College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 694C
Seminar in Energy Resources and Technology
Fall 2002 Ticket: 57564 Instructor: Larry Caretto

Midterm Exam Solutions

1. (20 points) An 800 MW coal-fired power plant that uses coal with a heating value of 13,500 Btu/lb has
an annual capacity factor of 85% and an annual average efficiency of 35%. As mined, the
coal used by the plant has trace amounts of mercury with an average mass fraction of
1.5x10-7. Coal cleaning processes before combustion remove 20% of the mercury. In the
combustion process, all the mercury is vaporized and leaves the combustion region with
the flue gases. Some of the mercury condenses on small particulate matter that is in the
flue gases and 30% of the mercury that enters the combustion process is removed by
fabric filter used as the particulate control device for the plant. What are the annual
emissions of mercury from this power plant?

The annual emissions of mercury, m

 Hg , are determined by the following equation:
 Hg  m
m  coal wHg (1  rcleaning )(1  rfilter )

In this equation, m  coal is the annual coal use in tons per year, wHg is the mass fraction of
mercury in the coal as mined, rcleaning = 20% is the fraction of mercury removed by coal cleaning
and rfilter = 30% is the fraction of mercury removed by the fabric filter. The mass flow rate of coal
is calculated from the required power output, P = 800 MW(e), the heat of combustion of the coal,
Qc, = 13,500 Btu(t)/lbm, the power plant efficiency, 0.35 MW(e)/MW(t), and the effective hours
at full capacity, Hcf. The effective hours at full capacity term is simply the product of the capacity
factor, and the total number of hours in the year.

PH cf
 coal 
m
Qc
For a capacity factor of 85% and a 365-day year, the H cf term is found as follows.

365 days 24 hours hours

H cf  85%  7,446
year day year

Substituting this value and the other date for the right-hand side terms used to compute the mass
flow rate of coal gives the following annual fuel use.

7,446 hr MW (t ) 3.412 x106 Btu(t ) lbm coal 4.302 x109 lbm coal
 coal  800 MW ( e)
m 
yr 0.35 MW ( e) MW (t )  hr 13,500 Btu(t ) yr
We can substitute this fuel rate into the equation for the mercury emission rate to obtain the
following result.

4.302 x109 lbm coal 1.5 x10 7 lbm Hg 361 lbm Hg

 Hg 
m (1  0.2)(1  0.3) 
yr lbm coal yr

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 m
 Hg = 361 lbm/yr.

2. (20 points) Table 8.3 on page 209 of the text shows various data on electric vehicles and their
batteries. (Assume that the vehicle mass data in this table are in metric tons.) Review the
data on the GM EV1 using lead-acid batteries and nickel-metal-hydride (NiMH) batteries
and answer the following questions.
(a)What is the specific power (maximum power per unit mass) of each battery type? How
does difference in specific power for the two battery types compare to the difference in
their specific energy (in kWh/kg)?

The data from the table for this vehicle with the two battery types are shown in the table below. The
specific power and specific energy are computed from the data on the battery mass and the battery energy
and power.

lead-acid NiMH
Maximum power (kW) 102 102
Vehicle mass (t) 1.4 1.32
Battery mass (kg) 595 521
Battery energy storage (kWh) 18.7 26.4
Vehicle range (km) 89-153 121-209
Specific energy (Whr/kg) 31.4 50.7
Specific power (W/kg) 171.4 195.8

The NiMH battery has about 1.14 times the specific power of the lead-acid battery, but its specific
energy is about 1.61 times that of the lead-acid battery.

(b)What battery property accounts for the difference in range between the lead-acid and
the NiMH batteries?

The greater specific energy for the NiMh battery accounts for the extended range. It is the
additional stored energy, not the increase in power, that provides the range. The maximum and
minimum ranges for the NiMh batteries are about 1.36 times the corresponding values for the
lead-acid batteries. Although this is less than the factor of 1.61 for the increase in specific energy
it is greater than the factor of 1.14 for the increase in specific power.

(c)How would the vehicle performance be changed if new batteries were developed that
(i)had three times the specific power of a NiMH battery?

The battery with three times the specific power could provide greater acceleration, hill-
climbing ability, and top speed. Alternatively, the mass of the battery pack could be reduced
and the overall vehicle mass would be reduced. This could increase the vehicle range. The
final design would likely be a compromise between the increase in vehicle power and the
increase in vehicle range.

(ii)had three times the specific energy of a NiMH battery?

Tripling the specific energy could not provide greater acceleration, hill-climbing ability, and
top speed. However, it could provide a significant increase in the vehicle range.

(iii)had both three times the specific energy and three times the specific power of a
NiMH battery?

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 A battery with these characteristics could be used to build a vehicle that had a significant
increase in range while still maintaining good power performance. With these improvements,
a battery-powered car could be competitive in performance and range with a car powered by
a combustion engine. Such a battery-powered car could still cost more and recharging the
batteries would take significantly longer that filling a fuel tank.

3. (15 points) Some power plants located along rivers or by the ocean use once-through cooling. In this
approach, water from the river (or ocean) is used as the coolant for the condenser and the
warm water leaving the condenser is simply returned to the original water body. If a
nuclear power plant with a peak power output of 1000 MW has an efficiency of 33%, what
flow rate of cooling water into the condenser is required if the cooling water enters the
condenser at 20oC and leaves at 40oC?

The heat rejected, Qout, in the condenser is simply the difference between the heat input, Q in and
the work done, W. However, the heat input is equal to the word done divided by the efficiency, .
Combining these two principles and using the work and heat rates gives the following relationship
between the power output, P, and the heat rejected in the condenser, Q 
out .

P 1 
Q out  Q in  P   P  P
 

The heat rejected by the condenser heats the coolant, with a mass flow rate of m c , and a
temperature rise, T = 40 C – 20 C = 20 C, according to the following equation, where cp is the
o o o

heat capacity of the coolant.

1 
Q out  P m
 c c p T


Using the data supplied in the problem we can solve the equations above for the coolant mass
flow rate in terms of the power plant output. In the calculation below we use a value of
4184 kJ/kg-C for the heat capacity of liquid water

P 1   1000 MW kg  C 106 J 1  .33 kg

c 
m   2.43x104
c p T  20 C 4184 J MW  s .33 s

m
 c = 2.43x104 kg/s

4. (20 points) (a) Compute the average energy rate per unit area for the months of January and July for
a solar collector located in Los Angeles. (This should be the energy that is actually
available as useful heat transfer to water.) Use the following data. The collector
captures 80% of the incoming solar irradiance and it is used to heat water to 135 F. In
January the average daily solar irradiance is 890 BTU/ft 2/day over five hours of bright
sunshine and the average air temperature during these five hours is 60 F. During July,
the average daily solar irradiance is 2,260 Btu/ft 2/day over ten hours of bright sunshine
and the average air temperature during these hours is 85 F. The heat transfer
coefficient between the collector and the air is 1.0 Btu/hr/ft2/F.

We can use equation 7.2 on page 159 of the text for these computations.

q  I  U (Tc  Ta )

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 In this equation q is the net heat flux to the water being heated; this is the term that we are trying
to find. The other terms are is the fraction of the incoming solar irradiance, I, that is absorbed
by the collector, U is the overall heat transfer coefficient, T c, is the temperature of the water being
heated and Ta is the ambient temperature. Using the data provided in the problem we can
compute the value of q for January and June as follows.

890 Btu 1.0 Btu 5 hr 337 Btu

q January  80%  (135 F  60 F )  2
ft  day hr  ft  F day
2 2
ft  day

2,260 Btu 1.0 Btu 10 hr 1,308 Btu

q July  80%  (135 F  85 F )  2
ft  day hr  ft  F day
2 2
ft  day

(b) What is the average efficiency of the solar collector in January and July?

The efficiency, , is the ratio of the heat delivered to the water, q, divided by the solar irradiance,
I. The efficiencies in January and July are found from the results above.

337 Btu 1,308 Btu

qJanuary ft 2  day q July ft 2  day
 January    37.9%  July    57.9%
I January 890 Btu I July 2,260 Btu
ft 2  day ft 2  day

(c) How many gallons of water could be heated in January and July, per unit collector area
per day, if the desired temperature increase for the water was 60 F?

 , where Q  m
 , can heat a mass flow rate of water, m
The total heat transfer rate, Q  c p T .
Here cp is the heat capacity of water and T is the temperature increase. The relationship
between mass flow rate and volume flow rate, V , is m   V , where  is the density of water.

Since q = Q , we can relate the volume flow rate and the heat flux, q, as follows.
A

Q m c p T Vc p T V q
q    
A A A A  c p T

We use the density and heat capacity of water as 8.34 pounds per gallon and 1 Btu/lb m-F along
with the given temperature rise, T, of 60 F to compute the volume of water that can be heated
per unit area of collector.

 V  q 337Btu gal lbm  F 1 gallons

 A   January  2
c T ft  day 8.34 lb 1 Btu 60 F
 0.673 2
ft  day
  January p m

 V  q July 1,308Btu gal lbm  F 1 gallons

    2  2.61 2
 A  July c p T ft  day 8.34 lbm 1 Btu 60 F ft  day

5. (25 points) This problem raises additional questions about the power plant described in problem one.
The data for that problem gave the maximum capacity as 800 MW, the heating value of the
coal as 13,500 Btu/lbm, the annual capacity factor as 85%, and an annual average efficiency
as 35%. The ultimate analysis of the coal used in that plant (on a weight basis) is as

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 follows: Carbon 82%, Hydrogen 10%, Sulfur 1.5%, Nitrogen 1.0%, Oxygen 1%, Mineral
matter: 4.5%.

(a) If all the sulfur in the coal is converted to sulfur dioxide, what is the emission rate in
pounds of SO2 per Btu before any controls to remove SO2?

We can use the same equation that was used for the solution to problem 9.2 on the September
m
 SO 2
18 homework solutions. The ratio of SO2 emissions to the heat input rate of the fuel, was

Q fuel
found, in that problem, from the following equation.

M SO2
wS m fuel
m SO2 m SO2 MS wS M SO2
  
Q fuel Qc m fuel Qc m fuel Qc M S

Substituting the problem data gives.

m
 SO2 w M SO2 lbm coal  0.015 lbm S  64.064 lbm SO2
 S 

Q fuel Qc M S 13,500 Btu  lbm coal  32.065 lbm S

m
 SO2 lb SO2
 2.15 x10 6 m

Q fuel Btu

(b) The plant has an allocation of 35,000 tons/year of SO2 under the Phase II of the Acid
Rain program of the 1990 Clean Air Act Amendments. Calculate the amount of
allocations it will have to purchase or have available to sell for the operating
conditions given in the problem.

From problem one we know that the annual fuel use will be 4.302x10 9 lbm of coal. We can
calculate the SO2 emission rate directly from the fuel use and the weight fraction of sulfur in the
fuel.

M SO2 0.015 lbm S 4.302x109 lbm coal 64.064 lbm SO2 1 ton SO2 tons SO2
 SO2  wS
m  coal 
m  62,444
MS lbm coal yr 32.065 lbm S 2000 lbm SO2 yr

Since the emissions are greater than the allowance, the plant operator will have to purchase
allowances to make up the difference (62,444 – 35,000 tons per year).

Thus, the plant will have to purchase

27,444 tons per year of allowances.

(c) If the coal is burned with 25% excess air, what is the mole fraction of SO 2 in the dry
exhaust gases?

We have to find the values of w, v, x, y, and z in the general fuel formula C xHySzOwNv. These are
found from the ultimate analysis and the atomic weight of the various elements using equations
[45] from the combustion notes as shown below.

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 wt%C wt%H wt%S wt%N wt%O
x= y= z= v= w=
MC MH MS MN MO

Substituting problem data and atomic weights gives the following calculations.

82 10 1.5 1 1
x= y= z= v= w=
12.0107 1.00794 32.065 14.0067 15.9994

Completing the calculations and substituting the results into the fuel formula gives C xHySzOwNv as
C6.827H9.921S0.0468O0.0625N0.0714. In burning one mole of this fuel, the exhaust gases will have z =
0.0468 moles of SO2. The total dry moles, D, that result from combustion of one mole of fuel, with
25% excess air ( = 1.25) are found from equations [2], [13], and [15] from the combustion notes.

y w
A= x + +z- [2]
4 2

v
D= x+ z + +  A Bd - A [13]
2

Bd  4.7742 and Bw  4.7742 + 7.67597  [15]

Substituting the values we have for x, y, z, and w into equation [3] for A gives the following result.

9.921 0.0625
A = 6.872 + + 0.0468 -  9.323
4 2

We can now compute the total dry moles from the known values of x, z, v, , A, and Bd.

0.0714
D = 6.872 + 0.0468 + + 1.25( 9.323)( 4.7742) - 9.323  53.224
2

So the mole fraction of SO2 is 0.0468/53.224 = 0.000879

6. (40 points) The attachment is a summary of an Energy Information Agency study on the potential use
of biomass fuels for electricity generation (http://www.eia.doe.gov/oiaf/analysispaper/).
Review this summary and answer the following questions.
(a) How significant will biomass fuels become in the period of used in this study?
The summary notes that 3 quads (3x1015 Btu) of biomass energy were used in 2000. It calculates
that the amount of biomass fuel that would be available at a cost competitive with coal is only
0.234 quads in 2020. There is a conflict here. This conclusion indicates that the present use of
biomass fuel is not based on the economics of competition with coal-fired generation. From an
energy standpoint, then, there appears to be no growth in biomass fuel availability. However, the
study forecasts that the generation capacity will increase from 6.6 GW in 2000 to 10.4 GW in
2020. Assuming that the capacity factors are the same in both years, this represents a 10.4/6.6-1
= 58% increase in biomass energy between 2000 and 2020. The projections for US energy use
from the lecture notes for the first class show an increase from 382 to 612 quads between 1999
and 2020. This is a 612/382-1 = 60% increase in total energy use. Thus, the forecast for
biomass energy use, implied by the increase in generating capacity, shows that this energy
source will retain about the same proportion of total energy use over the forecast period.

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(b) Give your explanation for the shapes of the curves for fuel availability as a function of
cost.
The lack of any increase in availability as a function of cost implies that there is some other
constraint on supply that is not affected by cost. For sources that are residues, the obvious limit
is the total amount of residues produced. There is no clear indication of why there should be a
limit on energy crops. Could not new crops be planted if there was a market for them at a higher
price? Perhaps this limit represents the total forecast production of the crops listed in the
summary as energy crops.
(c) What are the major conclusions of the study?
The summary presents the following conclusions: (1) the maximum available biomass energy in
2020 without special incentives, 7.1 quads, occurs only if the price of competing fuels increases
to $7 per million Btu; (2) with no changes in law or policy the biomass power generating capacity
will grow from 6.6 to 10.4 GW between 2000 and 2020; (3) increases to (a) 12.3 GW or (b) 70
GW are possible if (a) there are lower costs for biomass fuels or (b) if there is a mandated use of
20% renewable fuels.
(d) Do you think that these conclusions are reasonable?
The first conclusion is consistent with the figure shown. In particular, the following levels would
be available: agricultural residues, 2.35 quads; forestry residues, 2.1 quads; energy crops, 1.5
quads, and urban wood waste and mill residues, 1.25 quads. This is a total of 7.1 quads. What
is not mentioned in the summary is that most of this maximum amount, 6.7 quads, is available at
if the price rises to $4 per million Btu. This comes from the following components: agricultural
residues, 2.35 quads; forestry residues, 1.75 quads; energy crops, 1.5 quads, and urban wood
waste and mill residues, 1.1 quads.

The forecast increase in generating capacity (6.6 to 10.4 GW) seems reasonable. It is consistent
with the overall forecasts in energy growth. However, this forecast assumes that there are
special incentives for the use of biomass fuel that go beyond the economic cost comparison with
coal. If biomass fuel use were determined by this cost comparison alone, there would be a
decrease in biomass fuel use.

If the maximum fuel amount of 7.1 quads became available, and the capacity factors of biomass
generating plants did not change, the generating capacity would increase by a factor of 7.1/3-1 =
2.37. This should produce a generating future generating capacity of 2.37(6.6GW) = 12.6 GW; it
is not clear how the forecast generating capacity of 70 GW for 2020 can be established.

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