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3 years ago

What is the difference between a refrigeration cycle and a heat pump cycle?

Engineering

1 answer:

sukhopar [10]3 years ago

5 0

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

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At 120 KTAS, what is a good angle of bank to use to generate a level standard rate turn for instrument flight conditions

wlad13 [49]

Answer:

The correct answer will be "18 degrees AOB".

Explanation:

The given value is:

KTAS = 120

For determining bank angles through standard rate switches, as we understand

⇒ $10 \ percent \ of KTAS +\frac{1}{2} \ of \ the \ number$

On substituting the given value of KTAS, we get

⇒ $10 \ percent \ of 120+\frac{1}{2} \ of \ the \ number$

⇒ $12+6$

⇒ $18 \ degree \ angle \ of \ bank$

3 0

4 years ago

What are the well-known effects of electricity

Sever21 [200]

Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

3 0

3 years ago

Read 2 more answers

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

6 0

3 years ago

The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate t

Ostrovityanka [42]

Q)The shear flow at a point is given by q = 1 of 2 An I-beam has a flange width b = 200 mm , height h = 200 mm , web thickness tw = 8 mm, and flange thickness t = 12 mm, For the given point The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate the value of Q. Express your answer with appropriate units to three significant figures.( <u>I have also attached the diagram for better understanding)</u>

Answer:

Q = 56400 mm^3

Explanation:

<u>I have given the explanation in the attached file below.</u>

6 0

3 years ago

How to make a police car

S_A_V [24]

search it and you will get on internet

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2 years ago