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3 years ago

9. An orifice tube restriction is often indicated by:

Engineering

1 answer:

andriy [413]3 years ago

3 0

Answer:

I believe 80% the answer is low gage pressure.

Explanation:

while naturally a restriction of an orifice like placing your thumb on a water hose would cause a high pressure output, however in an entire system if the proper amount of fluid is not making its way through the system then there would be inadequate system pressure. I hope my train of thought is correct though. let me know either way

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A Service Schedule is...

VikaD [51]

Answer:

option c

Explanation:

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3 years ago

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To prevent drainage of the transmission fluid from the converter when the

-BARSIC- [3]

Answer
D I think
Explanation

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3 years ago

An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

Brilliant_brown [7]

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

$COP = \frac{T_{L}}{T_{H}-T_{L}}$

$COP = \frac{293.15\,K}{308.15\,K-293.15\,K}$

$COP = 19.543$

Lastly, the power required to operate the air conditioning system is:

$\dot W = \frac{\dot Q_{L}}{COP}$

$\dot W = \frac{32\,kW}{19.543}$

$\dot W = 1.637\,kW$

Hence, the answer is E.

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3 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

4 years ago

FAST PLLZZ!! Ideally, the backrest is tilted back slightly, so when you turn the wheel your shoulders are _______ the seat.

exis [7]

Answer:

touching

Explanation:

The backrest of the seat should be tilted back ever so slightly, and when turning the steering wheel your shoulders should remain in contact with the seat – rather than hunched forward.

8 0

3 years ago