Answer:
The rate of fuel required to drive the air conditioner [tex]Q_h = 6.061 kW[/tex]
The flow rate of the cold air is [tex]\r m = 0.30765 kg/s[/tex]
Explanation:
From this question we are told that
The efficiency is [tex]\eta = 33[/tex]% = 0.33
Temperature for the hot day is [tex]T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)[/tex]
Temperature after cooling is [tex]T_c = 5^oC = 278K[/tex]
The input power is [tex]P_{in} = 2kW[/tex]
The rate of fuel required to drive the air conditioner can be mathematically represented as
[tex]Q_h = \frac{P_{in}}{\eta}[/tex]
[tex]= \frac{2}{0.33} = 6.061 kW[/tex]
From the question the air condition is assumed to be half as a Carnot refrigeration unit
This can be Mathematically interpreted in terms of COP(coefficient of performance) as
[tex]\beta_{air} = 0.5 \beta[/tex]
where [tex]\beta[/tex] denotes COP and is mathematically represented as
[tex]\beta = \frac{Q_c}{P_{in}}[/tex]
= > [tex]Q_c = \beta P_{in}[/tex]
Where [tex]Q_c[/tex] is the rate of flue being burned for cold air to flow
Now if the COP of a Carnot refrigerator is having this value
[tex]\beta_{Carnot } = \frac{T_c}{T_h - T_c}[/tex]
[tex]= \frac{278}{308-278}[/tex]
[tex]\beta_{Carnot} = 9.267\\[/tex]
Then
[tex]\beta_{air} = 0.5 * 9.2667[/tex]
[tex]= 4.6333[/tex]
Now substituting the value of [tex]\beta[/tex] to solve for [tex]Q_c[/tex]
[tex]Q_c = \beta P_{in}[/tex]
[tex]= 4.6333 *2[/tex]
[tex]9.2667kW[/tex]
The equation for the rate of fuel being burned for the cold air to flow
[tex]Q_c = \r mc_p \Delta T[/tex]
Making the flow rate of the cold air
[tex]\r m = \frac{Q_c}{c_p \Delta T}[/tex]
[tex]= \frac{9.2667}{1.004}* (308 - 278)[/tex]
[tex]= 0.30765 kg/s[/tex]
