Pg. 7.1
SOLUTIONS FOR CHAPTER 7
7.1 From (1.9),
mg / m3=ppm x mol wt
24.465 (at 1 atm and 25 oC)
7.2 70% efficient scrubber, find S emission rate:
600 MWe
600/0.38=1579 MWt
7.3 If all S converted to SO2 and now using a 90% efficient scrubber:
7.4 70% scrubber, 0.6 lb SO2/106 Btu in, find % S allowable:
a.
Pg. 7.2
7.5 Compliance coal:
7.6 Air Quality Index:
_________________________________________________________
Pollutant Day 1 Day 2 Day 3
_________________________________________________________
O3, 1-hr (ppm) 0.15 0.22 0.12
CO, 8-hr (ppm) 12 15 8
PM2.5, 24-hr (µg/m3) 130 150 10
PM10
, 24-hr (µg/m3) 180 300 100
SO2, 24-hr (ppm) 0.12 0.20 0.05
NO2, 1-hr (ppm) 0.4 0.7 0.1
___________________________________________________________
7.7 8 hrs of CO at 50 ppm, from (7.6):
7.8 Tractor pull at 436 ppm CO:
7.9 RH to produce HCHO:
RO •+O2→HO2•+R’ CHO (7.19)
Pg. 7.3
7.10 RH = propene = CH2=CH-CH3 = C3H6 so, R = C3H5
so the sequence of reactions (7.16) to (7.19) is:
7.11 A 20-µm particle blown to 8000 m. From (7.24) its settling velocity is
7.12 Residence time for 10-µm particle, unit density, at 1000m:
7.13 Settling velocity and Reynolds numbers:
a. 1 µm:
v = d2
ρ
g
18
η
=(1x10−6m)2x 106g / m3 x 9.80m/s2
18 x 0.0172g/m – s =3.2x10−5m / s
Pg. 7.4
7.14 Finding the percentage by weight of oxygen and the fraction (by weight) of oxygenate
needed to provide 2% oxygen to the resulting blend of gasoline.
b. Methyl tertiary butyl ether (MTBE), CH3OC(CH3)3
c. Ethyl tertiary butyl ether (ETBE), CH3CH2OC(CH3)3
d. Tertiary amyl methyl ether (TAME), CH3CH2C(CH3)2OCH3
7.15 Ethanol fraction CH3CH2OH (sg = 0.791) in gasoline (sg = 0.739) to give 2% O2:
Pg. 7.5
7.16 The CAFE fuel efficiency for flex-fuel cars that get:
a. 18 mpg on gasoline and 12 mpg on ethanol.
b. 22 mpg on gasoline and 15 mpg on ethanol
c. 27 mpg on gasoline and 18 mpg on ethanol
7.17 At 25 miles/100 ft3 of natural gas, 0.823 gallons of gasoline equivalents per 100 ft3,
and each equivalent gallon counting as 0.15 gallons of gasoline gives a CAFÉ rating of
7.18 The “break-even” price of E85 with gasoline at $3.50/gallon:
7.19 On an energy-content basis, the cheapest would be:
a. E85 at $2/gallon or gasoline at $3/gallon
Pg. 7.6
7.20 With a 15-gallon fuel tank and 25 mpg on gasoline.
a. E10 = 0.10 x 75,670 Btu/gal + 0.90 x 115,400 Btu/gal = 111,427 Btu/gal
7.21 A 45-mpg PHEV, 30-mile/day on electricity at 0.25 kWh/mile; 50 mi/d, 5 days per
week, 2 days @ 25 mi /d .
b. On an annual basis:
c. At $3000 for batteries:
Pg. 7.7
7.22 From Figure 7.28 the well-to-wheels CO2/mile are:
11.2 kg CO2/gal x 1000 g/kg
7.23 At 0.25 kWh/mi from a 60%-efficient NGCC plant with a 96%-efficient grid, 14.4
gC/MJ of n. gas and 1 kWh = 3.6 MJ:
a. The EV carbon emissions would be
7.24 With 5.5 hr/day of sun, 17%-efficient PVs, 75% dc-ac, 0.25 kWh/mile, 30 mi/day:
Pg. 7.8
7.25 A 50%-efficient SOFC, 50% into electricity, 20% into useful heat, compared to 30%-
efficient grid electricity and an 80% efficient boiler:
7.26 NG CHP versus separated systems; Natural gas 14.4 gC/MJ, grid 175 gC/kWh. The
joule equivalent of one kWh of electricity is 3.6 MJ.
a. CHP with 36% electrical efficiency and 40% thermal efficiency versus an 85%-
efficient gas boiler for heat and the grid for electricity.
b. CHP with 50% electrical & 20% thermal efficiency vs a 280 gC/kWh, coal-fired
power plant for electricity and an 80% efficient gas-fired boiler for heat.
Pg. 7.9
7.27 Power plants emitting 0.39 x 1012
g particulates from 685 M tons coal with a heat content
of 10,000 Btu/kWh while generating 1400 billion kWh/yr.
7.28 Derivation for the dry adiabatic lapse rate:
plugged into (1) gives:
Pg. 7.10
7.29 Plotting the data, extending from ground level to crossing with ambient profile at the
adiabatic lapse rate, and extending from the stack height gives:
600
800
plume risemixing depth
7.30 From Problem 7.29, projection from the ground at 22oC crosses ambient at 500m.
Need the windspeed at 250 m (halfway up) using (7.46) and Table 7.6 for Class C,
7.31 Below the knee, the plume is fanning which suggests a stable atmosphere, which could be
7.32 H=50m, overcast so Class D, A at 1.2km, B at 1.4km.
Pg. 7.11
7.33 Bonfire emits 20g/s CO, wind 2 m/s, H=6m, distance = 400m. Table 7.7, clear night,
stability classification = F
Cx, 0
( )
=Q
π
u
σ
y
σ
z
exp −H2
2
σ
z
2
(7.49)
7.34 H=100m, Q=1.2g/s per MW, uH=4m/s, uAnemometer=3+m/s, C<365µg/m3.
7.35 Atmospheric conditions, stack height, and groundlevel concentration restrictions same as
Prob. 7.34 so that:
Pg. 7.12
7.36 H = 100m, ua = 4 m/s, Q = 80g/s, clear summer day so Class B:
First, find the windspeed at the effective stack height using (7.46) and Table 7.7:
uH=ua
H
za
p
=4m / s 100
10
0.15
=5.65m / s
a. At 2 km, Table 7.9: σy = 290 m, σz = 234 m
7.37 For class C, notice from (7.47) and (7.48) with Table 7.9 and f =0
σ
y
σ
z
=ax0.894
cxd+f=104x0.894
61x0.911 =1.7x−0.017 ≈fairly constant ≈k (about 1.7)
So, assume σy = k σz , then from (7.49)
Pg. 7.13
7.38 Find the effective stack height of the Sudbury stack:
130 C 20 m/s
15.2m 380m
10 C
o o
8 m/s