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An air-conditioning system is used to maintain a house at 70°F when the temperature outside is 100°F. The house is gaining heat through the walls and the windows at a rate of 800 Btu/min, and the heat generation rate within the house from people, lights, and appliances amounts to 100 Btu/min. Determine the minimum power input required for this air- conditioning system.

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gbenewaeternity

Answer:

Minimum power output required = 1.1977 hp

Explanation:

Given Data:

Temperature outside = 100°F.

House temperature =  70°F

Rate of heat gain(Qw) =  800 Btu/min

Generation rate within(Ql) = 100 Btu/min.

Converting the outside temperature 100°F from fahrenheit to ranking, we have;

1°F = 460R

Therefore,

100°F = 460+100

 To     = 560 R

Converting the house temperature 70°F from fahrenheit to ranking, we have;

1°F = 460R

Therefore,

70°F = 460+70

    Th      = 530 R

Consider the equation for coefficient of performance (COP) of refrigerator in terms of temperature;

COP =Th/(To-Th)

        = 530/(560-530)

        = 530/30

        = 17.66

Consider the equation for coefficient of performance (COP) of refrigerator;

COP = Desired output/required input

         =  Q/Wnet

          = Ql + Qw/ Wnet

Substituting into the formula, we have;

17.667 = (100 + 800)/Wnet

17.667 = 900/Wnet

Wnet = 900/17.667

         = 50.94 Btu/min.

Converting from Btu/min. to hp, we have;

1 hp = 42.53 Btu/min.

Therefore,

50.94 Btu/min =  50.94 / 42.53

                         = 1.1977 hp =

Therefore, minimum power output required = 1.1977 hp

Lanuel

Given the following data:

  • Outside temperature = 100°F.
  • House temperature =  70°F.
  • Rate of heat gain =  800 Btu/min.
  • Heat generation rate = 100 Btu/min.

The conversion of temperature.

We would convert the value of the temperatures in Fahrenheit to Rankine.

Note: 1°F = 460R

Conversion:

  • Outside temperature = 100°F = [tex]460+100[/tex] = 560R
  • House temperature =  70°F = [tex]460+70[/tex] = 530R

To calculate the minimum power input that is required for this air- conditioning system:

The coefficient of performance (COP)

In Science, the coefficient of performance (COP) is a mathematical expression that is used to show the relationship between the power output of an air-conditioning system and the power input of its compressor.

Mathematically, the coefficient of performance (COP) is given by the formula:

[tex]COP =\frac{T_h}{T_o-T_h}[/tex]

Substituting the given parameters into the formula, we have;

[tex]COP =\frac{530}{560-530}\\ \\ COP =\frac{530}{30}[/tex]

COP = 17.66

For the power input:

[tex]COP = \frac{E_o}{E_i} \\ \\ COP = \frac{Q_l + Q_w}{Q_{net}}\\ \\ 17.67 = \frac{100+800}{Q_{net}}\\ \\ Q_{net}=\frac{900}{17.67} \\ \\ Q_{net}=50.93\;Btu/min[/tex]

Conversion:

1 hp = 42.53 Btu/min.

X hp = 50.93 Btu/min.

Cross-multiplying, we have:

[tex]X=\frac{50.93}{42.53} [/tex]

X = 1.1975 hp

Read more on power input here: https://brainly.com/question/16188638

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