Suggested languages for you:

Americas

Europe

Q41P

Expert-verified
Found in: Page 606

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# An air conditioner operating between ${\mathbf{93}}{\mathbf{°}}{\mathbf{F}}$and ${\mathbf{70}}{\mathbf{°}}{\mathbf{F}}$ is rated at ${\mathbf{4000}}{\mathbf{Btu}}{\mathbf{/}}{\mathbf{h}}$ cooling capacity. Its coefficient of performance is ${\mathbf{27}}{\mathbf{%}}$ of that of a Carnot refrigerator operating between the same two temperatures. What horsepower is required of the air conditioner motor?

The power of the air conditioner motor is |0.25hp .

See the step by step solution

## Step 1: The given data

Temperatures at which the air conditioner operates are ${\mathrm{T}}_{\mathrm{L}}=70°\mathrm{F}=294\mathrm{K}$ and ${\mathrm{T}}_{\mathrm{H}}=93°\mathrm{F}=307\mathrm{K}$ .

The cooling capacity of the air conditioner, $\frac{{\mathrm{Q}}_{\mathrm{L}}}{\mathrm{t}}=4000\mathrm{Btu}/\mathrm{h}$

Coefficient of performance of Carnot refrigerator, $\mathrm{K}=0.27{\mathrm{K}}_{\mathrm{c}}$

## Step 2: Understanding the concept of the Carnot refrigerator

The coefficient of performance of the Carnot refrigerator is given. Using this, we can find the coefficient of performance of the air conditioner. Once we get that, using the relation between work done, coefficient of performance and heat absorbed we find the horsepower of the motor.

Formulae:

Coefficient of performance of Carnot refrigerator,

${\mathrm{K}}_{\mathrm{c}}=\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}$ (1)

The work done per cycle of the Carnot engine,

$\mathrm{W}=\frac{\mathrm{Q}}{\mathrm{K}}$ (2)

## Step 3: Calculation of power of air conditioner

From the given value of the coefficient of performance and using equation (1), we can get the coefficient of performance of the air conditioner as given:

$\mathrm{K}=0.27×\left(\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}}\right)\phantom{\rule{0ex}{0ex}}=0.27×\left(\frac{294\mathrm{K}}{307\mathrm{K}-294\mathrm{K}}\right)\phantom{\rule{0ex}{0ex}}=0.27×23\phantom{\rule{0ex}{0ex}}=6.21$

Dividing t into both sides of equation (2), we can get the power of the air conditioner as given:

$\frac{\mathrm{W}}{\mathrm{t}}=\frac{{\mathrm{Q}}_{\mathrm{L}}/\mathrm{t}}{\mathrm{K}}\phantom{\rule{0ex}{0ex}}=\frac{40000\mathrm{Btu}/\mathrm{h}}{6.21}\phantom{\rule{0ex}{0ex}}=643\mathrm{Btu}/\mathrm{h}\phantom{\rule{0ex}{0ex}}=643\mathrm{Btu}/\mathrm{h}×\left(\frac{0.0003929\mathrm{hp}}{1\mathrm{Btu}/\mathrm{h}}\right)\phantom{\rule{0ex}{0ex}}=0.25\mathrm{hp}$

Hence, the value of the required power is 0.25 hp