A piston–cylinder device initially contains 0.6 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 300 kPa is required to move it. Heat is now slowly transferred to the steam until the volume becomes 1.5 m3. Use the data from the steam tables.

a) Determine the final temperature.
b) Determine the work done during this process.
c) Determine the total heat transfer .

Respuesta :

Answer:

T2 = 662°c Q(in) = 911kj W(out) = 180kj

Explanation:

1. Firstly, we find the mass from the initial volume and specific volume from steam table A-5 for given pressure and state.

M = V(1)/a(1) = 0.6/0.7183(kg)

Where a(1) = specific volume

M = 0.835kg

The work is determined from the final pressure and the volume change that occurs when it is reached.

W = P2(V2 - V1)

W = P2V2

= 300×0.6kg = 180kg

Therefore, to get the final temperature and final pressure, the final temperature is determined fro steam table A-6 using interpolation.

T2 = T1* + T2*-T1*/a2*-a1* (a2-a1*)

T2 = 600 + 700-600/1.4958-1.34139 (1.43746-1.34139)

T2 = 662°c

2. The final internal energy can be determined in the same way just like the initial ones were done. From A-5 on steam table given the pressure

The heat Q(in) = ∆u + W(out)

= m(U2-U1) + W(out)

= 0.835(3412.3-2536.8)kj

+ 180

Q(in) = 911kj