Question #196806

Specific volume 7.40 m3 of air per minute at 31˚C DBT and 18.5˚C WBT is passed over the cooling

coil whose surface temperature is 4.4˚C. The coil cooling capacity is 3.56 tons of refrigeration

under the given condition of air. Determine DBT and WBT of the air leaving the cooling coil.


Expert's answer

Mass flow rate = 7.40 m3 /min = 1.23 *7.40 =9.102 kg/min = 0.1517 kg/s

Heat observed by the coil from air

qˉ=12.50.1517=82.39947\bar{q}= \frac{12.5}{0.1517}=-82.39947 The negative sign shows heat is absorbed.

q=cp[T2T1]DBT    82.39947=1.005[T231]DBTq=c_p[T_2-T_1]_{DBT} \implies -82.39947 =1.005[T_2-31]_{DBT}

[T2]DBT=50.989520C[T_2]_{DBT}=-50.98952 ^0C

qˉ=12.50.1517=82.39947\bar{q}= \frac{12.5}{0.1517}=-82.39947 The negative sign shows heat is absorbed.

q=cp[T2T1]WBT    82.39947=1.005[T218.5]WBTq=c_p[T_2-T_1]_{WBT} \implies -82.39947 =1.005[T_2-18.5]_{WBT}

[T2]WBT=63.489520C[T_2]_{WBT}=--63.48952 ^0C


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