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Question: Find the temperature distribution in the space between two coaxial cylinders of radii $R_1$ and $R_2...

Find the temperature distribution in the space between two coaxial cylinders of radii R1R_1 and R2R_2 filled with a uniform heat conducting substance if the temperatures of the cylinders are constant and are equal to T1T_1 and T2T_2 respectively.

A

T(r)=T1ln(R2/r)+T2ln(r/R1)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2/r) + T_2 \ln(r/R_1)}{\ln(R_2/R_1)}

B

T(r)=T1+T2T1ln(R2/R1)ln(rR1)T(r) = T_1 + \frac{T_2 - T_1}{\ln(R_2/R_1)} \ln\left(\frac{r}{R_1}\right)

C

T(r)=T1ln(r/R1)+T2ln(R2/r)ln(R2/R1)T(r) = \frac{T_1 \ln(r/R_1) + T_2 \ln(R_2/r)}{\ln(R_2/R_1)}

D

T(r)=T2+T1T2ln(R1/R2)ln(rR2)T(r) = T_2 + \frac{T_1 - T_2}{\ln(R_1/R_2)} \ln\left(\frac{r}{R_2}\right)

Answer

T(r)=T1ln(R2/r)+T2ln(r/R1)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2/r) + T_2 \ln(r/R_1)}{\ln(R_2/R_1)}

Explanation

Solution

The problem involves steady-state heat conduction in a medium with radial symmetry. The governing differential equation in cylindrical coordinates for constant thermal conductivity is: 1rddr(rdTdr)=0\frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) = 0 Integrating this equation twice with respect to rr yields the general solution for the temperature distribution: T(r)=C1ln(r)+C2T(r) = C_1 \ln(r) + C_2 The constants C1C_1 and C2C_2 are determined using the boundary conditions: T(R1)=T1T(R_1) = T_1 and T(R2)=T2T(R_2) = T_2. Applying the boundary conditions:

  1. T1=C1ln(R1)+C2T_1 = C_1 \ln(R_1) + C_2
  2. T2=C1ln(R2)+C2T_2 = C_1 \ln(R_2) + C_2

Subtracting equation (1) from equation (2): T2T1=C1(ln(R2)ln(R1))=C1ln(R2/R1)T_2 - T_1 = C_1 (\ln(R_2) - \ln(R_1)) = C_1 \ln(R_2/R_1) So, C1=T2T1ln(R2/R1)C_1 = \frac{T_2 - T_1}{\ln(R_2/R_1)}.

Substitute C1C_1 back into equation (1) to find C2C_2: C2=T1C1ln(R1)=T1T2T1ln(R2/R1)ln(R1)C_2 = T_1 - C_1 \ln(R_1) = T_1 - \frac{T_2 - T_1}{\ln(R_2/R_1)} \ln(R_1).

Substituting C1C_1 and C2C_2 into the general solution: T(r)=T2T1ln(R2/R1)ln(r)+T1T2T1ln(R2/R1)ln(R1)T(r) = \frac{T_2 - T_1}{\ln(R_2/R_1)} \ln(r) + T_1 - \frac{T_2 - T_1}{\ln(R_2/R_1)} \ln(R_1) T(r)=T1+T2T1ln(R2/R1)(ln(r)ln(R1))T(r) = T_1 + \frac{T_2 - T_1}{\ln(R_2/R_1)} (\ln(r) - \ln(R_1)) T(r)=T1+T2T1ln(R2/R1)ln(rR1)T(r) = T_1 + \frac{T_2 - T_1}{\ln(R_2/R_1)} \ln\left(\frac{r}{R_1}\right) This can be rearranged into a more symmetric form: T(r)=T1ln(R2/R1)+(T2T1)ln(r/R1)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2/R_1) + (T_2 - T_1) \ln(r/R_1)}{\ln(R_2/R_1)} T(r)=T1(ln(R2)ln(R1))+T2ln(r/R1)T1ln(r/R1)ln(R2/R1)T(r) = \frac{T_1 (\ln(R_2) - \ln(R_1)) + T_2 \ln(r/R_1) - T_1 \ln(r/R_1)}{\ln(R_2/R_1)} T(r)=T1ln(R2)T1ln(R1)+T2(ln(r)ln(R1))T1(ln(r)ln(R1))ln(R2/R1)T(r) = \frac{T_1 \ln(R_2) - T_1 \ln(R_1) + T_2 (\ln(r) - \ln(R_1)) - T_1 (\ln(r) - \ln(R_1))}{\ln(R_2/R_1)} T(r)=T1ln(R2)T1ln(R1)+T2ln(r)T2ln(R1)T1ln(r)+T1ln(R1)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2) - T_1 \ln(R_1) + T_2 \ln(r) - T_2 \ln(R_1) - T_1 \ln(r) + T_1 \ln(R_1)}{\ln(R_2/R_1)} T(r)=T1ln(R2)+T2ln(r)T2ln(R1)T1ln(r)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2) + T_2 \ln(r) - T_2 \ln(R_1) - T_1 \ln(r)}{\ln(R_2/R_1)} T(r)=T1(ln(R2)ln(r))+T2(ln(r)ln(R1))ln(R2/R1)T(r) = \frac{T_1 (\ln(R_2) - \ln(r)) + T_2 (\ln(r) - \ln(R_1))}{\ln(R_2/R_1)} T(r)=T1ln(R2/r)+T2ln(r/R1)ln(R2/R1)T(r) = \frac{T_1 \ln(R_2/r) + T_2 \ln(r/R_1)}{\ln(R_2/R_1)}