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there will be the answer
Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K
![\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}](https://tex.z-dn.net/?f=%5CDelta%20%5Cbar%7Bt%7D_%7Ba%7D%20%3D%5Cfrac%7Bt_%7BA_%7B1%7D%7D%2Bt_%7BA_%7B2%7D%7D%7D%7B2%7D%20-%20%5Cfrac%7Bt_%7BB_%7B1%7D%7D%2Bt_%7BB_%7B2%7D%7D%7D%7B2%7D)
Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;
![\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K](https://tex.z-dn.net/?f=%5CDelta%20%5Cbar%7Bt%7D_%7Ba%7D%20%3D%5Cfrac%7B1100%2B733.33%7D%7B2%7D%20-%20%5Cfrac%7B300%2B850%7D%7B2%7D%20%3D%20341%5Ctfrac%7B2%7D%7B3%7D%20%5C%2C%20K%20%3D%20341.67%20%5C%2C%20K)
Hence, from;
, we have
5912500 = 90 × A × 341.67
![A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B5912500%20%20%7D%7B90%20%5Ctimes%20341.67%7D%20%3D%20192.3%20%5C%2C%20m%5E2)
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
Gross building area
Explanation:
The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.
There must be a photo for me to answer!
Answer:
first one, rest are okay ig.....
Explanation: