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julia.julia

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answer:

Pressure of the closed feed water heaterP4=1MPa

Take the extracted pressure from the second turbine P11=0.3MPa

Operating pressure of open feed water heater=0.3MPa

State7: from the superheated water

P7=10MPa and t7=600degree C

H7=3625.8kJ/kg

S7=6.9045kJ/kg-k

Since state 2 is fixed so we know that

S7=s8=6.9045kJ/kg.K and P8=1.8Mpa

H8=3047.77kJ/kg

The state exit of the first turbine is the same as a exit of the first turbine.

State9:from the saturtrated water pressure table

P9=1MPa and s9=s7=6.9045kJ/kg-K

S9=sf+x3(sfx)

6.9045=2.1381+(x9)(4.4470)

X9=0.9895

Then

H9=hf=x9(hfg)

=762.51+(0.98)(2014.6)

=2736.818kJ/kg

State 10:

Steam is superheated state so

From the superheated water

P10 =1MPa and T10=550 degree C

H10=3588.85Kj/kg

S10=7.89765kJ/kg-K

State11:

Steam is superheated state so

From the superheated water

P11=0.3MPa and s11=s10=7.89765kJ/kg-K

H11=3093.325kJ/kg

We know that

From the saturated water table

P1=P12=7.5KPa and S12=S10=7.89765kJ/kg-K

H11=3093.325kJ/kg

We know that

From the saturated water table

P1=P12=7.5KPa and s12=s10=7.89765kJ/kg-K

The quality at state 6 is found to be X12=0.95

H12=hf+x12(hfx)

=168.75+(0.95)(2405.3)

=2453.785kJ/kg

And

State1:

H1@7.5KPa=hf=168.75kJ/kg

V1=0.001008M cube/kg

State2:

H2=h1+v1(P2-P1)

=168.75=(0.001008)(0.3-0.0075)

=168.75kJ/kg

State 3:

Since liquid leaving the open feed water heater at state 9is saturated liquid at 0.3MPa.

Thenh3=561047kJ/kg

The specific enthalpy at the exit of the second pump is h4=h3=v3(P4-p3)

=561.47+(0.001073)(10-0.3)

561.48kj/kg

The specific enthalpy of the feed water heater existing of the closed fed water heater at 10MPa and 550 degree C

Then

H5=hf=vf(P5-P sat @550 degree C)

=884.233kJ/kg

By applying the mass balance equation closed feed water heater

Y1=h3-h4/h5-h4

=884.233-561.48/3047.77-561.48

=0.1298

Similarly for open feed water heater

Yii=(1-y)(h2)+y1(h3-h2)/(h2-h11)

=(0.8702)(168.75)+(0.1298)(561.48-168.25)/168.75-3093.325

=0.0675

Work developed by the turbine per unit mass entering

Wt1=(h7-hg)=(1-y1)(hg-h9)

=(3625.8-3047.77)+(0.8702)(3093.325-2453.785)

=944.564kJ/kg

First pump work

Wp1=(1-y1-yii)(h1-h3)

=(0.8027)(168.75-561.47)

=315kJ/kg

Second pump work is

Wp2=h4-h3

=561.48-561.47

=0.01kJ/kg

Heat input is

Qtn=(h7-h6)+(1-y1)(h10-h9)

=(3625.8-884.233)+(0.8702)(3588.85-2736.818)

=3483.00kJ/kg

So here we needed 400 MW power so that vary the mass flow rate according to the need of the demand,

So here we required 400 MW so mass flow rate is

M=W/Wn+Wt2-Wp1-Wp2

=400MW(3600s/h)/822.521+944.527-315-0.01

=991.700k.kg/h

Required power output is

P=W*m

=1452.038kJ/kg*991.2*10 cubic kg/h/3600

=399.99MW

=400MW.