The heat transfer by conduction through a thick cylinder (Q) is given by (where T1 = Higher temperature, T2 = Lower temperature, r1 = Inside radius, r2 = Outside radius, $$l$$ = Length of cylinder and k = Thermal conductivity)
A. $${\text{Q}} = \frac{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$
B. $${\text{Q}} = \frac{{2.3\log \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}$$
C. $${\text{Q}} = \frac{{2\pi \left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3l{\text{k}}\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$
D. $${\text{Q}} = \frac{{2\pi l{\text{k}}}}{{2.3\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$
Answer: Option A
Solution (By Examveda Team)
The heat transfer by conduction through a thick cylinder (Q) is given by$${\text{Q}} = \frac{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$
Where T1 = Higher temperature, T2 = Lower temperature, r1 = Inside radius, r2 = Outside radius, $$l$$ = Length of cylinder and k = Thermal conductivity.
This Question Belongs to Mechanical Engineering >> Heat And Mass Transfer
Join The Discussion