Answer:
The effective overall heat transfer coefficient for the combination, ![U = 0.19 Btu/hr-ft^2 -F](https://tex.z-dn.net/?f=U%20%3D%200.19%20Btu%2Fhr-ft%5E2%20-F)
Explanation:
According to the question, the total area includes the areas of the wall, window and door.
Total area, A = 100 * 10 = 1000 ft²
Area of the window, ![A_{wind} = 80 ft^2](https://tex.z-dn.net/?f=A_%7Bwind%7D%20%3D%2080%20ft%5E2)
Area of the door, ![A_{d} = 8 * 2.5 = 20 ft^2](https://tex.z-dn.net/?f=A_%7Bd%7D%20%3D%208%20%2A%202.5%20%3D%2020%20ft%5E2)
Area of the wall, ![A_{wall} = A - (A_{d} + A_{wind} )](https://tex.z-dn.net/?f=A_%7Bwall%7D%20%3D%20A%20-%20%28A_%7Bd%7D%20%2B%20A_%7Bwind%7D%20%29)
![A_{wall} = 1000 - ( 80 + 20)\\A_{wall} = 900 ft^2](https://tex.z-dn.net/?f=A_%7Bwall%7D%20%3D%201000%20-%20%28%2080%20%2B%2020%29%5C%5CA_%7Bwall%7D%20%3D%20900%20ft%5E2)
<u>Heat transfer coefficients:</u>
For wall, ![h_{wall} = 0.2Btu/hr-ft^2-F](https://tex.z-dn.net/?f=h_%7Bwall%7D%20%3D%200.2Btu%2Fhr-ft%5E2-F)
For window, ![h_{wind} = 0.1Btu/hr-ft^2-F](https://tex.z-dn.net/?f=h_%7Bwind%7D%20%3D%200.1Btu%2Fhr-ft%5E2-F)
For door, ![h_{d} = 0.1Btu/hr-ft^2-F](https://tex.z-dn.net/?f=h_%7Bd%7D%20%3D%200.1Btu%2Fhr-ft%5E2-F)
To get the overall heat - transfer coefficient for the combination:
![AU = A_{wall} h_{wall} + A_{wind} h_{wind} + A_{d} h_{d}\\1000U = (900*0.2) + (80*0.1) + (20*0.1)\\1000U = 190\\U = 190/1000\\U = 0.19 Btu/hr-ft^2 -F](https://tex.z-dn.net/?f=AU%20%3D%20A_%7Bwall%7D%20h_%7Bwall%7D%20%2B%20A_%7Bwind%7D%20h_%7Bwind%7D%20%2B%20A_%7Bd%7D%20h_%7Bd%7D%5C%5C1000U%20%3D%20%28900%2A0.2%29%20%2B%20%2880%2A0.1%29%20%2B%20%2820%2A0.1%29%5C%5C1000U%20%3D%20190%5C%5CU%20%3D%20190%2F1000%5C%5CU%20%3D%200.19%20Btu%2Fhr-ft%5E2%20-F)
Answer:
Hans Christian Ørsted
One of the most important discoveries relating to current was made accidentally by Hans Christian Ørsted in 1820, when, while preparing a lecture, he witnessed the current in a wire disturbing the needle of a magnetic compass.
Explanation:
is that what you were looking for
Answer:
(a) 561.12 W/ m² (b) 196.39 MW
Explanation:
Solution
(a) Determine the energy and power of the wave per unit area
The energy per unit are of the wave is defined as:
E = 1 /16ρgH²
= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²
=3927. 83 J/m²
Thus,
The power of the wave per unit area is,
P = E/ t
= 3927. 83 J/m² / 7 s = 561.12 W/ m²
(b) The average and work power output of a wave power plant
W = E * л * A
= 3927. 83 J/m² * 0.35 * 1 *10^6 m²
= 1374.74 MJ
Then,
The power produced by the wave for one km²
P = P * л * A
= 5612.12 W/m² * 0.35 * 1* 10^6 m²
=196.39 MW