Training Convection & Dimensionless Numbers Nusselt Number and Forced Convection Correlations
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Nusselt Number and Forced Convection Correlations

24 min Convection & Dimensionless Numbers

Nusselt Number and Forced Convection Correlations

The Nusselt number is the dimensionless convection coefficient. All convection correlations express $Nu$ as a function of $Re$ and $Pr$, which we then convert to $h$.

Nusselt Number

$$Nu = \frac{hL_c}{k_f}$$

Ratio of convective to conductive heat transfer. $k_f$ = fluid thermal conductivity, $L_c$ = characteristic length. Once you know $Nu$, solve for $h$:

$$h = \frac{Nu \cdot k_f}{L_c}$$

Key Forced Convection Correlations

Internal flow, laminar ($Re_D < 2300$), fully developed:

$$Nu_D = 3.66 \quad(\text{constant } T_s) \qquad Nu_D = 4.36 \quad(\text{constant } q'')$$

Internal flow, turbulent ($Re_D > 10{,}000$): Dittus–Boelter:

$$Nu_D = 0.023\,Re_D^{0.8}\,Pr^{n}$$

$n = 0.4$ (heating), $n = 0.3$ (cooling). Valid for $0.6 \le Pr \le 160$, $L/D > 10$.

External flow, flat plate, laminar ($Re_x < 5 \times 10^5$):

$$Nu_x = 0.332\,Re_x^{1/2}\,Pr^{1/3}$$

External flow, flat plate, turbulent:

$$Nu_x = 0.0296\,Re_x^{4/5}\,Pr^{1/3}$$

Example 1 — Turbulent Pipe Flow (Dittus–Boelter)

Water ($k = 0.6$, $Pr = 7$, $\nu = 10^{-6}$) flows at 1.5 m/s through a 3 cm heated pipe. Find $h$.

$Re = 1.5 \times 0.03 / 10^{-6} = 45{,}000$ (turbulent)

$Nu = 0.023 \times 45{,}000^{0.8} \times 7^{0.4}$

$45{,}000^{0.8} = e^{0.8 \ln 45000} = e^{0.8 \times 10.714} = e^{8.571} = 5277$

$7^{0.4} = e^{0.4 \times 1.946} = e^{0.778} = 2.177$

$Nu = 0.023 \times 5277 \times 2.177 = 264.3$

$$h = \frac{264.3 \times 0.6}{0.03} = 5{,}286 \text{ W/(m²\cdot K)}$$

Example 2 — Laminar Pipe

Oil flows slowly through a 1 cm pipe at constant surface temperature. $Re = 500$, $k_f = 0.14$ W/(m·K). Find $h$.

Laminar, constant $T_s$: $Nu = 3.66$

$$h = \frac{3.66 \times 0.14}{0.01} = 51.2 \text{ W/(m²\cdot K)}$$

Example 3 — Flat Plate

Air at 20 m/s flows over a 50 cm heated plate. $\nu = 1.5 \times 10^{-5}$, $k = 0.026$, $Pr = 0.71$. Find average $h$ (laminar regime).

$Re_L = 20 \times 0.5 / (1.5 \times 10^{-5}) = 666{,}667$ (past transition)

For average $Nu$ over a plate with transition, use mixed correlation, but the first 75 cm is laminar. At $x = 0.5$ m: $Re_x = 666{,}667$. Since this exceeds $5 \times 10^5$, use turbulent at this point:

$Nu = 0.0296 \times 666{,}667^{0.8} \times 0.71^{1/3}$

$666{,}667^{0.8} = 59{,}340$

$0.71^{1/3} = 0.893$

$Nu = 0.0296 \times 59{,}340 \times 0.893 = 1569$

$h = 1569 \times 0.026 / 0.5 = 81.6$ W/(m²·K)

Practice Problems

1. Water in a 5 cm pipe, $Re = 80{,}000$, $Pr = 5$, $k = 0.62$. Find $h$ using Dittus–Boelter (heating).
2. Oil in a 2 cm pipe, $Re = 800$, constant $q''$. $k_f = 0.14$. Find $h$.
3. Air over a plate, $Re_x = 10^5$, $Pr = 0.71$, $k = 0.026$, $x = 0.3$ m. Find local $h$.
4. If $Re$ doubles in a turbulent pipe, by what factor does $Nu$ change? ($Nu \propto Re^{0.8}$)
5. A pipe has $Nu = 200$, $D = 4$ cm, $k_f = 0.6$. Find $h$.
6. Why is $Nu$ for turbulent flow much higher than for laminar?
Show Answer Key

1. $Nu = 0.023 \times 80{,}000^{0.8} \times 5^{0.4} = 0.023 \times 8710 \times 1.904 = 381.6$. $h = 381.6 \times 0.62/0.05 = 4732$ W/(m²·K)

2. Laminar, constant $q''$: $Nu = 4.36$. $h = 4.36 \times 0.14/0.02 = 30.5$ W/(m²·K)

3. $Nu = 0.332 \times (10^5)^{0.5} \times 0.71^{1/3} = 0.332 \times 316.2 \times 0.893 = 93.7$. $h = 93.7 \times 0.026/0.3 = 8.1$ W/(m²·K)

4. $2^{0.8} = 1.74$, so $Nu$ increases by a factor of 1.74.

5. $h = 200 \times 0.6 / 0.04 = 3000$ W/(m²·K)

6. Turbulence enhances mixing, which dramatically increases heat transfer between the wall and bulk fluid.

Reynolds and Prandtl Numbers Natural Convection and Grashof/Rayleigh Numbers