Question

Can you solve the question ?

Find out about the wall construction of the cabins of submarine, the range of ambient conditions under which they operate, typical heat transfer coefficients on the inner and outer surfaces of the wall, and the heat generation rates inside. Determine the size of the heating and airconditioning system that will be able to maintain the cabin at 200C at all times for a submarine capable of carrying 70 people.

Answer 1

Since no particular class of submarine is mentioned, we will take a look at the Balao Class Submarine which was used in World War 2 by the United States Navy and allies.

The typical cross-section of the wall of a Balao Class Submarine is as shown below:

Heat travel direction T(water)= 70 T(inside) = 20 C 1/4 inch Aluminium 7/8 inch Steel - 1 inch Fibreglass insulation

It is usually estimated that an average human being dissipates around 90-100 W of energy. For the safer side, we will assume it to be 100 W.

There are 70 humans on board and thus, 70 x 100 = 7000 W = 7 kW of heat energy generated inside the Submarine just by the human beings.

There are many systems and computers and machines on a submarine. Although an exact figure is always classified, as evident in many websites, the closest approximation I could find was around 2500 W (all systems combined - including generators, computers, sonar systems, tubelights, bulbs, emergency systems etc).

Thus, a net heat load of 9500 W can be expected.

Now, this is just the amoung of heat that is produced internally. We will have to investigate the amount of heat that leaves the submarine due to the temperature difference between the inside and the outside.

Although the submarine is not perfectly cylindrical, we consider an average outer diameter of 7 m.

Also, the Balao Class submarines had a length of about 95 m.

There are a total of 3 thermal resistances applicable.

• Internal Convection resistance (R1)
• Conduction resistance through Steel (R2)
• Conduction resistance through Fibreglass insulation (R3)
• Conduction resistance through Aluminium (R4)
• External Convection resistance (R5)

--------------

Internal Convection resistance (R1)

Thermal Resistance = $R1=\frac{1}{h_{internal}A}$

where

• h(internal) = Internal heat convection coefficient = 14.1956 W/m2K (from the internet)
• A = Internal Surface Area = $\prod DL+\left (\frac{\prod D^{2}}{4} \right )\times 2=\prod \times 6.946\times 95+\left (\frac{\prod 6.946^{2}}{4} \right )\times 2=2148.83 m^{2}$
  • D = Inner Dia = Outer Dia - thickness = 7 - 0.054 = 6.946 m
    • Outer Dia = 7 m
    • Thickness = 2.125 inches = 0.054 m
  • L = Length = 95 m

Thus,

R1 = 3.2783e-5K/W hinternal A hits 14.1956 x 2148.83

-----

Conduction resistance through Steel (R2)

Thermal Resistance = $R2=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{steel}L}$

where

• k(steel) = 14.4 W/mK (Stainless Steel - from the internet)
• L = Length = 95 m
• r(i) = Inner radius = 6.946 / 2 = 3.473 m
• r(o) = Outer radius = Inner radius + thickness of steel = 3.473 + 0.022 = 3.495 m
  • Thickness of steel = 0.875 inches = 0.022 m

Thus, $R2=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{steel}L}=\frac{ln\left (\frac{3.495}{3.473} \right )}{2\prod \times 14.4\times 95}=7.3465e-7\: K/W$

-----

Conduction resistance through Fibreglass insulation (R3)

Thermal Resistance = $R3=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{fibreglass-insulation}L}$

where

• k(fibreglass-insulation) = 0.043 W/mK (from the internet)
• L = Length = 95 m
• r(i) = Inner radius = 6.946 / 2 = 3.495 m
• r(o) = Outer radius = Inner radius + thickness of steel = 3.495 + 0.0254 = 3.5204 m
  • Thickness of steel = 1 inches = 0.0254 m

Thus, $R3=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{fibreglass-insulation}L}=\frac{ln\left (\frac{3.5204}{3.495} \right )}{2\prod \times 0.043\times 95}=2.8213e-4\: K/W$

----

Conduction resistance through Aluminium (R4)

Thermal Resistance = $R4=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{Al}L}$

where

• k(Al) = 170 W/mK (from the internet)
• L = Length = 95 m
• r(i) = Inner radius = 6.946 / 2 = 3.5204 m
• r(o) = Outer radius = Inner radius + thickness of steel = 3.5204 + 0.00635 = 3.52675 m
  • Thickness of steel = 0.25 inches = 0.00635 m

Thus, $R4=\frac{ln\left (\frac{ro}{ri} \right )}{2\prod k_{Al}L}=\frac{ln\left (\frac{3.52675}{3.5204} \right )}{2\prod \times 170\times 95}=1.7759e-8\: K/W$

-----

External Convection resistance (R5)

Thermal Resistance = $R5=\frac{1}{h_{external}A}$

where

• h(external) = External heat convection coefficient = 454.261 W/m2K (from the internet)
• A = Internal Surface Area = $\prod DL+\left (\frac{\prod D^{2}}{4} \right )\times 2=\prod \times 7\times 95+\left (\frac{\prod \times 7^{2}}{4} \right )\times 2=2166.128m^{2}$
  • D = Outer Dia = 7 m
  • L = Length = 95 m

Thus, $R5=\frac{1}{h_{external}A}=\frac{1}{454.261\times 2166.128}=1.0163e-6 \: K/W$

------------------------------------------------------------

Thus, net heat Transfer via the walls is:

$Q=\frac{T_{inside}-T_{water}}{R1+R2+R3+R4+R5}$

We have all the required values, thus,

$Q=\frac{20-7}{3.2783e-5+7.3465e-7+2.8213e-4+1.7759e-8+1.0163e-6}$

=> Q = 41050.685 W

----------------------------------------------------------------

Now, the Submarine is generating 9500 W of heat energy and 41050.685 W of heat energy is leaving the system. Thus, the submarine will soon turn very cold if adequate heating system is not maintained.

Thus, our heating system's capacity should be 41050.685 - 9500 = 31550.685 W.

We need to have a heater with a minimum Heating effect of 31550.685 W to be able to survive on this Submarine.

---------------------------------------------------------------

Note: The data provided here is only indicative. The actual heating/cooling load depends on the actual design data of the Balao Class Submarine which remains classified. The data that I have used here is available in the public domain and is in very vague form. Thus, the heating load is purely indicative and not a reflection of heating load of the actual Submarine.

---------------------------------------------------------------

Kindly upvote if you are satisfied with my efforts. :)