Question

A submarine is submerged and cruising at 10 m/s (~ 20 knots). If the engines are supplying 1.67 MW of power to propel the submarine at cruising speed, estimate the average heat transfer coefficien

Answer 1

An estimate the average heat transfer coefficient is 2755 W/m^2K

Step-by-step explanation:

The full question is:

"A submarine is submerged and cruising at 10 m/s (~ 20 knots). If the engines are supplying 1.67 MW of power to propel the submarine at cruising speed, estimate the average heat transfer coefficient between the submarine and sea if the surface area of the sub is 3140 m^2. The properties of seawater are approximately given by water at 10 C, density = 1000 kg/m^3, cp = 4189 J/kg-K, viscosity = 1225 • 10^-6 kg/m-s, k = 0.59 W/m-K and Pr = 8.81. (5200 W/m2-K)"

For this estimation we can approximate the surface area of the submarine to be the one of a sphere, thus we have

`A = 4\pi r^2 \\ A = 4\pi (D/2)^2 \\ A = \pi D^2`

So its diameter is approximately

`D = \sqrt{\cfrac{A}\pi}`

Thus the Reynolds number is:

`Re = \cfrac{\rho v D}{\mu}`

Replacing values

`Re = \cfrac{1000 \cfrac{kg}{m^3 }* 10 \cfrac ms * \cfrac{3140}{\pi}m}{1225 * 10^(-6) \cfrac{kg}{m s}}\\Re= 8.15913*10^9`

Since we have a very high Reynolds number we know the flow is turbulent thus we can use a correlation like Dittus-Boelter

`N_u= 0.023 Re^(0.8)Pr^(0.4)`

Where the Nussel number can be written as

`N_u= \cfrac{HD}{k}`

So we have

`\cfrac{HD}{k}=0.023 Re^(0.8)Pr^(0.4)\\H=0.023\cfrac KD Re^(0.8)Pr^(0.4)`

Replacing values we get

`H=0.023\cfrac {0.59 \cfrac{W}{mK}}{\cfrac{3140}\pi m} (8.16*10^9)^(0.8)(8.81)^(0.4)\\H=2755.13 \cfrac{W}{m^2 K}`

So we get an estimate the average heat transfer coefficient as 2755 W/m^2K