Power Plant’s Emissions A coal-fired power plant produces a net electricity output of 1 GW. A key problem faced by the power plant is the emission of sulfur as a combustion by-product. The coal contains 1% sulfur, and it is known to react with rainwater in the atmosphere to produce sulfuric

Question

Power Plant’s Emissions

A coal-fired power plant produces a net electricity output of 1 GW. A key problem faced by the power plant is the emission of sulfur as a combustion by-product. The coal contains 1% sulfur, and it is known to react with rainwater
in the atmosphere to produce sulfuric acid—the main ingredient of acid rain. The power plant is equipped with a scrubber system that removes 96% of the sulfur from the exhaust gas. See Figure 7.23. (a) By using the plant’s real efficiency of 32%, calculate the plant’s daily fuel consumption.
(b) How much sulfur escapes the scrubber and is released into the environment each day?

Approach
We will consider a 1-s time interval during which the power plant draws heat, produces work, and exhausts some sulfur into the atmosphere. The plant’s real efficiency relates “what we get” relative to “what we paid,” and by using
Equation (7.15),

[latex] \eta =\frac{W_{t}-W_{P}}{Q_{sg}} \approx \frac{W_{t}}{Q_{sg}} [/latex] (7.15)

the heat supplied to the plant by the fuel can be determined.

Table 7.3

Table 7.3 Heating Values for Certain Fuels*

		Heating Value, H
Type	Fuel	MJ/kg	Btu/lbm
Gas	Natural gas Propane	47 46	[latex]20.2×10^{3}[/latex] [latex]19.8×10^{3}[/latex]
Liquid	Gasoline Diesel Fuel oil	45 43 42	[latex]19.3×10^{3}[/latex] [latex]18.5×10^{3}[/latex] [latex]18.0×10^{3}[/latex]
Solid	Coal Wood	30 20	[latex]12.9×10^{3}[/latex] [latex]8.6×10^{3}[/latex]
*The numerical values are representative, and values for specific fuels could vary with their chemical composition.

lists the heating value of coal as 30 MJ/kg. The given information for the sulfur percentage and the scrubber’s efficiency can be used to determine the amount of sulfur released each day.

Accepted Answer

(a) By using Equation (7.5), the plant produces

[latex] W_{t}=(1 GW) (1 s) \longleftarrow \left[P_{avg}=\frac{W}{\Delta t} ight] [/latex]

[latex]=1 GW . s [/latex]

[latex]=1\left(\frac{GJ}{\cancel{s}} ight) (\cancel{s})[/latex]
[latex]= 1 GJ [/latex]

of electrical energy each second. The amount of heat that must be supplied to the steam generator by burning coal is

[latex] Q_{sg}=\frac{1 G}{0.32} \longleftarrow \left[\eta =\frac{W_{t}}{Q_{sg}} ight] [/latex]

[latex]= 3.125 GJ [/latex]

where we have neglected the small amount of pump work, [latex]W_p[/latex], in
Equation (7.15). By using Equation (7.6),

Q = mH (7.6)

the mass of fuel consumed each second is

[latex] m_{coal}=\frac{3125MJ}{30MJ/kg} \longleftarrow \left[Q=mH ight] [/latex]

[latex]=104.2(\cancel{MJ})\left(\frac{kg}{\cancel{MJ}} ight) [/latex]

[latex]=104.2 kg [/latex]

where we have converted the SI prefix on the numerical value for [latex] Q_{sg}[/latex] so that the numerator and denominator are dimensionally consistent. Over the course of a day, the plant must therefore use

[latex] m_{coal}=\left(104.2\frac{kg}{s} ight)\left(60\frac{s}{min} ight) \left(60\frac{min}{h} ight)\left(24\frac{h}{day} ight) [/latex]

[latex]=9.000 imes 10^6\left(\frac{kg}{\cancel{s}} ight)\left(\frac{\cancel{s}}{\cancel{min}} ight)\left(\frac{\cancel{min}}{\cancel{h}} ight) \left(\frac{\cancel{h}}{day} ight) [/latex]

[latex]=9.000 imes 10^6 \frac{kg}{day} [/latex]

(b) With the coal containing 1% sulfur and a scrubber efficiency of 96%,

[latex]m_{sulfur} = (0.01) (1 - 0.96) \left(9.000 imes 10^6 \frac{kg}{day} ight) [/latex]

[latex]= 3.600 imes 10^3\frac{kg}{day} [/latex]

is the daily release of sulfur.
Discussion
Because each quantity contains a large power-of-ten exponent, we apply
SI prefixes from Table 3.3

Table 3.3 Order-of-Magnitude Prefixes in the SI

Name Symbol Multiplicative Factor
tera	T	[latex]1,000,000,000,000 = 10^{12}[/latex]
giga	G	[latex]1,000,000,000 = 10^9[/latex]
mega	M	[latex]1,000,000 = 10^6[/latex]
kilo	k	[latex]1000 = 10^3[/latex]
hecto	h	[latex]100 = 10^2[/latex]
deca	da	[latex]10 = 10^1[/latex]
deci	d	[latex]0.1 = 10^{-1}[/latex]
centi	c	[latex]0.01 = 10^{-2}[/latex]
milli	m	[latex]0.001 = 10^{-3}[/latex]
micro	[latex]μ[/latex]	[latex]0.000,001 = 10^{-6}[/latex]
nano	n	[latex]0.000,000,001 = 10^{-9}[/latex]
pico	p	[latex]0.000,000,000,001 = 10^{-12}[/latex]

to represent the final results more concisely. Coal typically contains about 0.524% sulfur by weight; so the emission from a power plant could be less or more depending on the type of coal used. Although power plants in the United States are equipped with scrubber systems that remove a large fraction of the sulfur (but not all of it) from the exhaust stack, not all countries use such scrubber systems. As a result, the large-scale pollution effects of fossil-fuel consumption can indeed be high. Mechanical engineers have an active role to play in the public policy arena by balancing environmental issues with the need for an abundant and inexpensive supply of electricity.

[latex]m_{coal} =9.000 \frac{Gg }{day} [/latex]

[latex]m_{sulfur} =3.600 \frac{Mg }{day} [/latex]

Question 7.12: Power Plant’s Emissions A coal-fired power plant produces a ...

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