Possible error in textbook question? (Heat transfer)

Question

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. (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)} $
$R_H = 0.598899$ $(6 s.f.)$
$(\frac{T_{out} - T_w}{T_{in} - T_w})=exp(\frac{-\pi{d}L}{m{C_p}{R_H}})$
$ln(\frac{70-110}{25-110})=\frac{-\pi(0.4)(0.9)}{m(4.2)(0.598899)}$
$-0.753772={-0.449624\over m}$
$m = 0.596499$ $kg\cdot s^{-1} = 2147.40$ $kg\cdot h$

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?

Textbook:
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.

Answer 1

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
$m*(T_{out}-T_{in})*C_P=\frac{T_w-T_{out}}{R_H}*A$
$m = \frac{(T_w-T_{out})*A}{R_H*(T_{out}-T_{in})*C_P}=\frac{(110-70)*{\pi}(0.4)(0.9)}{0.5995*(70-25)*4.2} = 0.399265kg{\cdot}s^{-1}=1440kg{\cdot}h^{-1}$ (3 s.f.)

Answer 2

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: $$Q=mC_p(T_{out}-T_{in})=\frac{\pi dL}{R_H}(\Delta T)_{LM}$$

Answer 3

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