A piston-cylinder device initially contains 50L of liquid water at 10 oC and 300

Question

A piston-cylinder device initially contains 50L of liquid water at 10 oC and 300 kPa.

Heat is added to the water at constant pressure until the temperature reaches 250 oC.

Determine the following:

a) the mass of the water

b) the volume after heating in m3

c) the enthalpy change after heating in kJ

Now the water is compressed in an isothermal process until half the mass is in the liquid form.

d) What is the final volume?

e) What is the final pressure?

f) What is the enthalpy change in the water for the 2-step process?

Explanation / Answer

a) for the mass you're going to need the density at the intial.

1000L = 1m^3

volume = 50L = 50Lm^3/1000L = 0.05m^3

using thermodynamics tables it can be seen that the water is in the compressed region therefore density= 1000kg/m^3

mass = density * Volume = 1000 * 0.05 = 50kg

b)pressure is constant and at 250C with 300Kpa the water is in the superheated region. this is a closed system so mass is also constant

using the tables it is found that
specific volume: 0.79645 m^3/kg
Internal energy: 2728 kJ/kg
Enthalpy: 2967.9 kJ/kg
Entropy: 7.5180 kJ/kg*K
Volume= Specific volume * mass = 39.8225 m^3 = 39822.5 L

c)

since the water is initially in compressed form h=hf @10 degrees = 42.02 kJ/kg

h after heateing = 2967.9kJ/kg

h = m(2967.9-42.02)= 50(2967.9-42.02) = 146,294kJ

d)Temperature remains the same
quality of water x = 0.5 because half the the mass is liquid.
since the water is now a mixture it is also at saturation temperature 250 oC.
specifiv volume v = vf +x*vfg ; where vf is specific volume at saturated liq and vfg is the difference between vg and vf, where vg is the specific volume at saturated vapor
enthapy: h = hf + x*hfg

v= 0.025668 m^3/kg
h=1884.46 kJ/kg

Volume: v*mass = 0.025668*50 = 1.2834 m^3 = 1283.4L

e)since water is a mixture at Tsat = 250 oC, the water is also at Psat = 3976.2kPa

f) h = (hfinal - hinitial)mass = (1884.46-42.02)50 = 92122 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.