# Conductive heat transfer water through concrete to water

So I've been asked to do a calculation on heat loss from a floating swimming pool. Strange and not something you're likely to come across on an everyday basis.

I have already calculated the evaporation loss, convection loss and radiative loss. The last part I am unsure of is the conductive loss to the surrounding water.

The pool will be heated to 25°C and I am to take the surrounding water temperature as 3°C. My first point of call was the typical:

$$Q=k A \frac{\Delta T}{d}$$

Since this i have been asked to double check my calculation as the value of 61.9kW was seen to be on the low side.

My questions are as follows;

1. Does the fact that I am transferring heat from water to water affect the calculation?
2. The fabric build up is 200mm insulation (@ 0.025w/m$$^2$$ U-value) and 300mm concrete (@ 1.3W/m$$^2$$) do I need to take this as a single fabric element, or can i work each out individually and sum the totals?

Any help anyone can offer on this is greatly appreciated.

• What is a "floating swimming pool"? – Bob D Feb 27 at 19:51
• This depends more on how your full model is working. On the sides of this vessel, the heat loss due to conduction in the walls will also depend on the convection of the water on both sides, since the heat loss through the layers would be in series with each other. – JMac Feb 27 at 19:58
• Also, (1) will the heater in the floating swimming pool stay on and theoretically maintain the temperature of the pool and (2) can we assume the surrounding water is so massive that it is an ideal heat sink (no change in temperature)?. In other words, are you calculating the heat loss rate at steady state conditions? – Bob D Feb 27 at 20:00
• You need to know the density and heat capacity of the insulator and brick. If you set all the parameters, I calculate the heat flux. Does the brick come in with water in the pool? – Alex Trounev Feb 27 at 21:28
• A floating swimming pool is a something that has been adopted in Nordic countries to offer a safe place to swim within a reservoir. these swimming pools do not tend to be heated. They tend to be mounted upon height adjustable legs that rise and fall with the water level. – Abul Lais Mar 1 at 9:11

We use the heat equation:

$$\rho c_p\frac {\partial T}{\partial t}=\lambda \nabla^2 T$$

Set for brick wall thickness $$L_b=0.3$$ and for insulator $$L_i=0.2$$ (use SI):

brick $$\rho =2000, c_p=800,\lambda =1.35$$ for $$0\le x\le L_b$$

insulator $$\rho =100, c_p=1260, \lambda =0.025$$ for $$L_b

Boundary conditions for $$T(t,x)$$

$$T(t,0)=3, T(t,0.5)=25, t>0$$

Initial conditions: $$T(0,x)=3$$

In fig. shows the temperature profile $$T(t,x)$$ and heat flux $$q=-\lambda \nabla T$$ from the pool. Temperature profile is set for a week, heat flux reaches $$q_m=-59.3889$$. 