Download Energy, work and power of the body

Energy, work and power of the body
For basal conditions there are about 25% of body's
energy used by skeletal muscles and heart, 19%used
by the brain, 10%used by the kidneys, 27% used by
the liver, spleen. The body uses food energy to operate
its various organs, maintain constant body temperature
and do external work, small percentage 5% of food
energy excreted in the feces and urine any energy left
over is stored as body fat.
Conservation of energy in the human body
A thermodynamic system exchange energy with its
surroundings by mean of heat and work. When heat is
added to the system and work is done by it ,both Q, W
are positive, when heat is transferred out of the system
and work is done on it Q, W are both negative.
The energy of the human body is controlled by the first
law of thermodynamics for a closed system.
∆U: changed in stored energy
∆Q: heat lost or gained
∆W: work done by the body.
It's an important to determine the rates for the
energy changes in a small time ∆t:
U Q W


t
t
t
Where
∆U/∆t: catabolic rate
∆Q/∆t: heat loss or absorption rate
∆W/∆t: mechanical power
Basal Metabolic Rate (BMR): The amount of energy
needed to perform a minimal body function (breathing,
pumping blood through the arteries under resting
conditions). BMR depends on thyroid function. A person
of an overactive thyroid has a higher BMR than with
normal thyroid function.
Since the energy used for metabolism becomes heat
and dissipated from the skin so it related to surface area
or the mass of the body.
BMR depends on the temperature of the body, if a
patient has temp. 40° C or 3 above normal, BMR is about
30% greater than normal.
Example: suppose you wish to lose 4.54 Kg through
physical activity or by dieting.(a). How long would
you have to work at an activity of 15 Kcal/min to lose
4.54 Kg of fat?(b) If you normally use 500 Kcal/day,
how long must you diet at 2000 Kcal/day to lose 4.54
Kg?
Solution:
(a) from the table below there are 9.3 Kcal/g of fat
T(min) (15 Kcal/min)=(4.54*103g)*9.3 Kcal/g
T
=4.2*104 Kcal/(15Kcal/min)
= 2810 min =47 hours.
(b)
T
energy of 4.54Kg fat
diet per day
4.2  104 Kcal

 84 day
5  102 Kcal / day
Work and power:
Chemical energy stored in the body is converted into
external mechanical work as well as in life-preserving
functions.
Mechanical work is usually defined by Δw=F. Δx
where F is the force on the same line of displacement
x, or it can be also written as:
(Δw= F Δx cos Ө) where Ө is the angle between F and
the direction of movement.
The power is work per unit time.
P= Δw/Δt = F Δx / Δt = F V where V is the velocity
Power P=∆W/∆t = F∆x/∆t
= FV
Note: When the body doing an external work this
requires a conversion of stored energy U to work W and
like any machine no conversion possible without any
heat loss.
The effieciency of conversion of U to W :
€=ΔW/ ΔU
An example : When a man is climbing a hill or walking
up stair we calculate the work done by multiplying a
person weight (m g) by vertical distance (h). when a man
walking at aconstant speed on level surface the force act
in the direction perpendicular to his motion, thus external
work done by him appears to be zero. However, his
muscles are done as internal work which appear as heat
and causes a rise in muscle temp.
Heat losses from the body
The human body core temperature is constant around
37.1 ± 1.0 oC. The hypothalamus of the brain contains
body thermostat to keep temp. close to normal value. If
the core temp. rises, the hypothalamus initiates sweating
and vasodilatation which increases skin temp. if the skin
temp. drops, the thermo- receptors on the skin inform
hypothalamus and it initiate shivering, which causes an
increase in the core temp. the difference between energy
radiated by the body and the energy absorbed from
surrounding can be calculated by:
Transfer of heat by radiation, evaporation, convection,
conduction. All objects regardless on their temp. emit
electromagnetic radiation, the amount of energy emitted
by the body is proportional to the absolute temp. raised
to the fourth power. The body also receives radiant
energy from surrounding objects. The amount of heat
difference between the energy radiated by the body and
the energy absorbed from the surrounding can be
calculated from the equation:
H r  k r AR e(TS  TW )
Hr energy loss or gain
Ar surface area of emitting radiation
e emissivity of surface
TS skin temp.
Tw surrounding temp.
Kr constant= 5kcal/m2hr.°C
The heat loss due to convection Hc is:
H C  K C AC (TS  Ta )
KC const. depend on the movement of air
AC surface area
Ta
air temp.
When the wind const. =2.3 Kcal/m3 hr.C, Ta=25 °C,
TS=34°C, AC=1.2m2 . The nude body loses about 25 Kcal
/hr by convection or when the air is moving, the constant
Kc increase according to the equation :
K C  10.45  V  10 V
Where V is wind speed in m/sec.
Homework :
Calculate the convective heat loss per hour for a nude
standing in a 5 m/sec wind. Assume TS=33°, Ta=10°,
AC=1.2 m2. if wind speed 2.23 m/sec find the still air
temp. that would produce the same heat loss. Hent
kc=27.8