---
title: "Laminar forced convection over a heated flat plate"
subject: "Heat Transfer"
chapter: 3
key: "semester_5/heat_transfer:laminar-flow-over-heated-plate-air"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/laminar-flow-over-heated-plate-air
site: BCT Engineering
---

# Laminar forced convection over a heated flat plate

Heat Transfer · Chapter 3: Convection Heat Transfer

- Asked in: year not given, 8 marks
- Video: [Convective Heat Transfer over a Flat Plate - Example Problem](https://www.youtube.com/watch?v=P2eN2EApGEY) — LearnChemE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/laminar-flow-over-heated-plate-air
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-3

## Question

Air at 20 °C flows at 3 m/s parallel to a flat plate 1 m long and 0.6 m wide whose surface is at 100 °C. Using properties at the film temperature (nu = 18.97 x 10^-6 m^2/s, k = 0.02896 W/m K, Pr = 0.7202), determine (a) the Reynolds number at the trailing edge and the type of flow, (b) the average heat transfer coefficient, (c) the heat lost by the plate, and (d) the hydrodynamic and thermal boundary-layer thickness at the trailing edge. Use Nu_L = 0.664 Re^0.5 Pr^(1/3) and delta = 5x/sqrt(Re_x).

## Short answer

Using laminar correlation Nu_L = 0.664 Re^0.5 Pr^(1/3) and boundary layer formulas: Re_L = 1.58e5 (laminar), average h = 6.85 W/m^2K, heat loss Q = 329 W, hydrodynamic thickness delta = 12.6 mm, thermal thickness delta_t = 14.0 mm.

## Answer

Film temperature $T_f = (100+20)/2 = 60$ °C, the temperature at which the given properties apply.

#### (a) Reynolds number

$$
Re_L = \frac{u_\infty L}{\nu} = \frac{3\times1}{18.97\times10^{-6}} = 1.58\times10^5
$$

This is less than $5\times10^5$, so the flow over the whole plate is **laminar** and the laminar correlation is valid.

#### (b) Average heat transfer coefficient

$$
\begin{aligned}
Nu_L &= 0.664\,Re_L^{0.5}\,Pr^{1/3}\\
&= 0.664\,(1.58\times10^5)^{0.5}\,(0.7202)^{1/3}\\
&= 0.664\,(397.7)(0.8962) = 236.7
\end{aligned}
$$

$$
\bar h = \frac{Nu_L\,k}{L} = \frac{236.7\times0.02896}{1} = 6.85\ \text{W/m}^2\text{K}
$$

#### (c) Heat loss

$$
Q = \bar h A (T_s-T_\infty) = 6.85\times(1\times0.6)\times(100-20) = 329\ \text{W}
$$

(Heat is lost from one side. For both faces the value doubles.)

#### (d) Boundary layer thickness at $x = 1$ m

$$
\delta = \frac{5x}{\sqrt{Re_x}} = \frac{5(1)}{397.7} = 0.01257\ \text{m} = 12.6\ \text{mm}
$$

$$
\delta_t = \frac{\delta}{Pr^{1/3}} = \frac{12.57}{0.8962} = 14.0\ \text{mm}
$$

Since $Pr<1$ the thermal layer is slightly thicker than the velocity layer.

**Answer:** $Re_L = 1.58\times10^5$ (laminar); $\bar h = 6.85\ \text{W/m}^2$K; $Q = 329$ W; $\delta = 12.6$ mm, $\delta_t = 14.0$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Re_L = 1.58 x 10^5 means:
   A. Laminar over the whole plate
   B. Turbulent over the whole plate
   C. Transition at mid-plate
   D. Cannot be decided
   Answer: A. It is below 5 x 10^5. (Key point: Flow type)

2. Average h comes from:
   A. h = Nu k x
   B. h = Nu_L k / L with Nu = 0.664 Re^0.5 Pr^(1/3)
   C. h = Nu L / k
   D. h = 0.664 Re Pr
   Answer: B. Nu_L = hL/k, so h = Nu k/L = 6.85 W/m^2K. (Key point: Correlation)

3. Heat lost from one side of the plate is:
   A. About 658 W
   B. About 6.85 W
   C. About 329 W
   D. About 82 W
   Answer: C. Q = 6.85 x 0.6 x 80 = 329 W. (Key point: Heat loss)

4. Why is delta_t slightly larger than delta here?
   A. Pr > 1
   B. Flow is turbulent
   C. Plate is hotter than air
   D. Pr < 1, so delta_t = delta/Pr^(1/3)
   Answer: D. With Pr = 0.72 the ratio is 0.896, giving 14.0 mm vs 12.6 mm. (Key point: Thermal layer)

## Related in Heat Transfer

- [Natural convection from a vertical plate](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Anatural-convection-vertical-plate-heat-loss) — Heat Transfer, chapter 3
- [Natural and forced convection compared](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Afree-versus-forced-convection) — Heat Transfer, chapter 3
- [Dimensional analysis of forced convection](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Abuckingham-pi-forced-convection) — Heat Transfer, chapter 3
- [Heat loss from a cylinder in cross flow](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Across-flow-air-cylinder-heat-loss) — Heat Transfer, chapter 3
- [Condensation of steam on a vertical plate](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Afilm-condensation-vertical-plate-numerical) — Heat Transfer, chapter 5
- [Newton's law of cooling and convection coefficient](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Anewtons-law-cooling-convection-coefficient) — Heat Transfer, chapter 3

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