Uncover valuable information and solutions with IDNLearn.com's extensive Q&A platform. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
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.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
