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otez555 [7]
3 years ago
14

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio

n Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.
Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
c. If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Engineering
1 answer:
Mama L [17]3 years ago
3 0

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

P_{incandescent} = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

P_{fluorescent} = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

P_{saved} = P_{incandescent}  - P_{fluorescent}

P_{saved} = 750 - 180

P_{saved}  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

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Answer:

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Explanation:

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Using the given mass m, molar mass M, we can get the following equation:  

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To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

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4 years ago
Why might construction crews want to install pipes before the foundation is poured
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The answer is choice C

Explanation:

As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .

So it will be useful for the construction crews to  connect the pipes to the sewer lines before the foundation is poured.

But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured  and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,

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4 years ago
A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

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Explanation:

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    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

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            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

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                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

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          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

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                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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