Question

Saturated liquid water at 60 bar is heated until the quality is 43%.

Calculate the amount of heat (kJ/kg) that needs to add into the system.

Answer #1

State 1:

Saturated liquid water at 60 bar. From steam table following properties can be found:

v_{1} = 0.001319
m^{3}/kg h_{1} = 1213.32
kJ/kg, T_{1} = 275.64 ^{o}C

Now this sytem is heated utill the quality is 43%

State 2: P_{2} = 60 bar, T_{2} = 275.64
^{o}C and x = 0.43

h_{2} =(1-x)h_{l}+xh_{v} = (0.57 x
1213.32)+(0.43 x 2784.33) = 1888.85 kJ/kg

Since the process is a constant pressure process, the heat transfer will be equalt to change in enthalpy, thus

Q = h_{2} - h_{1} = 1888.85 - 1213.32 = 675.53
kJ/kg

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