Question

1.) Saturated liquid water flowing at 3kg/s is initialy at 30 bar.

a.) this steam is throttled through a valve to 9 bar. What is the vapor fraction of the steam after the valve?

b.) afterwards, 6.0 MW is added to the steam. What is the final temperature of the steam?

What is the entropy change for this heating, in kW/K?

Answer 1

a) Throttling is an isenthalpic process.

From steam table:

At 30 bar,

Tsat = 233.858 C

Hl = 1008.37 kJ/kg ; Hv = 2803.26 kJ/kg

Sl = 2.0944 kJ/kg-K ; Sv = 6.6212 kJ/kg-K

As steam is saturated liquid, H1 = Hl = 1008.37 kJ/kg

After throttling, H2 = H1 = 1008.37 kJ/kg

Now, from steam table at 9 bar, Tsat = 175.358 C

Hl = 742.725 kJ/kg ; Hv = 2773.04 kJ/kg

H2 = xHv + (1-x)Hl

x = quality of steam or vapour fraction

.: 1008.37 = x*2773.04 + (1-x)*742.725

.: x = 0.1308

Similarly,

S2= xSv + (1-x)Sl

.: S2 = 2.6865 kJ/kg-K

b) 6 MW is added to the steam.

By energy balance:

Q = m(H3 - H2) ....H3 is the enthalpy after addition of heat

.: 6000 kJ/s = 3 kg/s*(H3 - 1008.37) kJ/kg

.: H3 = 3008.37 kJ/kg

At 9 bar, ethalpy for saturated vapour = 2773.04 kJ/kg < 3008.37 kJ/kg

This implies that steam is superheated.

Look for H = 3008.37 kJ/kg at 9 bar in the superheated steam table and find the corresponding temperature.

we get, T = 278.5 C

entropy S3 = 7.095 kJ/kg-K

Change in entropy = m(S3 - S2) = 3 kg/s *(7.095 - 2.6865) kJ/kg = 13.2255 kW/K