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hw 5 thermo.pdf

# hw 5 thermo.pdf - 4.28 A diffuser shown in Fig P428 has air...

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Unformatted text preview: 4.28 A diffuser, shown in Fig. P428, has air entering at 100 kPa, 300 K, with a velocity of 200 ads. The inlet cross-sectional area of the diffuser is 100 1111112. At the exit, the area is 860 1111112= and the exit velocity is 20 m-"s. Determine the exit pressure and temperature of the air. Solution: Continuity Eq.4.3: 151i = Aiviwi = the = Aeves've, Energyr Eq.[per unit mass ﬂow) 4.13: bi + éviz = he + éVez he — h1- =% x2002s'1000 —% x202..-'1000 = 19.8 III-"kg .- 19.8 kl-"kg Te = Ti + [he — base]: = 300 + 1.004 kJs’kg—K = 319.72 It Now use the continuityr equation and the ideal gas law Aeve , Aeve v“ 2 Vi [AM] : (RTi'IIPi) [AM : RTE?“ P _P E AiVi _10G[319.72] 100x200 _123 921:1) e‘ i T- AeVe ‘ 300 360x20 _ ' " Inlet Exit Hi V Low v LO“? P, A H1 P, A 4.64 A condenser (cooler) receives 0.05 kg-"s R-410a at 2000 kPa. 80°C and cools it to 10°C. Assume the exit properties are as for saturated liquid with the same T. ‘What cooling capacity (kW) must the condenser have? Solution: CV. R-4lﬂa condenser. Steady state single ﬂow= heat transfer out and no work. Energyr Eq. 4.12: 1i: h] = til hz + flout Inlet state: Table 3.4.2 111 = 343.22 kl-‘kg. Exit state: Table 3.4.] 1:12 = 73.2] kJH'kg (compressed liquid) To be more accurate we could have added [P-Psat}v to the hz. Process: Neglect kinetic and potential energj,r changes. Cooling capacityr is taken as the heat transfer out i.e. positive out so (gout = 151011- 112) = 0.05 kg.-"s (343.22 — 73.21)kJ.-1cg = 13.5 kw 4.85 A compressor receives 0.05 kgz‘s R—410a at 200 kPa, -20°C and 0.1 kgfs 11-4103 at 400 kPa, 00C. The exit ﬂow is at 1000I kPa, 60°C as shown in Fig. P435. Assume it is adiabatic, neglect kinetic clergies and ﬁnd the required power input. CV. whole compressor steady, 2 inlets= 1 exit, no heat transfer Q = 0 Continuity Eq.4.9: 1111 + 1112 = ﬁ13 = 0.05 + 0.1 = 0.15 ngS Eﬂﬁl'gy liq-4.10: 121111111 '1' ﬁlzhz — WC 2 1113113 Table B32: 111: 278.72 ka'kg, 3 2 h2 = 290.42 kakg - 'WC Table B32: 113 = 335.75 kakg t. ‘77,: = 0.05 x 278.72 + 0.1 x 290.42 — 0.15 x 335.75 = —7.385 kW Power input is 7.4 kW 4.130 A 1-m3 tank contains ammonia at 150 kPa, 25°C. The tank is attached to a line ﬂowing ammonia at 1200 kPa, 60°C. The valve is opened, and mass ﬂows in until the tank is half full of liquid, by volume at 25°C. Calculate the heat transferred ﬁ'om the tank during this process. Solution: CV. Tank. Transient process as ﬂow comes in. State 1 Table 3.2.2 intelpolate behween 2.0 ”C and 30°C: v1 = 0.9552 m31kg; u1 = 1380.6 kykg m1: var] = 110.9552 = 1.041r kg State 2: 0.5 m3 liquid and 0.5 1113 vapor ﬁ'om Table B.2.1 at 25°C vf= 0.001658 m3fkg; vg = 0.12313 mgi’kg mm22 = 0510001658 = 301.57 kg, 111W.P2 = 051012813 = 3.902 kg 1112 = 305.47 kg, 12 = mVAPAzJ'lIIIE = 0.01277, From continuity equation m1- : m2 - m1= 304.42 kg Table 13.2.1: “2 = 296.6 + 0.0125"?I x 1038.4 = 309.9I kJﬂcg State inlet: Table B.2.2 hi = 1553.3 ldikg Energy Eq.4.16: ch + mihi = 1112112 - 1111111 QCV = 305.47 >5 309.9 - 1.047 X 1380.6 - 304.42 K 1553.3 = —379 636 kJ 4.139 A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat is transferred to the water thereby raising its pressure and temperature. At a pressure of 2 MPa a safety valve opens and saturated vapor at 2 MPa ﬂows out. The process continues, maintaining 2 MPa inside until the qualityr in the tank is 90%, then stops. Determine the total mass of water that ﬂowed out and the total heat transfer. Solution: CV. Tan]:= no work but heat transfer in and ﬂow sat vap out. Denoting State 1: initial state, State 2: valve opens= State 3: ﬁnal state. Continuity Eq.: m3 — n11 = — 111e Energy EQ-i 1113113 — 1111111 = — ”Jab: + 1Q3 State 1 Table 3.1.2: v1 = Vf+ lefg = 0.001043 + 0.01x1.69296 = 0.01297 m3tkg u1 = uf+ xlufg = 417.33 + 0.01x2088.72 = 438.22 kJrkg m =‘Wv =0.2n13f 0.01792 1:13:11 =11.13k 1 1 a a State 3 (2MPa): V3 = Vf'l' X3Vfg = 0.001177 + G.9)<D.09845 = 0.8973 msi’kg u3 = Uf+ X311fg = 906.42 + O.9)<1693.84 = 2430.88 kakg 1113 = v43 = 0.2 111311008973 n13t'kg]= 2.23 kg Exit state (2MPa): h,3 = hg = 2799.51 ka'kg Hence: me = 1111 —Jn3 = 11.13 kg— 2.23 kg = 8.90 kg Applying the lst law between state 1 and state 3 1Q3 = I113113 — n11111+ mehe = 2.23 x 2430.88 — 11.13 x 438.22 + 8.90 x 2799.51 = 25 459 k] = 25.45 MJ ...
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