Heat Transfer Example Problems May 2026

Let’s compute resistances per unit length:

The outside air convection is the bottleneck. Insulating the pipe would dramatically reduce heat loss. Problem 5: Lumped Capacitance – Transient Cooling Scenario: A copper sphere (diameter ( D = 0.02 , \text{m} )) at ( T_i = 200^\circ\text{C} ) is suddenly placed in air at ( T_\infty = 25^\circ\text{C} ) with ( h = 20 , \text{W/m}^2\text{K} ). Copper properties: ( \rho = 8933 , \text{kg/m}^3 ), ( c_p = 385 , \text{J/kg·K} ), ( k = 401 , \text{W/m·K} ). Check if lumped capacitance is valid. If yes, find the time to reach ( 100^\circ\text{C} ).

[ \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}) ] heat transfer example problems

Try modifying the numbers: add a contact resistance, change the emissivity, or switch to a different fluid. That’s where the real learning happens.

For a cylindrical system: [ \frac{Q}{L} = \frac{T_{hot} - T_{cold}}{\frac{1}{h_i (2\pi r_1)} + \frac{\ln(r_2/r_1)}{2\pi k} + \frac{1}{h_o (2\pi r_2)}} ] Let’s compute resistances per unit length: The outside

[ R_{conv,o} = \frac{1}{10 \cdot 2\pi \cdot 0.06} = \frac{1}{3.7699} = 0.2653 , \text{m·K/W} ]

First, compute the thermal resistances per unit area: [ R_A = \frac{0.2}{1.2} = 0.1667 , \text{m²·K/W} ] [ R_B = \frac{0.1}{0.15} = 0.6667 , \text{m²·K/W} ] [ R_{total} = 0.1667 + 0.6667 = 0.8334 , \text{m²·K/W} ] Copper properties: ( \rho = 8933 , \text{kg/m}^3

Heat transfer is the backbone of countless engineering applications—from designing a CPU cooler to building a power plant. But theory can only take you so far. To truly understand conduction, convection, and radiation, you have to work through the numbers.