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trasher [3.6K]
3 years ago
8

In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gall

on of ethanol. If ethanol can provide 76,330 Btu per gallon, what is the net energy yield for one gallon of ethanol?
Engineering
1 answer:
V125BC [204]3 years ago
4 0

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield (\% e), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy (E_{g}), measured in Btu, and the energy needed to produce the ethanol (E_{p}), measured in Btu, divided by the energy needed to produce the ethanol. That is:

\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\% (1)

If we know that E_{g} = 76330\,Btu and E_{p} = 74488\,Btu, then the net energy yield of 1 gallon of ethanol:

\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%

\%e = 2.473\,\%

The energy yield for one gallon of ethanol is 2.473 %.

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A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t
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Answer:

a)Δs = 834 mm

b)V=1122 mm/s

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Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

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At t= 4 s

s = 15t^3 - 3t\ mm

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We know that velocity V

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At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

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a=\dfrac{d^2s}{dt^2}

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3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
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Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

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3 years ago
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