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3 years ago

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A certain heat pump produces 200 kW of heating for a 293 K heated zone while only using 75 kW of power and a heat source at 273

K. Calculate the COP of this device as well as the theoretical maximum COP

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

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Answer:

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

Explanation:

Given that

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

We know that

Rex=ρvx/μ

So

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

$Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}$

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$Nu_x=C.x^{0.8}=C.x^{4/5}$

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Nux=hx/K

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The average value of h given as

$\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}$

$\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}$

$\bar{h}=\dfrac{5m}{4}L^{-1/5}$ ---------1

The value of local heat transfer coefficient at x=L

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From 1 and 2 we can say that

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

8 0

3 years ago

A heat pump designer claims to have an air-source heat pump whose coefficient of performance is 1.8 when heating a building whos

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Answer:

The claim is valid.

Explanation:

Let assume that heat pump is reversible. The coefficient of performance for the heat pump is:

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The claim is valid as real heat pumps have lower coefficients of performance.

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3 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

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b) Maximum shear stress in BC

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3 years ago

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

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Now put all the given values in the above equation, we get the final velocity of the ball.

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$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

7 0

3 years ago