Answer to Question #196806 in Mechanical Engineering for muhammed mudhasir

Question

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.

Answer 1

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

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

$q=c_p[T_2-T_1]_{DBT} \implies -82.39947 =1.005[T_2-31]_{DBT}$

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

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

$q=c_p[T_2-T_1]_{WBT} \implies -82.39947 =1.005[T_2-18.5]_{WBT}$

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