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Water is the working fluid in an ideal Rankine cycle.Saturated vapor enters the turbine at 16...

Water is the working fluid in an ideal Rankine cycle.Saturated vapor enters the turbine at 16 MPa, and thecondenser pressure is 8 kPa. The mass flow rate of steam enteringthe turbine is 120 kg/s. Determine
(a) the net power developed, in kW.
(b) the rate of heat transfer to the steam passing through theboiler, in kW.
(c) the thermal efficiency.
(d) the mass flow rate of condenser cooling water, in kg/s, ifthe cooling water undergoes a temperature increse of 18 degrees Cwith negligible pressure change in passing through thecondenser.
Please fill the table :
States p (kPa) volume T(K) h s q(kW) w(kW)
state 1-2
state 2-3
state 3-4
state 4-1
0 0
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Answer #1
Concepts and reason

Mass flow rate:

It is the mass of the substance which is passing per unit time. It is denoted as (т)
and its SI unit is kg/s
.

Heat transfer:

It is a branch of thermal engineering which deals with the generation, conversion, and transfer of thermal energy in the form of heat across the boundary of a system. The movement of thermal energy is due to the difference in the temperature between system and surroundings.

Ideal Rankine Cycle:

It is a thermodynamic cycle in which heat addition and rejection takes place in constant pressure process and isentropic expansion and compression takes place. Ideal cycle does not include the irreversibility in the processes. Cycle uses working fluid in both liquid and vapor phases.

Thermal efficiency:

Thermal efficiency of a Rankine cycle is the ratio of net work output from the system to the heat supplied to the boiler.

Fundamentals

The schematic diagram of an Ideal Rankine cycle is shown as in Figure.

Boiler
Turbine
Wout
Qin
Ритp
Win
Condenser
Qout
Figure

Show the Т-s
diagram of the Ideal Rankine cycle as in Figure(2)
.

1
Qin
Wout
4
Win
2
3
Qout
S
Figure 2

Here, work output from the turbine is out
, work input to the pump is W
in
, heat output from the condenser is out
,heat input to boiler is in
.

Turbine (Isentropic expansion process 1-2):

The superheated steam from the boiler is at higher temperature and pressure. Work is produced by the expansion of the steam in the turbine. Steam at low pressure is transferred to the condesor. Neglect the heat transfer with the surroundings.

The mass and energy balance in the turbine is given as follows:

rbine = mh-h)

Here, work output from the turbine is out urbine
, mass flow rate is т
, enthalpy at state 1 is , and enthalpy at state 2 is .

Condenser (Constant pressure heat rejection process 2-3):

In this process the steam enters the condenser from the turbine is condensed to liquid state at constant pressure in the condenser.

The mass energy balance in the condenser is given as follows:

О
— т(h - h)
Рout

Here, heat output from the condenser is out
, enthalpy at state 3 is , and enthalpy at state 2 is .

Pump (Isentropic compression process 3-4):

In the Isentropic compression process, the working fluid (water) is pumped into boiler at high pressure from the condenser at low pressure. In this process, no heat is transferred to the surroundings.

The mass and energy balance in the pump is given as follows:

|(W,) m(-)
npump

Here, work input to the pump is (W)
inpmp
and enthalpy at state 4is .

Boiler (Constant pressure heat addition process 4-1):

The high pressure saturated liquid water enters the boiler and is converted into steam by addition of heat, conversion takes place at constant pressure.

The mass and energy balance in the boiler is given as follows:

Qu=m(h-h,)

Here, heat input to boiler is in
.

The formula to calculate the thermal efficiency of the cycle is given as follows:

(W
net Jout
thermal
in

Here, thermal efficiency of the cycle is thermal
and net work output from the cycle is (W2
net out
.

The net work output from the cycle is calculated as follows:

(W (W (W),
in pump
out turbine
net lout

From the tables of “Saturated steam”, select the properties of water at pressure of 16 MPa
and dryness fraction of saturated vapor of x=1
.

Specific enthalpy at state 1,h, 2580.6 kJ/kg

Specific volume at state 1, v 0.00931 m/kg

Entropy at state 1,s, 5.2455 kJ/kg K

Temperature at state 1, T, = 347.4 °C

Since, the process 1-2 is an isentropic, thenthe entropy at state 1 is equal to entropy at state 2.

From the tables of “Saturated pressure table of water”, selectthe properties of water at pressure of .

Specific enthalpy of fluid, h, 173.88 kJ/kg

Enthalpy of vaporization, he 2403.1 kJ/kg

Specific entropy of fluid, = 0.5926 kJ/kg-K

Entropy of vaporization, - 7.6361 kJ/kg K

Specific volume of fluid, - 0.0010084 m/kg|

Specific volume of vapor, 18.099 m/kg

Calculate the dryness fraction at state 2 .

Substitute 5.2455 kJ/kg K
for , 0.5926 kJ/kg K
for , and for .

|(5.2455 kJ/kg K) = (0.5926 kJ/kg K)+x,(7.6361 kJ/kg-K)
5.2455 kJ/kg K-0.5926 kJ/kg K
7.6361 kJ/kg K
x 0.6093

Calculatespecific volume at state 2:

2= ,+x (
ニッ +(v。-)

Substitute for , 0.0010084 m3/kg
for , and for .

Calculate specific enthalpy at state 2(42)
.

Substitutefor , for х,
, and for .

h(173.88 kJ/kg)+(0.6093) (2403.1 kJ/kg)
h2 1638.159 kJ/kg

From the tables of “Saturated water”, select the properties of water at pressure of and dryness fraction of saturated liquid of .

Specific enthalpy at state 3,h 173.88 kJ/kg

Specific volume at state 3,

Calculate enthalpy at state 4.

,hv(P-P

Substitute for , for , for , and for .

Calculate the rate of heat output from the condenser (One
our
.

О
— т(h - h)
Рout

Substitute for , for , and for .

(120 kg/s) (1638.159 kJ/kg-173.88 kl/kg)
Qa 175713.48 kW

Show the properties of all states as in Table .

(a)

Write the mass and energy balance in the turbine.

rbine = mh-h)

Substitute for , 2580.6 kJ/kg
for , and 1638.159 kJ/kg
for .

(120 kg/s)(2580.6 kJ/kg-1638.159 kJ/kg)
out turbine
|(Wuthin113092.92 kW
out turbine

The mass and energy balance in the pump is given as follows,

|(W,) m(-)
npump

Substitute for , for , and for .

Calculate the net power developed .

(W .),
+(W)
net
out urbine
ритp

Substitute 113092.92 kW
for out urbine
, and -1935.6 kW
for (W)
inpmp
.

Wht (113092.92 kW)+(-1935.6 kW)
net
= 1 11157.32 kW
net

(b)

Calculate the rate of heat transfer to the steam passing through the boiler.

Qu=m(h-h,)

Substitute for , for , and for .

Q(120 kg/s) (2580.6 kJ/kg-190.01 kJ/kg)
Qn 286870.8 kW
=

(c)

Calculate the thermal efficiency.

W
net
thermal
in

Substitute 111157.32 kW
for W
net
and 286870.8 kW
for in
.

(111157.32 kWx100-
nthermal
286870.8 kW
hermal = 0.3874x 100
T thermal = 38.74%

(d)

Write the relation to calculate the mass flow rate of condenser cooling waterт.
сooling
.

(e,(-)m
т
=
cooling
steam
water
|(4-)

Here, specific heat capacity of water is (e)
water
, and increase in temperature is .

Substitute |4.18 kJ/kg°C
for (e)
water
, -18 °C
for , for steam
, for , and for .

(4.18 kJ/kg°C)(18 °C)=-(120 kg/s)(1638.159 kJ/kg-173.88 kJ/kg)
2335.37 kg/s
m
cooling
т
соoling

List out the calculated values as shown in Table (2)
.

W (kW)
(kJ/kg K)(kW)
p (kPa)
(m3/kg)T (K)h (kJ/kg)
States
1/
113092.92
State
16000
0.00931
620.4
2580.6
5.2455
0
(1)
175713.4

[Part d.2]

Ans: Part a

The net power developed is111157.32 kW
.

Part b

The rate of heat transfer to the steam passing through the boiler (Q.)
in
is286870.8 kW
.

Part c

The thermal efficiency Ithermal
is 38.74%
.

Part d.1

The mass flow rate of condenser cooling water т.
сooling
is .

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