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Eduardwww [97]
3 years ago
15

A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.80 MW of power, a

nd the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to useful energy that can power the community).
Assuming sunlight has a constant intensity of 1 180 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation?
Engineering
1 answer:
Greeley [361]3 years ago
6 0

Given Information:  

Output power required = Pout = 2.80 MW

Efficiency = η = 30%

Intensity = I = 1180 W/m²

Required Information:  

Effective area = A = ?

Answer:  

Effective area = A = 7.907x10³ m²

Step-by-step explanation:  

A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%

As we know efficiency is given by

η = Pout/Pin

Where Pout is the output power and Pin is the input power.

Pin = Pout/η

Pin = 2.80x10⁶/0.30

Pin = 9.33x10⁶ W

The effective area of a perfectly absorbing surface used in such an installation can be found using

A = Pin/I

Where I is the in Intensity of the sunlight in W/m²

A = 9.33x10⁶/1180

A = 7.907x10³ m²

Therefore, the effective area of the absorbing surface would be 7.907x10³ m².

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Gas chromatography separates compounds depending on their__________ . Benzene, m-xylene, and toluene have similar_________ , the
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Answer and Explanation:

Gas chromatography separates compounds depending on their **polarity and volatility**. Benzene, m-xylene, and toluene have similar **polarities**, therefore, the main basis for separation is **volatility**. The more volatile a component the ** higher its vapor pressure**, hence the more time it spends in the **gaseous mobile phase**, giving it a **shorter** retention time. Therefore, components of a liquid mixture will elute in order of **increasing boiling points/decreasing volatilities/increasing polarities with the stationary phase**.

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3 years ago
A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s
sergiy2304 [10]

Answer:

D_o=11.9inch

Explanation:

From the question we are told that:

Thickness T=0.5

Internal PressureP=2.2Ksi

Shear stress \sigma=12ksi

Elastic modulus \gamma= 35000

Generally the equation for shear stress is mathematically given by

 \sigma=\frac{P*r_1}{2*t}

Where

r_i=internal Radius

Therefore

 12=\frac{2.2*r_1}{2*0.5}

 r_i=5.45

Generally

 r_o=r_1+t

 r_o=5.45+0.5

 r_o=5.95

Generally the equation for outer diameter is mathematically given by

 D_o=2r_o

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Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

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7 0
3 years ago
P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

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3 years ago
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1 year ago
Read 2 more answers
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
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Answer: Weight on Mars = 0.02593N

Explanation:

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acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

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Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

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V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
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