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kirza4 [7]
2 years ago
15

Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power p

lant has an efficiency of 4.5 percent and a net power output of 150 kW, determine the average value of the required solar energy collection rate, in Btu/h.
Engineering
1 answer:
fgiga [73]2 years ago
3 0

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

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Which one of the following activities is not an example of incident coordination
Lady bird [3.3K]
Directing, ordering, or controlling
7 0
3 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
Ronny wants to calculate the mechanical advantage. He needs to determine the length of the effort arm and the length of the load
kakasveta [241]

Answer:

I hope it's helpful.

Explanation:

Simple Machines

Experiments focus on addressing areas pertaining to the relationships between effort force, load force, work, and mechanical advantage, such as: how simple machines change the force needed to lift a load; mechanical advantages relation to effort and load forces; how the relationship between the fulcrum, effort and load affect the force needed to lift a load; how mechanical advantage relates to effort and load forces and the length of effort and load arms.

Through investigations and models created with pulleys and levers, students find that work in physical terms is a force applied over a distance. Students also discover that while a simple machine may make work seem easier, in reality the amount of work does not decrease. Instead, machines make work seem easier by changing the direction of a force or by providing mechanical advantage as a ratio of load force to effort force.

Students examine how pulleys can be used alone or in combination affect the amount of force needed to lift a load in a bucket. Students find that a single pulley does not improve mechanical advantage, yet makes the effort applied to the load seem less because the pulley allows the effort to be applied in the direction of the force of gravity rather than against it. Students also discover that using two pulleys provides a mechanical advantage of 2, but that the effort must be applied over twice the distance in order to gain this mechanical advantage Thus the amount of work done on the load force remains the same.

Students conduct a series of experiments comparing the effects of changing load and effort force distances for the three classes of levers. Students discover that when the fulcrum is between the load and the effort (first class lever), moving the fulcrum closer to the load increases the length of the effort arm and decreases the length of the load arm. This change in fulcrum position results in an increase in mechanical advantage by decreasing the amount of effort force needed to lift the load. Thus, students will discover that mechanical advantage in levers can be determined either as the ratio of load force to effort force, or as the ratio of effort arm length to load arm length. Students then predict and test the effect of moving the fulcrum closer to the effort force. Students find that as the length of the effort arm decreases the amount of effort force required to lift the load increases.

Students explore how the position of the fulcrum and the length of the effort and load arms in a second-class lever affect mechanical advantage. A second-class lever is one in which the load is located between the fulcrum and the effort. In a second-class lever, moving the load changes the length of the load arm but has no effect on the length of the effort arm. As the effort arm is always longer than the load arm in this type of lever, mechanical advantage decreases as the length of the load arm approaches the length of the effort arm, yet will always be greater than 1 because the load must be located between the fulcrum and the effort.

Students then discover that the reverse is true when they create a third-class lever by placing the effort between the load and the fulcrum. Students discover that in the case of a third-class lever the effort arm is always shorter than the load arm, and thus the mechanical advantage will always be less than 1. Students also create a model of a third-class lever that is part of their daily life by modeling a human arm.

The CELL culminates with a performance assessment that asks students to apply their knowledge of simple machine design and mechanical advantage to create two machines, each with a mechanical advantage greater than 1.3. In doing so, students will demonstrate their understanding of the relationships between effort force, load force, pulleys, levers, mechanical advantage and work. The performance assessment will also provide students with an opportunity to hone their problem-solving skills as they test their knowledge.

Through this series of investigations students will come to understand that simple machines make work seem easier by changing the direction of an applied force as well as altering the mechanical advantage by afforded by using the machine.

Investigation focus:

Discover that simple machines make work seem easier by changing the force needed to lift a load.

Learn how effort and load forces affect the mechanical advantage of pulleys and levers.

8 0
2 years ago
Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125
Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0
1 year ago
The regulated voltage of an alternator is stated as 13.6 to 14.6 volts at 3000 rpm with the
lions [1.4K]

Answer:

  d)  1 volt​

Explanation:

The allowable range is 1 volt​. The allowed tolerance (deviation from nominal) depends on what the nominal voltage is.

5 0
3 years ago
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