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Artemon [7]
3 years ago
6

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
Physics
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}

in the question COP= \frac{60}{100} \times COP_{max}

⇒\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}

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Q_H= W

and T_H= 24° C= 297 K

1=\frac{60}{100}\times \frac{297}{297-T_C}

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3 years ago
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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
atroni [7]

(a) 2.79 rev/s^2

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

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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

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

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

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

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Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

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

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\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

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t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

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

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\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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3 years ago
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From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

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Generally the equation for the F north  is mathematically given as

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F_N=1.09090909*10^{-5}

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F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

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For more information on this visit

brainly.com/question/21811998

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alexgriva [62]

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