CEE4210-HW4-F2022-SOLUTION
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CEE 4210/6210 Renewable Energy Systems
Assignment 4 Fall 2022
SOLUTION
Question 1.
(10 points) Completion of curve-fitting exercise for observed wind data
In the wind unit Lectures 2.3 and 2.4 we reviewed the following wind data distribution that was
presented on Canvas:
Bin
Bin min
Bin max
Bin
avge
Hrs/yr
(mph)
(mph)
(mph)
1
0
0
0
40
2
0
2
1
65
3
2
4
3
301
4
4
6
5
482
5
6
8
7
669
6
8
10
9
833
7
10
12
11
884
8
12
14
13
895
9
14
16
15
902
10
16
18
17
811
11
18
20
19
726
12
20
22
21
568
13
22
24
23
499
14
24
26
25
354
15
26
28
27
270
16
28
30
29
195
17
30
none
33
266
Total:
8760
Complete this assignment by fitting the Rayleigh curve to this observed data set and producing a
modeled set of hourly output values.
In your response, include the predicted number of hours per year
for each of the bins 1 to 17, the average wind speed in mph, and a figure showing both modeled and
observed hours per year.
Hint: I did not teach a precise way to handle the last bin of wind speed above
30 mph, so slight variations in answers will be accepted for full credit, based on how you resolve this
question.
Solution:
The table is completed as follows with value of average wind speed using Excel Solver of
U
avg
= ~15.0
mph:
Bin max
CDF
Bin Prob
Mod
Observe
Error^2
1
hours
d
0
-
-
0
40
1600.0
2
0.0139
0.0139
122
65
3241.0
4
0.0545
0.0406
356
301
2992.3
6
0.1185
0.0640
561
482
6175.8
8
0.2009
0.0824
722
669
2772.8
10
0.2956
0.0947
830
833
11.3
12
0.3963
0.1006
882
884
5.4
14
0.4968
0.1006
881
895
195.7
16
0.5922
0.0954
836
902
4381.9
18
0.6787
0.0865
757
811
2879.4
20
0.7538
0.0751
658
726
4613.2
22
0.8166
0.0628
550
568
327.8
24
0.8671
0.0505
443
499
3161.8
26
0.9064
0.0393
344
354
98.8
28
0.9359
0.0295
258
270
136.5
30
0.9573
0.0214
188
195
55.4
Over
0.0427
374
266
11658.1
Comparing the modeled and observed hours per year gives the following graph:
Question 2.
(50 points) Biomass Combined Heat and Power (CHP) system
A biomass energy system is to be built up around a region of forest from which 2 tons of wood per acre
can be sustainably harvested each year (in other words, the rate of harvesting can be maintained
indefinitely without degrading the forest).
The average energy content of the wood is 18.6 Mmbtu/ton
(1 Mmbtu = 1 million Btu).
The wood is delivered by train car to the plant, and each car can carry 50
tons of wood.
The maximum number of cars that can enter the train yard for the plant in one train is 80
cars.
2
For calculating the economics of this system, use 25 years with a discount rate of 6%, and zero salvage
value at the end of the investment lifetime.
The CHP plant is based on a boiler that is 85% efficient and a generator that is 98% efficient.
The turbine
and process heater efficiency is based on the values in the table below.
In the steam cycle, water starts
at 50 kPa and is compressed to 6 MPa using a pump.
It is then heated until it is completely vaporized,
with the enthalpy and entropy values shown in the table below.
After that, the steam is expanded in a
turbine.
Part of the steam is partially expanded to 1 MPa and then flows to a process heater, where it
completely condenses and all the heat content is transferred to process heat to be used in industrial
processes, building heating, etc.
The other part of the steam that enters the turbine is fully expanded to
50 kPa and then completely condensed back to water, and the cycle begins again.
The working fluid that
leaves the process heater is also pumped up to 6 MPa, and then the two streams are combined in a
mixing chamber before being heated to make steam for the turbine.
The relevant tables are provided at
the end of this problem statement.
The value of specific volume for water is 0.00103 m
3
/kg at 50 kPa and 0.001127 m
3
/kg at 1 MPa.
The average behavior on a year-round basis that is the basis for evaluating the economics of the plant
are that a flow of 50 kg/s leaves the boiler and enters the turbine, and of this amount 30% is partially
expanded and sent to a process heater, and the remaining 70% is fully expanded.
All heat generated
from condensing steam in the process heater is transferred out of the system for heating or industrial
applications.
The rated capacity of the plant as an electricity production facility is 45 MW
electric
, so that
the plant can meet peaks in demand for electricity beyond the average output you have calculated.
The
capital cost of the plant is based on the rated capacity of the plant, and each kW costs $4500.
For operating cost, the cost of the wood is $32/ton, and all other operating and overhead costs amount
to $0.02/kWh.
The plant sells the heat it makes to customers for $7/Mmbtu, a figure recently obtained
from the rate that a district heating provider in Montpelier, VT.
Assume that the cycle is isentropic, i.e., there is no net decrease in entropy around the thermodynamic
cycle and all losses due to friction, heat loss from the equipment, etc., are negligible.
All heat generated
by the process heater is sold to customers so as to reduce the operating cost of the plant, and hence the
LCOE.
In other words:
LCOE
=
TotalCost
−
HeatSales
TotalkWh
.
Questions:
1)
As part of your solution, produce a table showing the enthalpy
h
in kJ/kg for each of the 8 states
in the thermodynamic cycle for this system.
Some of the values are already given to you in the
table below.
2)
What is the utilization of the system when operating in the average condition based on 50 kg/s
of flow through the boiler?
3)
What is the value of LCOE in $/kWh?
4)
Assuming continuous, sustainable harvesting of wood as described in the problem, how many
acres of forest are required to support the plant in its steady-state condition?
5)
If the wood is delivered in full 80-car trains, to the nearest 0.1 day, how frequently do the trains
arrive?
3
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Data tables:
Data for HW4 case study:
State:
Enthalpy (kj/kg)
Entropy
(kJ/kgK)
Pressure
(kPa)
1. Condensed water
340.54
1.0912
50
2. Compressed water
n/a
1.0912
6000
3. Saturated vapor
2784.60
5.8902
6000
4. Expanded water-vapor mix
n/a
5.8902
50
Data for calculating quality of steam:
Factor @ 50 kPa
Entropy s
Enthalpy h
(kJ/kg degK)
(kJ/kg)
Liquid
1.0912
340.54
Vapor
7.5931
2645.24
Difference
6.5019
2304.70
Data for calculating quality of steam at partial expansion 1000
kPa:
Factor @ 1000 kPa
Entropy s
Enthalpy h
(kJ/kg degK)
(kJ/kg)
Liquid
2.1381
762.51
Vapor
6.585
2777.1
Difference
4.4469
2014.59
4
Solution:
Refer to the diagram from lecture, and use the points as numbered (ignore values for s = 1.303, s =
6.070):
Part 1: Table of values.
Required summary of given and calculated enthalpy values: The table is presented first, and then the
calculation of the missing values follows below.
Enthalpies h1 to h8
Location
Enthalpy (kJ/kg)
h1
340.54
h2
346.67
h3
2784.60
h4
2462.33
h5
2041.62
h6
762.51
h7
768.15
h8
473.11
Calculations to derive missing enthalpy values: Begin with enthalpy h2 and the fully expanded steam.
Pump work to calculate enthalpy h2: We are given that the specific volume of water is 0.00103 m
3
/kg.
Therefore, multiplying by the change in pressure gives:
w
p
=
(
0.00103
) (
6000
−
50
)
=
6.13
kJ
/
kg
h
2
=
h
1
+
w
p
=
340.5
+
6.1
=
346.6
kJ
/
kg
Use of entropy to calculate h5: Since s5 = s3, we can use the quality of the steam to calculate h5.
Quality
x:
x
=
s
5
−
sf
∆ s
=
5.8902
−
1.0912
6.5019
=
0.738
5
h
5
=
h
1
+
x
(
h
g
−
h
1
)
=
h
1
+
x ×∆h
=
340.5
+
(
0.738
) (
2304.7
)
=
2041.6
kJ
/
kg
The calculation for the partially expanded steam follows a similar process, but the steam pressure
declines from 6000 kPa to 1000 kPa before being released from the turbine and passed to the process
heater.
w
p
=
(
0.001127
) (
6000
−
1000
)
=
5.6
kJ
/
kg
h
7
=
h
6
+
w
p
=
762.5
+
5.6
=
768.1
kJ
/
kg
Use of entropy to calculate h4: Since s4 = s3, we can use the quality of the steam to calculate h4.
Note
that because of the higher pressure the value of entropy for saturated liquid sf is higher than in the fully
expanded case.
Quality x:
x
=
s
4
−
sf
∆s
=
5.8902
−
2.1381
4.4469
=
0.844
h
4
=
h
6
+
x
(
h
g
−
h
6
)
=
h
6
+
x ×∆h
=
762.5
+
(
0.844
) (
2014.6
)
=
2462.3
kJ
/
kg
The remaining unknown enthalpy value is h8, the value leaving the exit chamber.
Based on the given
information, 50 kg/s leaves the mixing chamber, of which 35 kg/s arrives from the lower pressure and
the remaining 15 kg/s from the higher pressure.
Enthalpy h8 is calculated using a mass balance:
h
8
=
h
2
×m
2
+
h
7
×m
7
m
8
=
346.7
×
35
+
768.1
×
15
50
=
473.1
kJ
/
kg
Part 2. Utilization
With all enthalpy values known, it is now possible to calculate the utilization of the cycle.
Net power for
full and partial expansion of the steam are the following:
Full:
2784.6
−
2041.6
−
6.1
=
736.9
; power:
(
736.9
)
×
(
35
)
=
25,790
kW
Partial:
2784.6
−
2462.3
−
5.6
=
316.6
; power: (316.6) x 15 = ~4,749 kW
Total power:
25790
+
4749
=
30,539
kW
The process heater output is calculated based on the change in enthalpy h4 to h6:
2462.3
−
762.5
=
1699.8
kJ
/
kg
Rate of heat transfer:
∆h×m
6
=
(
1699.8
)
×
15
=
25497
kW
Heat input into boiler: 2784.6 - 473.1 = 2311.5 kJ/kg
Rate of heat transfer:
∆h×m
8
=
(
2311.5
)
×
50
=
115,574
kW
Utilization:
Heat
∈
¿
=
25497
+
30539
115574
=
56037
115574
=
0.485
ε
=
Heat out
+
Power
¿
Answer:
ε
=
48.5%
6
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Part 3. Calculation of LCOE
LCOE requires knowledge of both capital cost and operating cost.
Turning to capital cost, the required rated capacity is 45 MW
e
.
At $4.5 million per MW, the initial capital
cost is therefore
45
MW ×$
4.5
M
=
$
202.5
million initial cost.
This capital cost must be annualized:
(
$
202.5
M ,
6%
,
25
y
)
=
$
15.84
M
/
yr
Fuel cost is based on efficiency of combustion, the energy content of the wood, and the price per ton.
Since the boiler is 85% efficient, the rate of wood combustion is:
115574
0.85
=
135,970
kW
Energy content of wood per year is based on the number of hours per year and conversion of kWh to
Btu:
(
135970
)
×
(
8760
h
y
)
×
3412
btu
kWh
=
4.064
×
10
6
Mmbtu/year
Since wood has an energy content of 18.6 MMBtu per ton, the consumption per year is ~218,500
tons/year.
Accordingly, the cost at $32/ton is $7.00 million per year.
Lastly, the electricity output with the 98%-efficient generator is
30.5
MW ×
0.98
=
29.9
MW
and
the annual electricity production is
29.9
MW ×
8760
h
y
=
262.2
MillionkWh
/
yr
.
Therefore,
operating cost is
(
262.2
×
10
6
)
×
(
$
0.02
)
=
$
5.24
million per year.
The economic benefit of heat sales must be considered as well.
Ignoring losses in the heat transfer from
the cycle to the heat to be sold, the annual volume on ~25.5 MW of heat is:
(
25.5
)
(
8760
h
y
)
=
223,000
MWh
y
=
762,000
Mmbtu
/
y
On the basis of $7/mmbtu the value of this resource is ~$5.33 million per year.
Finally, LCOE for electricity is calculated based on net cost divided by generation:
LCOE
=
NetCost
kWh
=
$
15.84
M
+
$
7
M
+
$
5.24
M
−
$
5.33
M
262.2
MkWh
/
y
=
$
22.74
M
262.2
MkWh
/
y
=
$
0.0867
/
kWh
Answer: LCOE = $0.0867/kWh.
Part 4. Acres of forest required
Since the requirement is 218,500 tons of wood per year, and the rate of harvest is 2 tons per acre, the
area required is 109,250 acres.
7
Part 5. Arrival rate of trains
An 80-car train can carry 80 x 50 = 4,000 tons.
Since there are 218,500 tons consumed per year, the
number trains per year is:
218500
t
/
y
4000
t
/
train
=
54.6
trains
/
year
Then the number of trains per year is (365)/(54.6)=~6.7 days
Question 3
Small-scale hydropower (40 points)
Cornell University operates a hydro plant with Ossberger turbines rated at 1200 kW of maximum output.
Efficiency values of 85% for the turbine and 94% for the electricity generator can be used throughout.
The gross head is 42m but there are 7m of head losses that apply.
You can use a rounded density value
of 1,000 kg per m
3
of water throughout.
Model the flow as follows: The distribution of flow can be modeled with a Weibull distribution with scale
parameter σ = 131 and shape parameter
k
= 1.15.
Build your model with 10 cfs bins, in other words,
Bin1 is 0-10 cfs, Bin 2 is 10-20 cfs, etc.
Assume that at all times the first 10 cfs of water flow must be
allowed to bypass the plant to maintain flow in Fall Creek, and the second 10 cfs is not enough to
activate the turbine, so that output only starts with Bin 3 and flow in the range 20 <
Q
< 30 cfs.
Hint, be
sure to convert cfs to m
3
/s flow so that the units on power output are correct.
If the plant were built today, the cost would be $2,100/kW, and the investors would accept a long
payback of 5% over 50 years.
Operating cost is $0.02/kWh.
Questions: 1) What is the predicted kWh/year based on the given flow and turbine performance data?
2) In the case of a newly built hydropower plant, what is the LCOE in $/kWh?
3) In the case of the
Cornell plant, the cost of the plant is fully amortized, so only operating cost remains.
How much would it
cost per year for Cornell to operate the plant in this case, based on the output from part (1)?
Solution:
Solution:
Part 1. Set up a table of bin-by-bin probabilities.
For example, the probability that 20 cfs < Q < 30 cfs:
P
(
Q
<
20
)
=
1
−
exp
(
−
(
20
131
)
1.15
)
=
0.109
P
(
Q
<
30
)
=
1
−
exp
(
−
(
30
131
)
1.15
)
=
0.168
P
=
0.168
−
0.109
=
0.059
=
5.9%
Note also that the highest possible output is achieved in Bin 17 with average net flow of 155 cfs = 4.39
m
3
/s.
Calculate overall efficiency as that of the turbine and the generator, 0.85 x 0.94 = ~0.80.
Then
8
P
=
ρg ∆ zQε
=
1000
×
9.8
×
35
×
4.39
×
0.80
/
1000
=
1203
kW
=
1200
kW
.
Ideally, you would
cap the output at exactly 1200 kW although the increase to 1203 kW has a negligible effect.
The table includes flow in cfs and then converted to m/s:
Table:
Bin
Min cfs
Max cfs
CDF
Prob
h/yr
Avg net
cfs
Flow
m3/s
Power
kW
Output
MWh
1
0
10
0.051
0.051
443
0
0.00
0.0
0.0
2
10
20
0.109
0.058
510
5
0.14
0.0
0.0
3
20
30
0.168
0.059
516
15
0.42
116.4
60.1
4
30
40
0.226
0.058
506
25
0.71
194.0
98.3
5
40
50
0.281
0.056
489
35
0.99
271.7
132.8
6
50
60
0.335
0.053
467
45
1.27
349.3
163.1
7
60
70
0.385
0.051
443
55
1.56
426.9
189.0
8
70
80
0.433
0.048
418
65
1.84
504.5
210.7
9
80
90
0.478
0.045
392
75
2.12
582.1
228.4
10
90
100
0.520
0.042
367
85
2.41
659.7
242.2
11
100
110
0.559
0.039
343
95
2.69
737.3
252.6
12
110
120
0.595
0.036
319
105
2.97
815.0
259.9
13
120
130
0.629
0.034
296
115
3.26
892.6
264.3
14
130
140
0.660
0.031
274
125
3.54
970.2
266.2
15
140
150
0.689
0.029
254
135
3.82
1047.8
266.0
16
150
160
0.716
0.027
234
145
4.11
1125.4
263.8
17
160
170
0.741
0.284
2488
155
4.39
1200.0
2986.1
18
170
And up
1200.0
TOTAL:
5883.5
The total is ~5,883 MWh/yr or
~5.89 Million kWh
per year
.
Part 2. LCOE:
The initial capital cost is
1200
kW ×
$
2100
kW
=
$
2.52
million
.
Therefore, the annualized cost is
=-pmt(5%,50y,$2.52M) = ~$138,037.
Then the total annual cost and LCOE is:
TotCost
=
$
138
K
+
$
0.02
kWh
×
5.89
MkWh
=
$
255.7
K
LCOE
=
TotCost
TotkWh
=
$
255.7
K
5.89
MkWh
=
$
0.0435
/
kWh
Part 3. Operating cost for fully amortized plant:
9
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Since there is no longer any capital cost, we multiply output by $0.02:
TotCost
=
$
0.02
kWh
×
5.89
MkWh
=
$
117.8
K
10