The subject then returned to the hot environment, and exercise was resumed at the same rate for a further 30 minutes. The subject was then removed from the hot environment again to be weighed. The final period of 30 minutes was then commenced, with the subject being at rest in the cool environment. The subject was then weighed again.
All weights were calculated by weighing the subject 3 times, and calculating the mean. Aural temperature, and skin temperature at the calf, chest, and finger where recorded every 5 minutes during the experiment. The Douglas bag method was used to collect expired air in order to calculate oxygen consumption at STPD at the following times:
- during rest periods (1 and 4) – 2.5 - 7.5 minutes, 12.5 – 17.5 minutes, and 22.5 – 27.5 minutes,
- during exercise periods (2 and 3) – 3.5 – 6.5 minutes, 13.5 – 16.5 minutes, and 23.5 - 26.5 minutes.
RESULTS
Sample Calculations
Heat Balance = Heat production – Heat Loss
(M-W) – (H+E)
Heat production = M - W
M (metabolic rate) = total energy expenditure
W (work rate) = mechanical energy of exercise
Heat loss = H + E
H = C, K & R
H = heat lost or gained through conduction (C), convection (K), and radiation (R)
E = heat lost through evaporation (through skin surface and thought the respiratory tract)
Calculation of M
In order to calculate M, VO2 must first be calculated:
In order to calculate VO2, VE must first be calculated:
Calculation of VE
Volume of gas collected = 94.8
Collection period = 3 minutes
STPD factor = 0.9055
VE = (94.8/3) x 0.9055
= 28.61 l.min-1
Calculation of VO2
where 0.209 is the fractional O2 content of the inspired air, and FEO2 is the fractional O2 content of the expired air.
VE = 28.61 l.min-1
FEO2 = 0.161
VO2 = 28.61 x (0.0209 – 0.161)
= 1.40 l.min-1
Calculation of M
M (W.m-2) = VO2 x 21.1 x 1000
60 x AD
where VO2 = 1.40,
21.1 is the energy equivalence of 1l of oxygen. This is used to obtain the value in kJ.min-1
1000/60 is the number needed to multiply by to obtain the value in J.s-1
AD = subject’s surface area (used to obtain the value in W.m-2)
AD = 1.98 m2
M = 1.40 x 21.1 x 1000
60 x 1.98
= 248.47 W.m-2
Calculation of W
where 0.978 is a constant to take into account the circumference of the flywheel and the number of revolutions of the flywheel per revolution of the pedals
Load = 1.5 kP
Rpm = 60
AD = 1.98 m2
W = (1.5 x 60 x 0.978)
1.98
= 44.36 W.m-2
Calculation of H
In order to calculate H, Tskin (skin surface temperature) must first be calculated:
Calculation of Tskin
Tskin = (0.5 x Tchest) + (0.36 x Tcalf) + (0.14 x Tfingertip)
= (0.5 x 36.4) + (0.36 x 33.7) + (0.14 x 35.2)
= 35.3°C
Calculation of H
where Rd (thermal resistance of clothing) = 0.09
Tair = 39°C
Tskin = 35.3°C
AD = 1.98 m2
H = (35.3 – 39)
(0.09 x 1.98)
= -20.94 W.m-2
Calculation of E
In order to calculate E, evaporative fluid loss (g) must first be calculated:
Calculation of Evaporative fluid loss:
Evaporative fluid loss = (Subject weight pre – Subject weight post) + (Bottle weight pre – Bottle weight post) - (Weight of voided urine)
= (1000 x (74.02 – 73.98)) + (828 – 609) – (0)
= 259
Calculation of E
Evaporative fluid loss = 259
Time between weighings = 30 minutes
AD = 1.98 m2
E = (259 x 2.4 x 1000)
(60 x 30 x 1.98)
= 174.02 W.m-2
Calculation of Heat Balance
Heat Balance = Heat production – Heat Loss
(M-W) – (H+E)
Heat Balance = (248.47 – 44.36) – (-20.94 + 174.02)
= 51.03 W.m-2
Figures
Figure 1 – A graph to show heat balance against time
Figure 1 shows that time as increased, heat balance increased, peaked and then decreased. In the rest period 0-30 minutes, heat balance was fairly constant and stayed around 0. In the first period of exercise, 30- 60 minutes, heat balance increased dramatically and peaked at 103.19 W.m-2 at 45 minutes. It then decreased after this and went into negative value from 60 – 90 minutes (second period of exercise in the hot environment). The heat balance continues to decrease steadily throughout the final rest period (90 – 120 minutes).
Figure 2 – A graph to show aural temperature against time
Figure 2 shows that as time increased, aural temperature increased, then plateaued, before decreasing. In the rest period 0-30 minutes, aural temperature increased gradually. In the first period of exercise, 30- 60 minutes, heat balance increased to start with until it reached 37.50°C at 50 minutes, after which it stayed constant until the end of the second period of exercise in the hot environment (90 minutes). It was only when the subject was removed from the hot environment and the final rest period in the cool environment was commenced that the aural temperature began to come down.
Figure 3 – A graph to show VO2 against time
Figure 3 shows that as time increased, VO2 initially stayed fairly constant, then increased, before finally decreasing back to approximately its original level. In the rest period 0 - 30 minutes, VO2 remained steady but at the start of the first period of exercise, i.e. at 30minutes, VO2 increased to 1.4 l.min-1 and stayed at approximately this level during the whole exercise period in the hot environment (30 - 90). During the final rest period, 90 - 120 minutes, in the cool environment the VO2 decreased steadily.
DISCUSSION
The results which have been obtained are fairly good. This is shown in figures 1, 2 and 3 as they all correlate and portray what I would have expected. The conclusions that have been made are supported by scientific findings in this experiment. As shown in Figure 1 heat balance increases in a hot environment and goes back down when the subject returned to a cool environment at rest. The majority of heat lost was by sweating, this meant that heat balance increased. Figure 2 shows that the aural temperature increases in the hot environment. This was because heat gained from the surroundings and metabolism due to increased exercise was greater that heat lost.
Figure 3 shows that as the subject was moved into the hot environment and exercise was commenced the volume of oxygen consumption increased because of the greater demand of the body to produce energy (a process which requires oxygen).
The response to regulate body core temperature is related to negative feedback. Peripheral sensors detect a noticeable change in the external temperature and send signals to the central nervous system. From here signals are sent to the effector in order to create a response for instance vasodilation, which then has the effect of increasing the skin temperature in order to lose heat to the atmosphere.
In this experiment one of the errors was the relative humidity in the air. This posed a dilemma as the humidity was not constant throughout the 1 hour of exercise because of the enclosed environment in which the subjects were exercising. Therefore the subject’s sweating would have increased relative humidity in the chamber.
Another error in this experiment was that the body core temperature was taken aurally. However this is not the most effective and accurate method of reading the body core temperature. A more reliable method would have been to use an anal thermometer.
In conclusion, the results obtained in this experiment show that heat balance increases as the subject is moved into a hot environment and exercise is commenced. Aural temperature and the oxygen consumption also increased in relation to the heat balance.
REFERENCES
Schonbaum E, Lomax P. Thermoregulation: Physiology and Biochemistry. Pg 374 – 380. 1990.
Lomax P, Schonbaum E. Thermoregulation: Research and Clinical Application. Pg 256 – 289. 1989.
Mubeen Iqbal (SID:200242757) 29/10/07 -
