Exercícios de Termodinâmica Resolvidos Cap 3

Exercícios de Termodinâmica Resolvidos Cap 3

(Parte 5 de 6)

Sonntag, Borgnakke and Wylen

3.153E

A 35 ft3 rigid tank has air at 225 psia and ambient 600 R connected by a valve to a piston cylinder. The piston of area 1 ft2 requires 40 psia below it to float, Fig. P3.9. The valve is opened and the piston moves slowly 7 ft up and the valve is closed. During the process air temperature remains at 600 R. What is the final pressure in the tank? mA = PAVA RTA

= 225 × 35 × 144

53.34 × 600 = 35.433 lbm mB2 - mB1 = ∆VA vB

1 × 7 × 40 × 144

53.34 × 600 = 1.26 lbm mA2 = mA – (mB2 - mB1) = 35.433 – 1.26 = 34.173 lbm

PA2 = mA2RT VA

= 34.173 × 53.34 × 600

35 × 144 = 217 psia

Sonntag, Borgnakke and Wylen

3.154E

Give the phase and the missing properties of P, T, v and x. These may be a little more difficult if the appendix tables are used instead of the software. Solution:

aR-2 at T = 50 F, v = 0.6 ft3/lbm: Table F.9.1 v > vg

sup. vap. F.9.2 interpolate between sat. and sup. vap at 50 F.

P ≅ 98.73 + (0.6 - 0.5561)(80 -98.73)/(0.708 - 0.5561) = 93.3 lbf/in2

bH2O v = 2 ft3/lbm, x = 0.5: Table F.7.1
vf + vg = 4.3293 at 330 F,vf + vg = 3.80997 at 340 F.
linear interpolationT ≅ 336 F, P ≅ 113 lbf/in2
cH2O T = 150 F, v = 0.01632 ft3/lbm: Table F.7.1, v < vf

since vf is so small we find it approximately where vg = 4 ft3/lbm. compr. liquid P ≅ 500 lbf/in2

dNH3 T = 80 F, P = 13 lbf/in.2 Table F.8.1 P < Psat
sup. vap. interpolate between 10 and 15 psia:v = 26.97 ft3/lbm

v is not linear in P (more like 1/P) so computer table is more accurate.

eR-134a v = 0.08 ft3/lbm, x = 0.5: Table F.10.1
vf + vg = 0.1729 at 150 F,vf + vg = 0.1505 at 160 F.

since vf is so small we find it approximately where vg = 0.16 ft3/lbm. linear interpolation T ≅ 156 F, P ≅ 300 lbf/in2

Sonntag, Borgnakke and Wylen

3.155E

A pressure cooker (closed tank) contains water at 200 F with the liquid volume being 1/10 of the vapor volume. It is heated until the pressure reaches 300 lbf/in.2. Find the final temperature. Has the final state more or less vapor than the initial state?

Process:Constant volume and mass.
Vf = mf vf = Vg/10 = mgvg/10;Table F.7.1: vf = 0.01663, vg = 3.631

Solution:

x1 = mg mg + mf

= 10 mfvf / vg mf + 10 mfvf / vg = 10 vf 10 vf + vg

= 0.1663 0.1663 + 3.631 = 0.00492 v2 = v1 = 0.01663 + x1 × 3.615 = 0.1820 ft3/lbm P2, v2 ⇒ T2 = Tsat = 417.43 F

0.1820 = 0.01890 + x2 × 1.5286 x2 = 0.107 more vapor than state 1.

Sonntag, Borgnakke and Wylen

3.156E

Refrigerant-2 in a piston/cylinder arrangement is initially at 120 F, x = 1. It is then expanded in a process so that P = Cv−1 to a pressure of 30 lbf/in.2. Find the final temperature and specific volume.

Solution:

State 2: P2 = 30 lbf/in2 and on process line (equation).

v2 = v1P1

= 0.1924 × 274.6/30 = 1.761 ft3/lbm

Table F.9.2 between saturated at -1.71 F and 0 F:T2 ≅ -8.1 F

vP v

Sonntag, Borgnakke and Wylen

Compressiblity Factor

3.157E

A substance is at 70 F, 300 lbf/in.2 in a 10 ft3 tank. Estimate the mass from the compressibility chart if the substance is a) air, b) butane or c) propane. Solution:

Use Fig. D.1 for compressibility Z and table F.1 for critical properties m = PV ZRT =

530 ZR = 815.09 ZR

Air use nitrogenPc = 492 lbf/in.2; Tc = 227.2 R
Pr = 0.61;Tr = 2.3; Z = 0.98

m = PV

ZRT = 815.09 ZR = 815.09

0.98× 5.15 = 15.08 lbm

Butane Pc = 551 lbf/in.2; Tc = 765.4 R
Pr = 0.544;Tr = 0.692; Z = 0.09

m = PV

ZRT = 815.09 ZR = 815.09

0.09× 26.58 = 340.7 lbm

PropanePc = 616 lbf/in.2; Tc = 665.6 R
Pr = 0.487; Tr = 0.796;Z = 0.08

m = PV

ZRT = 815.09 ZR = 815.09

0.08× 35.04 = 290.8 lbm ln Pr

T = 2.0r a bc T = 0.7rT = 0.7r

Sonntag, Borgnakke and Wylen

(Parte 5 de 6)

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