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3 years ago

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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

, which cools the entire house to 20°C in 38 min. If the COP of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg·°C and cp = 1.0 kJ/kg·°C

2 answers:

Paul [167]3 years ago

6 0

Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

$Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W$

$COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W$

∴ Power drawn by the air conditioner is 1353.38 Watt

aev [14]3 years ago

4 0

Answer:

W_in = 1.353 KW

Explanation:

Given:

- Initial temperature of the house T_1 = 35°C

- Final temperature of the house T_2 = 20°C

- Time taken to cool the house dt = 38 min = 38×60 = 2280 s

- mass of air in the house m = 800 kg

- Specific heat at constant volume c_v = 0.72 kJ/kgK

- Specific heat at constant pressure c_p = 1.0 kJ/kgK

Find:

- Determine the power drawn by the air conditioner.

Solution:

- We will first compute the rate of heat removal from the room, we will use c_v due to a constant volume process, as follows:

Q_l = m*c_v*dT/dt

- In put values given:

Q_l = 800*0.72*(35-20) / 2280

Q_l = 3.7894 KW

- The relationship between the heat rejection and the COP of an air conditioner is given as:

COP = Q_l / W_in

W_in = Q_l / COP

W_in = 3.7894 / 2.8

W_in = 1.353 KW

- Hence the amount of power required for this process is 1.353 KW

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Also COP= $\frac{Q_H}{W}$

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$Q_H= W$

and T_H= 24° C= 297 K

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on calculating the above equation we get

$T_C$ =118.8 K

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

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The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

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$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

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The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

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Solving for t, we find

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$\alpha = \frac{\omega_f-\omega_i}{t}$

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$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

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The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

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brainly.com/question/21811998

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