Training Convection & Dimensionless Numbers Newton’s Law of Cooling and Convection Basics
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Newton’s Law of Cooling and Convection Basics

24 min Convection & Dimensionless Numbers

Convection is the transfer of heat between a solid surface and a moving fluid, governed by Newton’s law of cooling: q = hA(Ts − T∞). The convective heat-transfer coefficient h encapsulates the complex interplay of fluid velocity, viscosity, and thermal properties. Unlike conduction’s material constant k, the coefficient h depends on the flow conditions and must be determined from correlations, experiments, or computational fluid dynamics. Mastering this concept is the gateway to all convective heat-transfer design.

Newton’s Law of Cooling and Convection Basics

Convection is heat transfer between a surface and a moving fluid. Unlike conduction (which depends only on material properties), convection depends on fluid motion, viscosity, density, and geometry. The convection coefficient $h$ captures all of these effects in a single number.

Newton’s Law of Cooling

$$\dot{Q}_{\text{conv}} = hA_s(T_s - T_\infty)$$

$h$ = convection heat transfer coefficient (W/(m²·K)), $A_s$ = surface area (m²), $T_s$ = surface temperature, $T_\infty$ = bulk fluid temperature.

Typical Values of $h$

Free convection, air: $5$–$25$ • Forced convection, air: $25$–$250$ • Free convection, water: $50$–$1000$ • Forced convection, water: $100$–$20{,}000$ • Boiling water: $2500$–$100{,}000$ • Condensing steam: $5000$–$100{,}000$

Forced vs. Natural Convection

Forced convection: Fluid motion driven by external means (pump, fan, wind). Natural (free) convection: Fluid motion driven by buoyancy from temperature-induced density differences.

Example 1 — Heated Plate in Air

A vertical plate at 80°C is exposed to still air at 20°C. $h = 8$ W/(m²·K), $A = 0.5$ m². Find $\dot{Q}$.

  1. Apply Newton's law of cooling:
  2. $q = hA(T_s - T_\infty)$.
  3. $$\dot{Q} = hA(T_s - T_\infty) = 8 \times 0.5 \times (80 - 20) = 240 \text{ W}$$
Example 2 — Cooling a CPU

A CPU dissipates 95 W over a 4 cm × 4 cm area. The heatsink maintains $h = 150$ W/(m²·K) with forced air at 35°C. What is the surface temperature?

  1. $A = 0.04 \times 0.04 = 0.0016$ m²
  2. $$T_s = T_\infty + \frac{\dot{Q}}{hA} = 35 + \frac{95}{150 \times 0.0016} = 35 + 395.8 = 430.8\text{°C}$$
  3. Way too hot! This is why heatsinks use fins to dramatically increase the effective area $A$.
Example 3 — Required $h$

An electronic enclosure (surface area 0.3 m²) must reject 200 W with a maximum $T_s = 60$°C in 25°C air. What minimum $h$ is needed?

  1. $$h = \frac{\dot{Q}}{A(T_s - T_\infty)} = \frac{200}{0.3 \times 35} = 19.0 \text{ W/(m²\cdot K)}$$
  2. This is at the upper end of free convection in air — a small fan would ensure adequate cooling.

Practice Problems

1. A hot pipe surface ($T_s = 120$°C, $A = 2$ m²) in 20°C air, $h = 15$. Find $\dot{Q}$.
2. A water-cooled plate ($h = 5000$, $A = 0.1$ m²) with $\dot{Q} = 10$ kW. Find $T_s - T_\infty$.
3. Condensing steam ($h = 10{,}000$) on a 1 m² surface, $T_s = 99$°C, $T_{\text{steam}} = 100$°C. $\dot{Q}$?
4. Why is $h$ for water so much larger than for air?
5. If $h$ doubles, what happens to $\dot{Q}$ (all else equal)?
6. Free convection, air: $h \approx 10$, $A = 5$ m², $\dot{Q} = 1500$ W. Find $\Delta T$.
Show Answer Key

1. $\dot{Q} = 15 \times 2 \times 100 = 3000$ W

2. $\Delta T = 10{,}000/(5000 \times 0.1) = 20$°C

3. $\dot{Q} = 10{,}000 \times 1 \times 1 = 10{,}000$ W = 10 kW

4. Water has much higher thermal conductivity and density, which enhance heat transport.

5. $\dot{Q}$ doubles (linearly proportional).

6. $\Delta T = 1500/(10 \times 5) = 30$°C

🌬 Newton's Law of Cooling
q = hA(Tₛ − T∞)
Heat flux q″
Thermal resistance 1/(hA)
Reynolds and Prandtl Numbers