A power plant is connected to a supply of 150 psia steam at

A power plant is connected to a supply of 150 psia steam at 500 degree F. The load is constant and requires throttling to 100 psia between points 1 and 2 in order to hold the load. The turbine exhausts to 2 psia. In order to conserve energy, it is desired to bleed off a fraction of the steam at point 3 (which is at 30 psia) for feed water heating so that the throttle can be fully open without changing the load. The isentropic efficiency in both turbine stages is 86%. Determine (a) What is the maximum percentage of steam that can be bled without reducing the work output? (b) What is the maximum temperature to which returned water at 60 psia and 60 degree F can be heated?

Solution

The initial conditions for the entry of steam is given as T₁= 500 F = 260^oC

P₁= 150 psi = 1.034 MPa

Enthalpy of steam in that conditions is found out by consulting the steam tables for superheated steam and is given by

h₁= 2959.18 kJ/kg

It is given that the steam is throttled to a final pressure P₂= 150 psi = 0.689 MPa

Since throttling is isentropic process, the final entropy remains the same i.e. h₂= 2959.18 kJ/kg-K

The temperature corresponding to the entropy is T₂ = 250^oC

The entropy of steam for the same condition s₂= 7.04 kJ/kg-K

It is given that there is an expansion in the turbine. Expansion is supposed to be isentropic in nature. FInal entropy is s₃=7.04 kJ/kg-K

Temperature of steam for pressure of 30psi and entropy of 7.04 kJ/kg-K is T₃= 150^oC

Hence the maximum temperature to which the feed water can be heated to 150^oC = 302 F

On consulting the steam table, we find that the enthalpy for steam h₃= 2760 kJ/kg

Energy generated in turbine = change in enthalpy = 2959 - 2760 = 199 kJ/kg

But due to efficiency of 86%, work output = 171.4 kJ/kg

It is further expanded where entropy remains the same s₄ = 7.04 kJ/kg-K

Corresponding to that, dryness fraction is found to be x = 0.95

Enthalpy for that state h₄= 2569 kJ/kg

Work output in ideal conditions = 2760-2569 = 191 kJ/kg

But due to less efficiency, final work output = 164.26 kJ/kg

Hence, final work output = 171.4 + 164.26 = 335.26 kJ/kg

Part a asks us that what if the steam were not throttled but put into turbine, how much less steam would be used.

Let us find out the work output in that case.

h₁=2959 kJ/kg

s₁=6.899 kJ/kg-K

s₄=6.899kJ/kg-K

Then h₄=2524.144 kJ/kg

Hence ideal work output = 415kJ/kg

Real work output = 0.86 x 415 = 356.9 kJ/kg

Hence the fraction of steam that is used in case 2 is 94%

Hence 6% of the total steam can be extracted safely at 30 psi