estimating the surface area of a steam coil immersed in a tank

Question

I've been tasked to design a steam coil that will be immersed into a tank in order to maintain the fluid temperature above 25C (but not to exceed 40C), and I need to calculate its surface area & length. The fluid is a highly viscous oil additive and it's not agitated.

My approach was the calculate the total energy needed to heat the coil, using $Q = m.c_p\Delta T$. where m is the mass of fluid in the tank, $c_p$ is the specific heat capacity of the fluid, and $\Delta T$ is the difference between the initial temperature of fluid (25C) and what it should be maintained at (~35-40C).

Then I equated this energy to $UA\Delta T_m$, the energy delivered by the steam, where $U$ is the overall heat coefficient for steam, $A$ is the outer surface area of the coil and $\Delta T_m$ is the log mean temperature difference.

I am assuming this approach is correct, but I'm doubting my calculations because

1. It doesn't match with Immersion Coil Surface Area Calculator For Heating with Steam
2. It seems quite large

Apart from the temperatures I've mentioned above, these are the only parameters I've been provided with from the company:

• inner diameter of the tank = $3.35\,m$
• outer diameter of the tank = $3.6\,m$
• height of tank = $6.5\,m$
• density of the fluid = $850\,kg/m^3$
• viscosity of the fluid = $30, 000\,Cst$
• pressure at which steam enters the pipe = $150\,psi$
• Heat time of $2\,hours\,(7200s)$
• pipe diameter of $2'' (0.0508\,m)$

So here's the calculation I did:

$M_{fluid} = V * \rho = \frac{PI * 3.35^2 * 6.5}{4} * 850 = 49000\,kg$

Assuming the $C_p$ to be $1800\,J/kg.K$,

$Q = 49000 * 1800 * (40 - 25) = 1.323\,GJ$

I estimated $\Delta Tm$ to be $86$ using:

$\frac{(ΔT_1 - ΔT_2)}{ln(ΔT_1/ΔT_2)}$

With $T_1$ being the inlet steam - the final temperature of the fluid = 145 C.
With $T_2$ being the outlet 'steam' - initial temperature of the fluid = 45 C (I estimated the outlet 'steam' to be around 100.

Taking the $U factor$ to be 100 (estimated from this site),

A = 1.323 GJ / (100 * 86 * 7200) = $21m^2$.

And the corresponding length would be $\pi DL = A$ and $L = 131m$. Clearly something (or a few things) are wrong here since I don't think the tank can be that big so help would be appreciated on where I went wrong.

Answer 1

One thing to consider is that pipes used for heating like this are normally « finned » to increase the surface area - fins can be small discs fitted to the pipe or even wire loops - both increase the surface area massively and reduce the length required.

This article explains the finned pipe although they are collecting heat: The performance of a coiled finned-tube heat-exchanger submerged in a hot-water store: The effect of the exchanger's orientation.