Body heat dissipation and related water loss

Nico A.M. Schellart, Dept. of Med. Physics, AMC

Principles

The type of energy, e.g. heat, motion and radiation, produced by the body (nearly) all depends on chemical processes. With a constant body mass (no grow or weight loss) and in rest nearly all energy is transformed to heat, also the small amount of mechanical energy of heart and ventilation. All heat is transferred to the environment and there are various components of this heat transfer.

Calculation of he components can be performed by applying (semi-)empirical equations (curve fitting of experimental data) or by applying physical laws, as will be done here.

Table 1 summarises all components of heat release with their numerical values of a normal shaped man in rest and when performing heavy endurance sport (running). Some of the components are bidirectional, in other words heat can also be collected. This holds for radiation (sun bathing) and convection (as extreme example the hair dryer). The five components are dependent on biometric factors and to calculate numerically their contribution this factors should by defined. This is done in More Info.

Table 1

Heat release man in rest, 40 year, 75 kg, 175 height, body area 1.9 m^2
Components of release (vair < 0.15 m/s)	seated in rest	running 15 km/h
Cres, expiration of gas	1.2	41.8
Eres, evaporated water in expired air	9.3	332
Rskin, radiation	32.8	116.3
Cskin, convection along the skin	14.6	50.3^#
Eskin, perspiration and evaporation via skin	21.1*	685**
Total heat release	79	1226

^# lower limit (see More Info), *sweating 16 mL/h, ** sweating 2 L/h.

Two very small components, heat conduction and sound production (mainly heart) are neglected.

The dissipated heat of the runner is 15 times that of the seated person in rest. The runner has to dissipate this heat otherwise he will suffer from hyperthermia. A rough calculation yields a body temperature increase of 0.22 K/min. In rest, E_res and E_skin (for wind speed v < 0.15 m/s) together is 23.4 W, about ⅓ of the total, but during running ca. ¾. R_skin is 45 W, about ⅔ of the total loss of 79 Watt in rest. During running loss by radiation and convection become less important. Total water losses by evaporation and perspiration is 46 mL/h for the seated person in rest and 1342 mL/h for the runner (see More Info), 29 times more. The calculations show the enormous difference between both conditions.

Total loss in rest should be nearly the same as the basal energy expenditure, which is equal to:

male: BEE = 66.67 + 13.75W + 5H - 6.76A (kCal/day) (1a),

female: BEE = 665.1 + 9.56W +1.85H -4.68A, (1b)

where H is height (cm), W is weight (kg), A is age (year). With BEE_Watt = 0.0484 BEE_kCal/h, our model has an BEE of 82.4 Watt. The energy expenditure in rest and reclining is about 2% lower, but seated it is some 8% higher. All together, the calculations of heat losses and heat production are well in accordance.

The components can be calculated as follows. More Info gives details and the calculations of the values of Table 1.

Heat release by expiration

C_res = m·c_p·ΔT, (2)

with m the mass of gas expired per second, the temperature difference ΔT between inspired and expired gas and c_p the specific heat coefficient (the amount of energy to increase the temperature of one unit of mass of gas with 1^o C under constant pressure) of the expired gas.

Heat release by expired water vapor

E_res = m_H2O·ΔH_H2O, (3)

with m_H2O the mass of water vapor and ΔH_H2O the specific evaporation heat of water:

Heat release by radiation

The peak wavelength of the infrared radiated light by the human body is about 10 µm (calculated from the skin temperature with Wien’s law). The law of Stefan-Boltzmann says that a black emitting body radiating in an infinite space with 0 K as background temperature emits σ·A_body T⁴_body Watt. The constant σ is the constant of Stefan-Boltzmann, being 5.7 x 10^-8 W/(K⁴.m²). Extending the law for grey bodies (law of Kirchhoff) and a temperature > 0 K the equation becomes:

R_body =ε_body·σ·A_body (T⁴_body - T⁴_background)

≈ 0.5·ε_body·σ·A_body )T()T + 2T_background)³, (4)

where )T = T_body - T_wall and ε_body the emittance coefficient (0.8 <ε_body <1.0, and thin-clothed 0.95).

Convection along the skin

A difference in temperature always results in heat transport from a medium with a high temperature to a medium with a low temperature. Under many conditions, the human body releases heat to the surrounding air. (But the process can reverse, for instance by entering a warm room).The underlying mechanism is that the air particles colliding with the skin (the “wall”) obtain a larger momentum at the cost of the velocity of the Brownian motion of the skin particles. And so, the air particles increase their velocity, and consequently the boundary layer of air covering the skin obtains a higher temperature. This heated layer has a lower specific density than the cooler air at a larger distance. In rest, this difference causes a laminar ascend of the boundary layer. This is the process of heat release by convection. Heat release by convection is hard to calculate and various approaches can be found in literature (see Literature). With laminar convection the problem is easier than with turbulence, although still complicated.

A laminar gas flow has a Rayleigh number (Ra) between 10⁴ and 10⁸. Convection currents with a velocity v_air < 0.15 m/s (generally indoor) appear to be laminar since Ra is about 1.0x10⁶. Under the above conditions the refrigeration law of Newton (see Refrigeration) applies:

C_skin = α∙A∙ΔT, (5)

where α the heat convection coefficient, A the area of the body and ΔT the difference in temperature between skin and ambient air. The parameter α is 1.35∙(ΔT/H)^{1/ 4} at 1 bar and 20 ^oC, with H the effective height of the subject.

Perspiration and evaporation via the skin

Perspiration is the process of water evaporating though the skin driven by the vapor pressure difference in the outer skin and the lower vapor pressure of the surrounding air. Evaporation is the process of sweat evaporation from the skin surface. In rest with a low skin temperature there is no sweat production, but with a high skin temperature there is some sweat production. The release is calculated by:

E_skin = m·ΔH_{skin water,} (6)

where m is the loss of mass of liquid and ΔH the specific evaporation heat.

Application

Aerospace, altitude, and sports medicine. Occupational medicine for heavy exertion and extreme environmental conditions. Further air-conditioning (general, hospitals and commercial aviation) and clothing industry.

A typical application Dehydration during long flights is not caused by the cabin low humidity. The extra loss of water due to the humidity is only about 250 mL/day (see More Info). The actual reasons are a too low liquid intake by food, at all drinking to few and acute altitude diuresis.

More info

To clarify the equations the components of heat loss will be calculated in examples. First a human heat-model (standard subject) should be specified:

- 40 years, male.

- Body weight W = 75 kg, length H = 175 cm, body area A =0.007184∙W^0.425∙H^0.725 =1.90 m² .

- In rest, seated (on a poorly heat-conduction seat, indoor) and running during heavy endurance sport (running, 15 km/hour).

- Sweat production 0.26 in rest and 33 mL/min when running. Thin clothing.

- RMV (minute respiratory volume of inspiration) is 5.6 l/min in rest, in sitting position. During running 200 L/min.

- FI_N2/FE_N2 = 1.06 (N₂ fraction in inspired air/N₂ fraction in expired gas. From this ratio RMV_expiration (= RMV·FI_N2/FE_N2 ) is calculated.

- Temperature of expired air is 310 K, independent of the ambient air temperature (actually there is a small dependency).

- Temperature of the skin T_skin is 303 K (30 ^oC) at rest and 310 K when running.

- Ф_m = 88.2 W (seated some 10% more than lying). Ф_m is total metabolic power. Ф_m is age and sex dependent (see above).

- Air velocity: indoor 0.15 m/s: outdoor, produced by the runner, 4.17 m/s.

C_res, expiration of gas

C_res = {RMV_exp·ρ₀·(273/T)·p/60}·c_p,_air·ΔT, with: (2a)

- RMV_exp = (FI_N2/FE_N2)RMV = 1.06·RVM L/min. RVM is 5.6 and 200 L/min;

- ρ₀ = 1.29 kg/m³, the specific density of air at 273.15 K and 1 bar;

- T = 310 K, the temperature T of the expired gas is;

- p = 1 bar, the ambient pressure;

- 60, the conversion factor from minute to second;

- c_p,_air = 1.00 kJ·kg^-1K^-1, the specific heat capacity (at 0 ^oC);

- ΔT = 12 K (ambient temperature is 298 K).

After completing all values, 1.17 and 66.2 W is found for rest and running respectively.

A completely different, empirical approach (ref. 1) is :

C_res = 0.0014 Ф_m (307 -T_ambient), (2b)

where Ф_m the total metabolic power.

In a commercial aircraft, C_res is about 0.8² smaller since both ρ₀ and c_p,_air are reduced with 20%.

E_res, evaporated water in expired air

Starting from RMV_exp = RMV·FI_N2/FE_N2, considering the fraction of evaporated water vapor, correcting for temperature, considering seconds, the volume per second (m³/s) is found. Via the molecular volume of 22.4 m³ and the molecular weight of water (m_H2O), and ΔH_H2O (2260 kJ/kg), the evaporation in kg/s is found.

E_res = {[RMV·(FI_N2/FE_N2)·(FE_H2O - FI_H2O)·(273/310)/60]/22.4}m_H2O ·ΔH_H2O (3a)

Between brackets the volume of water vapor in m³/min at 273 K is found (FE_H2O = 0.0618 and FI_H2O = 0.003, only about 9% humidity, 310 K is body temperature). For the standard subject in rest E_res becomes 9.3 W and 332 W for the runner.

The water loss is 14.8 and 529 mL/h respectively.

There is no pressure dependency (mountaineering, diving) since the alveolar p_H2O is always 6.3 kPa. Consequently, in a commercial aircraft, E_res is about the same supposing FI_H2O = 0.003, which means very dry air. With a normal humidity (60%), water loss is about 30% less. The example shows that ventilatory water loss can be neglected in rest.

An empirical approach, modified after ref. 1, is:

E_res = 0.0173 Ф_m(5.87-p_H2O,ambient/1000). (3b)

R _body, Radiation

Subject in rest within a room Often, the body is surrounded at a finite distance by a wall. Whereas the body radiates in all directions to the walls, each small part of wall basically radiates to the rest of the wall and to the subject. A new parameter C is introduced to control these effects. It comprises the surface of the body and the wall, and the emittance factor of the body and the wall. Using the approximation of (3) the result is:

R _body ≈ C·σ·A_body )T()T + 2T _background)³/2, (4a)

The parameter C is defined as 1/C = 1/ε_body + (A_body /A_wall)(1/T_wall -1). The effective area of the sitting body A_s-skin is 1.33 m², (70% of A_skin,) and ,_wall = 0.9. With a room of 70 m³, C appears to be 5.4 x 10^-8 Wm^-2K^-1. With T_skin = 303 K and T_wall= 298 K the radiant power is 29.4 W/ m². In a similar way the radiant power of the chamber wall in the direction of the subject can be calculated. It amounts to 10.1 W/m². Since the absorption coefficient of the subject is about 0.65, the net dissipation is 29.4 – 4.8 = 24.6 W/m². For the sitting standard subject this finally yields R_skin = 32.8 W. When A_wall>>A_skin, in the definition of C, the second term at the right can be ignored.

Subject running outdoor Equation (4) with T_skin = 310 K yields 116.3 W.

A simple approach for most typical indoor conditions (ref. 1) is:

R’_skin = 4.7∙A∙ΔT (W) (4b).

This results in 31.3 W (sitting subject). For a large )T with a high body temperature this linear approximation is less accurate.

C_skin Convection along the skin

Subject in rest within a room The dependency of the heat convection coefficient α on the effective height H_e implies a dependency on body posture. For lying, sitting and standing the effective surface is some 65%, 70% ad 73% of total body area respectively, and H_e is some 17, 80 and 91% of actual height. This yields values for α of 2.7, 2.2 and 1.8. With ΔT = T_skin – T_air = 303 – 298 = 5 K and completing the equation C_skin = α∙A∙ΔT for reclining, seated and standing posture C_skin is 16.8, 14.6 and 12.5 W respectively, values closely together.

Subject running outdoor The runner has a ΔT of 12 K. His effective height is supposed to be 85% of the actual height and his effective surface 100%. This three changed values give an increase of a factor of 4.02 compared to standing in rest. Supposing that the factor α still applies for an air velocity of 15 km/h, than the convection loss of the runner is 50.3 W.

Air velocities v > 1 m/s give a substantial increase in C_skin: the factor of proportionality is (v∙p)^0.6. From this factor the wind-chill temperature factor can be calculated. The ratio of the air velocities yields a chill ratio of 7.35. This yields 92 W. The original value of 50.3 W is too low and the latter an upper limit. The factor α needs correction, since the thermal diffusivity (see Rayleigh number) is much higher in turbulent convection (the runner) than with laminar convection. The deviation from the upper limit depends on the clothing of the runner (extend of turbulence).

Foggy air augments C_skin.

Heat release by laminar convection is pressure dependent: C_{p bar} = p^¼C_{1 bar}. Consequently in aircrafts and at altitude it is less and in hyperbaric chambers (for hyperbaric oxygen treatment) it increases. The dependency on pressure can be clarified conceptually and only qualitatively as follows. With a higher pressure, there is a higher density and so more collisions with the wall. This effect augments the heat release. But a higher density means that in the gas there are also more collisions, reducing the “diffusion” of the heat. Also the flow of the convection behaves different. This all results in the exponent of ¼.

E_skin Perspiration and evaporation via the skin

E_skin = m·ΔH_{skin water.}

The mass (m) of sweat and perspirated skin water is 8.7 x 10^–6 kg/s (0.75 kg/day, supposing both contributions are the same at a skin temperature of 30 ^oC), and ΔH_{skin water} is 2428 kJ/kg (slightly higher than for pure water due to the dissolved minerals). The calculation yields E_skin = 21.1 W.

Supposing that 50% of the sweat production of the runner evaporates, perspiration and evaporation is 282 10^–6 kg/s ml is consequently E_skin = 685 W. With a large sweat production the calculation is actually more complicated since part of the sweat evaporates and the remaining part is cooled and drips off.

E_skin reduces with pressure (altitude) since the water particles make less collisions with the air particles which hampers their diffusion in the surrounding air. In an aircraft cabin, the dry air increases the water loss due to perspirated by some 30%, as holds for the water loss of E_res. Evaporation also increases some 30%, due to a higher convection.

Literature

1. ASHRAE, Fundamental Handbook, Ch. 8 Physiological Principles, Comfort and health. American Society of Heating, Refrigerating and Air-Conditioning Engineers, Atlanta, 1989.

2. Bernards J.A. and Bouwman L.N. Fysiologie van de mens. 5ft edition. Bohn, Scheltema & Holkema, 1988.

3. Polytechnisch Zakboekje, 1993, PBNA, Arnhem