The logarithmic mean temperature difference ($${{\text{t}}_{\text{m}}}$$) is given by (where $$\Delta {{\text{t}}_1}$$ and $$\Delta {{\text{t}}_2}$$ are temperature differences between the hot and cold fluids at entrance and exit)
A. $${{\text{t}}_{\text{m}}} = \frac{{\Delta {{\text{t}}_1} - \Delta {{\text{t}}_2}}}{{{{\log }_{\text{e}}}\left( {\frac{{\Delta {{\text{t}}_1}}}{{\Delta {{\text{t}}_2}}}} \right)}}$$
B. $${{\text{t}}_{\text{m}}} = \frac{{{{\log }_{\text{e}}}\left( {\frac{{\Delta {{\text{t}}_1}}}{{\Delta {{\text{t}}_2}}}} \right)}}{{\Delta {{\text{t}}_1} - \Delta {{\text{t}}_2}}}$$
C. $${{\text{t}}_{\text{m}}} = \left( {\Delta {{\text{t}}_1} - \Delta {{\text{t}}_2}} \right){\log _{\text{e}}}\left( {\frac{{\Delta {{\text{t}}_1}}}{{\Delta {{\text{t}}_2}}}} \right)$$
D. $${{\text{t}}_{\text{m}}} = \frac{{{{\log }_{\text{e}}}\left( {\Delta {{\text{t}}_1} - \Delta {{\text{t}}_2}} \right)}}{{\frac{{\Delta {{\text{t}}_1}}}{{\Delta {{\text{t}}_2}}}}}$$
Answer: Option A
Solution (By Examveda Team)
The logarithmic mean temperature difference ($${{\text{t}}_{\text{m}}}$$) is given by $${{\text{t}}_{\text{m}}} = \frac{{\Delta {{\text{t}}_1} - \Delta {{\text{t}}_2}}}{{{{\log }_{\text{e}}}\left( {\frac{{\Delta {{\text{t}}_1}}}{{\Delta {{\text{t}}_2}}}} \right)}}$$Where $$\Delta {{\text{t}}_1}$$ and $$\Delta {{\text{t}}_2}$$ are temperature differences between the hot and cold fluids at entrance and exit.
This Question Belongs to Mechanical Engineering >> Heat And Mass Transfer
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