Question

A hydraulic system operating at $80^\circ$C must be cooled to $45^\circ$C using a heat transfer oil. A water source of $30^\circ$C is available. The effectiveness of heat exchanger will be

• A. 0.7
• B. 0
• C. 1
• D. 0.5

Hint:

Effectiveness = (Actual temperature drop of hot fluid) $\div$ (Maximum possible drop). Always use inlet temperatures of both streams.

Answer 1

The Correct Option is A

Effectiveness ($\varepsilon$) of a heat exchanger is defined as:
\[ \varepsilon = \frac{\text{Actual heat transfer}}{\text{Maximum possible heat transfer}} \]
For a cooling process, the effectiveness is calculated using temperatures as:
\[ \varepsilon = \frac{T_{h,in} - T_{h,out}}{T_{h,in} - T_{c,in}} \]
Where:
$T_{h,in}$ = inlet temperature of the hot fluid = $80^\circ$C
$T_{h,out}$ = outlet temperature of the hot fluid = $45^\circ$C
$T_{c,in}$ = inlet temperature of the cold fluid = $30^\circ$C
Substitute the values into the formula:
\[ \varepsilon = \frac{80 - 45}{80 - 30} = \frac{35}{50} = 0.7 \]
Hence, the effectiveness of the heat exchanger is 0.7.