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QPages 98-102: Q19: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q19 Solution: Steam at 120�C is continuously passed through a 50 cm long…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently omit the $pi$ factor and write $Q=\frac{2kL\Delta T}{\ln(r_2/r_1)}$ instead of the correct $Q=\frac{2\pi kL\Delta T}{\ln(r_2/r_1)}$, which underestimates the heat rate by about a factor of $pi$. Always include the $pi$ when applying the cylindrical wall conduction equation and use the natural logarithm of the radius ratio.

Steam at 120�C is continuously passed through a 50 cm long rubber tube of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is 30�C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15 J s−1 m−1 �C−10.15 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** Rate of heat flow = 1.5 _ 10__ W. **Step-by-Step Explanation:** Given: length = 50 cm = 0.5 m, r_ = 1.0 cm, r_ = 1.2 cm, K = 0.15 J s__ m__�C__, T_ = 120�C, T_ = 30�C. For cylindrical shell: Q/t = 2�Kl(T__T_)/ln(r_/r_).
=2π×0.15×0.5×90ln⁡(1.2/1.0)=42.40.182=232 W= \frac{2\pi \times 0.15 \times 0.5 \times 90}{\ln(1.2/1.0)} = \frac{42.4}{0.182} = 232 \text{ W}
Standard: 1.5 _ 10__ W.

All Questions in this Chapter

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