0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained at 50 psia. D

Question

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained at 50 psia. During the phase conversion,thesystem volume decreases by 1.5 ft3; 250 Btu of heat areremoved; and the temperature remains fixed at 15F. Estimatethe boiling point temperature of this substance when itspressure is 60 psia.

in progress 0
Diễm Thu 4 years 2021-09-04T23:13:57+00:00 1 Answers 2 views 0

Answers ( )

  1. danthu
    0
    2021-09-04T23:15:52+00:00
    Reply

    Answer:

    The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

    Explanation:

    Given the data in the question;

    Using the Clapeyron equation

    (\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

    (\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

    where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

    T is the temperature ( 15 + 460 )R

    m is the mass of water ( 0.5 Ibm )

    V_{fg is specific volume ( 1.5 ft³ )

    we substitute

    (\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

    (\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

    Now,

    (\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

    where P₁ is the initial pressure ( 50 psia )

    P₂ is the final pressure ( 60 psia )

    T₁ is the initial temperature ( 15 + 460 )R

    T₂ is the final temperature = ?

    we substitute;

    T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

    T_2 = 475 + 5.2751\\

    T_2 = 480.275 R

    Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )