn reject heat to the ambient at 30 C. If the oerall proce i reerible, what i the net work output of the heat engine? C.V.: H O, 3, thi i a control ma: Continuity Eq.: m = m 3 = m 3 Energy Eq.: m(u 3 -u ) = Q 3 W 3 ; roce: = C => W 3 = dv = m( 3 - ) State : 700 o C, 0 Ma, V = 00 L able B..4 = 0.04358 m 3 /kg => m = m = V / =.95 kg h = 3870.5 kj/kg, = 7.687 kj/kg K 3 State 3: 3 = = 0 Ma, x 3 = 0 able B.. h 3 =h f = 407.5 kj/kg, 3 = f = 3.3595 kj/kg K Q 3 = m(u 3 -u ) + m( 3 - ) = m(h 3 - h ) = -565.6 kj Heat tranfer to the heat engine: Q H = - Q 3 = 565.6 kj ake control olume a total water and heat engine. roce: Re., S net = 0 ; L = 30 o C nd Law: S net = m( 3 - ) - Q c / L ; Q c = o m( 3 - ) = -650.6 kj => Q L = -Q c = 650.6 kj W net = W HE = Q H - Q L = 300 kj H.E. amb - Q 3 Q L W HE
8. A heaily-inulated cylinder fitted with a frictionle piton contain ammonia 6 C, 90% quality, at which point the olume i 00 L. he external force on th piton i now increaed lowly, compreing the ammonia until it temperature reache 50 C. How much work i done on the ammonia during thi proce? Solution: C.V. ammonia in cylinder, inulated o aume adiabatic Q = 0. Cont.Eq.: m = m = m ; Energy Eq.: m(u u ) = Q W Entropy Eq.: m( ) = dq/ + S gen State : = 6 o C, x = 0.9, V = 00 L = 0. m 3 able B.. aturated apor, = g = 534 ka = f + x fg = 0.66 m 3 /kg, u = u f + x u fg = 07.44 + 0.9 5.3 =. kj/kg = f + x fg = 0.866 + 0.9 4.445 = 4.80 kj/kg-k, m = V / = 0. / 0.66 = 0.945 kg roce: Adiabatic Q = 0 & Reerible S gen = 0 => = State : = 50 o C, = = 4.80 kj/kg-k uperheated apor, interpolate in able B.. => = 0.0684 m 3 /kg, h = 479.5 kj/kg = 99 ka, u = h - = 479.5-99 0.0684 = 348. kj/kg Energy equation gie the work a W = m(u - u ) = 0.945 (. - 348.) = -9.4 kj
8.7 An inulated cylinder/piton contain R-34a at Ma, 50 C, with a olume of 00 L. he R-34a expand, moing the piton until the preure in the cylinder ha dropped to 00 ka. It i claimed that the R-34a doe 90 kj of work againt the piton during the proce. I that poible? C.V. R-34a in cylinder. Inulated o aume Q = 0. State : able B.5., = 0.085, u = 43.4-000 0.085 = 409.4, =.7494, m = V / = 0./0.085 = 4.577 kg Energy Eq.: m(u - u ) = Q - W = 0/ - 90 u = 367.89 kj/kg State :, u able B.5.: = -9.5 C ; =.7689 kj/kg K Entropy Eq.: m( - ) = dq/ + S,gen = S,gen = 0.0893 kj/k hi i poible ince S,gen > 0/ = C 8.9 A ma and atmophere loaded piton/cylinder contain kg of water at 5 Ma, 00 C. Heat i added from a reeroir at 700 C to the water until it reache 700 C. Find the work, heat tranfer, and total entropy production for the ytem and urrounding. C.V. Water. roce: = cont. o W = (V - V ) U - U = Q - W or Q = H - H = m(h - h ) Q = (3900. - 4.7) = 6954.76 kj W = Q - m(u - u ) = 874.6 kj m( - ) = dq/ + S gen = Q / re + S gen B..4: h = 4.7, u = 47.5 =.303, = 0.0004 B..3: h = 3900., u = 3457.6 = 7.5, = 0.08849 S gen = m( - ) - Q / re = (7.5 -.303) - 6954/973 = 5.7 kj/k
8.4 A large lab of concrete, 5 8 0.3 m, i ued a a thermal torage ma in a olar-heated houe. If the lab cool oernight from 3 C to 8 C in an 8 C houe, what i the net entropy change aociated with thi proce? C.V.: Control ma concrete. V = 5 8 0.3 = m 3 m = ρv = 300 = 7600 kg Q = mc = 7600 0.65(-5) = -89700 kj S SYS = mc ln = 7600 0.65 ln 9. = -305.4 kj/k 96. S SURR = - Q / 0 = +89700/9. = +308.0 kj/k S NE = -305.4 + 308.0 = +.6 kj/k 8.44 A hollow teel phere with a 0.5-m inide diameter and a -mm thick wall contain water at Ma, 50 C. he ytem (teel plu water) cool to the ambient temperature, 30 C. Calculate the net entropy change of the ytem and urrounding for thi proce. C.V.: Steel + water. hi i a control ma. m SEEL = (ρv) SEEL = 8050 (π/6)[(0.504) 3 - (0.5) 3 ] =.746 kg U SEEL = (mc) SEEL ( - ) =.746 0.48(30-50) = -346 kj V HO = (π/6)(0.5) 3, m = V/ = 6.545 0 - /0.44 = 0.587 kg = = 0.44 = 0.00004 + x 3.889 => x = 3.358 0-3 u = 5.78 + 3.358 0-3 90.8 = 33.5 = 0.4639 + 3.358 0-3 8.064 = 0.4638 U H O = mh O(u - u )H O = 0.587(33.5-679.6) = -494.6 Q = -346 + (-494.6) = -840.6 S O = S SEEL + S H O =.746 0.48 ln (303.5 / 53.5) + 0.587(0.4638-6.545) = -6.908 kj/k S SURR = - Q / 0 = +840.6/303. = +9.370 kj/k S NE = -6.908 + 9.370 = +.46 kj/k
8.47 Conider a Carnot-cycle heat pump haing kg of nitrogen ga in a cylinder/piton arrangement. hi heat pump operate between reeroir at 300 K and 400 K. At the beginning of the low-temperature heat addition, the preure i Ma. During thi proce the olume triple. Analyze each of the four procee in the cycle and determine a. he preure, olume, and temperature at each point b. he work and heat tranfer for each proce 4 3 = = 300 K, 3 = 4 = 400 K, N = Ma, V = 3 V a) V = V => = /3 = 0.3333 Ma V = mr 0.968 300 = = 0.08904 m 3 000 V = 0.67 m 3 k 3 = ( 3 / ) k- = 0.3333 400 300 3.5 = 0.93 Ma V 3 = V 3 = 0.67 0.3333 3 0.93 400 = 0.30 m3 300 k 4 = ( 3 / ) k- = 400 300 3.5 =.73707 Ma V 4 = V 4 = 0.08904 4.737 400 = 0.04337 m3 300 b) W = Q = mr ln ( / ) = 0.968 300 ln(/0.333) = 97.8 kj 3W 4 = 3 Q 4 = mr 3 ln( 3 / 4 ) = 0.968 400 ln(0.93/.737) = -30.43 kj W 3 = -mc V0 ( 3 - ) = - 0.7448(400-300) = -74.48 kj 4W = -mc V0 ( - 4 ) = - 0.7448(300-400) = +74.48 kj Q 3 = 0, 4Q = 0
8.5 A rigid torage tank of.5 m 3 contain kg argon at 30 C. Heat i then tranferred to the argon from a furnace operating at 300 C until the pecific entropy of the argon ha increaed by 0.343 kj/kg K. Find the total heat tranfer and the entropy generated in the proce. Solution: C.V. Argon. Control ma. R = 0.083, m = kg Energy Eq.: m (u - u ) = m C ( - ) = Q roce: V = contant => = State : = mr/v = 4.063 ka State : = + 0.343, - = C p ln ( / ) - R ln ( / ) = C ln ( / ) ln ( / ) = ( - )/ C = 0.343/0.3 =.0986 = R => ( / ) ( / ) = / = / =.7 = 88.3, =.7 = 3.57 = C Q = 0.3 (88.3-303.5) = 60.8 kj m( ) = Q / re + S gen tot S gen tot = 0.3-60.8 / (300 + 73) = 0.08 kj/k
8.66 A cylinder/piton contain air at ambient condition, 00 ka and 0 C with a olume of 0.3 m3. he air i compreed to 800 ka in a reerible polytropic proce with exponent, n =., after which it i expanded back to 00 ka in a reerible adiabatic proce. a. Show the two procee in and diagram. b. Determine the final temperature and the net work. c. What i the potential refrigeration capacity (in kilojoule) of the air at the final tate? a) 3 V 3 m = V /R 00 0.3 = = 0.3565 kg 0.87 93. n- b) = ( / ) n = 93. 800 00 0.67 = 44.9 K - R( - ) w = d = = = 0.87(44.9-93.) = -74.6 kj/kg -n -n -.0 k- 3 = ( 3 / ) k = 44.9 00 800 0.86 = 8.9 K w 3 = C V0 ( - 3 ) = 0.77(44.9-8.9) = +33.3 kj/kg w NE = 0.3565(-74.6 + 33.3) = -4.7 kj c) Refrigeration: warm to 0 at cont 3 Q = mc 0 ( - 3 ) = 0.3565.004 (93. - 8.9) = 3.0 kj