CHE-312 · Lesson 7 of 12 · 23 min

Internal forced convection

Inside every pipe and tube at Northgate, a fluid flows and exchanges heat with the wall: district water in the transmission main, steam in the converters, glycol in the chilled-water loops, and oil in the heat-transfer systems of two industrial customers on the network. Internal flow differs from external flow in one essential way. The boundary layers grow from the walls until they meet at the centre of the tube, after which the flow is fully developed and confined. The fluid's mean temperature then changes along the tube as it gains or loses heat, so the temperature difference driving convection is not constant. This lesson develops the tools for internal flow: the mean temperature, entry lengths, the energy balance along a tube and the correlations for laminar and turbulent flow.

Hydrodynamic and thermal entry

Fluid entering a tube with a uniform velocity develops a boundary layer on the wall. In laminar flow the boundary layer fills the tube after a hydrodynamic entry length of about \(0.05\,\text{Re}_D D\), and the velocity profile becomes parabolic. Similarly, a thermal boundary layer grows from a heated or cooled wall and fills the tube after a thermal entry length of about \(0.05\,\text{Re}_D\text{Pr}\,D\). For gases these lengths are similar; for oils, with Prandtl numbers in the hundreds, the thermal entry length can be enormous, so the whole tube may lie in the developing region. In turbulent flow both entry lengths are short, roughly 10 to 60 diameters, and fully developed conditions are usually assumed.

The Reynolds number for a circular tube is \(\text{Re}_D = \rho u_m D/\mu = 4\dot m/(\pi D\mu)\), based on the mean velocity. Flow is laminar below about 2300 and fully turbulent above about 10,000, with a transitional range between in which predictions are uncertain. For non-circular ducts, the hydraulic diameter \(D_h = 4A_c/P\) is used in place of \(D\), which works well for turbulent flow and less well for laminar flow in sharp-cornered ducts.

Mean temperature and the energy balance

Because the temperature varies across the tube, a single representative value is needed. The mean (or bulk) temperature \(T_m\) is defined so that \(\dot m c_p T_m\) equals the rate at which the flow carries thermal energy; physically it is the temperature you would measure if you collected the fluid at that section in a cup and mixed it. An energy balance on the whole tube links the heat transferred to the change in mean temperature:

\[q = \dot m c_p\,(T_{m,o} - T_{m,i})\]

and an energy balance on a slice of length \(dx\) gives \(\dot m c_p\,dT_m = q_s''\,P\,dx\), where \(P\) is the perimeter. Two boundary conditions dominate practice. With a uniform surface heat flux, as from electric heating or a uniformly heated nuclear fuel pin, \(T_m\) rises linearly along the tube. With a uniform surface temperature, as in a tube surrounded by condensing steam or boiling liquid, the difference \(T_s - T_m\) decays exponentially:

\[\frac{T_s - T_{m,o}}{T_s - T_{m,i}} = \exp\!\Big(-\frac{P L \bar h}{\dot m c_p}\Big), \qquad q = \bar h A_s\,\Delta T_{\text{lm}}, \quad \Delta T_{\text{lm}} = \frac{\Delta T_o - \Delta T_i}{\ln(\Delta T_o/\Delta T_i)}\]

The log-mean temperature difference \(\Delta T_{\text{lm}}\) is the correct average driving force for a tube with a uniform wall temperature, and it reappears in heat exchanger analysis in Lesson 12. Using the arithmetic mean instead overestimates the heat rate when the inlet and outlet differences are very different. When the outside of the tube is not isothermal but exchanges heat with a surrounding fluid through insulation, as with a buried pipe, the same exponential applies with \(\bar h P\) replaced by the overall conductance per unit length, \(UP\) or \(1/R'\), and \(T_s\) replaced by the outside temperature.

Laminar flow

For fully developed laminar flow in a circular tube, the Nusselt number is a constant: \(\text{Nu}_D = 4.36\) for uniform surface heat flux and \(\text{Nu}_D = 3.66\) for uniform surface temperature. These exact results mean that \(h = \text{Nu}\,k/D\) is independent of velocity, and that small tubes give high coefficients. In the entry region the coefficient is higher, because the thermal boundary layer is thin. For a thermally developing flow with a uniform wall temperature, the Hausen correlation gives the average Nusselt number in terms of the Graetz number \(\text{Gz}_D = (D/L)\,\text{Re}_D\text{Pr}\):

\[\overline{\text{Nu}}_D = 3.66 + \frac{0.0668\,\text{Gz}_D}{1 + 0.04\,\text{Gz}_D^{2/3}}\]

For non-circular ducts, fully developed laminar Nusselt numbers are tabulated by shape; for a square duct with uniform wall temperature it is about 2.98, and for parallel plates with both walls at the same uniform temperature it is 7.54, based on the hydraulic diameter, which is twice the plate spacing.

Turbulent flow

For fully developed turbulent flow in smooth tubes, the classic Dittus-Boelter equation is

\[\text{Nu}_D = 0.023\,\text{Re}_D^{4/5}\,\text{Pr}^n, \qquad n = 0.4\ \text{(heating)},\ 0.3\ \text{(cooling)}\]

valid for \(\text{Re}_D > 10{,}000\), \(0.6 < \text{Pr} < 160\) and moderate temperature differences. Its errors can reach 25 percent. The Gnielinski correlation is more accurate, typically within 10 percent, and covers the transitional range from about 3000:

\[\text{Nu}_D = \frac{(f/8)(\text{Re}_D - 1000)\,\text{Pr}}{1 + 12.7\,(f/8)^{1/2}(\text{Pr}^{2/3} - 1)}, \qquad f = (0.790\ln\text{Re}_D - 1.64)^{-2}\]

where \(f\) is the Darcy friction factor, here from Petukhov's formula for smooth tubes; for rough tubes, the friction factor from the Colebrook equation or the Moody chart can be used, which accounts approximately for roughness raising heat transfer along with friction. The friction factor also gives the pressure drop, \(\Delta p = f\,(L/D)\,\rho u_m^2/2\), which sets the pumping power. The close link between friction and heat transfer is no accident: the same turbulent eddies carry momentum and heat, an idea known as the Reynolds analogy.

Turbulent heat transfer, friction and temperature drop in the district supply main at winter and summer flows, and a laminar oil stream heated by condensing steam in a tube, with entry effects.
# Internal forced convection: district water in a transmission main and oil heated by steam in a tube
import math
# Saturated liquid water (approximate table values): T (C), rho, cp, k, mu
WATER = [(20, 998.0, 4182, 0.598, 1.002e-3), (40, 992.0, 4179, 0.631, 0.653e-3), (60, 983.0, 4185, 0.654, 0.467e-3),
         (80, 972.0, 4197, 0.670, 0.355e-3), (100, 958.0, 4217, 0.679, 0.282e-3), (120, 943.0, 4244, 0.683, 0.232e-3)]
def water(T):
    for a, b in zip(WATER, WATER[1:]):
        if a[0] <= T <= b[0]:
            f = (T - a[0]) / (b[0] - a[0])
            return tuple(x + f * (y - x) for x, y in zip(a[1:], b[1:]))
    raise ValueError("temperature outside table")

def gnielinski(Re, Pr):
    f = (0.790 * math.log(Re) - 1.64) ** -2               # Petukhov friction factor, smooth tube
    return f, (f / 8) * (Re - 1000) * Pr / (1 + 12.7 * (f / 8) ** 0.5 * (Pr ** (2/3) - 1))

# 1. The 110 C supply main, DN200, at winter and summer flows
D, L = 0.2073, 1800.0
rho, cp, k, mu = water(110.0); Pr = mu * cp / k
UA_per_m = 49.4 / (110 - (-5))                           # buried-pipe value from Lesson 4, W/m.K
for V in (1.5, 0.2):
    m = rho * V * math.pi / 4 * D**2
    Re = rho * V * D / mu
    f, Nu_g = gnielinski(Re, Pr)
    Nu_db = 0.023 * Re**0.8 * Pr**0.4
    dp = f * L / D * rho * V**2 / 2
    T_out = -5 + (110 - (-5)) * math.exp(-UA_per_m * L / (m * cp))
    print(f"V = {V:.1f} m/s: m = {m:5.1f} kg/s, Re = {Re:.2e}, Nu Gnielinski {Nu_g:6.0f}, Dittus-Boelter {Nu_db:6.0f}, "
          f"h = {Nu_g*k/D:5.0f} W/m2.K")
    print(f"             friction f = {f:.4f}, pressure drop over {L/1000:.1f} km = {dp/1000:6.1f} kPa, "
          f"water arrives at {T_out:.2f} C")

# 2. Laminar heat-transfer oil heated by condensing steam in a 25 mm tube (constant surface temperature)
rho_o, cp_o, k_o, mu_o = 860.0, 2000.0, 0.13, 0.020    # approximate oil properties at the mean temperature
Do, mdot, Ts, Ti = 0.025, 0.10, 150.0, 40.0
Re_o = 4 * mdot / (math.pi * Do * mu_o); Pr_o = mu_o * cp_o / k_o
print(f"\noil: Re = {Re_o:.0f} (laminar), Pr = {Pr_o:.0f}, thermal entry length about {0.05*Re_o*Pr_o*Do:.0f} m")
for Lt in (3.0, 6.0, 12.0):
    Gz = Do / Lt * Re_o * Pr_o
    Nu = 3.66 + 0.0668 * Gz / (1 + 0.04 * Gz**(2/3))     # Hausen, combined entry (thermally developing)
    h = Nu * k_o / Do
    To = Ts - (Ts - Ti) * math.exp(-math.pi * Do * Lt * h / (mdot * cp_o))
    q = mdot * cp_o * (To - Ti)
    print(f"  tube {Lt:4.1f} m: Gz = {Gz:6.0f}, Nu = {Nu:5.2f}, h = {h:5.1f} W/m2.K, outlet {To:5.1f} C, duty {q/1000:4.1f} kW "
          f"(fully developed Nu = 3.66 would give {Ts-(Ts-Ti)*math.exp(-math.pi*Do*Lt*3.66*k_o/Do/(mdot*cp_o)):5.1f} C)")
Output
V = 1.5 m/s: m =  48.1 kg/s, Re = 1.15e+06, Nu Gnielinski   2216, Dittus-Boelter   1957, h =  7279 W/m2.K
             friction f = 0.0114, pressure drop over 1.8 km =  105.4 kPa, water arrives at 109.56 C
V = 0.2 m/s: m =   6.4 kg/s, Re = 1.53e+05, Nu Gnielinski    413, Dittus-Boelter    390, h =  1358 W/m2.K
             friction f = 0.0165, pressure drop over 1.8 km =    2.7 kPa, water arrives at 106.77 C

oil: Re = 255 (laminar), Pr = 308, thermal entry length about 98 m
  tube  3.0 m: Gz =    653, Nu = 14.54, h =  75.6 W/m2.K, outlet  49.4 C, duty  1.9 kW (fully developed Nu = 3.66 would give  42.4 C)
  tube  6.0 m: Gz =    326, Nu = 11.19, h =  58.2 W/m2.K, outlet  54.1 C, duty  2.8 kW (fully developed Nu = 3.66 would give  44.8 C)
  tube 12.0 m: Gz =    163, Nu =  8.63, h =  44.9 W/m2.K, outlet  61.0 C, duty  4.2 kW (fully developed Nu = 3.66 would give  49.4 C)

Worked problem

The supply main at winter flow

Water at 110 °C flows at 1.5 m/s in the DN200 supply main (inside diameter 207.3 mm). Properties: ρ = 951 kg/m³, \(c_p\) = 4231 J/kg·K, k = 0.681 W/m·K, μ = 2.57 × 10⁻⁴ kg/m·s, Pr = 1.60. Find the Reynolds number, the convection coefficient by Gnielinski and Dittus-Boelter, the pressure drop over 1.8 km of smooth pipe, and the temperature at the end of the buried line, whose loss conductance from Lesson 4 is 0.43 W/m·K to a −5 °C ground surface.

Show solution
  1. \(\text{Re}_D = 951(1.5)(0.2073)/2.57\times10^{-4} = 1.15\times10^6\): turbulent.
  2. \(f = (0.790\ln(1.15\times10^6) - 1.64)^{-2} = 0.0114\); Gnielinski \(\text{Nu}_D = 2216\), so \(h = 2216(0.681)/0.2073 = 7280\) W/m²·K. Dittus-Boelter gives \(\text{Nu}_D = 1957\), about 12 percent lower.
  3. \(\Delta p = 0.0114(1800/0.2073)(951)(1.5^2)/2 = 105\) kPa.
  4. \(\dot m = 48.1\) kg/s; \(T_{m,o} = -5 + 115\exp\!\big(-0.43(1800)/(48.1 \times 4231)\big) = 109.56\) °C.

Answer: At full flow the water loses less than half a kelvin over the whole line, and the water-side coefficient is far larger than the value of 5000 W/m²·K assumed in Lesson 2, confirming that it hardly matters. Real steel pipe is not smooth; a friction factor of about 0.014 to 0.016 is more typical for commercial steel at this Reynolds number, which would raise the pressure drop by a quarter or more.

At summer flow, when the network carries only domestic hot water loads, the velocity in the same main may fall to 0.2 m/s. The program shows the consequences: the Reynolds number drops to about 150,000, still turbulent; the pressure drop falls almost fifty-fold; and the water arriving at the hospital is now about 3 K cooler than it left the plant, because the same heat loss is spread over one-seventh of the flow. This is why district systems lose a larger fraction of their heat in summer, and why some utilities lower supply temperatures or use bypasses during low-load periods.

Worked problem

Laminar oil heated by condensing steam

One industrial customer heats a heat-transfer oil (ρ = 860 kg/m³, \(c_p\) = 2000 J/kg·K, k = 0.13 W/m·K, μ = 0.020 kg/m·s) from 40 °C in a 25 mm tube surrounded by condensing steam at 150 °C. The flow is 0.10 kg/s. Find the outlet temperature and duty for tube lengths of 3, 6 and 12 m.

Show solution
  1. \(\text{Re}_D = 4(0.10)/(\pi \times 0.025 \times 0.020) = 255\): laminar. \(\text{Pr} = 0.020(2000)/0.13 = 308\).
  2. Thermal entry length \(\approx 0.05(255)(308)(0.025) = 98\) m: the tube is entirely in the entry region.
  3. For 6 m: \(\text{Gz}_D = (0.025/6)(255)(308) = 326\); Hausen gives \(\overline{\text{Nu}}_D = 11.2\), so \(\bar h = 58.2\) W/m²·K.
  4. \(T_{m,o} = 150 - 110\exp\!\big(-\pi(0.025)(6)(58.2)/(0.10 \times 2000)\big) = 54.1\) °C; duty 2.8 kW.
  5. For 3 m: 49.4 °C, 1.9 kW; for 12 m: 61.0 °C, 4.2 kW.

Answer: Assuming fully developed flow with Nu = 3.66 would underpredict the outlet temperature badly, giving 44.8 °C for the 6 m tube. Laminar oil heaters are inherently poor, because the coefficient is low and the oil near the centre of the tube barely sees the wall. Designers use smaller tubes, twisted-tape inserts or static mixers to break up the core, or raise the flow into the turbulent range, and they keep the wall temperature low enough that the oil does not degrade at the hot wall.

A graph of temperature against distance along a tube from 0 to 12 metres. A horizontal line at 150 degrees Celsius shows the wall temperature set by condensing steam. The mean oil temperature rises from 40 degrees to about 61 degrees at 12 metres, rising faster near the inlet. A dashed curve below it, for a constant Nusselt number of 3.66, rises only to about 49 degrees.
The mean temperature of laminar oil heated by condensing steam; entry effects matter.

For chemical and process engineers in Canada, the same correlations underlie the heat exchanger design software used across the oil sands, refining and petrochemical industries in Alberta and elsewhere, where heavy, viscous streams often flow in the laminar or transitional range. Pressure vessels and piping for these services are designed and registered under provincial safety authorities using the CSA B51 code and the ASME codes it references.

Common mistakes

  • MistakeUsing the arithmetic mean temperature difference for a tube with a uniform wall temperature.

    InsteadUse the log-mean temperature difference, or the exponential solution.

  • MistakeAssuming fully developed laminar flow in viscous liquids.

    InsteadThe thermal entry length is 0.05 Re Pr D and can exceed the whole tube.

  • MistakeUsing Dittus-Boelter in transitional flow.

    InsteadIt requires Re above 10,000; Gnielinski covers lower Reynolds numbers.

  • MistakeUsing the outer diameter in Re_D.

    InsteadUse the inside diameter, or the hydraulic diameter for non-circular ducts.

  • MistakeForgetting that a smaller flow raises the temperature change.

    InsteadThe same heat loss over less flow gives a larger temperature drop.

Summary and key terms

  • Internal flow becomes fully developed after the entry lengths; for oils the thermal entry can be very long.
  • The mean temperature changes along the tube; q = ṁc_p(T_m,o − T_m,i).
  • With a uniform wall temperature, T_s − T_m decays exponentially and the log-mean difference applies.
  • Fully developed laminar flow: Nu = 4.36 (uniform flux) or 3.66 (uniform temperature); Hausen for entry.
  • Turbulent flow: Dittus-Boelter for estimates, Gnielinski for accuracy; friction and heat transfer go together.
  • The district main loses under half a kelvin in winter but about 3 K at summer flow.
Fully developed flow
Flow in which the velocity and dimensionless temperature profiles no longer change along the tube.
Entry length
The distance over which the boundary layers grow to fill the tube.
Mean temperature
The bulk temperature that represents the energy carried by the flow.
Hydraulic diameter
D_h = 4A_c/P, used for non-circular ducts.
Log-mean temperature difference
The appropriate average temperature difference for exponential temperature profiles.
Graetz number
Gz = (D/L) Re Pr, which governs thermal entry effects.
Dittus-Boelter equation
Nu = 0.023 Re^0.8 Pr^n for fully turbulent flow in smooth tubes.
Gnielinski correlation
A more accurate turbulent correlation using the friction factor.
Reynolds analogy
The link between momentum and heat transfer in turbulent flow.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. Water flows at 0.5 kg/s in a 25 mm tube with μ = 6.5 × 10⁻⁴ kg/m·s. Find Re_D.

    Show answer

    39,200; turbulent.

  2. Find h for fully developed laminar flow of water (k = 0.63 W/m·K) in a 10 mm tube with a uniform wall temperature.

    Show answer

    3.66(0.63)/0.010 = 231 W/m²·K.

  3. The inlet and outlet temperature differences are 60 K and 20 K. Find the log-mean difference.

    Show answer

    36.4 K.

  4. In the program, double the oil flow to 0.20 kg/s for the 6 m tube. What happens to the outlet temperature?

    Show answer

    It falls to about 49.4 °C, while the duty rises to about 3.7 kW: the higher flow raises h through the Graetz number, but each kilogram is heated less.

  5. Why does the district water arrive cooler at summer flow?

    Show answer

    The heat loss is similar but is spread over much less flow.

  6. Find the hydraulic diameter of a 20 mm by 40 mm rectangular duct.

    Show answer

    4(800)/120 = 26.7 mm.

Lesson quiz · 5 questions

Score 60% or more to complete this lesson.

Take the quiz