Question

Prior to making a polymer solution for coating plastic film a solvent is pumped into a jacketed carbon steel, Teflon®-lined vessel that is 3 m in diameter with a straight side of 4 m. The liquid depth is 3 m and the jacket extends 3 m up the vessel side. The vessel has an agitator installed. The impeller is a glass-lined hydrofoil that is 1210 mm in diameter with a power number of 0.39 operating at 125 RPM. The fluid has the following properties: Density 1000 kg / m3 Viscosity 0.001 Pa s Specific heat capacity 4180 J / kg °C Thermal conductivity 0.591 W / m °C The vessel’s carbon steel wall is 10 mm thick and the Teflon® lining is 1.5 mm thick. The heating medium in the jacket is condensing steam at 1 bar(a) pressure 100 deg C. The initial temperature of the solvent is 20 °C and final is 60 °C at which point the solid polymer pellets are added.
What is the process side heat transfer coefficient?
What is the overall heat transfer coefficient?

Answer 1

Power number = 0.39

$\Rightarrow P=0.39\rho N^{3}D^{5}$

$\Rightarrow P=0.39\left ( 1000 \right ) \left ( 125/60 \right )^{3}\left ( 3 \right )^{5}$

$\Rightarrow P=856933.5938\left ( Watts \right )$

Area of heat transfer, A

$A=2\left ( 3.14 \right )\left ( 1.5 \right )\left ( 3 \right )$

$\Rightarrow A=28.26\left ( m^{2} \right )$

By using heat balance

$P=UA\Delta T_{lm}$ ...........................................(2)

$\Delta T_{lm}=\frac{80-40}{ln\left ( \frac{80}{40} \right )}=57.71^{0}C$

From equation 2, we have

$856933.5938=U\left ( 28.26 \right )\left ( 57.71 \right )$

$\Rightarrow U=525.461\left ( \frac{W}{m^{2}C} \right )$

So, the overall heat transfer coefficienct is 525.461 (W / m²C)

Now

We can write

$UA\Delta T_{lm}=hA\Delta T$

$856933.5938=h\left ( 28.26 \right )\left ( 100-20 \right )$

$\Rightarrow h=379.04\left ( \frac{W}{m^{2}C} \right )$

So, the process heat transfer coefficienct is 379.04 (W / m²C)