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Angelina_Jolie [31]
3 years ago
15

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

f 2.5. How much power is required to drive the unit? For the same COP, how high a high temperature would a Carnot heat pump have, assuming the same low T?

Engineering
1 answer:
AysviL [449]3 years ago
7 0

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

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Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

                          v = +/- 0.323 m/s

- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

4 0
3 years ago
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