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Steam at 6 MPa, 400C is flowing in a pipe. Connected to this pipe through a valve is a tank of volume 0.4 m 3 . This tank initially contains saturated water vapor at 0.1MPa. The valve is opened and the tank fills with steam until the pressure is 6MPa, and then the valve is closed. The process takes place adiabatically. Determine the temperature in the tank right as the valve is closed.

Respuesta :

batolisis

Answer:

2400°C

Explanation:

Volume of tank = 0.4 m^3

steam pressure = 6 Mpa

Steam temperature = 40°C

Initial pressure of tank = 0.1 MPa

Final pressure of Tank = 6 Mpa

Calculate the temperature in the tank when the Pressure in Tank = 6Mpa

since the volume of the Tank is constant = 0.4 m^3

we will apply Gay-Lussac's Law

= [tex]\frac{T1}{P1} = \frac{T2}{P2}[/tex]  ------ ( 1 )

T1 =  40°c

P1 = 0.1 MPa

P2 = 6 Mpa

T2 = ?

From equation 1 above

T2 = ( T1 * P2 ) / P1

     = ( 40 * 6 ) / 0.1

     = 2400°C

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