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Question:
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 600lbf/in.2, 800∘F.
At the valve exit, corresponding to the turbine inlet, the pressure is 300lbf/in.2 At the turbine exit, the pressure is 5lbf/in.2
The work 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 super heated vapor, determine the temperature, in ∘F. If the state is a two-phase liquid–vapor mixture, determine the quality.
Answer / Explanation:
Given the following parameters,
Fluid Steam
At the valve inlet 1:
Temperature T1 = 800°F
Pressure P₁ = 600 lbf/in²
At the valve exit and turbine inlet 2:
Pressure P₂ = 300 lbf/in²
At the turbine exit 3:
Pressure P₃ = 5 lbf/in²
The work developed by the turbine w = 350 Btu per lb of steam flowing
While we are required to :
(1) Fix the state at the turbine exit
(2) If the state is super heated vapor, determine the temperature T₃, in ∘F.
(3) If the state is a two-phase liquid–vapor mixture, determine the quality X₃.
And assuming that:
Constant average value
Steady Flow
Neglecting kinetic and potential energy effect
Moving Forward:
Specific enthalpy of water from table A - 3E at T₁ = 800⁰F and
P₁ = 600 lbf/in₂ where
h₁ = 1407.6 Btu per lb
Hence, as the fluid is throttled, then
Δh = 0
Therefore,
h₂ = h₁ = 1407.6 Btu per lb
However, for turbine,
Energy equation can be defined as:
Q - W = Mout ( Hout + V²out/2 + gZout ) - Min ( Hin + V²in/2 + gZin )
Q - W = M₃ ( H₃ + V₃²/2 + g Z₃ ) - M₂ ( H₂ + V₂²/2 + g Z₂ )
Hence, as for neglecting kinetic and potential effect, we have:
Q -W = M₃ (H₃) - M₂ (H₂)
Also, taking note that as the system has no work done, we have:
-W = M₃ (H₃) - M₂ (H₂)
Recalling the basics from specific properties and mass balance, we have:
M₃ = M₂
Therefore:
-W = (H₃) - (H₂)
Therefore, the specific enthalpy at the turbine exit can be calculated as follows:
H₃ = - w + h₂
Inserting the values, we have:
- 350 + 1407.6 = 1057.6 Btu/lb
Recalling that specific enthalpy of water from table A - 3E at h₃ = 1057.6 Btu/lb and P₁ = 5 lbf/in²
We therefore at this point know that the state is in two phase zone
Hf = 130.17 Btu/lb
Hfg = 1000.83 Btu/lb
H₃ = Hf + XHfg
The quality of steam exiting the turbine can now theerfore be calculated thus:
X₃ = H₃ - Hf / Hfg
= 1057.6 - 130.17 / 1000 . 83
0.927
Therefore the quality of steam exiting the turbine is -X = 0.927
