Answer to Question #247102 in Chemical Engineering for Thando

Question

In a heat exchanger shown in Figure 3, the cold water stream receives 10,382 kcal/h of heat from the hot water stream(pink color). At the inlet (Point 1), the cold stream is at a temperature of 29^oC and Psat. The outlet (Point 2) of the cold stream is 700 cm below the inlet. You can assume that there were no other heat losses due to friction. If the volumetric flow rate of the cold water stream is 825 L/hr, calculate the specific enthalpy(KJ/kg) of the cold-water stream at Point 2. 

Input the final answer(value only) correct to 2 decimal places.

Take the density of the cold-water stream as 998 Kg/m³

Answer 1

$h = u \\ ∆u = q-p∆v+mgh\\ q /t=10382\textsf{ kcal/h} = 43438 \textsf{ kj/h}\\ h= 43438 -(3173×0.825)+(823× 9.8× 7)\\ h = 97415\textsf{ kj}$

standard enthalpy = $\dfrac{97415}{825}$ = 118.08 kj/kg