In certain HE, both the fluids have identical mass flow rate-spec

In certain HE, both the fluids have identical mass flow rate-spec
| In certain HE, both the fluids have identical mass flow rate-specific heat product. The hot fluid enters at 76°C and leaves at 47°C, and the cold fluid entering at 28°C leave at 55°C. The effectiveness of the HE is

A. 0.16

B. 0.58

C. 0.72

D. 1.0

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

Concept:

The heat exchanger effectiveness (ϵ) is defined as the ratio of actual heat transfer to the maximum possible heat transfer.

\(\epsilon = \frac{{{Q_{act}}}}{{{Q_{max}}}}\)

Qact = ṁh Cph (Th1 – Th2) = ṁc Ċpc (Tc2 – Tc1)

Fluid Capacity Rate, C : Ch = ṁh Cph, Cc = ṁCpc

Qmax = C(Th1 – Tc1) or Cc (Th1 – Tc1)

Qmax = Cmin (Th1 – Tc1)

\(\epsilon = \frac{{{C_h}\left( {{T_{h1}} - {T_{h2}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}} = \frac{{{C_c}\left( {{T_{c2}} - {T_{c1}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}}\)

Calculation:

Given:

Th1 = 76°C, Th2=47°C, Tc1 = 28°C, Tc2 = 55°C
Ch = Cc

\(\epsilon = \frac{{\left( {{T_{h1}} - {T_{h2}}} \right)}}{{\left( {{T_{h1}} - {T_{c1}}} \right)}} \)

\( \epsilon = \frac{{76 - 47}}{{76 - 28}} = \frac{{29}}{{48}} = 0.60 \simeq 0.58\)