Heat Transfer · Chapter Eight

Internal Forced Convection

Internal convection is governed by confinement: velocity and temperature profiles develop from the inlet until the walls control the entire cross-section.

A sound solution joins a bulk-flow energy balance to a regime- and boundary-condition-specific correlation, while pressure drop supplies the hydraulic cost.

Learning objectives

Define mean velocity, bulk mean temperature, and hydraulic diameter physically.
Distinguish hydrodynamic and thermal development and test entry-length assumptions.
Select laminar or turbulent correlations by Reynolds number, wall condition, and validity range.
Predict outlet temperature, wall temperature, heat rate, pressure drop, and pumping power.
Audit property temperatures, signs, dimensions, and limiting behavior.

8–1

Introduction

Internal flow is driven through a closed passage. For noncircular ducts use \(D_h=4A_c/P_w\), where \(P_w\) is wetted perimeter. The hydraulic diameter transfers established duct correlations; it does not make every noncircular velocity profile identical to a circular one.

Internal-flow regime and mass flow

\[Re_{D_h}=\frac{\rho V_mD_h}{\mu},\qquad \dot m=\rho V_mA_c\]

Laminar: \(Re<2300\); transition: \(2300\lesssim Re\lesssim4000\); turbulent: \(Re>4000\). Avoid interpolating casually through transition.

Problem-solving strategy

Identify passage geometry, wetted perimeter, wall thermal condition, and desired thermal and hydraulic outputs.
Estimate a representative bulk temperature, evaluate properties there, and evaluate wall viscosity separately if the correlation requires it.
Compute \(D_h\), \(V_m\), \(Re\), \(Pr\), and entry lengths.
Select a correlation matching regime, development, geometry, roughness, and heating or cooling.
Couple \(h\) to the bulk energy balance and audit units, directions, range limits, and pumping cost.

8–2

Mean Velocity and Mean Temperature

The mean velocity preserves mass flow. Bulk mean temperature preserves enthalpy flow: for nearly constant \(c_p\), \(\dot m c_pT_m=\int_A\rho u c_pT\,dA\). It is not generally the arithmetic area-average temperature because faster fluid carries more energy.

Driving temperature

Local convection uses \(q_s''=h_x(T_s-T_m)\). The wall-to-bulk difference, not wall-to-centerline or wall-to-inlet difference, is the thermodynamic driving force at a section.

Physical meaning

As the fluid is heated, \(T_m\) rises downstream and the available wall-to-fluid temperature difference may shrink. Constant heat flux and constant wall temperature therefore produce different temperature histories and different fully developed Nusselt constants.

8–3

The Entrance Region

Hydrodynamically developed means the normalized velocity profile no longer changes with \(x\); thermally developed means the normalized temperature-profile shape no longer changes. It does not mean velocity or bulk temperature is spatially constant.

Approximate entry lengths

\[\frac{L_h}{D}\approx0.05Re,\qquad \frac{L_t}{D}\approx0.05RePr\quad\text{(laminar)},\qquad L_h,L_t\sim10D\quad\text{(turbulent)}\]

These are screening estimates. A developing-flow correlation should be used when the heated length is not comfortably longer than the relevant entry length.

8–4

General Thermal Analysis

For steady flow with negligible kinetic and potential energy changes and no shaft work within the heated section, \(\dot Q=\dot m c_p(T_o-T_i)\). Constant heat flux makes \(T_m(x)\) linear when properties are constant; a constant wall temperature produces exponential approach and can never be crossed by the fluid in this model.

Two canonical wall conditions

\[T_m(x)=T_i+\frac{q_s''P_wx}{\dot mc_p}\quad(q_s''=\text{constant})\]\[\frac{T_s-T_o}{T_s-T_i}=\exp\!\left(-\frac{\bar hP_wL}{\dot mc_p}\right)\quad(T_s=\text{constant})\]

EXAMPLE 1Uniform-flux warming of calibration oil

Problem. Calibration oil enters a 25 mm diameter, 3.0 m heated tube at 25 °C with \(\dot m=0.080\) kg/s. The wall supplies \(q_s''=4.00\) kW/m². Given \(c_p=2100\) J/kg·K, \(k=0.145\) W/m·K, and fully developed laminar flow, find outlet bulk temperature and outlet wall temperature.

Strategy

Uniform heat flux gives a direct energy balance; the fully developed laminar constant-flux value \(Nu=4.36\) then fixes wall-to-bulk difference.

Givens

\(D=0.025\) m, \(L=3.0\) m, \(q_s''=4000\) W/m², \(\dot m=0.080\) kg/s.
\(T_i=25^\circ\)C, \(c_p=2100\) J/kgK, \(k=0.145\) W/mK.

Physical/model interpretation

Steady, single-phase, constant properties, negligible axial conduction and heat loss; positive heat enters the oil.

Numbered derivation steps

1. Heat input. \(A_s=\pi DL=0.2356\ {\rm m^2}\), so \(\dot Q=q_s''A_s=942\) W.

2. Bulk rise. \(T_o-T_i=\dot Q/(\dot mc_p)=942/[(0.080)(2100)]=5.61\) K, hence \(T_o=30.6^\circ\)C.

3. Coefficient. \(h=Nu\,k/D=(4.36)(0.145)/0.025=25.3\ {\rm W/m^2K}\).

4. Wall temperature. \(T_{s,o}-T_o=q_s''/h=4000/25.3=158\) K, so \(T_{s,o}=189^\circ\)C.

Unit/sign/limit checks

Units: W divided by kg/s·J/kgK gives K; W/m² divided by W/m²K gives K.

Signs: Positive flux raises both bulk and wall temperature.

Limit: The high wall temperature follows from oil's low \(h\); if \(q_s''\to0\), both temperature rises vanish.

Final answer: \(\boxed{T_o=30.6^\circ{\rm C},\quad T_{s,o}\approx189^\circ{\rm C}}\)

Engineering insight

The mild bulk rise hides a severe wall temperature. Real approval requires checking oil degradation, property variation, heater limits, and whether fully developed conditions actually exist.

EXAMPLE 2Water warming in a constant-temperature jacket

Problem. Water at 20 °C enters an 18 mm diameter, 2.0 m tube at 0.120 kg/s. A condensing-steam jacket maintains the inner wall at 80 °C. If the applicable average inside coefficient is 900 W/m²K and \(c_p=4180\) J/kgK, determine outlet temperature and heat rate.

Strategy

The fixed wall temperature and known average \(h\) call for the exponential bulk-temperature relation, followed by the fluid energy balance.

Givens

\(T_s=80^\circ\)C, \(T_i=20^\circ\)C, \(D=0.018\) m, \(L=2.0\) m.
\(\dot m=0.120\) kg/s, \(h=900\) W/m²K, \(c_p=4180\) J/kgK.

Physical/model interpretation

Steady single-phase heating, constant properties and wall temperature, negligible external loss. Heat into water is positive.

Numbered derivation steps

1. Form conductance. \(A_s=\pi DL=0.1131\ {\rm m^2}\), so \(hA_s=101.8\) W/K.

2. Form capacity rate. \(\dot mc_p=(0.120)(4180)=501.6\) W/K and \(hA_s/(\dot mc_p)=0.2030\).

3. Solve outlet temperature. \(T_o=80-(80-20)e^{-0.2030}=31.0^\circ\)C.

4. Energy audit. \(\dot Q=\dot mc_p(T_o-T_i)=(501.6)(11.0)=5.52\) kW.

Unit/sign/limit checks

Units: \(hA/(\dot mc_p)\) is dimensionless and capacity rate times kelvin is W.

Signs: \(20

Limit: As \(L\to\infty\), the exponential tends to zero and \(T_o\to T_s\) without crossing it.

Final answer: \(\boxed{T_o=31.0^\circ{\rm C},\quad\dot Q=5.52\ {\rm kW\ to\ water}}\)

Engineering insight

Doubling area does not double heat indefinitely because the driving temperature shrinks downstream; the exponential model makes that diminishing return explicit.

8–5

Laminar Flow in Tubes

Exact fully developed circular-tube solutions give \(Nu=3.66\) for uniform wall temperature and \(Nu=4.36\) for uniform wall heat flux. These constants require both the stated boundary condition and thermal development. Developing laminar flow usually has a larger average coefficient and requires an entrance correlation, often with a wall-to-bulk viscosity correction.

Fully developed circular tube

\[Nu_D=3.66\ (T_s=\text{constant}),\qquad Nu_D=4.36\ (q_s''=\text{constant}),\qquad f_D=\frac{64}{Re}\]

Here \(f_D\) is the Darcy friction factor. Confusing it with the Fanning factor produces a fourfold pressure-drop error.

8–6

Turbulent Flow in Tubes

Turbulent mixing raises heat transfer and pressure loss. Dittus–Boelter is convenient for smooth, fully developed flow with modest property variation; Gnielinski covers a broader range and links \(Nu\) to friction factor. Roughness, strong viscosity changes, transition, or short tubes require a more suitable model.

Smooth-tube thermal and hydraulic relations

\[Nu=0.023Re^{0.8}Pr^n,\quad n=0.4\ {\rm(fluid\ heated)},\ n=0.3\ {\rm(fluid\ cooled)}\]\[f_D=(0.79\ln Re-1.64)^{-2},\quad \Delta P=f_D\frac{L}{D}\frac{\rho V_m^2}{2},\quad \dot W_{\rm fluid}=\dot V\Delta P\]

Dittus–Boelter: typically \(Re>10^4\), \(0.710\). Pump input is larger than \(\dot V\Delta P\) when efficiency is below one.

EXAMPLE 3Coolant-line thermal and pumping audit

Problem. Water coolant flows at 2.0 m/s through a smooth 30 mm diameter, 12 m tube whose wall cools the fluid. At the mean bulk temperature use \(\rho=995\) kg/m³, \(\mu=8.00\times10^{-4}\) Pa·s, \(k=0.620\) W/mK, \(Pr=5.4\). Estimate \(h\), frictional pressure drop, and hydraulic power.

Strategy

Check turbulent and length criteria, use Dittus–Boelter with the cooling exponent, then use a consistent Darcy friction factor for hydraulic cost.

Givens

\(V=2.0\) m/s, \(D=0.030\) m, \(L=12\) m; smooth tube.
Bulk properties supplied; fluid is being cooled, so \(n=0.3\).

Physical/model interpretation

Steady, incompressible, fully developed turbulent flow; fittings and elevation excluded. Pressure decreases in the flow direction.

Numbered derivation steps

1. Verify regime. \(Re=\rho VD/\mu=(995)(2.0)(0.030)/(8.00\times10^{-4})=7.46\times10^4\); \(L/D=400>10\).

2. Thermal result. \(Nu=0.023(7.46\times10^4)^{0.8}(5.4)^{0.3}=302\), so \(h=Nuk/D=6.24\times10^3\ {\rm W/m^2K}\).

3. Friction loss. \(f_D=(0.79\ln74600-1.64)^{-2}=0.0192\), hence \(\Delta P=(0.0192)(12/0.030)(995)(2.0^2/2)=15.3\) kPa.

4. Hydraulic power. \(\dot V=VA_c=2.0\pi(0.030)^2/4=1.414\times10^{-3}\ {\rm m^3/s}\); \(\dot W_{\rm fluid}=\dot V\Delta P=21.6\) W.

Unit/sign/limit checks

Units: In Darcy–Weisbach, dimensionless \(f_D(L/D)\) multiplies \(\rho V^2/2\), whose units are kg/(m·s²) = Pa. Then \(\dot V\Delta P\) has units m³/s·Pa = J/s = W, confirming both reported quantities.

Signs: Reported \(\Delta P=15.3\) kPa is a positive loss; static pressure drops along flow.

Limit: The valid \(Re\), \(Pr\), and \(L/D\) support the correlation; pump shaft power must exceed 21.6 W.

Final answer: \(\boxed{h=6.24\ {\rm kW/m^2K},\ \Delta P=15.3\ {\rm kPa},\ \dot W_{\rm fluid}=21.6\ {\rm W}}\)

Engineering insight

A complete loop design adds fittings, roughness, elevation, pump efficiency, fouling, and allowable erosion velocity; this calculation isolates only straight-tube behavior.

Chapter recap

Bulk quantities preserve mass and energy flow through nonuniform profiles.
Regime, entry length, geometry, wall condition, and property temperature control correlation selection.
Energy balances determine bulk-temperature change; correlations determine wall-to-bulk behavior.
Before moving on, be able to compute \(D_h\), \(Re\), entry lengths, \(Nu\), outlet temperature, pressure loss, and a specific audit.

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