Question

A thin copper tube of outer radius 0.5 cm carries a liquid flowing at T = 100°C. The copper tube loses heat according to Newton's law with constant of proportionality 3 × 10⁻³cal/cm²sec°C. The temperature of surrounding is 20°C. Now we coat a layer with thermal conductivity 2.8 × 10⁻³cal/cm°Csec. The layer is 0.5 cm thick. Assume that outer surface of layer loses heat with same constant of proportionality: (Take : ln2 = 0.7)

Answer 1

The rate of heat loss per centimeter of the tube is approximately 0.86 cal/s per cm.

Answer 2

Solution:

1. Determine the radii:

   • Inner surface of insulation (i.e. copper tube's outer surface):

     $r_1 = 0.5\,\text{cm}$

   • Outer surface of insulation:

     $r_2 = 0.5 + 0.5 = 1.0\,\text{cm}$

2. Thermal Resistances:

   Let the length $L = 1\,\text{cm}$ (per unit length).

   (a) Conduction resistance through the insulation:

   For a cylindrical shell:

   $$R_{\text{cond}} = \frac{\ln(r_2/r_1)}{2\pi k L}$$

   Using $k = 2.8\times10^{-3}\,\text{cal/(cm·°C·sec)}$ and $\ln(2)=0.7$:

   $$R_{\text{cond}} = \frac{0.7}{2\pi (2.8\times10^{-3})} \approx \frac{0.7}{6.28\times2.8\times10^{-3}}.$$

   Now,
   $6.28\times2.8\times10^{-3} \approx 0.017584$
   So,

   $$R_{\text{cond}} \approx \frac{0.7}{0.017584} \approx 39.8\,\text{sec·°C/cal}.$$

   (b) Convection resistance at the outer surface of the insulation:

   The convective resistance is given by:

   $$R_{\text{conv}} = \frac{1}{hA} = \frac{1}{h(2\pi r_2 L)}$$

   Here, $h = 3\times10^{-3}\,\text{cal/(cm}^2\text{·sec·°C)}$ and $r_2 = 1.0\,\text{cm}$:

   $$R_{\text{conv}} = \frac{1}{3\times10^{-3}\,(2\pi \times 1)} = \frac{1}{3\times10^{-3}\times6.28} \approx \frac{1}{0.01884} \approx 53.1\,\text{sec·°C/cal}.$$

3. Total Thermal Resistance:

   Since resistances are in series:

   $$R_{\text{total}} = R_{\text{cond}} + R_{\text{conv}} \approx 39.8 + 53.1 \approx 92.9\,\text{sec·°C/cal}.$$

4. Heat Loss Calculation:

   The liquid in the copper tube is at $T = 100\,°\text{C}$ and ambient is $20\,°\text{C}$. The temperature difference is:

   $$\Delta T = 100 - 20 = 80\,°\text{C}.$$

   The heat loss per unit length is given by:

   $$Q = \frac{\Delta T}{R_{\text{total}}} \approx \frac{80}{92.9} \approx 0.86 \,\text{cal/s per cm}.$$

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Explanation (Minimal):

• Compute conduction resistance $R_{\text{cond}}=\frac{\ln(1/0.5)}{2\pi (2.8\times10^{-3})} \approx 40$.
• Compute convection resistance $R_{\text{conv}}=\frac{1}{3\times10^{-3}(2\pi\times1)} \approx 53$.
• Total resistance $\approx 40+53=93$.
• Heat loss: $Q=\frac{80}{93}\approx 0.86\,\text{cal/s per cm}$.