Energy, Work and Power of Human Body PDF

By Dr/ Mohamed A. EL-Hag 1 Energy  The simplest definition of Energy is the ability to do work  There are many different forms of energy in the universe  Ex. Mechanical, thermal, chemical, electrical and nuclear energies By Dr/ Mohamed A. EL-Hag...

By Dr/ Mohamed A. EL-Hag 1 Energy  The simplest definition of Energy is the ability to do work  There are many different forms of energy in the universe  Ex. Mechanical, thermal, chemical, electrical and nuclear energies By Dr/ Mohamed A. EL-Hag 2 Conservation of Energy  “Energy never destroyed or created but it can transformed from one form to another”  Conservation of energy can also called the 1st law of thermodynamics By Dr/ Mohamed A. EL-Hag 3 Conservation of energy in the body  From a physics point of view, human body is considered to be an energy converter that should satisfy the law of conservation of energy Human body (Energy Input Energy Output Energy Converter) (Food energy + body fats) (Work done+heat loss) By Dr/ Mohamed A. EL-Hag 4  Conservation of energy within the body can be expressed as Change in stored energy in the body (i.e. food energy and body fats) = heat lost from the body + work done  Our bodies use food energy to  1) operate its organs  2) Maintain constant body temperature  3) Do work By Dr/ Mohamed A. EL-Hag 6  Units of Energy  In the international system (S.I.) Joule  In the French system erg  1 J = 107 erg  Joule and erg are the unit for measuring energy by physicists  The energy value stored in the food is measured in a special unit called calorie (cal). Therefore, calorie is unit used for measuring energy used by nutritionists)  1 cal = 4.18 J  1 Cal. = 1 Kcal. By Dr/ Mohamed A. EL-Hag 7 Energy changes in the body  Energy is produced in the body through the oxidation of glucose (the simplest form of sugar)  “The rate of oxidation of glucose within the body” is called the metabolic rate  Oxidation of one mole of glucose C6H12O6 (180 gm) produce 686 Kcal C6H12O6 + 6 O2 6 H2O + 6 CO2 + 686 Kcal (12x6) + (1x12) + (16x6) + 6(16x2) 6(1x2 + 16) + 6(12+ (16x2) + 686 Kcal To calculate this amount of energy in KJ 686 X 4.18 = 2867.48 K J By Dr/ Mohamed A. EL-Hag 8 some useful quantities for glucose metabolism:  The following parameters for glucose metabolism can be calculated 686 - The amount of Energy (Kcal) released per gm of fuel (glucose) = = 3.8 Kcal/gm 180 knowing that the volume of 1 mole of any gas (ex. O2 or CO2) at standard temperature (273 K) and pressure (76 cm Hg) (STP) is 22.4 Liter. The following parameters can be calculated 686 The amount of energy (Kcal) released per liter of Oxygen used = = 5.1 Kcal/L 6 𝑋 22.4 6 𝑋 22.4 Liters of O2 used per gm of fuel (glucose) = = 0.75 L/gm 180 - Simillarly 6 𝑋 22.4 Liters of CO2 produced per gm of fuel (glucose) = = 0.75 L/gm 180 Respiratory quotient Respiratory quotient is defined as “the ratio between the no. of moles of Co2 produced and the number of moles of O2 used” 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 𝑃𝑟𝑜𝑑𝑢𝑐𝑒𝑑 Respiratory quotient = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂2 𝑈𝑠𝑒𝑑  For normal person (person with normal metabolism) the no. of moles of CO2 produced 6 equals the no. of moles of O2 used). Respiratory quotient = = 1 6 - The metabolic rate increases with increasing the physical activities - The metabolic rate is temperature dependent. An increase in body by 1 0C will increase the metabolic rate by about 10%.  Therefore, a decrease of the body temp. by 3 0C will decrease the metabolic rate by about 30%. (that’s why patient’s temp. is lowered during heart surgery) The basal metabolic rate (BMR)  It is defined as “ The amount of energy needed to perform minimal body functions (i.e. to maintain the life)  BMR = 92 Kcal/ hr  92 10  4.18 3 BMR = = 107 J/s (Watt) 60  60  The BMR depends primarily on the thyroid function. Person with overactive thyroid (hyperthyroidism) will have higher BMR than normal thyroid person. While persons with underactive thyroid (hypothyroidism) will have lower BMR. By Dr/ Mohamed A. EL-Hag 12  To keep constant body weight for a person the consumed food should equal the energy output required for basal metabolism and work done by this person (physical activities) - Consuming little food will results in loss weight and starvation if continued for long period of time. - Consuming large amounts of food will lead to weight increase and obesity if continued for long period.  The energy value stored in 1 gm of fats is 9.3 Kcal and it is denoted as the maximum activity of fats  Loss of body weight is achieved by burning of the stored fats either through reducing the energy intake (diet) or increasing physical activities.  Example:  Suppose a person wish to lose 4.54 Kg through physical activity. How long would he has to work at an activity of 15 Kcal/min. knowing that the maximum activity of fat (energy stored in 1 gm of fat is 9.3 Kcal/gm)  Answer: 𝑓𝑎𝑡𝑠 𝑡𝑜 𝑏𝑒 𝑏𝑢𝑟𝑛𝑒𝑑 𝑋 𝑚𝑎𝑥.𝑎𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑡𝑠  Time of working (t) = 𝑎𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑟𝑎𝑡𝑒 4.54 𝑋 1000 𝑔𝑚 𝑋 9.3 𝐾𝑐𝑎𝑙/𝑔𝑚  = = 2810 min = 46.83 hr ~ 47 hr 15 𝐾𝑐𝑎𝑙/𝑚𝑖𝑛  Example:  Suppose a person wish to lose 4.54 Kg through reducing the food (energy) intake. if the person is normally uses 2500 Kcal/day. How long must he diet to 2000 Kcal/day.  Answer: 𝑓𝑎𝑡𝑠 𝑡𝑜 𝑏𝑒 𝑏𝑢𝑟𝑛𝑒𝑑 𝑋 𝑚𝑎𝑥.𝑎𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑡𝑠  Time of dieting = 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑡𝑎𝑘𝑒 4.54 𝑋 1000 𝑔𝑚 𝑋 9.3 𝐾𝑐𝑎𝑙/𝑔𝑚  = = 84 days (2500−2000) 𝐾𝑐𝑎𝑙/𝑑𝑎𝑦 Work  Chemical energy stored in the body is converted into external mechanical work as well as into life-preserving functions.  The Work done (W) by a person is the product of the force and displacement  W = Force X displacement F  W = F. X X  The unit of work is Joule  Which is the same unit of energy By Dr/ Mohamed A. EL-Hag 17  Power (P)  It is defined as “ The time rate of doing the work”  w 𝐹. 𝑋 𝑃= = = 𝐹. 𝑣 t 𝑡  J/s = Watt is the S.I unit for measuring power By Dr/ Mohamed A. EL-Hag 18  Ex: Calculate the work done by a person exerts a force of 10 N to move an object a distance of 2 m.  Answer:  F = 10 N X=2m W=FX  W= (10)( 2) = 20 J By Dr/ Mohamed A. EL-Hag 19  Ex. In the last example calculate the power if you know that the force was affected for 5 S.  Answer:  W = 20 J t=5S  P=W/t  P = 20 / 5 = 4 watt By Dr/ Mohamed A. EL-Hag 20  Example:  What is the energy required (work done) to walk 20 Km with velocity of 5 Km/hr. knowing that the time rate for consuming energy (Power) during walking at this velocity is 3.8 Kcal/min.  Answer: x 20  The time of walking = = = 4 hr = 4 X60 = 240 min. 𝛎 𝟓  Energy consumed on walking ( work done) = P. t  = 3.8 X 240 = 912 Kcal.  Example:  In the last example, what is the amount of food needed for walking. Knowing that each gm of food provide 5 Kcal of energy.  Answer: 𝐭𝐨𝐭𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐝 𝐢𝐧 𝐰𝐚𝐥𝐤𝐢𝐧𝐠  The amount of food needed = 𝐞𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐯𝐢𝐝𝐞𝐝 𝐩𝐞𝐫 𝐠𝐦 𝐨𝐟 𝐟𝐨𝐨𝐝 912  = = 182 gm 𝟓 The work done by a person to climb a hill or walking upstairs  The work done by a person to climb a hill or walk upstairs is given by  W = m.g.h  Where mg is the weight of the person  And h is the height of the hill By Dr/ Mohamed A. EL-Hag 23 Efficiency of the body as a machine  The efficiency of the body (E) as a machine is given by the equation Work done E= Energy consumed By Dr/ Mohamed A. EL-Hag 24 Example:  A man of mass 70 kg climbs stairs to a height of 15 m in 1.2 minutes. During this time, his average oxygen consumption is 3 liters per minute. During resting, he uses 0.255 liters per minute. What is his muscular efficiency during the climb? Assume 5 kcal is produced for each liter of oxygen consumed.. Answer:  The work performed (W) = mgh = (70)(9.8)(15) = 10.3 x 103 Joule  Oxygenconsumed due to climbing = (3- 0.255) x 1.2 = 3.29 liter of oxygen  Energy consumed in climbing = 3.29 x 5 = 16.45 kcal = 68.8 x 103 Joule Work done 10.3 × 103 𝐸= = = 0.149 Energy consumed 68.8 × 103 E ≈ 0.15 or 15%  Example  Calculate the energy consumed by a person of body mass 60 Kg to walk upstairs for the 3rd floor of height 10 m. knowing that his body efficiency is 15%. (use the acceleration of gravity = 10 m/s2)  Answer:  Work done (W) = mgh = 60 X 10 X 10 = 6000 J work done E= energy consumed work done 6000 Energy consumed = = = 40000 J = 40 KJ E 0.15  The efficiency of muscles in performing external work is at best 20%. Therefore, at least 80% of the energy consumed is converted into heat inside the body  If this heat were not eliminated, the body temperature would quickly rise to a dangerous level  During moderate physical activity, a 70-kg man may consume 260 Cal/hr. Of this amount, at least 208 Cal is converted to heat. If this heat remained within the body, the body temperature would rise by 3 ◦C/hr. By Dr/ Mohamed A. EL-Hag 28