B1 — Mass of oil heated per hour in a tube: 23-Chem-B1 Transport Phenomena, May 2018, Q3

Question 3: B1 — Mass of oil heated per hour in a tube (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Tube diameter, length	$D,\ L$	$0.01\ \text{m},\ 4.6\ \text{m}$
Inlet / outlet oil temperature	$T_{in},T_{out}$	$80\ \text{°C},\ 120\ \text{°C}$
Constant wall temperature	$T_s$	$165\ \text{°C}$
Specific heat	$c_p$	$2093.4\ \text{J/kg}\cdot\text{K}$
Thermal conductivity	$k$	$0.14365\ \text{W/m}\cdot\text{K}$
Viscosity (bulk 100 °C / wall 165 °C)	$\mu_b,\ \mu_s$	$4.80,\ 2.30\ \text{cP}$

Find. The mass flow rate $\dot m$ (kg/h) of oil that leaves at exactly $120\ \text{°C}$.

Check — corrected data line

The data line printed as “kinematic viscosity of water $=1.43651\times10^{-4}$ kW/m·K” is mislabeled: the units kW/m·K are those of a thermal conductivity, and $1.43651\times10^{-4}\ \text{kW/m}\cdot\text{K}=0.14365\ \text{W/m}\cdot\text{K}$ is a textbook value for hydrocarbon oil. It is read here as the thermal conductivity of the oil, which is what the convection calculation requires.

Fig. B1: Oil is heated in a constant-wall-temperature tube. The larger the flow, the less each kilogram is heated over the fixed length; the exit-temperature condition fixes the single flow rate that leaves at 120 °C.

Approach. Use the constant-wall-temperature energy balance to relate the exit temperature to the group $\dot m c_p$ and the tube conductance $hA$; because $h$ depends on flow through the laminar Sieder–Tate correlation — whose Graetz number $Re\,Pr\,D/L$ collapses to $4\dot m c_p/(\pi k L)$, eliminating the (unstated) density — the balance becomes one algebraic equation in $\dot m$.

Constant-wall-temperature energy balance. For a tube at fixed $T_s$, integrating $\dot m c_p\,dT=h(T_s-T)\,dA$ gives the exponential approach $$\frac{T_s-T_{out}}{T_s-T_{in}}=\exp\!\Big(-\frac{hA}{\dot m c_p}\Big)\ \Rightarrow\ \frac{hA}{\dot m c_p}=\ln\frac{T_s-T_{in}}{T_s-T_{out}}=\ln\frac{85}{45}=0.636.$$ Here $A=\pi D L=\pi(0.01)(4.6)=0.14451\ \text{m}^2$.
Choose the film correlation. Oil in a narrow tube runs laminar (confirmed as $Re\approx307$ below), and the tube is thermally developing, so the Sieder–Tate laminar correlation applies: $$Nu=\frac{hD}{k}=1.86\left(Re\,Pr\,\frac{D}{L}\right)^{1/3}\!\left(\frac{\mu_b}{\mu_s}\right)^{0.14}.$$
Eliminate density via the Graetz number. Writing $Re\,Pr=\dfrac{\rho V D}{\mu}\dfrac{c_p\mu}{k}=\dfrac{\rho V c_p D}{k}$ and $\dot m=\rho V\tfrac{\pi}{4}D^2$, the density cancels: $$Gz=Re\,Pr\,\frac{D}{L}=\frac{4\dot m c_p}{\pi k L}=\frac{4(2093.4)}{\pi(0.14365)(4.6)}\,\dot m=4034\,\dot m.$$ This is why no oil density is needed.
Property correction factor. At the bulk mean $100\ \text{°C}$, $\mu_b=4.80$ cP (table interpolation between 95 and 120 °C); at the wall $165\ \text{°C}$, $\mu_s=2.30$ cP. Hence $$\left(\frac{\mu_b}{\mu_s}\right)^{0.14}=\left(\frac{4.80}{2.30}\right)^{0.14}=1.109.$$
Assemble the balance as one equation in $\dot m$. With $h=Nu\,k/D$, $$\frac{hA}{\dot m c_p}=\frac{1.86\,(\mu_b/\mu_s)^{0.14}\,(k/D)\,A}{\dot m c_p}\,Gz^{1/3}=0.636.$$ Substituting $Gz=4034\,\dot m$ leaves $\dot m^{-2/3}$ on the left; solving, $$\boxed{\dot m\approx0.01158\ \text{kg/s}=41.7\ \text{kg/h}.}$$
Check the regime and coefficient. $Re=\dfrac{4\dot m}{\pi D\mu_b}=\dfrac{4(0.01158)}{\pi(0.01)(4.80\times10^{-3})}=307$ (laminar, $<2100$ &checkmark), $Gz=46.7$, $Nu=1.86(46.7)^{1/3}(1.109)=7.43$, and $h=Nu\,k/D=107\ \text{W/m}^2\text{K}$; back-substituting gives $hA/\dot m c_p=0.636$, closing the balance.

Quantity	Result
Required $\ln[(T_s-T_{in})/(T_s-T_{out})]$	$0.636$
Graetz number / Reynolds number	$Gz=46.7,\ Re=307$ (laminar)
Nusselt number / film coefficient	$Nu=7.43,\ h=107\ \text{W/m}^2\text{K}$
Oil heated per hour	$\dot m\approx41.7\ \text{kg/h}\ (0.01158\ \text{kg/s})$

Concept & Theory

Topic. Laminar internal-flow heating at constant wall temperature (thermally developing), solved for the throughput.

Key relations. Log-temperature approach $\tfrac{T_s-T_{out}}{T_s-T_{in}}=e^{-hA/\dot m c_p}$; Sieder–Tate laminar $Nu=1.86\,Gz^{1/3}(\mu_b/\mu_s)^{0.14}$; Graetz number $Gz=Re\,Pr\,D/L=4\dot m c_p/(\pi k L)$.

Why this works. The Graetz-number identity $Re\,Pr\,D/L=4\dot m c_p/(\pi k L)$ contains only $\dot m$, $c_p$, $k$, $L$ — not $\rho$ or $\mu$ — so the missing oil density never enters the answer, and the whole problem reduces to one equation in $\dot m$. The $(\mu_b/\mu_s)^{0.14}$ factor corrects for the steep viscosity drop near the hot wall, which thins the boundary layer and raises $h$.

Common pitfall. Using a simple LMTD with an assumed $h$, or applying a turbulent correlation — the flow is laminar here. Also using arithmetic $\Delta T$ instead of the log approach, or forgetting that $\mu_s$ is evaluated at the wall temperature (165 °C), not the bulk.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 8 (laminar tube flow; Sieder–Tate, Eq. 8.57 and the constant-$T_s$ analysis); Geankoplis, Ch. 4.

Question 3 of 6: B1 — Mass of oil heated per hour in a tube

Question 3: B1 — Mass of oil heated per hour in a tube (25 marks)