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3 years ago

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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?

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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From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

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Explanation:

Let the volume of the system at any time be V.

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where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

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Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

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r = 2 inches/2 = 1 inch = 0.0254 m

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Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

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Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

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$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

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$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

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$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

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