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A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) as well as powering the car and other auxiliary equipment. On a hot (35°C) summer day, the air conditioner takes outside air in and cools it to 5°C, sending it into a duct using 2 kW of power input. It is assumed to be half as efficient as a Carnot refrigeration unit.

Tasks:
1. Find the extra rate of fuel (in kW) being burned just to drive the air conditioner unit.
2. Calculate the Coefficient of Performance (COP) of the air conditioner.
3. Determine the flow rate of cold air the air-conditioner unit can provide.

Answer :

Final answer:

The Coefficient of Performance (COP) of the air conditioner is calculated by taking half of the Carnot COP, obtained using provided temperatures (converted to Kelvin). The extra rate of fuel (kW) being burned to run the air conditioner is calculated as the power input divided by the car engine's thermal efficiency.

Explanation:

To calculate the extra rate of fuel (kW) being burned just to drive the air conditioner unit and also its Coefficient of Performance (COP), we can make use of the formulas for thermal efficiency and COP.

First, let's calculate the maximum Carnot COP, which is given by TC / (TH - TC), where TC is the cold temperature and TH is the hot temperature in Kelvin. Given the temperatures in Celsius, we convert them to Kelvin by adding 273.15. Therefore, Carnot COP = (5+273.15)/(35 - 5+273.15). The air conditioner is half as efficient as this, so COP of air conditioner = (Carnot COP)/2.

To find the extra rate of fuel burned, we know that the actual power used by the a/c unit equals to power input / thermal efficiency of car engine. As given, power input=2 kW and thermal efficiency=33% or 0.33. As a result, the extra rate of fuel being burned = 2 / 0.33 kW.

Learn more about Thermal Efficiency and Coefficient of Performance here:

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