Modern Engineering Thermodynamics

13.7 Rankine Cycle with Regeneration 471
and
h 5 = h 1 – ðh 1 – h 5s Þðη s Þ stage1
This equation allows us to calculate the mass fraction of vapor that must be removed from the prime mover and
added to the open loop regenerator to achieve a desired saturated liquid state at station 6.
Suppose we have a total of N regenerators (either open, closed, or any combination) in a system. Let the
fraction of vapor removed for the first regenerator be _m regen 1 = _m total = y 1 , the amount removed for the
second be y 2 , and so forth, with y N being fed to the Nth regenerator. Now, the general equation for thermal
efficiency is
η T =
_ W net
_Q boiler
=
_ Q boiler − j _Q condenser j
_Q boiler
= 1 − j _Q condenser j
_Q boiler
so that the thermal efficiency of a Rankine cycle with N regenerators can be written as
ðη T Þ Rankine cycle
with N regenerators
= 1 − j _Q condenser j
_Q boiler
= 1 − _m condenserðh in − h out Þ condenser
_m boiler ðh out − h in Þ boiler
= 1 − _m totalð1 − y 1 − y 2 − … − y N Þðh in − h out Þ condenser
_m total ðh out − h in Þ boiler
= 1 − ðh
in − h out Þ condenser
ð1 − y 1 − y 2 − … − y N Þ
ðh out − h in Þ boiler
Therefore, the thermal efficiency of a Rankine cycle with N open or closed regenerators can always be
determined from
ðη T Þ Rankine cycle
= 1 − ðh
in − h out Þ condenser
ð1 − y 1 − y 2 − … − y N Þ
(13.12a)
ðh out − h in Þ
with N regenerators
boiler
ðeither open or closedÞ
This equation provides a simple solution to a complex problem when multiple open or closed regenerators are
used. For the open and closed regenerators shown in Figure 13.20, Eq. (13.12a) reduces to
where
ðη T Þ Rankine cycle
with 1 regenerator
ðeither open or closedÞ
= 1 − h
2 − h 3
ð1 − yÞ
h 7 − h 6
(13.12b)
h 2 = h 5 – ðh 5 – h 2s Þðη s Þ stage2
h 5 = h 1 – ðh 1 – h 5s Þðη s Þ stage1
h 7 = h 6 + v 6 ðp 7 – p 6 Þ/ðη s Þ pump2
The only property in Eq. (13.12a) that is directly affected by the presence of the regenerators is the boiler inlet
enthalpy, (h in ) boiler . Notice that maximizing this value also maximizes the cycle’s thermal efficiency. Therefore,
an optimum set of regenerator flows exists that maximizes the cycle’s thermal efficiency. This is illustrated in the
next example problem.
EXAMPLE 13.6
A two-stage steam turbine receives dry saturated steam at 200. psia. It has an interstage pressure of 80.0 psia and a condenser
pressure of 1.00 psia. Determine
a. The isentropic Rankine cycle thermal efficiency of the system without regeneration present.
b. The isentropic Rankine cycle thermal efficiency of the system and the mass fraction of regeneration steam required with
an open loop boiler feedwater regenerator at a pressure of 80.0 psia.
(Continued )