Question

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and other auxiliary equipment. On a hot (35oC) summer day the A/C takes outside air in and cools it to 5oC sending it into a duct using 2 kW of power input and it is assumed to be half as good as a Carnot refrigeration unit. Find the rate of fuel (kW) being burned extra just to drive the A/C unit and its COP. Find the flow rate of cold air the A/C unit can provide.

Answer 1

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is  $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

    The efficiency is $\eta = 33$% = 0.33

   Temperature for the hot day is  $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

        Temperature after cooling is  $T_c = 5^oC = 278K$

      The input power is  $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

              $Q_h = \frac{P_{in}}{\eta}$

                    $= \frac{2}{0.33} = 6.061 kW$

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

             $\beta_{air} = 0.5 \beta$

where $\beta$  denotes COP and is mathematically represented as

                     $\beta = \frac{Q_c}{P_{in}}$

= >              $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if  the COP of a Carnot refrigerator is having this value

                $\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

                            $= \frac{278}{308-278}$

                            $\beta_{Carnot} = 9.267$

Then

     $\beta_{air} = 0.5 * 9.2667$

            $= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

                             $Q_c = \beta P_{in}$

                                  $= 4.6333 *2$

                                  $9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

                       $Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

                       $\r m = \frac{Q_c}{c_p \Delta T}$

                            $= \frac{9.2667}{1.004}* (308 - 278)$

                            $= 0.30765 kg/s$