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2-lbm of water at 500 psia initially fill the 1.5-ft3 left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300°F. Determine the final pressure of water, in psia, and the total internal energy, in Btu, at the final state.

Respuesta :

Answer:

a) pressure = 66.985 psia

b) internal energy 920.56  Btu

Explanation:

Given data:

left chamber characterisitics

2lbm water, 500 psia, 1.5 ft^3

right chamber

volume = 1.5 ft^3

when partitioned ruptured, the final volume will be

[tex]V_2 = 1.5 + 1.5 = 3.0 ft^3[/tex]

specific volume will be  [tex]= \frac{V_2}{m} = \frac{3}{2} = 1.5 ft^3/lbm[/tex]

at [tex]T_2 =  300 F[/tex], From saturated water tables

[tex]v_f = 0.1745 ft^3/lbm[/tex]

[tex]v_{fg} = 6.4536 ft^3/lbm[/tex]

[tex]v_G = 6.471 FT^3/lbm[/tex]

[tex]v_2 = v_f + x_2 v_{fg}[/tex]

[tex]1.5 = 0.01745 + x_2(6.4536}[/tex]

[tex]x_2 = 0.229[/tex]

the final pressure at temperature 300 F

[tex]P_2  = P_{sat} = 66.985 psia[/tex]

internal energy at fnal stage

[tex]U_2 = m (u_f \  @300F + X_2 U_{FG}\  @ 300 F)[/tex]

[tex]U_2 = 2( 269.51 + 0.229(830.45))[/tex]

[tex]U_2 = 920.56 Btu[/tex]

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