I am dealing with either (a) a severe misunderstanding of the question, or (b) a wrong textbook answer.

Here is question verbatim:

A jacketed vessel is used to heat a water stream using steam available at $110°C$, as shown in Figure $10.10$. The steam temperature remains unchanged as its latent heat is removed. The jacketed vessel has a circular cross section with internal diameter $0.4 m$ and height $0.9 m$. It is fitted with an appropriate stirring device such that its content is well mixed. The water-side resistance for heat transfer is $0.4545 m^2 °C/kW$, while the steam-side resistance is $0.14 m^2 °C/kW$. The vessel wall is $2 mm$ thick and is made up of copper with a thermal conductivity of $0.4 kW/m °C$. What is the maximum hourly flow rate of water that this vessel can handle to provide hot water at the specified outlet temperature? The specific heat capacity of water is $4.2 kJ/kg °C$ and heat transfer through the top and bottom surfaces of the vessel may be assumed to be negligible. enter image description here (a) $2880 kg/h$
(b) $720 kg/h$
(c) $1440 kg/h$ (this is the textbook answer)
(d) $2160 kg/h$

Here's my working:

$2mm = 0.002m$
$R_H=0.4545+0.14+ {ln({r_o\over r_i})\over 2L{\pi}K} $
$R_H = 0.5945 +{ln({0.4 \div 2 +0.002\over 0.4 \div 2})\over 2(0.9){\pi}(0.4)} $
$R_H = 0.598899$ $(6 s.f.)$
$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$
$-0.753772={-0.449624\over m}$
$m = 0.596499$ $kg\cdot s^{-1} = 2147.40$ $kg\cdot h$

Seeing as my answer is totally different from any of the presented options, I likely made a total logic problem, so what was my mistake?

Khan, S. A., Farooq, S., Swee, K. Y., Jangam, S. V., & Wee, E. C. L. (2021). Heat Transfer II. Chemical Engineering: Visualizing Foundational Concepts (Rev. ed., pp 134-135). Department of Chemical and Biomolecular Engineering, National University of Singapore.

  • 1
    $\begingroup$ Your answer doesn't seem "totally different" from the presented options, your answer is the same as (D) if you had rounded 0.596499 kg/s to two digits. $\endgroup$ Nov 14 at 17:00
  • $\begingroup$ Yeah, but it's totally different from Option C, 1440 Kg/h. That's the textbook answer. $\endgroup$ Nov 14 at 17:02
  • $\begingroup$ I don't have a textbook in front of me to go over the formulas you're using, but your $R_H$ equation seems to have unit inconsistencies. The first two terms are given in $m.m.C/kW$ while the third term looks to be in $C/kW$ judging by the numbers from the problem you were plugging in. So it's possible that this is throwing off your $R_H$ value. Units look like they match properly for the equation with the exponential, though. You could work backward and solve for $m\cdot R_H$ to see what the textbook expects $R_H$ to be, to try to figure out the above unit issue. $\endgroup$ Nov 14 at 17:53
  • $\begingroup$ Your calculation of the wall resistance is incorrect. It should be 0.002/0.4. $\endgroup$ Nov 14 at 17:57
  • $\begingroup$ Please include a proper citation for the textbook. $\endgroup$ Nov 17 at 19:41

3 Answers 3


I figured out the problem with my working. I used the following equation earlier: $(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$
This was the wrong equation to use because it assumes that there is a temperature gradient within the jacketed vessel, which isn't the case because it is well-mixed and thus the internal temperature can be assumed to be constant (in this case, at $70°C$).

Correct working:

$R_H = 0.4545+0.14+\frac{{\Delta}X_W}{K} = 0.5945+\frac{0.002}{0.4} = 0.5995$ (Thanks @Chet Miller)
Rate of heat transfer from hot water to cold water in the vessel = rate of heat transfer across the jacketed vessel wall from hot steam to water
Let $m =$ volumetric flow rate of cold water into the vessel
Let $A =$ area of vessel wall
$m = \frac{(T_w-T_{out})*A}{R_H*(T_{out}-T_{in})*C_P}=\frac{(110-70)*{\pi}(0.4)(0.9)}{0.5995*(70-25)*4.2} = 0.399265kg{\cdot}s^{-1}=1440kg{\cdot}h^{-1}$ (3 s.f.)

  • 1
    $\begingroup$ This is not done correctly. The right hand side should be using the log-mean temperature difference (59.7 C), not the wall temperature minus the outlet temperature. The latter is not the average driving force of heat transfer. $\endgroup$ Nov 16 at 12:04
  • $\begingroup$ Sorry, I don't get it. I've never encountered 'log-mean' before and it's not in my textbook. Admittedly that the reason may be that I'm only in my first semester as a year 1 student. Care to elaborate? $\endgroup$ Nov 16 at 13:52
  • $\begingroup$ See my added Answer $\endgroup$ Nov 16 at 14:37

The log-mean temperature difference arises from your original equation: $$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$$or$$\ln{\left(\frac{T_{out} - T_w}{T_{in} - T_w}\right)}=\frac{-\pi{d}L}{m{C_p}{R_H}}$$This can be arranged to $$Q=mC_p(T_{out}-T_{in})=\frac{\pi dL}{R_H}\frac{[(T_w-T_{in})-(T_w-T_{out})]}{\ln{\left(\frac{T_{w} - T_{in}}{T_{w} - T_{out}}\right)}}$$where Q is the heat load and where the log-mean temperature difference is given by:$$(\Delta T)_{LM}=\frac{[(T_w-T_{in})-(T_w-T_{out})]}{\ln{\left(\frac{T_{w} - T_{in}}{T_{w} - T_{out}}\right)}}$$Then we get: $$Q=mC_p(T_{out}-T_{in})=\frac{\pi dL}{R_H}(\Delta T)_{LM}$$

  • $\begingroup$ The reason I was given as to not use that formula for this question is that this formula assumes that the liquid isn't subject to bulk motion from stirring, which would give a temperature gradient in the liquid (cooler nearer to the intake, hotter nearer to the outlet). However the liquid in the solution is well-mixed (see diagram) which gives roughly even temperature distribution in the liquid. $\endgroup$ Nov 16 at 23:57
  • $\begingroup$ Good point. That explains it. $\endgroup$ Nov 17 at 0:12

Overall resistance $R= 0.4545+0.14+\frac{0.002}{.4}=0.5995\ (m^2C)/kW$

Heat Transfer coefficient $U=\frac{1}{R}=1.6681\ kW/(m^2C)$

log mean temperature difference $=\frac{85-40}{\ln{(85/40)}}=59.7\ C$

Area $=(0.4)\pi (0.9)=1.131\ m^2)$

Heat Load = (1.6681)(1.131)(56.35)=112.6 kW

So $$\dot{m}C (70-25)=106.3$$

and $$\dot{m}=.5959\ kg/sec=2145\ kg/hr$$


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