[ R_{conv,o} = \frac{1}{10 \cdot 2\pi \cdot 0.06} = \frac{1}{3.7699} = 0.2653 , \text{m·K/W} ]
In this post, we’ll walk through five example problems covering the three core modes of heat transfer. No fluff, just step-by-step solutions with practical insights.
[ R_{total} = 0.03183 + 0.00193 + 0.2653 = 0.2991 , \text{m·K/W} ] heat transfer example problems
[ R_{conv,i} = \frac{1}{100 \cdot 2\pi \cdot 0.05} = \frac{1}{31.416} = 0.03183 , \text{m·K/W} ]
Radiation dominates at high temperatures. Even with a 200 K difference, over 3 kW is transferred. Problem 4: Overall Heat Transfer Coefficient (Conduction + Convection) Scenario: A steam pipe (inner radius ( r_1 = 0.05 , \text{m} ), outer radius ( r_2 = 0.06 , \text{m} )) has ( k = 15 , \text{W/m·K} ). Inside: steam at ( T_{hot} = 200^\circ\text{C} ) with ( h_i = 100 , \text{W/m}^2\text{K} ). Outside: room air at ( T_{cold} = 25^\circ\text{C} ) with ( h_o = 10 , \text{W/m}^2\text{K} ). Find the heat loss per unit length ( Q/L ). [ R_{conv,o} = \frac{1}{10 \cdot 2\pi \cdot 0
[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h A_s}{\rho V c_p} t\right) ] For a sphere: ( A_s/V = 6/D ). [ \frac{100 - 25}{200 - 25} = \exp\left(-\frac{20 \cdot 6}{8933 \cdot 0.02 \cdot 385} t\right) ] [ \frac{75}{175} = 0.4286 = \exp(-0.001744 \cdot t) ] [ \ln(0.4286) = -0.8473 = -0.001744 , t ] [ t \approx 486 , \text{seconds} , (\approx 8.1 , \text{minutes}) ]
Newton’s law of cooling: [ Q = h , A , (T_s - T_\infty) ] [ 600 = h \cdot 0.5 \cdot (80 - 20) ] [ 600 = h \cdot 0.5 \cdot 60 = h \cdot 30 ] [ h = 20 , \text{W/m}^2\text{·K} ] Even with a 200 K difference, over 3 kW is transferred
Small, highly conductive objects reach thermal equilibrium very quickly. Final Thoughts These five examples cover the fundamentals: conduction through composites, convection from surfaces, radiation between black bodies, combined modes in cylinders, and transient cooling. The key to mastering heat transfer is not memorizing formulas—it’s understanding when to apply which resistance, and how simplifying assumptions (like lumped capacitance) can save hours of work.