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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
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;
Hence, from;
, we have
5912500 = 90 × A × 341.67
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C
