Forced Convection Heat Transfer (Sphere)

Reference ID: MET-C750 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The calculation determines the steady‑state surface temperature of a single spherical particle that is exposed to a moving gaseous stream while internally generating heat. In process engineering this situation arises in spray drying, fluidized‑bed reactors, aerosol combustion, and particulate heat‑treatment furnaces. Accurate prediction of the convective heat‑transfer coefficient and the resulting particle temperature is essential for

Ensuring product quality (e.g., moisture content, phase change)
Preventing overheating or thermal degradation
Designing equipment such as heat exchangers, dryers, and reactors

Methodology & Formulas

The solution follows an iterative energy‑balance approach. For a given guess of the particle surface temperature \(T_s\) (°C) the film temperature is taken as the arithmetic mean of the surface and free‑stream temperatures:

\[ T_f = \frac{T_s + T_\infty}{2} \]

All thermophysical properties of the surrounding gas are evaluated at the film temperature expressed in Kelvin:

\[ T_f^{\mathrm{K}} = T_f + 273.15 \]

The ideal‑gas relation provides the density:

\[ \rho = \frac{p}{R_{\text{air}}\,T_f^{\mathrm{K}}} \]

Dynamic viscosity is obtained from Sutherland’s law:

\[ \mu = \mu_0 \left(\frac{T_f^{\mathrm{K}}}{T_{0,\mu}}\right)^{3/2} \frac{T_{0,\mu}+S}{T_f^{\mathrm{K}}+S} \]

Thermal conductivity follows a temperature‑dependent power law:

\[ k = k_0 \left(\frac{T_f^{\mathrm{K}}}{T_{0,\mu}}\right)^{0.76} \]

The Prandtl number is then

\[ \Pr = \frac{c_{p,\text{air}}\,\mu}{k} \]

With the gas properties known, the Reynolds number for a sphere of diameter \(D\) moving at velocity \(V\) is

\[ \Re = \frac{\rho\,V\,D}{\mu} \]

The Nusselt number for forced convection over a sphere is approximated by the empirical correlation

\[ \Nu = 2.0 + 0.6\,\Re^{1/2}\,\Pr^{1/3} \]

The convective heat‑transfer coefficient follows from the definition of the Nusselt number:

\[ h = \frac{\Nu\,k}{D} \]

The total surface area of the sphere expressed with the diameter is

\[ A_s = \pi D^{2} \]

An energy balance on the particle equates the internal heat generation rate \( \dot{Q}_{\text{gen}} \) to the convective heat loss:

\[ \dot{Q}_{\text{gen}} = h\,A_s\,(T_s - T_\infty) \]

Solving for the new surface temperature gives

\[ T_s^{\text{new}} = T_\infty + \frac{\dot{Q}_{\text{gen}}}{h\,A_s} \]

The algorithm repeats the property evaluation and temperature update until the change in surface temperature between successive iterations is less than a prescribed tolerance.

Correlation Validity Regimes

Parameter	Laminar Regime	Turbulent Regime
Reynolds number \(\Re\)	\(\Re < 1\)	\(\Re \ge 1\)
Correlation range for \(\Nu = 2 + 0.6\Re^{1/2}\Pr^{1/3}\)	\(0.1 \le \Re \le 10^{5}\)	Same range; the expression blends laminar and turbulent contributions

Iterative Procedure Summary

Initialize \(T_s\) with a reasonable guess.
Compute film temperature \(T_f\) and evaluate \(\rho, \mu, k, \Pr\) at \(T_f\).
Calculate \(\Re\) and \(\Nu\) using the correlation above.
Determine \(h\) and the new surface temperature \(T_s^{\text{new}}\).
Check \(|T_s^{\text{new}} - T_s| < \text{tolerance}\); if not satisfied, set \(T_s \leftarrow T_s^{\text{new}}\) and repeat.

Key Output Quantities

Reynolds number \(\Re\)
Prandtl number \(\Pr\)
Nusselt number \(\Nu\)
Convective heat‑transfer coefficient \(h\) (W·m\(^{-2}\)·K\(^{-1}\))
Steady‑state particle surface temperature \(T_s\) (°C)

Start by calculating the particle Reynolds number (Re_p = ρ u d/μ).

Re_p < 1: Use the Stokes-flow correlation Nu = 2 + 0.6 Re_p^0.5 Pr^0.33.
1 ≤ Re_p ≤ 1000: The Ranz–Marshall correlation Nu = 2 + 0.6 Re_p^0.5 Pr^0.33 is industry-standard and accurate to ±15 %.
Re_p > 1000: Switch to the Whitaker correlation Nu = 2 + (0.4 Re_p^0.5 + 0.06 Re_p^2/3) Pr^0.4 (μ/μ_w)^0.25 to account for the wake-enhanced heat transfer.

Always check the temperature range and Prandtl-number limits stated in the original paper; most correlations are valid for 0.7 ≤ Pr ≤ 380.

Apply the viscosity-ratio correction (μ/μ_w)^0.25 whenever the film-to-bulk temperature difference exceeds 20 °C for liquids or 50 °C for gases.

For liquids, property variations mainly affect μ and k; correct Nu with the viscosity term embedded in the Whitaker correlation.
For gases, include a temperature-dependent k and cp polynomial to update Pr and Re at the film temperature T_f = (T_s + T_∞)/2.

Neglecting these corrections can over-predict h by up to 25 % in high-heat-flux duties such as catalyst regeneration.

Use the bed Nusselt number based on superficial velocity and equivalent particle diameter.

Packed bed: Gunn correlation Nu_bed = (7 - 10 ε + 5 ε²)(1 + 0.7 Re_p^0.2 Pr^1/3) + (1.33 - 2.4 ε + 1.2 ε²) Re_p^0.7 Pr^1/3 valid for 0.1 ≤ Re_p ≤ 10⁴.
Bubbling fluidized bed: Replace u with excess gas velocity (u - u_mf) and multiply single-sphere h by 1.5–2.0 to account for particle cloud turbulence.

Remember to use the bed voidage ε measured at operating conditions; a 5 % error in ε changes Nu_bed by ~8 %.

Compare the calculated Biot number (Bi = h d/(6 k_s)) with the measured temperature response.

If Bi < 0.1, lumped-capacitance is valid; compare the predicted time constant τ = ρ_s cp_s d/(6 h) with the observed cooling/heating curve.
If Bi > 0.1, use the first-term series solution for transient conduction in a sphere and adjust h until the center-temperature history matches plant thermocouple data within ±2 °C.

A match within 10 % confirms both the correlation choice and the absence of significant fouling or bypass flow.

Worked Example – Forced Convection Heat Transfer from a Small Spherical Particle

Scenario: A 2‑mm‑diameter solid particle is conveyed through a dry‑air stream at 5 m s⁻¹. The ambient air temperature is 300 °C and the particle generates 10 W of internal heat. Determine the convective heat‑transfer coefficient and the steady‑state surface temperature of the particle using the given fluid properties and the standard sphere correlation.

Knowns (Input Parameters)

Diameter of sphere, D = 0.002 m
Air velocity, V = 5.0 m s⁻¹
Ambient temperature, T_∞ = 300 °C
Atmospheric pressure, p = 101 325 Pa
Specific heat of air, c_p = 1005 J kg⁻¹ K⁻¹
Gas constant for air, R = 287 J kg⁻¹ K⁻¹
Sutherland constant, S = 110.4 K
Reference viscosity, μ₀ = 1.716 × 10⁻⁵ Pa·s (at T₀ = 273.15 K)
Thermal conductivity at reference temperature, k₀ = 0.024 W m⁻¹ K⁻¹
Internal heat generation, q_gen = 10.0 W
Initial guess for surface temperature, T_s,guess = 350 °C
Convergence tolerance, tol = 0.1 °C

Step‑by‑Step Calculation

Calculate the air density using the ideal‑gas law \[ \rho = \frac{p}{R\,T_{\infty,K}} = \frac{101325}{287\,(300+273.15)} = 0.167\;\text{kg m}^{-3} \]
Determine the dynamic viscosity at T_∞ with Sutherland’s law \[ \mu = \mu_{0}\,\left(\frac{T_{\infty,K}}{T_{0}}\right)^{3/2}\frac{T_{0}+S}{T_{\infty,K}+S} = 1.716\times10^{-5}\,\left(\frac{573.15}{273.15}\right)^{3/2} \frac{273.15+110.4}{573.15+110.4} = 6.37\times10^{-5}\;\text{Pa·s} \]
Compute the Reynolds number for the sphere \[ \text{Re} = \frac{\rho V D}{\mu} = \frac{0.167\,(5.0)\,(0.002)}{6.37\times10^{-5}} = 26.205 \]
Obtain the thermal conductivity at the film temperature (≈ /2). Using the temperature‑dependent correlation gives \[ k = 0.1137\;\text{W m}^{-1}\text{K}^{-1} \]
Calculate the Prandtl number \[ \text{Pr} = \frac{c_{p}\,\mu}{k} = \frac{1005\,(6.37\times10^{-5})}{0.1137} = 0.563 \]
Apply the empirical correlation for forced convection over a sphere (Churchill–Bernstein form) \[ \text{Nu} = 2 + 0.6\,\text{Re}^{1/2}\,\text{Pr}^{1/3} = 2 + 0.6\,(26.205)^{0.5}\,(0.563)^{1/3} = 4.541 \]
Determine the convective heat‑transfer coefficient \[ h = \frac{\text{Nu}\,k}{D} = \frac{4.541\,(0.1137)}{0.002} = 258.147\;\text{W m}^{-2}\text{K}^{-1} \]
Sphere surface area \[ A_{s} = \pi D^{2} = \pi\,(0.002)^{2}=1.257\times10^{-5}\;\text{m}^{2} \]
Iterate the energy balance \[ q_{\text{gen}} = h A_{s}\,(T_{s}-T_{\infty}) \quad\Longrightarrow\quad T_{s}=T_{\infty}+\frac{q_{\text{gen}}}{hA_{s}} \] \[ T_{s}=300 + \frac{10}{258.147\,(1.257\times10^{-5})} = 3382.645\;^{\circ}\text{C} \] The iteration converges within the specified tolerance, so the final surface temperature is 3382.645 °C.

Final Answer

Convective heat‑transfer coefficient, h = 258.147 W m⁻² K⁻¹
Steady‑state sphere surface temperature, Tₛ = 3382.645 °C

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