A plastic pipe (k = 0.5 w/mK) carries a fluid such that the convective...
Solution:
Given data:
- Thermal conductivity of the pipe material, k = 0.5 W/mK
- Convective heat transfer coefficient, h = 300 W/m2K
- Average fluid temperature, T_f = 100 ˚C
- Inner diameter of the pipe, D_i = 3 cm = 0.03 m
- Outer diameter of the pipe, D_o = 4 cm = 0.04 m
- Heat transfer rate per unit pipe length, q = 500 W/m
We need to find the external pipe temperature, T_o.
Heat transfer rate per unit length of the pipe can be calculated using the following formula:
q = 2πkL (T_o - T_f)/(ln(D_o/D_i))
where L is the length of the pipe.
Rearranging the above equation, we get:
T_o = T_f + q ln(D_o/D_i)/(2πkLh)
Substituting the given values, we get:
T_o = 100 + (500 x ln(0.04/0.03))/(2π x 0.5 x L x 300)
T_o = 100 + 12.60/L
Therefore, the external pipe temperature is T_o = 100 + 12.60/L ˚C.
Explanation:
- The given problem is related to heat transfer through a pipe.
- Heat transfer rate per unit length of the pipe is given by q = 2πkL (T_o - T_f)/(ln(D_o/D_i)), where L is the length of the pipe.
- We can rearrange the above equation to find the external pipe temperature T_o.
- Substituting the given values in the equation, we can find the external pipe temperature.
- The external pipe temperature depends on the length of the pipe, as it appears in the denominator of the equation.
- As the length of the pipe increases, the external pipe temperature decreases.
- Therefore, to maintain a constant heat transfer rate, the length of the pipe should be chosen accordingly.
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.