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nexus9112 [7]
3 years ago
13

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 30 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.
Engineering
1 answer:
ipn [44]3 years ago
3 0

Answer:

\dot W = 1.667\, kW

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW

The power drawn by the air conditioner is:

\dot W = \frac{\dot Q_{L}}{COP_{R}} \\\dot W = \frac {4.667\, kW}{2.8}\\\dot W = 1.667\, kW

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The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published
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Answer:

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The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

  From the question we are told that

        The  temperature is  T =  7200K

        The diameter of the ball is  d = 1.5 km = 1.5 *1000 = 1500m

       Hence the radius  == \frac{1500}{2} = 750m

 The total energy radiated can be mathematically represented as

                         E = \sigma A T^4

Where \sigma is the Stefan-Boltzmann constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}

            A is the area of a sphere  = \pi d^2  = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2

 Substituting values we have

                    E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4

                        =1.077*10^{15} W

Now the state of the energy is mathematically represented as

                           Rate  \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4

                                                            = 5.67*10^{-8} * 7200^2

                                                            = 1.523747635*10^9 W/m^2

A sketch illustrating the b part of the question is shown on the first uploaded image

     looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

      from the diagram x can be calculated as follows

                      x = \sqrt{40^2 + 16^2}

                        = 43.0813 Km

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

                              G = \frac{E}{4 \pi r^2}

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

                            G= \frac{1.077*10^15}{4 \pi (431081.3^2)}

                                G =46.177\ kW/m^2

Generally the amount of energy absorbed can be mathematically represented as

                            Amount \ of \ energy \ absorbed \ (E_B ) = G * t

Where t is the time taken

       Therefore     E_B = 46.177 *10 = 461.77 KJ

       

                         

                       

             

   

6 0
3 years ago
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