Question

A rigid tank has a volume of **0.01 m3**. It
initially contains saturated water at a temperature of **200
oC** and a quality of **0.4**. The top of the
tank contains a pressure regulating valve which maintains the vapor
at constant pressure. This system undergoes a process where it is
heated until all the liquid vaporizes. How much heat in (kJ) is
required? You may assume there is no pressure drop in the exit
line.

Answer #1

V = 0.01 m3 = 10 L

sat water at 200°C, x = 0.4

Q = ? until x = 1

find H

this is deffinetively an isobaric process

so

Q = m*Cp*dT

since this is at equilibrium

Q = Hf-Hi

Hf = saturated vapor , 1.0 quality

Hi = sat. liquid, 0.4 quality

a)

step 1, find Pressure saturation

at T = 200°C, P = 1.5538 MPA (from thermo tables)

H of liquid = 852.45 kJ/kg

H of gas = 2793.2 kJ/kg

Hfg = 1940.7 kJ/kg

We need the mixed Enthalpy:

H = x*Hfg + Hf

H = 0.4*1940.7 + 852.45 + = 1628.73kJ/kg

then

we need mass of system

vf = 0.001157 ---> 0.01 m3

mass = vf/V = 0.01 m3 /0.001157 m3 / kg *= 8.6430 kg

Q = m(Hf-Hi)

substitute data

Q = 8.6430* (2793.2 -1628.73) = 10064.51421 kJ

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