Forced convection in internal flow

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1 Errata

• Slide 12 reverses the temperature ratios in the logarithm and exponential expressions. Use the integrated balance

  ln⁡(Δ𝑇𝑒Δ𝑇𝑖)=−𝐴𝑠¯¯¯¯ℎ𝐿˙𝑚𝐶𝑝$$\ln\left(\frac{\Delta T_e}{\Delta T_i}\right) =-\frac{A_s\bar h_L}{\dot m C_p}$$

  Rearranging gives the temperature ratio

  𝑇𝑠−𝑇𝑒𝑇𝑠−𝑇𝑖=exp⁡(−𝐴𝑠¯¯¯¯ℎ𝐿˙𝑚𝐶𝑝)$$\frac{T_s-T_e}{T_s-T_i} =\exp\left(-\frac{A_s\bar h_L}{\dot m C_p}\right)$$

• Correction to the earlier worked solution for slide 22: pair the stated inlet velocity with density at the inlet, 77 degrees Celsius. Using a mean-temperature density for that velocity underestimates mass flow. The corrected wall heat flux is 159 W m−2$\mathrm{W\,m^{-2}}$ and exit wall temperature is 132.73 degrees Celsius. The PDF question itself does not print the incorrect density or answer.

• Correction to the earlier worked solution for slide 23: use cooling-water transport properties near its bulk mean of 30 degrees Celsius, with inlet density at 20 degrees Celsius for tube counting. The previous near-70-degree property set gave 79 tubes at 1.58 m. The corrected estimate is 78 tubes at 1.92 m. The PDF question itself does not print those properties or answers.

2 Reading list

• Çengel and Ghajar (2015), chapter 8.
• Bird et al. (2006).

Use 𝑞𝑠$q_s$ for inward wall heat flux, ˙𝑄$\dot Q$ for heat rate, and 𝑇𝑖,𝑇𝑒$T_i,T_e$ for inlet and outlet bulk temperatures.

3 Choose an internal-flow model

Given conditions	Model and checks
First establish flow and wall conditions	Use diameter in Reynolds number; distinguish prescribed wall temperature from prescribed wall flux.
Laminar and fully developed	Use Nu = 3.66 for fixed wall temperature, or Nu = 48/11 for fixed wall flux.
Laminar with thermal or combined entry	Use the length-average entrance correlation for its stated wall condition; see Eq. 20.
Turbulent flow	Check the range and mean-bulk property rule of Eq. 24 or the chosen alternative.
Outlet temperature or tube length	Use the bulk energy balance and appropriate local or average coefficient; include all resistances when using an overall coefficient.

4 Convective heat transfer: internal flow

4.1 Mean velocity, temperature and coefficient

Bulk temperature

The mixing-cup temperature that gives the same enthalpy flow as the actual fluid temperature profile. For constant properties it is weighted by the local mass flow, rather than being a simple area average.

Use 𝑞$q$ for heat flux in W m−2$\mathrm{W\,m^{-2}}$ and ˙𝑄$\dot Q$ for heat rate in W, as in earlier topics. The source’s 𝑄$Q$ means a rate. Its 𝑇𝑜$T_o$ and 𝑇𝑒$T_e$ both mean outlet mean temperature; here use 𝑇𝑒$T_e$, with 𝑇𝑖$T_i$ for inlet. 𝑥$x$ is axial distance in m, synonymous with the source’s 𝑧$z$, 𝑟$r$ is radial distance and 𝑅 =𝐷/2$R=D/2$ the circular-tube radius. The flow area 𝐴𝑐$A_c$ and wall area 𝐴𝑠$A_s$ are in m2$\mathrm{m^2}$. The wetted perimeter 𝑃$P$ is in m, not pressure.

For constant density 𝜌$\rho$ in kg m−3$\mathrm{kg\,m^{-3}}$, the area-mean axial velocity 𝑢𝑚$u_m$ in m s−1$\mathrm{m\,s^{-1}}$ and the mass flow ˙𝑚$\dot m$ in kg s−1$\mathrm{kg\,s^{-1}}$ are

𝑢𝑚=1𝐴𝑐∫𝐴𝑐𝑢𝑥 𝑑𝐴$$u_m=\frac{1}{A_c}\int_{A_c}u_x\,dA$$

˙𝑚=𝜌𝑢𝑚𝐴𝑐$$\dot m=\rho u_m A_c$$(1)

The mixing-cup or bulk temperature 𝑇𝑚$T_m$ is defined by matching enthalpy flow. If 𝐶𝑝$C_p$ is constant, in J kg−1 K−1$\mathrm{J\,kg^{-1}\,K^{-1}}$,

˙𝑚𝐶𝑝𝑇𝑚=∫𝐴𝑐𝜌𝑢𝑥𝐶𝑝𝑇 𝑑𝐴𝑇𝑚=1𝑢𝑚𝐴𝑐∫𝐴𝑐𝑢𝑥𝑇 𝑑𝐴$$\begin{gathered} \dot m C_pT_m=\int_{A_c}\rho u_x C_pT\,dA\\ T_m=\frac{1}{u_m A_c}\int_{A_c}u_x T\,dA \end{gathered}$$(2)

It is the temperature measured by catching and mixing the stream, rather than a plain area average. Fast fluid near the axis contributes more enthalpy than nearly stationary fluid near the wall. For varying properties use enthalpy itself in the definition instead of assuming 𝐶𝑝𝑇$C_pT$.

Figure 1: Developed laminar velocity and a schematic heating temperature profile from the tube axis to its wall. Radial position is vertical; velocity and temperature use separate horizontal scales. Velocity vanishes at the wall and peaks on the axis; temperature is highest at the heated wall.

Define the local coefficient by

𝑞𝑠(𝑥)=ℎ(𝑥)[𝑇𝑠(𝑥)−𝑇𝑚(𝑥)],$$q_s(x)=h(x)[T_s(x)-T_m(x)],$$

where 𝑞𝑠$q_s$ is positive heat entering the fluid. The subscript 𝑠$s$ identifies the wall; its inward sign differs from an outward radial Fourier flux. ℎ$h$ is in W m−2 K−1$\mathrm{W\,m^{-2}\,K^{-1}}$. Conductivity 𝑘$k$ is in W m−1 K−1$\mathrm{W\,m^{-1}\,K^{-1}}$, and Nu𝐷 =ℎ𝐷/𝑘$\mathrm{Nu}_D=hD/k$ is dimensionless. Temperature differences in K and degrees Celsius have the same numerical value.

For constant fluid properties, what does the mixing-cup temperature preserve?

• The temperature at the pipe axis.
• The simple area-average temperature.
• The wall temperature.
• The energy carried by the flowing fluid, accounting for how velocity and temperature vary across the pipe. ✓

Explanation

The bulk temperature is defined by the stream’s enthalpy flow. Faster fluid carries more mass and contributes more to this mean than nearly stationary fluid near the wall. A simple area average generally gives a different value.

4.2 Thermal analysis of a pipe slice

At steady state, neglect kinetic and potential energy changes, shaft work, axial conduction and heat storage. For an incompressible liquid or ideal gas with approximately constant 𝐶𝑝$C_p$, the whole-pipe balance is

˙𝑄=˙𝑚𝐶𝑝(𝑇𝑒−𝑇𝑖)$$\dot Q=\dot m C_p(T_e-T_i)$$(3)

This conservation statement holds for either wall condition. A slice of length Δ𝑥$\Delta x$ has wall area 𝑃Δ𝑥$P\Delta x$. Equating wall input to the enthalpy rise and taking the zero-length limit gives

˙𝑚𝐶𝑝Δ𝑇𝑚=𝑞𝑠𝑃Δ𝑥˙𝑚𝐶𝑝𝑑𝑇𝑚𝑑𝑥=𝑃𝑞𝑠=𝑃ℎ(𝑥)[𝑇𝑠(𝑥)−𝑇𝑚(𝑥)]$$\begin{gathered} \dot m C_p\Delta T_m=q_sP\Delta x\\ \dot m C_p\frac{dT_m}{dx}=Pq_s =Ph(x)[T_s(x)-T_m(x)] \end{gathered}$$(4)

More perimeter or a larger coefficient heats the stream faster. A larger capacity flow rate ˙𝑚𝐶𝑝$\dot m C_p$, in W K−1$\mathrm{W\,K^{-1}}$, slows its temperature rise. Integration requires a wall condition.

A stream has capacity flow rate ˙𝑚𝐶𝑝 =200 W K−1$\dot mC_p=200\ \mathrm{W\,K^{-1}}$ and receives 1000$1000$ W without losses. What is its bulk-temperature rise?

• 5$5$ K. ✓
• 0.2$0.2$ K.
• 200$200$ K.
• 1000$1000$ K.

Explanation

The heat supplied raises the stream’s bulk temperature:

𝑇𝑒−𝑇𝑖=˙𝑄˙𝑚𝐶𝑝=1000200=5 K$$T_e-T_i=\frac{\dot Q}{\dot mC_p}=\frac{1000}{200}=5\ \mathrm K$$

The bulk-temperature rise is 5 K. This energy balance applies to either wall condition under the stated assumptions.

4.3 Limiting case 1: constant surface heat flux

Uniform electrical heating or imposed irradiation can approximate constant 𝑞𝑠$q_s$. With constant 𝑃$P$ and ˙𝑚𝐶𝑝$\dot m C_p$, integrating Eq. 4 gives

𝑇𝑚(𝑥)=𝑇𝑖+𝑃𝑞𝑠𝑥˙𝑚𝐶𝑝$$T_m(x)=T_i+\frac{Pq_s x}{\dot m C_p}$$

𝑇𝑠(𝑥)=𝑇𝑚(𝑥)+𝑞𝑠ℎ(𝑥)$$T_s(x)=T_m(x)+\frac{q_s}{h(x)}$$(5)

The bulk temperature is linear even in the entrance region. The wall is linear only where ℎ$h$ is constant. Its hottest point during heating is normally the outlet, so check outlet metal temperature for burnout or coking. A fired heater’s real flux varies along and around the tube; this idealisation does not represent peak-flux maldistribution.

To calculate ℎ$h$, use the cylindrical energy equation from the conduction topic. Assume steady, laminar, axisymmetric flow, constant properties, an already developed axial velocity, negligible axial conduction and negligible volumetric generation or viscous dissipation. Then

𝑢𝑥∂𝑇∂𝑥=𝛼1𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)$$u_x\frac{\partial T}{\partial x} =\alpha\frac{1}{r}\frac{\partial}{\partial r} \left(r\frac{\partial T}{\partial r}\right)$$(6)

The thermal diffusivity is

𝛼=𝑘𝜌𝐶𝑝$$\alpha=\frac{k}{\rho C_p}$$

Péclet number

The ratio of advection to diffusion over the same length 𝐿$L$ and speed 𝑈$U$. For heat,

Pe𝐿=𝑈𝐿𝛼=Re𝐿Pr.$$\mathrm{Pe}_L=\frac{UL}{\alpha}=\mathrm{Re}_L\mathrm{Pr}.$$

For a species,

Pe𝑚,𝐿=𝑈𝐿𝐷𝐴𝐵=Re𝐿Sc.$$\mathrm{Pe}_{m,L}=\frac{UL}{D_{AB}}=\mathrm{Re}_L\mathrm{Sc}.$$

The diffusivity must match the transported quantity. See the thermal scale estimate and the species scale estimate.

4.4 Checking neglected transport terms

For Pe𝐿 ≪1$\mathrm{Pe}_L\ll1$, diffusion is faster than advection over the chosen length. At Pe𝐿 ∼1$\mathrm{Pe}_L\sim1$, both terms must be retained. For Pe𝐿 ≫1$\mathrm{Pe}_L\gg1$, advection dominates and diffusion in that direction can often be neglected away from end regions. This comparison does not remove diffusion across a tube. With 𝑈 =1 m s−1$U=1\ \mathrm{m\,s^{-1}}$, 𝐿 =0.10 m$L=0.10\ \mathrm m$ and 𝛼 =2 ×10−5 m2 s−1$\alpha=2\times10^{-5}\ \mathrm{m^2\,s^{-1}}$, Pe𝐿 =5000$\mathrm{Pe}_L=5000$. If 𝐷 =0.01 m$D=0.01\ \mathrm m$, the diameter-based value is only 500. Use

Pe𝐿=Re𝐷Pr𝐿𝐷$$\mathrm{Pe}_L=\mathrm{Re}_D\mathrm{Pr}\frac{L}{D}$$

for the axial test, rather than substituting Re𝐷Pr$\mathrm{Re}_D\mathrm{Pr}$ alone.

Brinkman number

The dimensionless ratio

Br=𝜇𝑈2𝑘Δ𝑇∗,$$\mathrm{Br}=\frac{\mu U^2}{k\Delta T_*},$$

comparing viscous heat generation with conduction over the same transverse scale. Here 𝜇$\mu$ and 𝑘$k$ are fluid properties, 𝑈$U$ is the specified velocity scale, and Δ𝑇∗ >0$\Delta T_*>0$ is a characteristic temperature difference.

Small Br supports neglecting viscous heating; values of order one or larger require that assumption to be checked. For 𝜇 =1 Pa s$\mu=1\ \mathrm{Pa\,s}$, 𝑈 =1 m s−1$U=1\ \mathrm{m\,s^{-1}}$, 𝑘 =0.2 W m−1 K−1$k=0.2\ \mathrm{W\,m^{-1}\,K^{-1}}$ and Δ𝑇∗ =10 K$\Delta T_*=10\ \mathrm K$, Br =0.5$\mathrm{Br}=0.5$. That model cannot dismiss viscous heating as small. A zero wall-to-bulk temperature difference cannot be used as the denominator; select a nonzero physical scale or retain the dissipation term in the energy balance.

𝛼$\alpha$ is thermal diffusivity in m2 s−1$\mathrm{m^2\,s^{-1}}$. Steadiness removes the time derivative, axisymmetry removes angular diffusion, and developed axial flow removes radial and angular advection. Axial conduction relative to axial advection scales as

1Re Pr(𝐿𝐷)$$\frac{1}{\mathrm{Re}\,\mathrm{Pr}\left(\frac{L}{D}\right)}$$

and is below one percent when that product exceeds 100. Low Péclet numbers need another model. Viscous heating matters when the Brinkman number

Br=𝜇𝑢2𝑚𝑘∣Δ𝑇∣$$\mathrm{Br}=\frac{\mu u_m^2}{k|\Delta T|}$$

approaches one; it can matter in polymer melts even though it is negligible in the air exercise.

The Hagen-Poiseuille profile supplied by the momentum prerequisite is

𝑢𝑥(𝑟)=2𝑢𝑚[1−(𝑟𝑅)2]$$u_x(r)=2u_m\left[1-\left(\frac{r}{R}\right)^2\right]$$(7)

Its area average is 𝑢𝑚$u_m$, so centreline speed is twice the mean. The inlet condition 𝑇(𝑟,0) =𝑇𝑖$T(r,0)=T_i$ determines the entrance solution. In the thermally developed constant-flux region the temperature profile translates upward without changing shape. Consequently

∂𝑇∂𝑥=𝑑𝑇𝑚𝑑𝑥$$\frac{\partial T}{\partial x}=\frac{dT_m}{dx}$$

is constant across the radius and

∂2𝑇∂𝑥2=0.$$\frac{\partial^2 T}{\partial x^2}=0.$$

Constant wall flux alone does not establish this in the entrance. Radial symmetry gives ∂𝑇/∂𝑟 =0$\partial T/\partial r=0$ at 𝑟 =0$r=0$, and the inward wall flux gives

𝑘∂𝑇∂𝑟=𝑞𝑠$$k\frac{\partial T}{\partial r}=q_s$$

at 𝑟 =𝑅$r=R$.

Using 𝑃 =2𝜋𝑅$P=2\pi R$ and ˙𝑚 =𝜌𝑢𝑚𝜋𝑅2$\dot m=\rho u_m\pi R^2$ in Eq. 4, then substituting Eq. 7 into Eq. 6, gives

∂𝑇∂𝑥=2𝑞𝑠𝜌𝐶𝑝𝑢𝑚𝑅1𝑟𝑑𝑑𝑟(𝑟𝑑𝑇𝑑𝑟)=4𝑞𝑠𝑘𝑅[1−(𝑟𝑅)2]$$\begin{gathered} \frac{\partial T}{\partial x}=\frac{2q_s}{\rho C_pu_mR}\\ \frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) =\frac{4q_s}{kR}\left[1-\left(\frac{r}{R}\right)^2\right] \end{gathered}$$(8)

Multiply by 𝑟$r$ and integrate twice. A term 𝐶1/𝑟$C_1/r$ in the gradient would be singular on the axis, so regularity and symmetry set 𝐶1 =0$C_1=0$:

𝑟𝑑𝑇𝑑𝑟=4𝑞𝑠𝑘𝑅(𝑟22−𝑟44𝑅2)𝑇=4𝑞𝑠𝑘𝑅(𝑟24−𝑟416𝑅2)+𝐶2$$\begin{gathered} r\frac{dT}{dr}=\frac{4q_s}{kR} \left(\frac{r^2}{2}-\frac{r^4}{4R^2}\right)\\ T=\frac{4q_s}{kR} \left(\frac{r^2}{4}-\frac{r^4}{16R^2}\right)+C_2 \end{gathered}$$(9)

Set 𝑇(𝑅,𝑥) =𝑇𝑠(𝑥)$T(R,x)=T_s(x)$ to fix 𝐶2$C_2$. With 𝜂 =𝑟/𝑅$\eta=r/R$,

𝑇−𝑇𝑠=4𝑞𝑠𝑅𝑘(𝜂24−𝜂416−316)$$T-T_s=\frac{4q_sR}{k} \left(\frac{\eta^2}{4}-\frac{\eta^4}{16}-\frac{3}{16}\right)$$(10)

The profile is quartic because a parabolic velocity carries energy. Apply the velocity-weighted mean Eq. 2, not an area mean:

𝑇𝑚−𝑇𝑠=4∫10𝜂(1−𝜂2)(𝑇−𝑇𝑠) 𝑑𝜂∫10𝜂(1−𝜂2)(𝜂24−𝜂416−316)𝑑𝜂=−11384𝑇𝑚−𝑇𝑠=−11𝑞𝑠𝐷48𝑘$$\begin{gathered} T_m-T_s=4\int_0^1\eta(1-\eta^2)(T-T_s)\,d\eta\\ \int_0^1\eta(1-\eta^2) \left(\frac{\eta^2}{4}-\frac{\eta^4}{16}-\frac{3}{16}\right)d\eta =-\frac{11}{384}\\ T_m-T_s=-\frac{11q_sD}{48k} \end{gathered}$$(11)

For heating 𝑇𝑚 <𝑇𝑠$T_m<T_s$, as the sign requires. Equating this difference to 𝑞𝑠/ℎ$q_s/h$ from Newton’s definition yields

ℎ=48𝑘11𝐷,Nu𝐷=4811=4.3636…$$h=\frac{48k}{11D},\qquad \mathrm{Nu}_D=\frac{48}{11}=4.3636\ldots$$(12)

Nu is independent of Re and Pr under these assumptions. The dimensional ℎ$h$ still depends on 𝑘/𝐷$k/D$. Flow rate changes the axial temperature slope rather than the developed profile’s shape. Thus constant flux and constant ℎ$h$ give parallel wall and bulk temperature lines with a fixed gap.

A pipe has uniform wall heat flux, constant perimeter and constant capacity flow rate. Its thermal entrance region has a changing ℎ(𝑥)$h(x)$. Which statement follows?

• Both wall and bulk temperatures must be constant.
• The bulk temperature must bend whenever the wall temperature bends.
• Bulk temperature changes linearly along the pipe, but the difference between wall and bulk temperatures can change. ✓
• The wall-to-bulk gap must be zero.

Explanation

A balance on a short pipe section gives

𝑑𝑇𝑚𝑑𝑥=𝑃𝑞𝑠˙𝑚𝐶𝑝$$\frac{dT_m}{dx}=\frac{Pq_s}{\dot mC_p}$$

The right-hand side is constant, so bulk temperature increases linearly with distance. The wall-to-bulk temperature difference is

𝑇𝑠(𝑥)−𝑇𝑚(𝑥)=𝑞𝑠ℎ(𝑥)$$T_s(x)-T_m(x)=\frac{q_s}{h(x)}$$

This difference changes as the heat transfer coefficient changes near the inlet.

Figure 2: Comparison of wall and bulk temperatures under the two wall conditions.

Wall and bulk temperatures during heating. The entrance-region wall curve is schematic.

4.5 Limiting case 2: constant surface temperature

A pure vapour condensing or a fluid boiling at nearly fixed pressure can provide an approximately isothermal source. A much larger capacity flow rate on the other side also approximates it. Finite wall and film resistance means the tube’s inner surface need not equal the source temperature, as the condenser exercise accounts for through 𝑈$U$.

For a fixed inner-wall 𝑇𝑠$T_s$, define Δ𝑇 =𝑇𝑠 −𝑇𝑚$\Delta T=T_s-T_m$. Then

𝑑Δ𝑇𝑑𝑥=−𝑑𝑇𝑚𝑑𝑥.$$\frac{d\Delta T}{dx}=-\frac{dT_m}{dx}.$$

Substitution in Eq. 4 and separation give

𝑑Δ𝑇𝑑𝑥=−𝑃ℎ(𝑥)˙𝑚𝐶𝑝Δ𝑇𝑑Δ𝑇Δ𝑇=−𝑃˙𝑚𝐶𝑝ℎ(𝑥) 𝑑𝑥$$\begin{gathered} \frac{d\Delta T}{dx}=-\frac{Ph(x)}{\dot m C_p}\Delta T\\ \frac{d\Delta T}{\Delta T}=-\frac{P}{\dot m C_p}h(x)\,dx \end{gathered}$$(13)

Number of transfer units

For heat exchange with a fixed wall temperature, the number of transfer units is

NTU𝐻=¯¯¯¯ℎ𝐴𝑠˙𝑚𝐶𝑝.$$\mathrm{NTU}_H=\frac{\bar h A_s}{\dot m C_p}.$$

It compares surface transfer conductance with stream heat-capacity rate. For a species with fixed wall concentration on a consistent concentration basis,

NTU𝑚=¯¯¯¯𝑘𝑐𝐴𝑠𝑄,$$\mathrm{NTU}_m=\frac{\bar k_c A_s}{Q},$$

where 𝑄$Q$ is volumetric flow rate. Both are dimensionless and describe the exponential approach to the wall state in these constant-property models.

Small NTU gives little approach to the wall state; large NTU leaves a small driving difference. NTU 0.1 leaves 𝑒−0.1 =0.905$e^{-0.1}=0.905$ of the inlet difference; NTU 3 leaves 𝑒−3 =0.050$e^{-3}=0.050$. This supports selecting the heated length. NTU alone is not the effectiveness of an arbitrary two-stream exchanger, whose capacity-rate ratio and arrangement also matter. The mass-transfer example uses the corresponding concentration difference.

The driving force decays exponentially if ℎ$h$ is constant. Define its length-average and integrate from inlet to exit:

¯¯¯¯ℎ𝐿=1𝐿∫𝐿0ℎ(𝑥) 𝑑𝑥ln⁡Δ𝑇𝑒Δ𝑇𝑖=−𝑃˙𝑚𝐶𝑝∫𝐿0ℎ(𝑥) 𝑑𝑥=−𝐴𝑠¯¯¯¯ℎ𝐿˙𝑚𝐶𝑝𝑇𝑒=𝑇𝑠−(𝑇𝑠−𝑇𝑖)𝑒−𝐴𝑠¯¯¯¯ℎ𝐿˙𝑚𝐶𝑝$$\begin{gathered} \bar h_L=\frac{1}{L}\int_0^Lh(x)\,dx\\ \ln\frac{\Delta T_e}{\Delta T_i} =-\frac{P}{\dot m C_p}\int_0^Lh(x)\,dx =-\frac{A_s\bar h_L}{\dot m C_p}\\ T_e=T_s-(T_s-T_i)e^{-\frac{A_s\bar h_L}{\dot m C_p}} \end{gathered}$$(14)

For heating the exit difference is smaller, so the left side must be negative to match the right. The corrected outlet-over-inlet form above also handles variable ℎ$h$ exactly through its integral. The ratio

𝐴𝑠¯¯¯¯ℎ𝐿˙𝑚𝐶𝑝$$\frac{A_s\bar h_L}{\dot m C_p}$$

is the number of transfer units. At a value of 3 the remaining difference is about 5 percent of its inlet value; further length gives diminishing temperature change.

Use Eq. 3 and 𝑇𝑒 −𝑇𝑖 =Δ𝑇𝑖 −Δ𝑇𝑒$T_e-T_i=\Delta T_i-\Delta T_e$ to eliminate ˙𝑚𝐶𝑝$\dot m C_p$ from Eq. 14:

˙𝑄=𝐴𝑠¯¯¯¯ℎ𝐿Δ𝑇ln⁡Δ𝑇ln⁡=Δ𝑇𝑖−Δ𝑇𝑒ln⁡(Δ𝑇𝑖Δ𝑇𝑒)$$\begin{gathered} \dot Q=A_s\bar h_L\Delta T_{\ln}\\ \Delta T_{\ln}= \frac{\Delta T_i-\Delta T_e}{\ln\left(\frac{\Delta T_i}{\Delta T_e}\right)} \end{gathered}$$(15)

For heating both differences are positive. Swapping inlet and exit leaves the mean unchanged, and equal differences have the continuous limit of their common value. For cooling, signed differences are both negative with this wall-minus-fluid convention, so ˙𝑄$\dot Q$ and the signed log mean are negative; the heat-loss magnitude uses positive fluid-minus-wall differences. A crossing of zero is outside this single isothermal-wall model. A positive logarithmic mean assumes positive heating differences. An arithmetic mean overestimates the positive log mean. The latter follows from integration rather than an arbitrary choice of averaging.

A fluid enters colder than a constant-temperature wall. With positive finite ℎ$h$, no sources and the stated pipe balance, how does its bulk temperature behave?

• It crosses above the wall temperature after enough length.
• It approaches the wall temperature from below as the driving difference decreases. ✓
• It rises linearly without limit.
• It immediately equals wall temperature at the inlet.

Explanation

As the fluid warms, the wall-to-fluid temperature difference becomes smaller. For constant coefficient,

𝑇𝑠−𝑇𝑚(𝑥)𝑇𝑠−𝑇𝑖=exp⁡(−ℎ𝑃𝑥˙𝑚𝐶𝑝)$$\frac{T_s-T_m(x)}{T_s-T_i}=\exp\left(-\frac{hPx}{\dot mC_p}\right)$$

This ratio stays positive at any finite heated length. The fluid approaches the wall temperature from below. A varying positive coefficient changes the rate of approach, but does not make the fluid hotter than the wall in this model.

4.6 Developed constant-temperature flow

Graetz number

For internal thermal development,

Gz𝑥=Re𝐷Pr𝐷𝑥=𝑈𝐷2𝛼𝑥,$$\mathrm{Gz}_x= \mathrm{Re}_D\mathrm{Pr}\frac{D}{x}=\frac{U D^2}{\alpha x},$$

using mean speed 𝑈$U$, tube diameter 𝐷$D$ and distance 𝑥 >0$x>0$ from the start of heating. Its inverse

𝑥𝐷Re𝐷Pr$$\frac{x}{D\mathrm{Re}_D\mathrm{Pr}}$$

is the dimensionless axial distance used in entrance solutions. Replace 𝛼$\alpha$ by 𝐷𝐴𝐵$D_{AB}$ for species development.

Large Gz means little transverse diffusion time has elapsed during passage to 𝑥$x$, so entry effects can persist. At Gz of order one, transverse diffusion has acted over a distance comparable to the diameter during that passage. Small Gz supports a developed scalar profile under the stated laminar assumptions. Use an entrance-region solution while the profile is developing, rather than a fully developed constant Nusselt number. For Re𝐷Pr =1000$\mathrm{Re}_D\mathrm{Pr}=1000$ and 𝑥/𝐷 =10$x/D=10$, Gz𝑥 =100$\mathrm{Gz}_x=100$; at 𝑥/𝐷 =1000$x/D=1000$ it is one. A developed velocity profile alone does not imply a developed temperature profile. With 𝐷 =2𝑅$D=2R$, the coordinate below is 𝜉 =2/Gz𝑥$\xi=2/\mathrm{Gz}_x$; check factors of two before using a chart.

The same Eq. 6 and parabolic velocity now have 𝑇(𝑅,𝑥) =𝑇𝑠$T(R,x)=T_s$, 𝑇(𝑟,0) =𝑇𝑖$T(r,0)=T_i$ and axis symmetry. The axial gradient varies along the tube, so the constant-flux two-integration solution does not apply. The Graetz solution expands the temperature in radial eigenfunctions. To see its form, let

Θ=𝑇𝑠−𝑇𝑇𝑠−𝑇𝑖$$\Theta=\frac{T_s-T}{T_s-T_i}$$

and

𝜉=𝛼𝑥2𝑢𝑚𝑅2.$$\xi=\frac{\alpha x}{2u_mR^2}.$$

Substitution of

Θ=𝐹(𝜂)𝑒−𝜆2𝜉$$\Theta=F(\eta)e^{-\lambda^2\xi}$$

gives

1𝜂𝑑𝑑𝜂(𝜂𝑑𝐹𝑑𝜂)+𝜆2(1−𝜂2)𝐹=0𝐹′(0)=0,𝐹(1)=0$$\begin{gathered} \frac{1}{\eta}\frac{d}{d\eta}\left(\eta\frac{dF}{d\eta}\right) +\lambda^2(1-\eta^2)F=0\\ F'(0)=0,\qquad F(1)=0 \end{gathered}$$(16)

The inlet determines expansion coefficients. Far downstream the slowest mode dominates and its first eigenvalue is 𝜆0 ≃2.7044$\lambda_0\simeq2.7044$. For that mode

𝑑Δ𝑇𝑚𝑑𝑥=−𝜆20𝛼Δ𝑇𝑚2𝑢𝑚𝑅2.$$\frac{d\Delta T_m}{dx}=-\frac{\lambda_0^2\alpha\Delta T_m}{2u_mR^2}.$$

The slice balance independently gives

−2ℎΔ𝑇𝑚𝜌𝐶𝑝𝑢𝑚𝑅.$$-\frac{2h\Delta T_m}{\rho C_pu_mR}.$$

Equating these rates gives

Nu𝐷=𝜆202≃3.657≃3.66$$\mathrm{Nu}_D=\frac{\lambda_0^2}{2}\simeq3.657\simeq3.66$$(17)

Both developed Nu values are local asymptotes, independent of pipe length. The average over an entire short tube depends on its entrance. A length ratio 𝐿/𝐷 >60$L/D>60$ can make entrance effects small for gases near Pr = 1, but is not a universal laminar guarantee for high-Pr fluids. The imposed-flux condition has a roughly 19 percent greater developed Nu than the fixed-temperature case. These are different normalized profile shapes, not a universal claim that one arbitrary heater always transfers more heat.

For fully developed laminar flow in a circular tube with constant properties, which Nusselt values match the wall conditions?

• Both wall conditions give 3.66$3.66$.
• Fixed wall temperature gives 4.36$4.36$; fixed wall flux gives 3.66$3.66$.
• Either value can be used whenever the Reynolds number is below 2300.
• Fixed wall temperature gives 3.66$3.66$; fixed wall flux gives 4.36$4.36$. ✓

Explanation

The wall conditions produce different normalized temperature profiles. The developed circular-tube limits are 3.66 for fixed wall temperature and 4.36 for fixed wall heat flux. Entrance effects and other duct shapes need their own results.

4.7 Geometry and hydraulic diameter

Hydraulic diameter

The length 𝐷ℎ =4𝐴𝑐/𝑃$D_h=4A_c/P$ used to characterise duct flow, where 𝐴𝑐$A_c$ is the flow area and 𝑃$P$ is the wetted perimeter. It equals the diameter for a circular pipe.

A force balance gives Δ𝑝𝐴𝑐 =𝜏𝑤𝑃𝐿$\Delta p A_c=\tau_wPL$, with pressure drop Δ𝑝$\Delta p$ and mean wall shear 𝜏𝑤$\tau_w$ in Pa. Matching a round tube’s

𝜏𝑤=Δ𝑝𝐷4𝐿$$\tau_w=\frac{\Delta p D}{4L}$$

defines

𝐷ℎ=4𝐴𝑐𝑃$$D_h=\frac{4A_c}{P}$$

Re=𝜌𝑢𝑚𝐷ℎ𝜇=𝑢𝑚𝐷ℎ𝜈$$\mathrm{Re}=\frac{\rho u_mD_h}{\mu} =\frac{u_mD_h}{\nu}$$(18)

The Prandtl number remains a fluid-property ratio:

Pr=𝐶𝑝𝜇𝑘=𝜈𝛼$$\mathrm{Pr}=\frac{C_p\mu}{k}=\frac{\nu}{\alpha}$$

For a rectangle,

𝐷ℎ=2𝑎𝑏𝑎+𝑏;$$D_h=\frac{2ab}{a+b};$$

for wide parallel plates this tends to twice the gap. Hydraulic diameter does not remove laminar shape dependence. The circular-tube Nu values above apply only to circular tubes.

4.8 Example: comparing development lengths

For laminar flow, use the slide estimates

𝐿ℎ≃0.0575Re𝐷$$L_h\simeq0.0575\mathrm{Re}D$$

The thermal entrance-length estimate is

𝐿𝑡≃0.0575Re Pr𝐷$$L_t\simeq0.0575\mathrm{Re}\,\mathrm{Pr}D$$(19)

Dividing gives 𝐿𝑡/𝐿ℎ ≃Pr$L_t/L_h\simeq\mathrm{Pr}$. Oils develop thermally later than hydrodynamically, whereas gases near Pr =1$\mathrm{Pr}=1$ have comparable lengths. Check the heated length against 𝐿𝑡$L_t$ before using developed Nu.

A laminar fluid has Pr =100$\mathrm{Pr}=100$. Its velocity profile has developed, but heating has only recently begun. What should be checked before using a developed thermal Nusselt value?

• Whether the fluid has travelled far enough from the start of heating for its temperature profile to develop. ✓
• Only whether the pipe has a circular cross-section.
• Whether Prandtl number can be replaced by one.
• Nothing, because velocity development guarantees thermal development.

Explanation

For the laminar entry-length estimates,

𝐿𝑡𝐿ℎ≃Pr=100$$\frac{L_t}{L_h}\simeq\mathrm{Pr}=100$$

The temperature profile can need about 100 times as much length to develop as the velocity profile. Check the distance from where heating begins. If it is too short, use a correlation that includes thermal entry effects.

4.9 Laminar entrance correlations

Check Reynolds number and both entrance lengths before choosing a coefficient. The following correlations give the length-average ¯¯¯¯ℎ𝐿$\bar h_L$ over a heated or cooled length 𝐿$L$ in a circular tube with constant wall temperature. They are not the local exit coefficient and do not apply to uniform wall flux. Use properties at the mean bulk temperature unless a wall viscosity is shown.

For developed velocity but developing temperature, slide 19 uses

Gz𝐿=Re𝐷Pr𝐷𝐿¯¯¯¯¯¯¯¯Nu𝐷=3.66+0.065Gz𝐿1+0.04Gz2/3𝐿$$\begin{gathered} \mathrm{Gz}_L=\mathrm{Re}_D\mathrm{Pr}\frac{D}{L}\\ \overline{\mathrm{Nu}}_D= 3.66+\frac{0.065\mathrm{Gz}_L}{1+0.04\mathrm{Gz}_L^{2/3}} \end{gathered}$$(20)

Use laminar flow, Re𝐷 <2300$\mathrm{Re}_D<2300$, negligible axial conduction and modest property variation. The constant 3.66$3.66$ supplies the developed limit as Gz𝐿 →0$\mathrm{Gz}_L\to0$. The slide’s rounded coefficient 0.065 is retained here and in Workshop 7; the common Hausen version uses 0.0668. These are slightly different fits, so state which one you use (Çengel and Ghajar 2015).

For appreciable viscosity variation or a combined velocity and thermal entrance, the laminar Sieder-Tate estimate is

¯¯¯¯¯¯¯¯Nu𝐷=1.86Gz1/3𝐿(𝜇𝑏𝜇𝑠)0.14$$\overline{\mathrm{Nu}}_D= 1.86\mathrm{Gz}_L^{1/3} \left(\frac{\mu_b}{\mu_s}\right)^{0.14}$$(21)

Use Re𝐷 <2300$\mathrm{Re}_D<2300$, approximately 0.48 ≤Pr ≤16700$0.48\le\mathrm{Pr}\le16700$ and an entry-dominated length, Gz𝐿 ≳10$\mathrm{Gz}_L\gtrsim10$. Evaluate all properties at the mean bulk temperature except 𝜇𝑠$\mu_s$, which uses the wall temperature. This is an empirical viscosity correction, not permission to ignore large changes in all properties. Do not extrapolate the entry-only expression to a long tube where it predicts a value below the developed constant-temperature limit. The laminar entrance correlations and their property rules are tabulated in Çengel and Ghajar (2015).

Flow and wall condition	Coefficient to use
Developed laminar velocity and temperature; constant wall temperature	Local developed Nu𝐷 =3.66$\mathrm{Nu}_D=3.66$
Developed laminar velocity and temperature; constant wall flux	Local developed Nu𝐷 =48/11$\mathrm{Nu}_D=48/11$
Developed laminar velocity; developing temperature; constant wall temperature	Length-average Eq. 20
Combined laminar entry, or a significant wall-to-bulk viscosity difference	Length-average Eq. 21 within its stated range
Turbulent flow	Select a turbulent correlation and check its range

For example, Re𝐷 =1000$\mathrm{Re}_D=1000$, Pr =7$\mathrm{Pr}=7$, 𝐷 =0.010 m$D=0.010\ \mathrm m$ and 𝐿 =0.50 m$L=0.50\ \mathrm m$ give Gz𝐿 =140$\mathrm{Gz}_L=140$ and ¯¯¯¯¯¯¯¯Nu𝐷 =8.04$\overline{\mathrm{Nu}}_D=8.04$. With 𝑘 =0.60 W m−1 K−1$k=0.60\ \mathrm{W\,m^{-1}\,K^{-1}}$, ¯¯¯¯ℎ𝐿 =482 W m−2 K−1$\bar h_L=482\ \mathrm{W\,m^{-2}\,K^{-1}}$. Using 3.66 over this whole length would miss the entrance enhancement. Put the average coefficient into the integrated isothermal-wall balance, rather than a local wall-temperature formula.

Workshop 7’s oil pipeline uses the same thermal-entry model with developed inlet velocity.

5 Turbulent flow and the friction analogy

Fanning friction factor

The dimensionless factor

𝑓𝐹=𝜏𝑤12𝜌𝑢2𝑚$$f_F=\frac{\tau_w}{\frac{1}{2}\rho u_m^2}$$

for internal flow, using mean speed 𝑢𝑚$u_m$. It has the same wall-shear normalization as the Skin friction coefficient. The Darcy factor is 𝑓𝐷 =4𝑓𝐹$f_D=4f_F$.

Darcy friction factor

The dimensionless pressure-loss factor in

Δ𝑝=𝑓𝐷𝐿𝐷𝜌𝑢2𝑚2$$\Delta p=f_D\frac{L}{D}\frac{\rho u_m^2}{2}$$

for fully developed flow in a straight constant-diameter pipe. It is four times the Fanning factor. Entrance, fitting, elevation and acceleration contributions require separate treatment.

For example, 𝑓𝐹 =0.005$f_F=0.005$ means 𝑓𝐷 =0.020$f_D=0.020$. Inserting 0.005 into a Darcy pressure-loss formula underpredicts that loss by four. The heat-transfer analogy below uses the same convention as its correlation.

Turbulent eddies transport momentum and energy, requiring empirical closure rather than the laminar analytical solution. The Chilton-Colburn analogy from topic 3 uses the Fanning factor 𝑓𝐹 =𝐶𝑓$f_F=C_f$. The Stanton number is

St=ℎ𝜌𝑢𝑚𝐶𝑝=NuRe Pr.$$\mathrm{St}=\frac{h}{\rho u_mC_p}=\frac{\mathrm{Nu}}{\mathrm{Re}\,\mathrm{Pr}}.$$

The analogy gives

𝑓𝐹2=St Pr2/3,𝑓𝐷=4𝑓𝐹Nu=𝑓𝐷8Re Pr1/3$$\begin{gathered} \frac{f_F}{2}=\mathrm{St}\,\mathrm{Pr}^{2/3},\qquad f_D=4f_F\\ \mathrm{Nu}=\frac{f_D}{8}\mathrm{Re}\,\mathrm{Pr}^{1/3} \end{gathered}$$(22)

The distinction matters by a factor of four. Circular laminar flow has 𝑓𝐷 =64/Re$f_D=64/\mathrm{Re}$ and 𝑓𝐹 =16/Re$f_F=16/\mathrm{Re}$. Inserting the source’s smooth turbulent fit 𝑓𝐷 =0.184Re−0.2$f_D=0.184\mathrm{Re}^{-0.2}$ gives

Nu=0.023Re0.8Pr1/3$$\mathrm{Nu}=0.023\mathrm{Re}^{0.8}\mathrm{Pr}^{1/3}$$(23)

The coefficient is 0.184/8$0.184/8$ and the Re exponent is 1 −0.2$1-0.2$; the Pr exponent comes from the analogy. It remains empirical because the friction fit and analogy are empirical. The source specifies 0.7 ≤Pr ≤160$0.7\le\mathrm{Pr}\le160$, Re above 10000 and developed smooth-pipe flow, approximately 𝐿/𝐷 >10$L/D>10$. Roughness increases pressure loss without an identical heat-transfer increase, so a rough-pipe friction factor cannot be inserted without checking applicability.

5.1 Correlations used in the classroom examples

For developed, smooth-tube turbulent flow, the Dittus-Boelter estimate is

Nu=0.023Re0.8Pr𝑛𝑛=0.4 for fluid heating𝑛=0.3 for fluid cooling$$\begin{gathered} \mathrm{Nu}=0.023\mathrm{Re}^{0.8}\mathrm{Pr}^{n}\\ n=0.4\ \text{for fluid heating}\\ n=0.3\ \text{for fluid cooling} \end{gathered}$$(24)

Use bulk mean properties, Re >10000$\mathrm{Re}>10000$, 0.7 ≲Pr ≲160$0.7\lesssim\mathrm{Pr}\lesssim160$, modest property variation and 𝐿/𝐷 >10$L/D>10$. The classroom examples state whether developed laminar or turbulent flow applies.

Electrically heated air tube

Air (1 atm, 77 °C$77\ \degree\mathrm{C}$) enters a 5 mm I.D. tube with a bulk average velocity of 2.5 m/s$2.5\ \mathrm{m/s}$. The velocity profile is developed and the thermal profile is “developing”. The tube length is 1.0 m, and a constant (uniform) heat flux is imposed at the tube surface on the air over the entire length. The bulk average exit temperature of the air stream is 127 °C$127\ \degree\mathrm{C}$. Determine:

• The heat transfer coefficient at the exit (ℎ𝐿$h_L$)
• The heat flux imposed at the surface of the tube (𝑞𝑠$q_s$)
• The surface temperature (𝑇𝑠$T_s$) of the tube at the exit

Use ideal-gas air with 𝑅air =287.05 J/(kg K)$R_{\mathrm{air}}=287.05\ \mathrm{J/(kg\,K)}$ and 𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$. At the mean bulk temperature of 375.15 K, use 𝜇 =2.1805 ×10−5 Pa s$\mu=2.1805\times10^{-5}\ \mathrm{Pa\,s}$, 𝑘 =0.03184 W/(m K)$k=0.03184\ \mathrm{W/(m\,K)}$ and 𝐶𝑝 =1011.5 J/(kg K)$C_p=1011.5\ \mathrm{J/(kg\,K)}$.

Solution

Figure 3: Heat enters through the tube wall while air flows from inlet Ti to outlet Te. In this example the inward wall flux is uniform.

Data

Quantity	Value
Inside diameter	𝐷 =0.0050 m$D=0.0050\ \mathrm m$
Heated length	𝐿 =1.0 m$L=1.0\ \mathrm m$
Inlet mean speed	𝑢𝑚 =2.5 m s−1$u_m=2.5\ \mathrm{m\,s^{-1}}$
Inlet bulk temperature	𝑇𝑖 =77 °C$T_i=77\ \degree\mathrm C$
Outlet bulk temperature	𝑇𝑒 =127 °C$T_e=127\ \degree\mathrm C$
Inlet air density	𝜌𝑖 =1.00810 kg m−3$\rho_i=1.00810\ \mathrm{kg\,m^{-3}}$
Dynamic viscosity	𝜇 =2.1805 ×10−5 Pa s$\mu=2.1805\times10^{-5}\ \mathrm{Pa\,s}$
Air conductivity	𝑘 =0.03184 W m−1 K−1$k=0.03184\ \mathrm{W\,m^{-1}\,K^{-1}}$
Specific heat capacity	𝐶𝑝 =1011.5 J kg−1 K−1$C_p=1011.5\ \mathrm{J\,kg^{-1}\,K^{-1}}$

The wall heat flux is uniform, and the inlet velocity profile is developed. The mean bulk temperature is 102 degrees Celsius or 375.15 K. Use transport properties near 375 K, but pair inlet velocity with inlet density:

𝜌𝑖=𝑃𝑅air𝑇𝑖=101325287.05×350.15=1.00810 kg m−3.$$\begin{aligned} \rho_i&=\frac{P}{R_{\mathrm{air}}T_i}\\ &=\frac{101325}{287.05\times350.15}\\ &=1.00810\ \mathrm{kg\,m^{-3}}. \end{aligned}$$

Assumptions

• Steady state.
• Constant viscosity, conductivity and heat capacity; ideal-gas density varies with temperature at approximately constant pressure.
• Heat supplied at the wall enters the air.
• Axial conduction, viscous heating and kinetic and potential energy changes are neglected.
• The tube is circular and the flow remains laminar.

Calculations

First check the flow regime and whether the temperature profile has developed by the exit. Using the diameter in the Reynolds number, as defined in Eq. 18, gives

Re𝐷=𝜌𝑖𝑢𝑚,𝑖𝐷𝜇=1.00810(2.5)(0.0050)2.1805×10−5=577.91$$\mathrm{Re}_D=\frac{\rho_i u_{m,i}D}{\mu} =\frac{1.00810(2.5)(0.0050)}{2.1805\times10^{-5}}=577.91$$

The Prandtl number is

Pr=𝐶𝑝𝜇𝑘=1011.5(2.1805×10−5)0.03184=0.69271$$\mathrm{Pr}=\frac{C_p\mu}{k} =\frac{1011.5(2.1805\times10^{-5})}{0.03184}=0.69271$$

The flow is laminar. The thermal-entry estimate from Eq. 19 is

𝐿𝑡≃0.0575Re𝐷Pr𝐷=0.0575(577.91)(0.69271)(0.0050)=0.1151 m$$L_t\simeq0.0575\mathrm{Re}_D\mathrm{Pr}D =0.0575(577.91)(0.69271)(0.0050)=0.1151\ \mathrm m$$

The temperature profile develops after heating starts, but the exit at 1.0 m$1.0\ \mathrm m$ lies well beyond this estimated entrance length. Use the developed local constant-flux result from Eq. 12, Nu𝐷 =48/11$\mathrm{Nu}_D=48/11$:

ℎ𝐿=Nu𝐷𝑘𝐷=48(0.03184)11(0.0050)=27.79 W m−2 K−1$$h_L=\frac{\mathrm{Nu}_Dk}{D} =\frac{48(0.03184)}{11(0.0050)} =27.79\ \mathrm{W\,m^{-2}\,K^{-1}}$$(25)

The total energy balance determines the wall heat flux independently of the local entrance coefficient. First calculate the air mass flow:

˙𝑚=𝜌𝑖𝑢𝑚,𝑖𝜋𝐷24=1.00810(2.5)𝜋(0.0050)24=4.94851×10−5 kg s−1$$\dot m=\rho_i u_{m,i}\frac{\pi D^2}{4} =1.00810(2.5)\frac{\pi(0.0050)^2}{4} =4.94851\times10^{-5}\ \mathrm{kg\,s^{-1}}$$

The enthalpy rise is the total heat input:

˙𝑄=˙𝑚𝐶𝑝(𝑇𝑒−𝑇𝑖)=(4.94851×10−5)(1011.5)(127−77)=2.5027 W$$\dot Q=\dot mC_p(T_e-T_i) =(4.94851\times10^{-5})(1011.5)(127-77) =2.5027\ \mathrm W$$

Since the flux is uniform over area 𝜋𝐷𝐿$\pi DL$,

𝑞𝑠=˙𝑄𝜋𝐷𝐿=2.5027𝜋(0.0050)(1.0)=159.33 W m−2$$q_s=\frac{\dot Q}{\pi DL} =\frac{2.5027}{\pi(0.0050)(1.0)} =159.33\ \mathrm{W\,m^{-2}}$$(26)

The local wall relation in Eq. 5,

𝑞𝑠=ℎ𝐿[𝑇𝑠(𝐿)−𝑇𝑒],$$q_s=h_L[T_s(L)-T_e],$$

then gives

𝑇𝑠(𝐿)=𝑇𝑒+𝑞𝑠ℎ𝐿=127+159.3327.79=132.73 °C$$T_s(L)=T_e+\frac{q_s}{h_L} =127+\frac{159.33}{27.79} =132.73\ \degree\mathrm{C}$$(27)

The wall is 5.73 K$5.73\ \mathrm K$ hotter than the exit air, as required for inward heat flow. Near the inlet the coefficient is larger, so the same heat flux needs a smaller wall-to-bulk difference. The developed coefficient must not be applied over that entrance region. The large axial Peclet product

Re Pr𝐿𝐷≃8.01×104$$\mathrm{Re}\,\mathrm{Pr}\frac{L}{D}\simeq8.01\times10^4$$

also supports neglecting axial conduction. At the same volume flow and temperature rise, replacing air by water would require a much larger heat input because water has a much larger volumetric heat capacity.

Answer

At the exit, ℎ𝐿 ≃27.8 W m−2 K−1$h_L\simeq27.8\ \mathrm{W\,m^{-2}\,K^{-1}}$. The imposed heat flux is 𝑞𝑠 ≃159 W m−2$q_s\simeq159\ \mathrm{W\,m^{-2}}$, and the exit wall temperature is 𝑇𝑠(𝐿) ≃132.7 °C$T_s(L)\simeq132.7\ \degree\mathrm{C}$.

Steam condenser tube sizing

A horizontal tubular condenser is used to condense saturated steam at 80 °C$80\ \degree\mathrm{C}$. The condenser is a shell and tube one with brass tubes (𝑘 =110 W/(m⋅K)$k=110\ \mathrm{W/(m\cdot K)}$, 1.59 cm OD, 1.34 cm ID). Steam is outside of the tubes (ℎ =10760 W/(m2⋅K)$h=10760\ \mathrm{W/(m^2\cdot K)}$) and cooling water enters the tube at 20 °C$20\ \degree\mathrm{C}$ with a velocity of 1.4 m/s$1.4\ \mathrm{m/s}$. Water leaves the tubes at 40 °C$40\ \degree\mathrm{C}$. If the rate of cooling water supply is 55000 kg/h and the latent heat of condensation of steam at 80 °C$80\ \degree\mathrm{C}$ is 2304 kJ/kg, calculate:

• The number of tubes required
• The length of each tube

Use inlet water density 𝜌𝑖 =998.2 kg/m3$\rho_i=998.2\ \mathrm{kg/m^3}$ at 20 degrees Celsius. At the bulk mean of 30 degrees Celsius, use 𝜇 =7.98 ×10−4 Pa s$\mu=7.98\times10^{-4}\ \mathrm{Pa\,s}$, 𝑘 =0.615 W/(m K)$k=0.615\ \mathrm{W/(m\,K)}$ and 𝐶𝑝 =4180 J/(kg K)$C_p=4180\ \mathrm{J/(kg\,K)}$. Treat 1.4 m/s as the nominal inlet speed; after rounding the number of tubes up, use the exact total mass flow to calculate the duty and Reynolds number of each tube.

Solution

Figure 4: Water is heated as it passes through a condenser tube. The heat flux is inward; the wall temperature and local flux need not be uniform. The steam film and brass wall are included in the overall coefficient.

Data

Quantity	Value
Steam temperature	𝑇steam =80 °C$T_{\mathrm{steam}}=80\ \degree\mathrm C$
Outer diameter	𝐷𝑜 =0.0159 m$D_o=0.0159\ \mathrm m$
Inner diameter	𝐷𝑖 =0.0134 m$D_i=0.0134\ \mathrm m$
Brass conductivity	𝑘𝑤 =110 W m−1 K−1$k_w=110\ \mathrm{W\,m^{-1}\,K^{-1}}$
Outside coefficient	ℎ𝑜 =10760 W m−2 K−1$h_o=10760\ \mathrm{W\,m^{-2}\,K^{-1}}$
Water inlet temperature	𝑇𝑖 =20 °C$T_i=20\ \degree\mathrm C$
Water outlet temperature	𝑇𝑒 =40 °C$T_e=40\ \degree\mathrm C$
Nominal inlet water speed	𝑢𝑚 =1.4 m s−1$u_m=1.4\ \mathrm{m\,s^{-1}}$
Total water mass flow	˙𝑚 =55000 kg h−1$\dot m=55000\ \mathrm{kg\,h^{-1}}$
Inlet water density	𝜌𝑖 =998.2 kg m−3$\rho_i=998.2\ \mathrm{kg\,m^{-3}}$
Dynamic viscosity	𝜇 =7.98 ×10−4 Pa s$\mu=7.98\times10^{-4}\ \mathrm{Pa\,s}$
Water conductivity	𝑘 =0.615 W m−1 K−1$k=0.615\ \mathrm{W\,m^{-1}\,K^{-1}}$
Specific heat capacity	𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$

The rounded water properties are consistent with the inlet and bulk mean states. See the IAPWS liquid-water property release. The inlet density determines flow area; mean-temperature viscosity, conductivity and heat capacity determine heat transfer.

Assumptions

• Steady state.
• Equal flow through parallel tubes.
• Constant properties and heat transfer coefficients.
• Developed turbulent flow in smooth tubes.
• Clean walls, with fouling resistance neglected.
• Negligible heat loss to the surroundings.

Calculations

The nominal inlet speed first determines a trial mass flow per tube:

˙𝑚trial=𝜌𝑖𝑢𝑚,𝑖𝜋𝐷2𝑖4=998.2(1.4)𝜋(0.0134)24=0.19708 kg s−1$$\dot m_{\mathrm{trial}}=\rho_i u_{m,i}\frac{\pi D_i^2}{4} =998.2(1.4)\frac{\pi(0.0134)^2}{4} =0.19708\ \mathrm{kg\,s^{-1}}$$

Divide total flow by this value and round up:

𝑁∗=˙𝑚total˙𝑚trial=55000/36000.19708=77.52$$N_* =\frac{\dot m_{\mathrm{total}}}{\dot m_{\mathrm{trial}}} =\frac{55000/3600}{0.19708}=77.52$$(28)

Use 78 tubes. The exact mass flow per tube is then

˙𝑚tube=550003600(78)=0.195869 kg s−1.$$\dot m_{\mathrm{tube}}=\frac{55000}{3600(78)} =0.195869\ \mathrm{kg\,s^{-1}}.$$

This gives an inlet speed of 1.3914 m s−1$1.3914\ \mathrm{m\,s^{-1}}$, close to the nominal value. Use this mass flow throughout the remaining calculation.

The inside coefficient follows from the mass flow and mean-temperature properties:

Re𝑖=4˙𝑚tube𝜋𝐷𝑖𝜇=4(0.195869)𝜋(0.0134)(7.98×10−4)=23322$$\mathrm{Re}_i=\frac{4\dot m_{\mathrm{tube}}}{\pi D_i\mu} =\frac{4(0.195869)}{\pi(0.0134)(7.98\times10^{-4})}=23322$$Pr=𝐶𝑝𝜇𝑘=4180(7.98×10−4)0.615=5.4238$$\mathrm{Pr}=\frac{C_p\mu}{k} =\frac{4180(7.98\times10^{-4})}{0.615}=5.4238$$

The flow is turbulent. Since the water is heated, use 𝑛 =0.4$n=0.4$ in the Dittus-Boelter relation from Eq. 24:

Nu𝑖=0.023Re0.8𝑖Pr0.4=0.023(23322)0.8(5.4238)0.4=141.14$$\mathrm{Nu}_i=0.023\mathrm{Re}_i^{0.8}\mathrm{Pr}^{0.4} =0.023(23322)^{0.8}(5.4238)^{0.4}=141.14$$

Convert Nusselt number using the inside diameter:

ℎ𝑖=Nu𝑖𝑘𝐷𝑖=141.14(0.615)0.0134=6478 W m−2 K−1$$h_i=\frac{\mathrm{Nu}_ik}{D_i} =\frac{141.14(0.615)}{0.0134} =6478\ \mathrm{W\,m^{-2}\,K^{-1}}$$(29)

Heat passes through the steam film, the brass wall and the water film in series. The steam temperature is fixed, but the inner wall is cooler. Using the cylindrical conduction resistance, the overall coefficient on inside area satisfies

1𝑈𝑖=𝐷𝑖𝐷𝑜ℎ𝑜+𝐷𝑖ln⁡(𝐷𝑜𝐷𝑖)2𝑘𝑤+1ℎ𝑖=0.01340.0159(10760)+0.0134ln⁡(0.01590.0134)2(110)+16478=2.4311×10−4 m2 K W−1$$\begin{aligned} \frac{1}{U_i} &=\frac{D_i}{D_oh_o} +\frac{D_i\ln\left(\frac{D_o}{D_i}\right)}{2k_w}+\frac1{h_i}\\ &=\frac{0.0134}{0.0159(10760)} +\frac{0.0134\ln\left(\frac{0.0159}{0.0134}\right)}{2(110)}+\frac1{6478}\\ &=2.4311\times10^{-4}\ \mathrm{m^2\,K\,W^{-1}} \end{aligned}$$(30)

Thus 𝑈𝑖 =4113 W m−2 K−1$U_i=4113\ \mathrm{W\,m^{-2}\,K^{-1}}$. The water film supplies about 63.5 percent of the total resistance, the steam film 32.2 percent and the brass wall 4.3 percent. Changing the wall conductivity alone would therefore have little effect. Area bases must remain consistent, with 𝑈𝑖𝐴𝑖 =𝑈𝑜𝐴𝑜$U_iA_i=U_oA_o$.

The driving differences are 80 −20 =60 K$80-20=60\ \mathrm K$ at the water inlet and 80 −40 =40 K$80-40=40\ \mathrm K$ at the outlet. Their logarithmic mean, defined in Eq. 15, is

Δ𝑇ln⁡=Δ𝑇𝑖−Δ𝑇𝑒ln⁡(Δ𝑇𝑖Δ𝑇𝑒)=60−40ln⁡(6040)=49.326 K$$\Delta T_{\ln}=\frac{\Delta T_i-\Delta T_e} {\ln\left(\frac{\Delta T_i}{\Delta T_e}\right)} =\frac{60-40}{\ln\left(\frac{60}{40}\right)}=49.326\ \mathrm K$$

The per-tube heat duty follows from the water enthalpy rise:

˙𝑄tube=˙𝑚tube𝐶𝑝(𝑇𝑒−𝑇𝑖)=0.195869(4180)(40−20)=16375 W$$\dot Q_{\mathrm{tube}}=\dot m_{\mathrm{tube}}C_p(T_e-T_i) =0.195869(4180)(40-20)=16375\ \mathrm W$$

Set this equal to 𝑈𝑖𝜋𝐷𝑖𝐿Δ𝑇ln⁡$U_i\pi D_iL\Delta T_{\ln}$ and solve for length:

𝐿=˙𝑄tube𝑈𝑖𝜋𝐷𝑖Δ𝑇ln⁡=163754113𝜋(0.0134)(49.326)=1.92 m$$L=\frac{\dot Q_{\mathrm{tube}}}{U_i\pi D_i\Delta T_{\ln}} =\frac{16375}{4113\pi(0.0134)(49.326)}=1.92\ \mathrm m$$(31)

The ratio 𝐿/𝐷𝑖 ≃143$L/D_i\simeq143$ supports the developed turbulent approximation. The ideal isothermal steam source needs no flow-arrangement correction to the logarithmic mean. The arithmetic mean 50 K$50\ \mathrm K$ would overestimate the driving force by about 1.37 percent.

As a steam-side check, the total water duty and the stated latent heat give

˙𝑄total=550003600(4180)(40−20)=1.2772×106 W$$\dot Q_{\mathrm{total}}=\frac{55000}{3600}(4180)(40-20) =1.2772\times10^6\ \mathrm W$$˙𝑚steam=˙𝑄totalℎ𝑓𝑔=1.2772×1062.304×106=0.5544 kg s−1≃1996 kg h−1$$\dot m_{\mathrm{steam}}=\frac{\dot Q_{\mathrm{total}}}{h_{fg}} =\frac{1.2772\times10^6}{2.304\times10^6} =0.5544\ \mathrm{kg\,s^{-1}}\simeq1996\ \mathrm{kg\,h^{-1}}$$

The latent heat checks the steam requirement; it does not determine tube count. These are clean-tube estimates using the stated correlation. A practical design also needs allowances for fouling, pressure drop and available tube sizes.

Answer

Using properties at the stated water temperatures, provide 78 parallel tubes of approximately 1.92 m$1.92\ \mathrm m$ heated length each.

6 Additional examples

Which pipe profile develops first?

Oil enters a 16 mm diameter pipe with Re =900$\mathrm{Re}=900$ and Pr =60$\mathrm{Pr}=60$. Estimate the hydrodynamic and thermal entrance lengths. Is a 5 m pipe long enough for both profiles to develop? Assume a uniform velocity and temperature at the inlet and heating beginning there.

Solution

Data

• Diameter 𝐷 =0.016 m$D=0.016\ \mathrm m$
• Pipe length 𝐿 =5.0 m$L=5.0\ \mathrm m$
• Reynolds number Re =900$\mathrm{Re}=900$
• Prandtl number Pr =60$\mathrm{Pr}=60$

Assumptions

• Uniform inlet velocity and temperature.
• Heating begins at the pipe inlet.
• Constant properties.

Calculations

The Reynolds number is below 2300, so use the laminar estimates in Eq. 19, converting the diameter to 0.016 m:

𝐿ℎ=0.0575(900)(0.016)=0.828 m𝐿𝑡=0.0575(900)(60)(0.016)=49.7 m$$L_h=0.0575(900)(0.016)=0.828\ \mathrm m \qquad L_t=0.0575(900)(60)(0.016)=49.7\ \mathrm m$$

The velocity profile develops within the pipe, but the temperature profile does not. The ratio 𝐿𝑡/𝐿ℎ =Pr =60$L_t/L_h=\mathrm{Pr}=60$ expresses the slower thermal diffusion relative to momentum diffusion in this oil. A heat transfer calculation for the 5 m pipe needs a thermal-entry correlation. A developed velocity profile alone does not justify a developed Nusselt number.

If heating starts after a long unheated approach, measure thermal entry from the start of heating. The two entrance lengths then have different origins.

Answer

The hydrodynamic entrance length is 𝐿ℎ =0.828 m$L_h=0.828\ \mathrm m$ and the thermal entrance length is 𝐿𝑡 =49.7 m$L_t=49.7\ \mathrm m$. The 5.0 m$5.0\ \mathrm m$ pipe permits velocity development but remains thermally developing.

Compare constant wall temperature and constant wall heat flux

Water enters a 2.0 m$2.0\ \mathrm m$ heated tube at 20 °C$20\ \degree\mathrm{C}$. The inside diameter is 0.020 m$0.020\ \mathrm m$, mass flow is 0.020 kg s−1$0.020\ \mathrm{kg\,s^{-1}}$ and 𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$. Adopt a constant effective coefficient ℎ =400 W m−2 K−1$h=400\ \mathrm{W\,m^{-2}\,K^{-1}}$ to isolate the effect of the wall condition. Neglect axial conduction and heat loss.

Find the outlet temperature and total heat input for two separate cases: a wall maintained at 80 °C$80\ \degree\mathrm{C}$, and a uniform inward wall heat flux of 10000 W m−2$10000\ \mathrm{W\,m^{-2}}$. Explain how the wall-to-bulk temperature difference changes with distance in each case.

Solution

Figure 5: The same flowing stream is heated in two separate cases. One fixes the wall temperature; the other fixes inward heat flux. The energy balance relates inlet and outlet bulk temperatures in both cases.

Data

Quantity	Value
Tube length	𝐿 =2.0 m$L=2.0\ \mathrm m$
Diameter	𝐷 =0.020 m$D=0.020\ \mathrm m$
Mass flow	˙𝑚 =0.020 kg s−1$\dot m=0.020\ \mathrm{kg\,s^{-1}}$
Inlet temperature	𝑇𝑖 =20 °C$T_i=20\ \degree\mathrm C$
Specific heat capacity	𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Supplied effective coefficient	ℎ =400 W m−2 K−1$h=400\ \mathrm{W\,m^{-2}\,K^{-1}}$
Wall temperature in the first case	𝑇𝑠 =80 °C$T_s=80\ \degree\mathrm C$
Wall heat flux in the second case	𝑞𝑠 =10000 W m−2$q_s=10000\ \mathrm{W\,m^{-2}}$

Assumptions

• Steady state.
• Constant fluid properties.
• Axial conduction and heat loss are neglected.

Calculations

The heated area is 𝐴𝑠 =𝜋𝐷𝐿 =0.12566$A_s=\pi DL=0.12566$ m² and the stream heat-capacity rate is ˙𝑚𝐶𝑝 =83.6$\dot mC_p=83.6$ W K⁻¹.

For constant wall temperature, Eq. 14 gives

ℎ𝐴𝑠˙𝑚𝐶𝑝=400[𝜋(0.020)(2.0)]0.020(4180)=0.6013$$\frac{hA_s}{\dot mC_p}=\frac{400[\pi(0.020)(2.0)]}{0.020(4180)}=0.6013$$𝑇𝑒=𝑇𝑠−(𝑇𝑠−𝑇𝑖)𝑒−ℎ𝐴𝑠˙𝑚𝐶𝑝=80−(80−20)𝑒−0.6013=47.11 °C$$T_e=T_s-(T_s-T_i)e^{-\frac{hA_s}{\dot m C_p}} =80-(80-20)e^{-0.6013}=47.11\ \degree\mathrm{C}$$˙𝑄=83.6(47.11−20)=2267 W$$\dot Q=83.6(47.11-20)=2267\ \mathrm W$$

The driving difference decreases from 60 K to 32.89 K. Even with a constant coefficient, the heat flux falls from 24000 to 13155 W m⁻². Fully developed flow does not require a constant temperature difference.

For constant wall heat flux, the overall balance instead gives

˙𝑄=𝑞𝑠𝐴𝑠=10000[𝜋(0.020)(2.0)]=1257 W$$\dot Q=q_s A_s=10000[\pi(0.020)(2.0)]=1257\ \mathrm W$$𝑇𝑒=𝑇𝑖+˙𝑄˙𝑚𝐶𝑝=20+125783.6=35.03 °C$$T_e=T_i+\frac{\dot Q}{\dot m C_p}=20+\frac{1257}{83.6}=35.03\ \degree\mathrm{C}$$𝑇𝑠−𝑇𝑚=𝑞𝑠ℎ=25.0 K$$T_s-T_m=\frac{q_s}{h}=25.0\ \mathrm K$$

The bulk temperature rises linearly, and the wall temperature rises with it, from 45.0 °C$45.0\ \degree\mathrm{C}$ to 60.03 °C$60.03\ \degree\mathrm{C}$. Their difference stays constant under the stipulated constant coefficient. In a real entrance region, ℎ(𝑥)$h(x)$ varies and a constant imposed flux need not give a constant difference. The two cases prescribe different heat inputs, so their duties need not agree.

Answer

For fixed wall temperature, 𝑇𝑒 =47.1 °C$T_e=47.1\ \degree\mathrm{C}$ and ˙𝑄 =2.27 ×103 W$\dot Q=2.27\times10^3\ \mathrm W$, with a decreasing wall-to-bulk difference. For fixed heat flux, 𝑇𝑒 =35.0 °C$T_e=35.0\ \degree\mathrm{C}$ and ˙𝑄 =1.26 ×103 W$\dot Q=1.26\times10^3\ \mathrm W$, with a constant 25.0 K$25.0\ \mathrm K$ wall-to-bulk difference.

7 What you should now be able to calculate

Distinguish velocity and thermal entry, calculate outlet and wall temperatures, and size tubes using consistent mass flow and an overall resistance.

Practice: Workshop 7 for water heating and developing oil flow.

8 Report an error

Email Ranjeet Utikar with the title Forced convection in internal flow, the URL /heat-transport/internal-forced-convection/, and the equation or passage to check.

References

Bird, R. Byron, Warren E. Stewart, and Edwin N. Lightfoot. 2006. Transport Phenomena. Revised 2nd. John Wiley & Sons. https://www.wiley-vch.de/en/areas-interest/engineering/transport-phenomena-978-0-470-11539-8.

Çengel, Yunus A., and Afshin J. Ghajar. 2015. Heat and Mass Transfer: Fundamentals and Applications. 5th ed. McGraw-Hill Education. https://glencoe.mheducation.com/sites/0073398187/information_center_view0/application_chapters.html.

Shah, Milinkumar. n.d. “CHEN3001-TP: Heat Transport 4.” Curtin University. Unpublished.