Question
Download Solution PDFThe effectiveness of a counter-flow heat exchanger has been estimated as 0.25. Hot gases enter at 200°C and leave at 75°C. Cooling air enters at 40°C. The temperature of the air leaving the unit will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Heat exchanger effectiveness:
\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}}\)
\({{\rm{Q}}_{{\rm{actul}}}} = {{\rm{\dot m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right) = {{\rm{\dot m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)
\({{\rm{Q}}_{{\rm{max}}}} = {{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)\)
\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{c}}1}}} \right)}} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} \;-\; {{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{c}}1}}} \right)}}\)
Calculation:
Given:
ϵ = 0.25, Th1 = 200°C, Th2 = 75°C, Tc1 = 40°C
Heat capacities of gases and air are not given, so for calculating cold fluid outlet temperature, we have to assume the heat capacity of cold air is minimum.
Because then only, the effectiveness formula includes cold air outlet temperature.
\(ϵ =\frac{\left( {{T}_{ce}}-{{T}_{ci}} \right)}{\left( {{T}_{hi}}-{{T}_{ci}} \right)}\)
\(0.25=\frac{{{T}_{ce}}-40}{200-40}\)
Tce - 40 = 40
Tce = 80°CLast updated on May 28, 2025
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