A two-fluid heat exchanger is pictured in the diagram. The two fluids are flowin

Question

A two-fluid heat exchanger is pictured in the diagram. The two fluids are flowing in opposite directions. This situation is referred to as counterflow. The m(dot)Cp value of the product for fluid 1 is twice the m(dot)Cp product for fluid 2. Suppose that fluid 2 enters the heat exchanger with a temperature of 95.3 C and leaves at a temperature of 47.1 C. With respect to fluid 1, its inlet temperature is 19.1 C. What is the exit temperature of fluid 1? There is no heat transfer at the external surfaces of the heat exchanger.

A two-fluid heat exchanger is pictured in the diagram. The two fluids are flowing in opposite directions. This situation is referred to as counterflow. The m(dot)Cp value of the product for fluid 1 is twice the m(dot)Cp product for fluid 2. Suppose that fluid 2 enters the heat exchanger with a temperature of 95.3 C and leaves at a temperature of 47.1 C. With respect to fluid 1, its inlet temperature is 19.1 C. What is the exit temperature of fluid 1? There is no heat transfer at the external surfaces of the heat exchanger.

Explanation / Answer

Rate of heat transffered by 2

Q2 = m(dot)2*Cp2*(Th,out - Th,in)

Q2 = m(dot)2*Cp2*(95.3 - 47.1) = m(dot)2*Cp2*48.2

Rate of heat recieved by 1

Q1 = m(dot)1*Cp1*(Tc,out - Tc,in)

Q1 = m(dot)1*Cp1*(Tc,out - 19.1)

Given

m(dot)1*Cp1 = 2*m(dot)2*Cp2 ------(1)

Since no heat trnasfer at external surfaces

Rate of heat transffered by 2 = Rate of heat recieved by 1

m(dot)1*Cp1*(Tc,out - 19.1) = m(dot)2*Cp2*48.2

From 1

2*m(dot)2*Cp2*(Tc,out - 19.1) = m(dot)2*Cp2*48.2

(Tc,out - 19.1) = 24.1

Tc,out = 43.2 deg

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