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.

The rate of fuel required to drive the air conditioner

The flow rate of the cold air is  

Explanation:

From this question we are told that

    The efficiency is % = 0.33

   Temperature for the hot day is  

        Temperature after cooling is  

      The input power is  

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

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

where  denotes COP and is mathematically represented as

= >              

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

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

Then

Now substituting the value of to solve for

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

Making the flow rate of the cold air

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the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: