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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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Answer:

well what do you wanna talk about friend?

Explanation:

7 0
3 years ago
A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret
andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = \frac{8000 X 10^-^3}{0.7}

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = \frac{0.002}{5}

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0
3 years ago
Air pressure is higher above an airfoil.<br> true or false
attashe74 [19]

Answer: true

Explanation:

it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However

6 0
3 years ago
The aluminum rod AB (G 5 27 GPa) is bonded to the brass rod BD (G 5 39 GPa). Knowing that portion CD of the brass rod is hollow
Temka [501]

Answer:

Qcd=0.01507rad

QT= 0.10509rad

Explanation:

The full details of the procedure and answer is attached.

7 0
3 years ago
For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals
kvv77 [185]

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   Pr=\dfrac{\nu }{\alpha }

  Where \nu is the molecular diffusivity of momentum

             \alpha is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    Pr=\left (\dfrac{\delta }{\delta _t}\right )^3

Where \delta is the hydrodynamic boundary layer thickness and \delta_t is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0
3 years ago
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