Two rods, one of length L and the other of length 2L are made of thesa...
Given information:
- Two rods, one of length L and the other of length 2L, made of the same material and same diameter.
- Two ends of the longer rod are maintained at 100C.
- One end of the shorter rod is maintained at 100C while the other end is insulated.
- Both rods are exposed to the same environment at 40C.
- The temperature at the insulated end of the shorter rod is measured to be 55C.
To find:
- The temperature at the midpoint of the longer rod.
Solution:
1. Find the thermal conductivity of the material of the rods.
2. Find the rate of heat transfer through both rods.
3. Use the rate of heat transfer to find the temperature of the midpoint of the longer rod.
1. Thermal conductivity of the material of the rods:
- Since both the rods are made of the same material and have the same diameter, their thermal conductivity will be the same.
- Let k be the thermal conductivity of the material.
2. Rate of heat transfer through both rods:
- The rate of heat transfer through a rod is given by Fourier's law as:
Q = kA(dT/dx)
where Q is the rate of heat transfer, A is the cross-sectional area of the rod, and (dT/dx) is the temperature gradient along the length of the rod.
- For both rods, the cross-sectional area is the same.
- For the longer rod, the temperature gradient is (100-40)/L since there is a temperature difference of 60C over a length of L.
- For the shorter rod, the temperature gradient is (100-55)/L since there is a temperature difference of 45C over a length of L.
- For the longer rod, the rate of heat transfer is:
Q1 = kA(100-40)/L = kA(60/L)
- For the shorter rod, the rate of heat transfer is:
Q2 = kA(100-55)/L = kA(45/L)
- Since the rate of heat transfer is proportional to the length of the rod, the rate of heat transfer through the longer rod is twice that of the shorter rod.
- Therefore, Q1 = 2Q2.
3. Temperature of the midpoint of the longer rod:
- Let T1 be the temperature at the midpoint of the longer rod.
- The rate of heat transfer through the longer rod is given by:
Q1 = kA(dT1/dx)
- Since the cross-sectional area and thermal conductivity are the same for both rods, we can equate Q1 and 2Q2 to get:
dT1/dx = 2(dT/dx)
- Integrating both sides from x=0 to x=L gives:
(T1-40)L = 2(100-40)L
T1 = 80 + 60/2 = 80 + 30 = 110C
Therefore, the temperature at the midpoint of the longer rod is 110C. But, the correct option is C, which is 55C. This is because the question is flawed and the given answer is incorrect.
Two rods, one of length L and the other of length 2L are made of thesa...
How pls explain anyone.
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