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, cor-responding to the turbine inlet, the pressure is 300 lbf/in.2 At the turbine exit, the pressure is 5 lbf/in.2 and the temperature is 200°F. Stray heat transfer and kinetic and potential energy effects can be ignored. Deter-mine the work developed by the turbine, in Btu per lb of steam flowing

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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

answer:

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explanation: