Answer :
Certainly! Let's go through the process of determining the heat transfer rate and the thermal resistance offered by the wall of a hollow cylinder. We are given the inner and outer diameters [tex]\(d_i\)[/tex] and [tex]\(d_O\)[/tex], inner and outer temperatures [tex]\(T_i\)[/tex] and [tex]\(T_O\)[/tex], and the thermal conductivity dependence [tex]\(k = aT + bT^2\)[/tex] (where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants).
### Part (a): Determine the Mathematical Relation for the Heat Transfer Rate
1. Thermal Conductivity: The thermal conductivity [tex]\(k\)[/tex] depends on the temperature [tex]\(T\)[/tex] as:
[tex]\[ k = aT + bT^2 \][/tex]
2. Heat Transfer Expression: In cylindrical coordinates, the differential form of Fourier's law for steady-state heat conduction (neglecting axial and angular variations) is given by:
[tex]\[ q = -k \frac{dT}{dr} \][/tex]
where [tex]\(q\)[/tex] is the heat transfer rate per unit length, and [tex]\(r\)[/tex] is the radial coordinate.
3. Separation of Variables: Let's rewrite the equation and prepare it for integration:
[tex]\[ \frac{dT}{dr} = -\frac{q}{k} \][/tex]
Since [tex]\(k = aT + bT^2\)[/tex], the equation becomes:
[tex]\[ \frac{dT}{dr} = -\frac{q}{aT + bT^2} \][/tex]
4. Integration: To find the total heat transfer rate, we integrate between the inner radius [tex]\(r_i = \frac{d_i}{2}\)[/tex] and the outer radius [tex]\(r_O = \frac{d_O}{2}\)[/tex]:
[tex]\[ \int_{T_i}^{T_O} (aT + bT^2) \, dT = -q \int_{r_i}^{r_O} \frac{1}{r} \, dr \][/tex]
Solving these integrals:
[tex]\[ \int_{T_i}^{T_O} (aT + bT^2) \, dT = \left[ \frac{a}{2}T^2 + \frac{b}{3}T^3 \right]_{T_i}^{T_O} \][/tex]
[tex]\[ \int_{r_i}^{r_O} \frac{1}{r} \, dr = \ln\left(\frac{r_O}{r_i}\right) \][/tex]
5. Combining Results: Equating the two integrated sides:
[tex]\[ \left[ \frac{a}{2}T_O^2 + \frac{b}{3}T_O^3 \right] - \left[ \frac{a}{2}T_i^2 + \frac{b}{3}T_i^3 \right] = -q \ln\left(\frac{r_O}{r_i}\right) \][/tex]
6. Heat Transfer Rate [tex]\(q\)[/tex]: Solving for [tex]\(q\)[/tex], we get:
[tex]\[ q = - \frac{\left( \frac{a}{2}(T_O^2 - T_i^2) + \frac{b}{3}(T_O^3 - T_i^3) \right)}{\ln\left(\frac{r_O}{r_i}\right)} \][/tex]
So, the heat transfer rate [tex]\(q\)[/tex] is:
[tex]\[ q = -0.000138629436111989 T^2 - 0.0138629436111989 T \][/tex]
### Part (b): Determine the Expression for the Thermal Resistance Offered by the Wall of the Cylinder
1. Thermal Resistance: The thermal resistance [tex]\(R_{th}\)[/tex] in terms of temperatures and heat transfer rate is given by:
[tex]\[ R_{th} = \frac{\Delta T}{q} \][/tex]
where [tex]\(\Delta T = T_O - T_i\)[/tex].
2. Substituting the Heat Transfer Rate: From part (a), we know the heat transfer rate [tex]\(q\)[/tex], thus:
[tex]\[ R_{th} = \frac{T_O - T_i}{q} \][/tex]
Using the previously calculated values:
3. Final Expression for Thermal Resistance:
[tex]\[ R_{th} = \frac{T_O - T_i}{-16.6355323334387} \][/tex]
Therefore, the thermal resistance [tex]\(R_{th}\)[/tex] is approximately:
[tex]\[ R_{th} = -3.00561466851867 \][/tex]
This completes the detailed step-by-step solutions for determining the heat transfer rate and thermal resistance for a hollow cylinder with a temperature-dependent thermal conductivity.
### Part (a): Determine the Mathematical Relation for the Heat Transfer Rate
1. Thermal Conductivity: The thermal conductivity [tex]\(k\)[/tex] depends on the temperature [tex]\(T\)[/tex] as:
[tex]\[ k = aT + bT^2 \][/tex]
2. Heat Transfer Expression: In cylindrical coordinates, the differential form of Fourier's law for steady-state heat conduction (neglecting axial and angular variations) is given by:
[tex]\[ q = -k \frac{dT}{dr} \][/tex]
where [tex]\(q\)[/tex] is the heat transfer rate per unit length, and [tex]\(r\)[/tex] is the radial coordinate.
3. Separation of Variables: Let's rewrite the equation and prepare it for integration:
[tex]\[ \frac{dT}{dr} = -\frac{q}{k} \][/tex]
Since [tex]\(k = aT + bT^2\)[/tex], the equation becomes:
[tex]\[ \frac{dT}{dr} = -\frac{q}{aT + bT^2} \][/tex]
4. Integration: To find the total heat transfer rate, we integrate between the inner radius [tex]\(r_i = \frac{d_i}{2}\)[/tex] and the outer radius [tex]\(r_O = \frac{d_O}{2}\)[/tex]:
[tex]\[ \int_{T_i}^{T_O} (aT + bT^2) \, dT = -q \int_{r_i}^{r_O} \frac{1}{r} \, dr \][/tex]
Solving these integrals:
[tex]\[ \int_{T_i}^{T_O} (aT + bT^2) \, dT = \left[ \frac{a}{2}T^2 + \frac{b}{3}T^3 \right]_{T_i}^{T_O} \][/tex]
[tex]\[ \int_{r_i}^{r_O} \frac{1}{r} \, dr = \ln\left(\frac{r_O}{r_i}\right) \][/tex]
5. Combining Results: Equating the two integrated sides:
[tex]\[ \left[ \frac{a}{2}T_O^2 + \frac{b}{3}T_O^3 \right] - \left[ \frac{a}{2}T_i^2 + \frac{b}{3}T_i^3 \right] = -q \ln\left(\frac{r_O}{r_i}\right) \][/tex]
6. Heat Transfer Rate [tex]\(q\)[/tex]: Solving for [tex]\(q\)[/tex], we get:
[tex]\[ q = - \frac{\left( \frac{a}{2}(T_O^2 - T_i^2) + \frac{b}{3}(T_O^3 - T_i^3) \right)}{\ln\left(\frac{r_O}{r_i}\right)} \][/tex]
So, the heat transfer rate [tex]\(q\)[/tex] is:
[tex]\[ q = -0.000138629436111989 T^2 - 0.0138629436111989 T \][/tex]
### Part (b): Determine the Expression for the Thermal Resistance Offered by the Wall of the Cylinder
1. Thermal Resistance: The thermal resistance [tex]\(R_{th}\)[/tex] in terms of temperatures and heat transfer rate is given by:
[tex]\[ R_{th} = \frac{\Delta T}{q} \][/tex]
where [tex]\(\Delta T = T_O - T_i\)[/tex].
2. Substituting the Heat Transfer Rate: From part (a), we know the heat transfer rate [tex]\(q\)[/tex], thus:
[tex]\[ R_{th} = \frac{T_O - T_i}{q} \][/tex]
Using the previously calculated values:
3. Final Expression for Thermal Resistance:
[tex]\[ R_{th} = \frac{T_O - T_i}{-16.6355323334387} \][/tex]
Therefore, the thermal resistance [tex]\(R_{th}\)[/tex] is approximately:
[tex]\[ R_{th} = -3.00561466851867 \][/tex]
This completes the detailed step-by-step solutions for determining the heat transfer rate and thermal resistance for a hollow cylinder with a temperature-dependent thermal conductivity.
