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____ [38]
2 years ago
15

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

r 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.
Engineering
1 answer:
fgiga [73]2 years ago
4 0

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

                         

                             

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