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MEM310 DESIGN PROJECT ASSIGNMENT
Nuclear Power Plant
Team 17
Emily Joyce
eaj62@drexel.edu – ID: 14424445
Lily Meuninck
lrm79@drexel.edu - ID: 14470244
Matthew Yturraspe
may53@drexel.edu – ID: 14425417
Submitted to Professor David L. Miller
on
2023 June 08
1
Summary
This study aims to design a nuclear power plant that can supply a net output of 1150MW of electric power and maximize efficiency by selecting the best pressure at which to operate the flash evaporator. The plant system includes a nuclear reactor, a flash evaporator, a turbine, a condenser, two pumps and a mixing chamber. We are given the following parameters: pressure at
state 1 is 107 atm, minimum quality at turbine exit is 86%, pump isentropic efficiency is 96 %, turbine isentropic efficiency is 86%, process water condenser exit temperature is 40
o
C, cooling water rise is 12
o
C, condenser efficiency is 93%, and electric generator efficiency is 96%. We conducted five trials each with unique simulated flash evaporator pressures. The best efficiency found for the plant is 41.2% running at a pressure of 0.35 MPa for the flash evaporator. 2
Table of Contents
Summary
2
List of Figures
4
System Definition / Problem Statement:
5
Solution
7
Discussions and Conclusions:
11
Recommendations For Future Analysis:
12
Appendices:
13
3
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List of Figures
Figure 1. Diagram of the devices and state points of the proposed nuclear power plant. 5
Table 1. Constraints
6
Figure 2. Temperature vs Entropy Diagram
6
Table 2. Pressure Relationships
7
Table 3. State Point Phases
7
Table 4. Entropy Assumptions
7
Figure 3. Example Test Calculations at P=0.35 MPa
7-9
Figure 4. Steam Quality at State 4a vs Trial Pressure Graph
8
Figure 5. Thermal Efficiency vs Trial Pressure Graph
8
Figure 6. Turbine Work Out vs Trial Pressure Graph
8
Table A.1. Equations Used
13
4
System Definition / Problem Statement
The Nuclear Power Plant, shown in Figure 1, contains water as a saturated liquid at point state 1. The water then flows through a flash evaporator at the teams selected trial pressure. The pressure is reduced by a constant enthalpy throttle valve and the total mass of the water is split into two parts. The portion of the water that is flashed into a saturated vapor flows to state point two while the rest of the mass flows as a saturated liquid to state point three. The saturated vapor at state two then enters the turbine, which generates shaft work stored in the electric generator. After point four, the saturated water mixture is cooled to 313 K in the condenser and flows to state point six. The saturated liquid then enters the low-pressure pump one, returning it to the pressure at state point two. The liquid then enters the mixing chamber, where it combines with the saturated liquid from state point three. The liquid then leaves the mixing chamber at the same
pressure as state point seven to reach state point eight. This liquid then is sent through the high-
pressure pump two to reach the pressure required for the nuclear reactor at state point five. The 5
water at state point five then goes through the nuclear reactor gaining thermal energy and returns to state point one. This entire system is held to the constraints, shown in Table 1, provided. Table 1
System Constraints
Quantity
Value
P1 (MPa)
10.8418MPa
Minimum Turbine Exit Quality
86%
Pump Isentropic Efficiency
96%
Turbine Isentropic Efficiency
86%
Process Water Condenser Exit Temperature (K)
313
Cooling Water Temperature rise (K)
12
Condenser Efficiency
93%
Electric Generator Efficiency
96%
This study aims to design a nuclear power plant that can supply a net output of 1150MW of electric power and maximize efficiency by selecting the best pressure at which to operate the flash evaporator. In order to do so we have to define: temperature, pressure, and entropy at the state points, overall flow rate of water through the nuclear reactor, heat transfer in the nuclear reactor, percent of the overall mass passing through the steam turbine, the power required for low-pressure pump 1, the power required for high-pressure pump 2, and the cooling water flow rate.
The relationship between the temperature and entropy throughout the Nuclear Power Plant is depicted by the T-s diagram shown in Figure 2. 6
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Figure 2.
T-s Diagram of the System
7
Solution
8
Through our given values (Table 1),
relationships (Table 2), and states (Table 3) we were able to
know that state points one, three, and six were all saturated
liquids, and state point two is a saturated vapor. The fact that at state point six the water is a saturated liquid allows us to assume the quality is zero (X
6
= 0). Given this assumption, and the temperature at state six, we are able to collect the entropy, enthalpy, and pressure for any test pressure. From there we are given the relationship between our test pressure at state point two and state points three, seven, and eight so we can take our given temperature to calculate the enthalpy and entropy. For state points seven, four, and five we calculated two different sets of values, a set for the isentropic and the actual values. The entropies at state points six and seven are going to be equal, however, the enthalpy will differ making the phase at state point seven a subcooled liquid. We then can calculate the enthalpy, entropy, and temperature at state point one because of the given pressure (P
1
= 10.8418MPa) and quality (X
1
=0). After this point, you can calculate the temperature, entropy, and enthalpy at state point two because you have the pressure,
the given pressure, and the quality of saturated vapor (X
2
= 1). This process is then repeated at state point three where the quality is zero because it is a saturated liquid. The isentropic values for state point four can then be calculated because we know the relationship between the pressures at state point four and state point six as well as the relationships between the entropies at state points four and six. The assumptions made for the relationships between the entropy values are shown in Table 4. After we have found the isentropic values at state point four, we can
find the actual values at state point four. The enthalpy, entropy, and temperature for state point eight can then be calculated. Lastly, the isentropic and actual values for state point five are calculated. 9
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The process described has been worked out below in Figure 3 in a sample test with a pressure of 0.35 MPa.
10
11
Figure 3.
Example Test Calculations at P=0.35 MPa
From our trials, we discovered that as the trial pressure we were testing rose so too did the thermal efficiency of the system; Figure 5 proves this correlation with a positive slope. We later calculated the work out of the turbine and found that it, as shown by Figure 6, also had a positive correlation with an increase in trial pressure. This proved to us that it was better to run our system at the highest possible pressure in the flash evaporator to get the most value out of the
system. Unfortunately, the quality of steam leaving the flash evaporator falls as the pressure rises. This leads to unsatisfactory turbine efficiency, which would be bad for the system overall, and it should thus be kept from going too high. Using this information as a guide, we then started
narrowing down how high we could get the pressure to be while still meeting the constraints of 12
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the system we have been contracted to design. The design constraints state that the turbine could not use steam with a quality less than 0.86 as it would impact the longevity and safety of the system. Figure 4 shows this relationship and plots our 0.35 MPa test pressure just above the critical boundary for the quality of the steam. This was then chosen as our optimal test pressure since it was able to deliver the greatest possible output within the given constraints. Figure 4. Steam Quality at State 4a vs Trial Pressure Graph
13
Figure 5. Thermal Efficiency vs Trial Pressure Graph
Figure 6. Turbine Work Out vs Trial Pressure Graph
14
15
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Discussions and Conclusions
The results from the trials help show that as the pressure in the flash evaporator rises, the thermal efficiency. This correlation is what led to the discovery that the maximum efficiency for the system occurs at 0.35 MPa. While a higher pressure would yield a more efficient system, design constraints on the quality of the steam make 0.35 MPa the ideal pressure to run the system. To make the system run at peak performance for this pressure, several operating conditions must also be met: The mass flow rate of water through the nuclear reactor should be held at or extremely near 104.62 Kg
s
, around 0.40% of this total mass should be making it through the steam turbine. Additionally, pumps one and two should be receiving 3.59
⋅
10
−
4
and
1.13
⋅
10
−
2
MW respectively in order to insure a proper flow rate. Following these specifications would only require a cooling water flow rate of 6.799489624 Kg
s
to run the system at maximum efficiency. 16
Recommendations for Future Analysis
To improve the overall nuclear plant efficiency, Either the work out of the system could or be increased or the overall energy put into the system by heat could be lowered. A higher work
output could be achieved by increasing the efficiency of the turbine. This would allow for a higher work output while keeping the inputs constant, which would lead to a direct improvement to the plant’s efficiency. One way to decrease the heat required by the system would be to decrease the enthalpy of reaction at state points one and five. Doing so would require the reactor to supply the water moving through it with less energy and thus the plant would need less heat to run smoothly.
To ensure that the profits of the Nuclear Power Plant are maximized, high-quality turbines should be installed to ensure high efficiency of the turbine and less overall maintenance to the system. Since liquid water content in steam increases as the quality decreases, it is not recommended to increase the pressure of the flash evaporator as this will lead to deterioration of the turbine’s blades. If the blades become eroded the efficiency of the system will drop and so too will profit margins. The steam turbine’s efficiency will need to be over 90 percent for it to last the longest in the Nuclear Power Plant we have designed. Another way to maximize profits could be by adding a reheat loop to the system. This would ensure that the water that enters the turbine will have high quality. This would also allow the flash evaporator to run at a higher pressure, which would lead to overall higher efficiency in the plant. This could be easily accomplished by harvesting some of the energy leaving the system by way of the condenser, and instead redirecting it to a reheat-coil. Lastly, in future analysis, we would need to take count any loss of mass, heat, or energy that is escaping the system by any non-intended means. This includes lack of insulation, sealant, and so 17
forth. Proper maintenance of the plant can ensure that these issues do not arise if given proper attention. 18
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Appendices
Table A.1. Equations Used
Actual Enthalpy: State Point 4
Actual Enthalpy: State Point 5
h
4
a
=
h
2
−
η
T
(
h
2
−
h
4
s
)
h
4a
: Actual enthalpy at state point four
h
2
: Enthalpy at state point two
h
T
: Efficiency of the turbine
h
4s: Isentropic enthalpy at state point four
h
5
a
=
h
8
+
v ∙
(
P
5
−
P
8
)
η
p
h
5a
: Actual enthalpy at state point five
h
8
: Enthalpy at state point eight
v: Specific volume
P
5
: Pressure at state point five
P
8
: Pressure at state point eight
η
p
: Pump efficiency of pump two
Actual Enthalpy: State Point 7
Enthalpy: State Point 8
h
7
a
=
h
6
+
v ∙
(
P
7
−
P
6
)
n
p
h
7a
: Actual enthalpy at state point seven
h
6
: Enthalpy at state point six
v: Specific volume
P
7
: Pressure at state point seven
P
6
: Pressure at state point six
η
p
: Pump efficiency
h
8
=
y∙h
7
+
(
1
−
y
)
∙h
3
h
8
: Enthalpy at state point eight
y: Percent of Overall Flow Passing through Turbine
h
7
: Enthalpy at state point seven
h
3
: Enthalpy at state point three
Work Into Pump 1
Work Into Pump 2
W
¿
, p
1
=
v
6
∙
(
P
7
−
P
6
)
η
p
W
in,p1
: Work Into Pump 1
v
6
: Specific volume at state point six
P
7
: Pressure at state point seven
P
6
: Pressure at state point six
η
p
: Pump efficiency of pump one
W
¿
, p
2
=
v
8
∙
(
P
5
−
P
8
)
η
p
W
in,p2
: Work Into Pump 2
v
8
: Specific volume at state point eight
P
5
: Pressure at state point five
P
8
: Pressure at state point eight
η
p
: Pump efficiency
Work Done by Turbine
Heat into Nuclear Reactor
W
T ,out
=
η
T
∙
(
h
4
−
h
2
)
W
T,out
: Work Done by Turbine
h
T
: Turbine Efficiency
h
4
: Enthalpy at state point four
h
2
: Enthalpy at state point two
Q
¿
,r
=
h
1
−
h
5
a
Q
in,r
: Heat into Nuclear Reactor
h
1
: Enthalpy at state point one
h
5a
: Actual Enthalpy at state five
19
Heat Out of Condenser
Net Work
Q
out , c
=
(
h
4
a
−
h
6
)
∙η
c
Q
out,c
: Heat Out of Condenser
h
4a
: Actual Enthalpy at state point four
h
6
: Enthalpy at state point six
h
c
: Condenser Efficiency
W
net
=
W
out
−
W
¿
W
net
= Net Work
W
out
: Work Out
W
in
: Work In
Thermal Efficiency
Mass Flow Rate Through Turbine
η
therm
=
W
net
Q
¿
h
therm
: Thermal Efficiency
W
net
: Net Work
Q
in
: Heat In
˙
m
4
=
Power
Turbine
W
T
˙
m
4
: Mass Flow Rate Through Turbine
Power
turbine
: Power from Turbine
W
T
: Work from Turbine
Percent of Overall Flow Passing Through Turbine
Total Mass Flow Rate
y
=
h
1
−
h
3
h
2
−
h
3
y: Percent of Overall Flow Passing Through Turbine
h
1
: Enthalpy at state point one
h
3
: Enthalpy at state point three
h
2
: Enthalpy at state point two
˙
m
tot
=
˙
m
4
y
˙
m
tot
: Total Mass Flow Rate
˙
m
4
: Mass Flow Rate Through Turbine
y: Percent of Overall Flow Passing Through Turbine
Mass Flow Rate: State 3
Heat through Cooling Water
˙
m
3
= ˙
m
tot
˙
m
3
: Mass Flow Rate Through state point three
˙
m
tot
: Total Mass Flow Rate
Q
w
=
Q
out ,c
∙
˙
m
4
Q
w
: Heat through Cooling Water
Q
out,c
: Heat out of Condenser
˙
m
4
: Mass Flow Rate Through Turbine
Mass Flow Rate: Cooling Water
˙
m
w
=
Q
w
4.18
∙ ΔT
cooling water
˙
m
w
: Mass Flow Rate of Cooling Water
Q
w
: Heat through Cooling Water
D
T
cooling water
: Change in temperature of the cooling water
20
Table B.1. Trial 1 (0.1 MPa)
Table B.2. Trial 2 (0.2 MPa)
Table B.2. Trial 3 (0.3 MPa)
21
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Table B.2. Trial 4 (0.35 MPa)
Table B.2. Trial 5 (0.4 MPa)
22
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23
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References
Darrow, Ken, et al. “Catalog of CHP Technologies, Section 4.”
CHP Technologies
, Mar. 2015, www.epa.gov/sites/default/files/2015-07/documents/catalog_of_chp_technologies_section
_4._technology_characterization_-_steam_turbines.pdf.
24
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Maps D Web design tutori. A MATH180: HW08-.
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What is the angle of rope 3?
Express your answer to two significant figures and include the appropriate units.
HA
?
2
O = 38
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ECO
5. AUTOMOTIVE. The power an
engine produces is called
horsepower. In mathematical
terms, one horsepower is the
power needed to move 550
pounds one foot in one
second, or the power needed
to move 33,000 pounds one
foot in one minute. Power, in
physics, is defined simply as
the rate of doing work. The
formula below gives the
horsepower at 5,252 radians
per second.
https://philkotse.com/toyota-corona-ior-sale-in-baguio/1991-for-sale-in-aid7017151
625T
1313
where H is the horsepower and T is the torque
a. Find the inverse of the model.
b. If a taxi produces a horsepower of 200, what is the torque it generates?
Solve here:
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