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Anettt [7]
3 years ago
5

An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat

e of heat transfer into the dwelling through the walls and roof is 30,000 Btu/h and the net power input to the air conditioner compressor is 3.3 hp.
Required:
a. Determine the coefficient of performance of the air conditioner.
b. Determine the power input required, in hp, and the coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures.
Engineering
1 answer:
IrinaVladis [17]3 years ago
3 0

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature T_L = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature T_H = 99° F = ( 99 + 460 )R = 559 R

Cooling Load Q_L = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

COP_{air-condition = Cooling Load Q_L  / power P

we substitute

COP_{air-condition = 30000 Btu/h / 8396.619 Btu/h

COP_{air-condition = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

Q_L / P_{required = T_L / ( T_H - T_L )

we substitute

30000 Btu/h / P_{required = 530 R / ( 559 R - 530 R )

30000 Btu/h / P_{required = 530 R / 29 R

we solve for P_{required

P_{required  = ( 30000 Btu/h × 29 R ) / 530 R

P_{required  = ( 870000 Btu/h / 530 )

P_{required  = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

P_{required  = ( 1641.5094 / 2544.43 ) hp

P_{required  = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

COP_{rev-air-condition = T_L / ( T_H - T_L )

we substitute

COP_{rev-air-condition = 530 R / ( 559 R - 530 R )

COP_{rev-air-condition = 530 R / 29 R

COP_{rev-air-condition = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

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A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
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Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

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  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

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Explanation:

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Learn more:

New technology brainly.com/question/5768621

#learnwithBrainly

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