Unformatted text preview:

BioenergeticsIn Chapter 8 (Thermodynamics), we find the maxi-mum theoretical efficiency of a heat engine, inturning heat to work1, to beηCarnot = T> − T<T> .Since the upper temperature (body internal tem-perature) is about 310 oK and the exterior tem-perature is—in cold weather—about 273 oK, theabsolute maximum efficiency is about 12%. In-cluding the effects of heat leaks, friction and theneed to have a reasonable power-to-weight ratio,the expected efficiency would be no better thanhalf that (and worse on hot days!). However, frommeasurement of work output and food input, wediscover the human being is about 25% efficient inconverting food energy to work. The conclusion isinescapable: living organisms are not heat en-gines.1. Energy input to living organismsHow does a living organism get its energy? As withall other things biological, Nature provides a smallmachine (actually, a series of small machines). Inthis case it is an energy-storing molecule calledadenosine tri-phosphate (ATP) that is constructedin the cellular energy factory (the mitochondrion, atubular organelle present in all cells) using theenergy from oxidation of sugar (glucose). The ATPmolecule gives up one of its phosphate groups to amolecule that needs to be promoted to a higherenergy state, thereby becoming adenosine bi-phos-phate (ADP). About 11.5 kcal/mol can be pro-vided by this maneuver, or about 0.5 eV perreaction2.The oxidation of food provides energy as shown inthe table below:Energy output from various foodsFood Kcal/ L(CO2)/ Kcal/gm(food) gm(food) L(O2)Lipid (fat) 9.3 1.39 4.7Protein 4.0 0.75 4.5Alcohol 7.1 0.97 4.9Sugar 3.8 0.74 5.1Carbohy-drate4.1 0.81 5.0The interesting column is the energy ouput perliter of oxygen consumed: the numbers are almostconstant. This is the reason why oxygen consump-tion can be used to measure basal metabolism, therate of energy consumption needed to just sustainthe processes of life, with no significant muscularor cognitive activity taking place3.Physics of the Human Body 45Chapter 5 Bioenergetics1. In principle stochiometric combustion of food with oxygen is 100% efficient in converting chemical energyto heat, so we neglect inefficiencies in the burners of heat engines.2. An electron volt is the energy gained by letting an electron move across a potential difference of one volt.Since the electron charge is 1.6×10-19 Coulombs, and since 1 Coulomb × 1 Volt = 1 Joule, 1eV is 1.6×10-19 J.3. As we shall see subsequently, the brain consumes significant amounts of energy during periods of intenseconcentration.2. Energy demand of organsWe can easily estimate the power required torun the human heart. Basically we use theenthalpy, which for a moving fluid ish = 12 ρ v2 + p + ρgz .The stream of blood leaving the heart has netpressure increment 60 mm (of Hg) or about8×104 dyne/cm2. When blood returns to theheart through the vena cava, its pressure is isbasically the diastolic (gauge) pressure of thebody and the speed of flow is roughly the sameas that of the exiting blood. The height is thesame so there is no change in potential energy.Therefore the amount of work done per unittime must beP = ∆p dVdt .The volume of the ventricle is about 100 cm3 soassuming 60 beats per minute (that is, one contrac-tion per second) we get P ≈ 8×104 × 100 ≈ 107erg ⁄ sec = 1 Watt .Larger hearts, faster beats or higher blood pressurescan double this. If we multiply by 4 or 5 to accountfor the inefficiency of converting chemical energyto work we find about 5-7 Watts or 90-120 Caloriesper day is needed to keep the heart pumping. Thisis only a few percent of our basal metabolic powerdemand.What about the energy demand of kidneys? Thethermodynamic potential of a concentration dif-ference across a permeable membrane4 is∆E = RT ln c2c1so that if the rate of transfer of material (in molesper unit time) is dνdt , the power needed isP = dνdt RT ln c2c1 .The logarithm of concentration ratio is a numberof order 1; the average flow through the kidneys is125 cm3/sec, or about 1/8 of a liter of fluid persecond. Assuming the material that must be eitherheld back (or eliminated into the bladder) is about0.1 molar in concentration, we getdνdt ≈ 0.0125 gm−mol ⁄ sec .Thus the average power demand is about 30 W, orabout 660 Cal/day. That is, the energy required tooperate our kidneys is a non-trivial fraction of thetotal energy budget of the human body.Suppose we apply a different method for estimatingenergy requirements, based on the measured bloodflow to various organs in the resting human being.46 Physics of the Human BodyEnergy demand of organs4. That is, the energy needed to maintain a concentration gradient against the process of diffusion in the op-posite direction.If we assume the energy requirements of eachorgan system in the table below is proportional tothe blood volume that circulates through it, thenwe find the results shown in the third column ofthe table:System Flow*(ml/min)Metabolism (Kcal/day)Liver, intestines,spleen1400 480Kidneys 1100 380Brain 750 260Heart 250 90Skeletal muscles 1200 410Skin 500 170Other organs 600 210TotalTotalTotalTotal 5800580058005800 2000200020002000*taken from McDonald’s Blood Flow in Arteries,4th ed. (Oxford U. Press, Inc. New York, 1988)Table 2.4, p. 36.We see that our estimate for the heart (based onwork output) is very close to that based on bloodflow through the coronary arteries feeding theheart muscle. Our estimate for the kidneys wassomewhat high, but considering that the concen-tration ratio c2 ⁄ c1 and the absolute concentrationof dissolved substances were only “guess-timates”,the agreement seems remarkably good.The energy demand of the resting brain, 12 W or260 Kcal/day seems rather small. Measurements ofblood flow to various parts of the brain5, as well asof its heat output, yield comparable numbers forenergy demand. However, when the brain is mostactive, as during periods of intense concentrationon difficult problems (for example, the homeworkin this course), its demand rises 20-fold. 3. Heating and coolingThere are three mechanisms of heat loss by thebody:a) radiation;b) conduction;c)convection/evaporation.Bodies radiate and absorb electromagnetic energy(black-body radiation) according to the Stefan-Boltzmann law,P = σ A T4where A is the radiating area, T the absolutetemperature (in oK), and the Stefan-Boltzmannconstant σ is6σ = 2π515 k4h3c2= 5.7×10−8 Watts ⁄ m2 ⁄


View Full Document

UVA PHYS 3040 - Bioenergetics

Download Bioenergetics
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Bioenergetics and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Bioenergetics 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?