Engineering, 23.03.2020 17:18 rederisbetter7637

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.

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Answer from: AnActualTrashcan

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

Answer from: Quest

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Answer from: Quest

you forgot the rest of the question sorry

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