Cooling the casualty - "Must Read"

Cooling with Ice and water facilities available
The quickest and least risk way of cooling a victim of heat stroke is whole body cooling, immersing them in a bath of ice cold water, and then monitor their temperature and symptoms until these have subsided.

Cooling when access to civilised facilities is not possible
If ice and cold water are not available in sufficient quantities for whole body immersion, then forced convection evaporative cooling should be used as described below.
Whole Body Evaporative Cooling - Technique

(1) When the victim is conscious all [ or most ] of their clothing should be removed and they should kneel on all fours on the ground. This position keeps most of the body's surface available for cooling.

(2) If the victim is unconscious then, in case of vomiting or breathing difficulties, the balance of risks suggests it is better to put the victim in the recovery position laying on the ground and to fan as much of the body surface as possible.

(3) As much of the body as possible should be covered in water. If water is in limited supply or some areas cannot easily be kept wet, then consider using thin cloth soaked in water and laid, held or tied on to the body.

(4) Vigorous fanning of the victim should start as soon as possible. Fanning should be from the side of the body so the air flows over both the upper and lower surfaces a bit like an aeroplane wing.

(5) Fanning from above on the back will not create effective air flow over the front of the body.

(6) The faster the air movement over the body the greater the cooling.


Extra Information

Medical specialists are clear that for heat stroke caused by exercise whole body cooling is the best way to get the body's core temperature down as quickly as possible.
A number of ways of cooling people have been measured and the most effective, with very few side effects, is putting the victim in a bath of ice cold water.

From the results of such cooling studies it is possible to work out approximately how quickly heat energy is lost - as shown in the table below.

Table of [approximate] heat loss

Cooling Method Temperature reduction Cooling time from 40 oC to 37 oC Approximate rate of heat loss

Ice bath immersion

0.2 oC per min 15-20 mins 1200 watts

Warm air and water spray

0.08 oC per min 35-45 mins 500 watts

Rest covered with cool wet towels

0.04 oC per min 70-90 mins 250 watts

The ice, cold water and equipment needed for cold water baths are usually available when incidents happen close to civilisation.
However in the wild the facilities needed are not likely to be available. When working in agriculture, or on industrial sites away from towns, similarly there may not be a supply of ice and or cold water.
When such help is not to hand the technique of evaporation cooling should be used where you spray the nude body with water and provide a rapid air flow with vigorous fanning .


Heat loss with air movement over the body
For the environmental conditions likely to be met in the field heat loss is roughly proportional to the air speed over the body.
The rate of heat removal [ in watts ] is roughly doubled at an air flow of 6 kph, compared to zero kph in the table below,
Note: 6 kph = 3.75 mile per hour = 5.5 feet per second = 1.7 metres per second.
This increase in cooling is approximately true for all the temperatures and relative humidities likely to be experienced.
ie 12 kps has three times the cooling of 0 kph and 18 kph 4 times as much.

Estimating the length of time that fanning has to go on for is very difficult. Rectal temperature measurement is the only sure way of knowing cooling has been successful. Without rectal temperature being available the return of mental capabilities and the disappearance of other symptoms are indications of cooling.


The table below shows the theoretical maximum evaporative cooling in still air. The table can only be used as a very rough guide.
If fanning/ air flow is used then cooling is increased as indicated above.

It is worth remembering that the average female weight is 50 kgs and that average female surface area is 70% of the average male.

Theoretical maximum cooling by evaporation for an average male of 70 kgs
Watts of heat loss for a nude body
with an average area of 1.8 sq meters

o F 80 85 90 95 98.6 100 105 110 115 120
o C 26.7 29.4 32.2 35.0 37.0 37.8 40.6 43.3 46.1 48.9
Relative 0% 1306 1306 1306 1306 1306 1306 1306 1306 1306 1306
Humidity 10% 1233 1220 1206 1189 1170 1148 1123 1095 1063 1027
20% 1160 1135 1105 1072 1033 990 940 884 820 749
Air speed 30% 1087 1049 1005 955 897 832 757 673 578 470
0 mph 40% 1015 964 905 838 761 674 574 462 335 192
0 kph 50% 942 878 805 721 625 516 392 251 92 -87
60% 869 792 704 604 489 358 209 40 -151 -365
70% 796 707 604 487 352 199 26 -171 -393 -644
80% 723 621 504 370 216 41 -157 -382 -636 -922
90% 651 536 404 253 80 -117 -340 -593 -879 -1201
100% 578 450 304 136 -56 -275 -523 -804 -1121 -1479


For those interested in why the cooling effect changes so much with humidity the following notes may be useful.

Notes on humidity effects and cooling.

1. Still air evaporation.
In the slightly over idealised view used for the table above, if there is no water vapour in the air, that is at 0% relative humidity, a nude body of 1.8 square meters area [ about 18 square feet ] could evaporate in one hour 1.88 litres of water [ about three and one third pints ] into still air .

2. Why that much ?
This figure comes from

  • the area available for evaporation over the whole body
  • the temperature of the water which is evaporating.
The warmer the water the faster it will evaporate. In this case the water is considered to be at body temperature of 37 oC.
1.88 litres evaporated in an hour would cause heat to be lost at a rate of about 1300 watts. If you look at the top line of the blue part of the table it is constant at all external temperatures because there is no effect from humidity.

3. The effect of humidity [ at any one temperature ]
If you look at the columns of figures in the table above a column gives you the cooling at one external temperature with increasing humidity. The cooling effect reduces as the humidity goes up because the amount of water that can evaporate per hour reduces.

4. The same relative humidity at different temperatures
If you look at the rows in the table the heat loss decreases for a particular relative humidity as the temperature goes up. Relative humidity is a measure of the water in the air as a percentage of the maximum water carrying capacity of air at the temperature concerned. As the air gets hotter it can carry more water per cubic meter before becoming saturated. So as the temperature goes up, at constant relative humidity, the weight of water vapour in the air goes up. With more water molecules in the surrounding air less water can evaporate and this effect reduces the amount of cooling possible.

5. Condensing effects
The red areas on the table above show that, at high temperatures and high relative humidity, heat is gained by the body. When the temperature of the water which is trying to evaporate is less than the surrounding environment and there is sufficient water vapour in the air for the water vapor in the air to want to condense, then water condenses on the skin instead of evaporating. Condensing water does the reverse of evaporating water, it releases the heat into the body instead of removing it.
This is a very dangerous situation.