<p>At steady state, a valve and steam turbine operate in series. The steam flowi

Question

<p>At steady state, a valve and steam turbine operate in series. The steam flowing through the valve undergoes a throttling process. At the valve inlet, the conditions are 600 lbf/in^2, 800 F. At the valve exit, corresponding to the turbine inlet, the pressure is 300 lbf/in^2. At the turbine exit, the pressure is 5 lbf/in^2. The power developed by the turbine is 350 Btu per lb of steam flowing. Stray heat transfer and kinetic and potential energy effects can be ignored. Fix the state at the turbine exit: If the state is superheated vapor, determine the temperature, in F. If the state is a two-phase liquid-vapor mixture, determine the quality.</p>

Explanation / Answer

for a throttling process enthalpy remains constant
h1 = h2 where state 1 corresponds to throttling valve inlet and state 2 corresponds to throttling valve exit

for given pressure of 600 psi T > Tsat so steam is in superheated state
and h1 corresponds to 600 psi and 800 F = 1408 BTU/lbm (from steam table)
hence h 2 = 1408 Btu/lbm
and P = 300 Psi
corrseponding to this T = 775 F
and since T > Tsat hence this is also in superheated state (from steam table)

now T2 = temp at inlet of turbine
turbine output = 350 Btu/lbm (given)
Turbine,out = (h2 - h3) (Btu/lbm)
350 = 1408- h3
h3 = 1058 Btu/lbm and P3 = 5 psi (given)
from steam table we see that corresponding to above values ,steam is not in superheated state.

so it will be in saturated region

h3 = hf + x3.hfg

1058 = 130.2 + x3 . 1000.8 (hf = 130.2 Btu/lbm and hg = 1131 Btu/lbm

x3 = 0.927 and hfg = hg- hf = 1000.8 Btu/lbm)

quality = 0.927

Since the exit state is a saturated mixture at 5 psi, the exit temperaturemust be the saturation temperature at this pressure, which is 162.2 F

Texit = 162.2 F

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