First of all, English is not my native language, excuse me for all the grammar mistakes I will make.

At the same temperature, water feels cooler than air. This is explained here. I know the temperature we feel is highly related to the rate of heat loss of our body. At water, we lose heat faster, and that's why it feels cooler. Knowing that, I wanted to calculate at which temperature the air has to be to feel the same as water at 11ºC. To do so, I need to calculate the rate of heat loss of our body in water at 11ºC.

How can I calculate it?

The person is completely naked and submerged into the water. His temperature is 37ºC. He does not move in the water. The water flowing past the person is moving at a V speed, for example 10m/s.


  • 1
    $\begingroup$ Please narrow your question. You are asking about natural convection in water. This is influenced by: 1) the clothes the person is wearing; 2) any water flowing past the person; 3) the difference between skin temperature and water temperature; 4) how much of the person's skin is in contact with water (i.e., are they completely submerged?); 5) whether or not the person is moving while in the water (are they swimming or treading water?). $\endgroup$ Apr 2, 2017 at 22:44
  • $\begingroup$ @DavidWhite Edited. $\endgroup$
    – J doeoeo
    Apr 2, 2017 at 22:57
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    $\begingroup$ Please provide a diagram of a naked man submerged in water. $\endgroup$ Apr 3, 2017 at 5:48
  • $\begingroup$ @electronpusher done. $\endgroup$
    – J doeoeo
    Apr 3, 2017 at 13:42
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    $\begingroup$ @JMac It isn't, but does it really matter? Anyway, if the air can affect the conditions of water in any way, we should just suppose there is no air outside $\endgroup$
    – J doeoeo
    Apr 3, 2017 at 15:05

1 Answer 1


Heat loss can be calculated using this equation. $$\frac{dQ}{dt}=hA(T_{surface}-T_{fluid})$$ The difference between the water and air is likely to be h, the heat transfer coefficient. So the equation for air temperature calculation will be, $$h_{air}A(37-T_{air})=h_{water}A(37-11)$$ Area will be canceled. The air temperature will be, $$T_{air}=37-\frac{h_{water}}{h_{air}}*26$$

For 10m/s water flow, the Reynolds number is $1 \times 10^{12}$, so it is turbulence flow. For 10m/s air flow, the reynolds number sis about $6 \times 10^{8}$. So both a turbulence flow.

The heat transfer coefficient can be obtained for the convective heat transfer. The air heat transfer coefficient is about $32W/m^2K$ , refer to http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html and the water heat transfer coefficient is between $300-6000W/m^2K$. By the way, even a strong person may not stand in a 10m/s flow current.

This gives -1095~-6795 degC. So you see it is impossible to achieve the same effect with air. It really because water can take away significant heat.


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