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Unformatted text preview: 3 .66 3 .69 A 1 3 tank is ﬁlled with a gas at room temperature 20°C and pressure 100 kPa.
How much mass is there if the gas is a) air, b) neon or c) propane '3 Solution: Use Table A2 to compare T and P to the critical T and P with
T=20°C=293.15K; P=100kPa€cZ PC forall
Air: T b“; TQM; Tom = 154.6 K so ideal gas; R: 0.287 kJJ'kg K
Neon: T >> Tc = 44.4 K so ideal gas; R = 0.41195 11ng K
Propane: T =1 Ta = 3'10 K, but P <41 PC = 4.25 MPa
so gas R = 0.18855 kJa’kg K All states are ideal gas states so the ideal gas law applies
PV = mRT PV 100><1 ‘9 mZEZozsa x 293.15 2 1'1” kg 100 x 1
0.41195 x 293.15 100 x]
0.18855 X29115 b) m = = 0.328 kg c) In = = 1.809 kg Is it reasonable to assume that at the given states the substance behaves as an
ideal gas? Solution:
a) Oxygen, 02 at 30°C, 3 MPa Ideal Gas (T >> To = 155 K from 131.2)
b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T )} To = 190 K from A.2) c) Water, H20 at 30°C, 3 MPa N0 compressed liquid P > PS,“ (B11) (1) R—134a at 30°C, 3 MPa N0 compressed liquid P > PS,“ (B51)
e) R—134a at 30°C, 100 kPa Ideal Gas P is low <1. Pm (35.1) 3.?2 3.88 A spherical helium balloon of 10 m in diameter is at ambient T and P, 15°C and
100 kPa. How much helium does it contain? It can lift a total mass that equals the mass of displaced atmospheric air. How much mass of the balloon fabric and cage
can then be lifted? We need to ﬁnd the masses and the balloon volume _E 3_E 3_ 3
V—ﬁD —610 —523.6m _ V_ PV_ 100 kPa x 523.6 m3
1? RT 2.0771 kakgK x 288 K mHE = pV = 87.5 kg m _Pv 10I01cPax523.ﬁm3 air RT = 0.28? kakgK x 288 K = 633 kg mliﬁ = mar—mHe = 633815 = 545.5 kg What is the percent error in speciﬁc volume if the ideal gas model is used to
represent the behavior of superheated ammonia at 40°C, 500 kPa'? What if the
generalized compressibility chart, Fig. D. 1, is used instead? Solution:
NH3 T =40°C = 313.15 K, Tc =405.5 K, PC = 11.35 MPa ﬁ'om TableAJ Table 13.2.2: v = 0.2923 m3mg RT _ 0.43819 X 313 Ideal gas: v = ? — 500 = 0.3056 msfkg 2:) 4.5% error _ _ 313.15 0.5
FigureDJ. Tr: 4055 =0.7'72, PI=—1135=0.044 3 2:11.97 ZRT 3
v = T = 0.2964 In fkg :b 1.4% error 3.92
Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z ﬁ'om Fig. D. 1 Ethylene Table A2: Tc = 282.4 K, PC = 5.04 MPa
Table AS: R = 0.2964 kag K The reduced temperature and pressure are: T 270 P 2.5
Tr=T =282_4=0.956, Pr=P =5_04=0.496 C C
Enter the chart with these coordinates and read: Z = 0.?6 anRT 2 k x 0.?6 x 0.2964 kak K x 270 K _ _ _ 3
V — P — 2500 kPa — 0.0487 n1 3.97
A new reﬁ‘igerant R125 is stored as a liquid at 20 “C with a small amount of vapor.
For a total of 1.5 kg R125 ﬁnd the pressure and the volume. Solution:
As there is no section B table use compressibility chart.
Table A2: R125 TC = 339.2 K PC = 3.62 MP3. Tr: TI'TC = 253.15 J“ 339.2 =0.746 We can read ﬁ‘om Figure D. 1 or a little more accurately interpolate from table
D.4 entries: Pr sat : 016; Zg : 086; 2f: 0.029 P=P P =0.16x3620kPa=579kPa rsat C qu = meﬁq RTFP = 0.029 xl.5 kg x 0.06927:Ir kakgK x 253.15 KI 579 kPa = 0.0013 1113 Z sat vapor ...
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 Summer '08
 Sherif

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