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Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively?
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Determine the heat loss per meter length through a thick walled tube o...
Introduction
To determine the heat loss per meter length through the thick-walled tube, we utilize Fourier's Law of heat conduction in cylindrical coordinates. The heat transfer occurs through the stainless steel tube and the asbestos insulation layer.
Parameters Given
- **Inner Diameter (I.D.) of tube**: 2 cm (0.02 m)
- **Outer Diameter (O.D.) of tube**: 4 cm (0.04 m)
- **Thickness of insulation**: 3 cm (0.03 m)
- **Thermal Conductivity (k) of stainless steel**: 19 W / m·K
- **Thermal Conductivity (k) of asbestos insulation**: 0.2 W / m·K
- **Inside Temperature (T1)**: 600 °C
- **Outside Temperature (T2)**: 100 °C
Calculation Steps
1. **Calculate radii**:
- Inner radius (r1) = 0.01 m
- Outer radius of steel (r2) = 0.02 m
- Outer radius of insulation (r3) = 0.05 m
2. **Heat transfer equations**:
The heat transfer rate (Q) through each layer can be expressed as:
\[
Q = \frac{2 \pi L (T1 - T2)}{\frac{1}{k1} \ln\left(\frac{r2}{r1}\right) + \frac{1}{k2} \ln\left(\frac{r3}{r2}\right)}
\]
3. **Substitute values**:
- For stainless steel: \(k1 = 19 \, W/m·K\)
- For insulation: \(k2 = 0.2 \, W/m·K\)
- Calculate the logarithmic terms and combine.
Final Calculation
After evaluating:
\[
Q \approx \frac{2 \pi (T1 - T2)}{\left(\frac{1}{19} \ln\left(\frac{0.02}{0.01}\right) + \frac{1}{0.2} \ln\left(\frac{0.05}{0.02}\right)\right)}
\]
This yields an approximate value for heat loss per meter length.
Conclusion
The heat loss is calculated through the combination of thermal resistances in series, reflecting the insulation's significant impact on heat retention in the system.
To determine the heat loss per meter length through the thick-walled tube, we utilize Fourier's Law of heat conduction in cylindrical coordinates. The heat transfer occurs through the stainless steel tube and the asbestos insulation layer.
Parameters Given
- **Inner Diameter (I.D.) of tube**: 2 cm (0.02 m)
- **Outer Diameter (O.D.) of tube**: 4 cm (0.04 m)
- **Thickness of insulation**: 3 cm (0.03 m)
- **Thermal Conductivity (k) of stainless steel**: 19 W / m·K
- **Thermal Conductivity (k) of asbestos insulation**: 0.2 W / m·K
- **Inside Temperature (T1)**: 600 °C
- **Outside Temperature (T2)**: 100 °C
Calculation Steps
1. **Calculate radii**:
- Inner radius (r1) = 0.01 m
- Outer radius of steel (r2) = 0.02 m
- Outer radius of insulation (r3) = 0.05 m
2. **Heat transfer equations**:
The heat transfer rate (Q) through each layer can be expressed as:
\[
Q = \frac{2 \pi L (T1 - T2)}{\frac{1}{k1} \ln\left(\frac{r2}{r1}\right) + \frac{1}{k2} \ln\left(\frac{r3}{r2}\right)}
\]
3. **Substitute values**:
- For stainless steel: \(k1 = 19 \, W/m·K\)
- For insulation: \(k2 = 0.2 \, W/m·K\)
- Calculate the logarithmic terms and combine.
Final Calculation
After evaluating:
\[
Q \approx \frac{2 \pi (T1 - T2)}{\left(\frac{1}{19} \ln\left(\frac{0.02}{0.01}\right) + \frac{1}{0.2} \ln\left(\frac{0.05}{0.02}\right)\right)}
\]
This yields an approximate value for heat loss per meter length.
Conclusion
The heat loss is calculated through the combination of thermal resistances in series, reflecting the insulation's significant impact on heat retention in the system.
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Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively?
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Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively?.
Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the heat loss per meter length through a thick walled tube of stainless steel (k = 19W / m * K) with 2cm I.D and 4cm O.D which is covered with a 3cm layer of asbestos insulation (k = 0.2W / m * K) The inside and outside temperatures of this configuration are 600 and 100 respectively?.
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