be the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out

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1 Chater 4 4. Energy Balances on Nonreactive Processes he general energy balance equation has the form Accumulation Inut Outut Heat added = + of Energy of Energy of Energy to System Work by done System Let E sys be the total energy (ternal + ketic + otential) of a system, m be the mass flow rate of the system ut stream, and m be the mass flow rates of the system ut stream, then d ( E E ) where + = m u U sys k, sys +, sys V + + gz m u V + + gz + Q W (3.-) U sys = system ternal energy E k,sys = system ketic energy E,sys = system otential energy u, u = ternal energies er unit mass of the system let and let streams V, V = average velocity of the system let and let streams Q = rate of heat added to the system W = rate of work done by the system he net rate of work done by the system can be written as W = W s + W f where W s = rate of shaft work = rate of work done by the system through a mechanical device (e.g., a um motor) W = rate of flow work = rate of work done by the system fluid at the let mus f rate of work done on the system fluid at the tlet P P Rate of work = Force distance = Force velocity time Rate of flow work done on the system fluid = P A V = P F 4-

2 Rate of flow work done by the system fluid = P A V = P F Eq. (3.-) becomes d ( E E ) V V + = m u + + gz m u + + gz + Q W s + P F P F (3.-) U sys k, sys +, sys he ternal energy can be combed with the flow work to give the enthaly ρf u + P F = ρf u P + ρ = ρf h In terms of enthalies h and h d ( E E ) U sys k, sys +, sys + = m h V + + gz m h V + + gz + Q W s (3.-3) he ternal energy and the enthaly can be related to the heat caacities where c = h, and c v = u v For constant values of c and c v h = c ( - ref ) and u = c v ( - ref ) For solid and liquid c c v If the system is at steady state with one let and one exit stream m = (4.-3) is simlified to m = m, equation h h + g(z z ) + ( ) Q W V V = s m m (4.-4) Q W Let = ( ) ( ), and q =, w = s be the heat added to the system and work done m m by the system, resectively, er unit mass flow rate. Equation (4.-4) becomes h + g z + V = q w (4.-5) 4-

3 his equation also alies to a system comrisg the fluid between any two ots along a streamle with a flow field. If these two ots are only fitesimal distance aart, the differential form of the energy equation is obtaed dh + gdz + VdV = δq δw (4.-6) he d() notation reresents a total or exact differential and alies to the change state roerties that are determed only by the itial and fal states of the roerties. he δ() notation reresents an exact differential and alies to the change roerties that deend uon the ath taken from the itial to the fal ot of the roerties. he forms of energy can be classified as either mechanical energy, associated with motion or osition, or thermal energy, associated with temerature. Mechanical energy is considered to be an energy form of higher quality than thermal energy sce it can be converted directly to useful work. Mechanical energy cludes otential energy, ketic energy, flow work, and shaft work. Internal energy and heat are thermal energy forms that cannot be converted directly to useful work. For systems that volve significant temerature changes, the mechanical energy terms are usually negligible comared with the thermal terms. In such cases the energy balance equation reduces to a heat or enthaly balance, i.e. dh = δq. Examle In a residential water heater, water at 60 o F flows at a constant 5 GPM to the 00 gallons tank and leaves at 3 GPM. Initially the tank has 0 gallons of 75 o F water it. he tank gas heater heats the tank contents at a constant rate of 800 Btu/m. Assume erfect mixg, determe the temerature of the discharge water after 0 m. of oeration. Water: c = c v = Btu/(lb.o F), density = 6.4 lb/ft 3. Unit conversion ft 3 = 7.48 gal. F o, o Q F, Solution Ste #: Defe the system. Ste #: Fd an equation that contas the temerature of the discharge water. he energy balance for the system gives the desired equation. 4-3

4 Ste #3: Aly the energy balance on the system with the reference temerature ref = 0 o F. Neglect the changes ketic and otential energies comared with the changes thermal energies. d (ρvc ) = ρf o c o - ρfc + Q d Q (V) = Fo o - F + ρc dv = 5-3 = => V = 0 + t Ste #4: Secify the itial condition for the differential equation. At t = 0, = 75 o F Ste #5: Solve the resultg equation and verify the solution. d V dv + = Fo o - F + Q ρc d (800)(7.48) (0 + t) + = (5)(60) (6.4)() d (t + 0) = d t = - ln 0t t + 0 = ln 0 t = => = t at t = 0 m., = 79. o F

5 Examle 4.- As art of a sace exeriment, a small strumentation ackage is release from a sace vehicle. It can be aroximate by a solid alumum shere, 0 cm diameter. If we take the surroundg sace to be 0 o K, how long will it take for the temerature of the ackage to change from 300 o K to 30 o K. Physical roerties of alumum: thermal conductivity k = W/m o K, emissivity ε = 0.05, density ρ = 70 kg/m 3, and heat caacity c = 798 J/kg o K. Solution q r o sur = 0 K R s Figure E4.- Heat leavg the shere by radiation. he heat flow from the shere by radiation is given by q r = Aεσ( s 4 sur 4 ) d Vρc = q q + q gen = q r = Aεσ 4 s = Aεσ 4 30 d = εσa ρc V t = εσa 3εσ t = t ρc V ρc R t = ρ (R / 3) c εσ (0./ 3) t = = s = 809 days 4-5

6 Human Body emerature Regulation Humans are homeotherms, or warm-blood animals, and can regulate body temerature rather than have it adjusted by the external environment. he average ternal temerature of the human body (core temerature) is mataed to with ±0.5 o C around the average temerature of 37 o C. he surface sk temerature is lower and under normal conditions is around 3 o C. he body exchanges heat with its ambient by surface radiation Q r (W), by conduction through direct contact with solid surfaces Q k (W), by surface convection heat transfer Q ku (W), and by surface evaoration energy conversion S = M h lg where M (kg/s) is the evaoration rate and h lg (J/kg) is the heat of evaoration for water. When ambient is at a temerature lower than the sk, heat flows of the body to the ambient. In addition to the surface heat losses, heat is loss through gas, liquid, and solid discharges. his heat loss is sustaed by the generation of heat by conversion of chemical bonds to thermal energy metabolic reactions, i.e. chemical reactions, and temorarily by the reducg the body temerature. When ambient is at a higher temerature, heat flows of the body by surface evaoration or sweatg sce the body must use its energy to break the hysical bonds of the liquid. he fat tissues have a thermal conductivity of one third of the other tissues, and therefore, act as an sulator. he blood flow to the surface is controlled by creasg it for heatg uroses (vasodilatation) or by decreasg it for reducg heat losses (vasoconstriction). he conversion of energy, blood flow rate, and sweatg are controlled by the nervous system feedback mechanisms and this control origates from the hyothalamus the bra. here are temerature sensors through the body that are heat-sensitive neurons, which send higher frequency signals to the bra as the temerature creases. Examle Consider a erson under a condition of hyothermia that is generatg a maximum heat under severe shiverg of 400 W. However, the total heat loss due to convection and radiation is 800 W. he body energy content (ρc v V) is assumed to be J/K. Determe how long it will take for the body temerature to dro by 0 o C. Solution Neglect any other energy loss, the heat balance for the body is lg lg lg de d = accumulated energy change (W) = ρcv V = q + q gen d = = 400 t =,50 0 =,500 s = 3.47 hr Kaviany, Prciles of Heat ransfer, Wiley, 00,

7 Examle Durg an emergency launch oeration, to fill a missile with RP-4 (a kerosene-based fuel), the ullage volume of the fuel storage tank is first ressurized with air from atmosheric ressure to a ressure of 50 sia. he air is available from large external storage tanks at high ressure (000 sia, 70 o F). his oeration is to be comleted as raidly as ossible. After the 50 sia ressure level is reached, the ma transfer valve is oened and fuel flows at a steady rate until the missile is loaded. It is necessary to mata a constant gas ressure of 50 sia side the fuel tank durg transfer. he fuel storage tank can be aroximated as a right circular cylder 40 ft tall and 0 ft diameter and is origally filled to 90% of caacity. ransfer of fuel to a residual volume of 0% must be comleted 8 mutes. What roblems would you anticiate if the let gas control valve were to malfunction and the gas sace above the fuel were to reach full storage ressure (000 sia)? (he fuel tank has been hydrostatically tested to 4000 sia.) (Ref.) Air 000 sia Air 50 sia to Missile Solution Model A: a. Identify any roblem if the air above the fuel reachs 000 sia Ste #: Defe the system. System = air side the storage tank at any time. Assume erfect mixg of the air so that the air temerature is uniform at any time. Ste #: Fd equation that contas the temerature of the system. 4-7

8 he temerature of air side the tank can be obtaed from the energy balance on the system. he temerature of air will crease with ressure and it might be high enough to ignite the fuel. Ste #3: Assume adiabatic oeration sce the air will be quickly ressurized from 4.7 sia to 000 sia. Durg this eriod the heat transfer from the air to the surroundgs (fuel and tank) is negligible. Energy balance d (nu ) = n h where n = moles of air the tank at any time and n = let molar flow rate of air dn = n, u = c v ( ref ), h = c ( ref ) Ste #4: Secify the boundary conditions for the differential equation. At = 4.7 sia, t = 0, = = 70 o F = = 530 o R At = 000 sia, = = 70 o F = 530 o R Ste #5: Solve the resultg equation and verify the solution. he energy balance on the system becomes d (nu ) = h dn Only the highest temerature when P = 000 sia is imortant, the time derivative can be cancelled. d(nu ) = h dn => d( nu ) = h dn => n u n u = h (n n ) Let ref = 0 o R then n c v n c v = (n n ) c Note c = c v + R for ideal gas where R = gas constant he ideal gas law can be alied with the condition V = V = V = constant 4-8

9 V R V R V V c = v R R c => ( - ) v c = c is the only unknown the above equation = c + ( ) v c = ( ) c c v c k = k + ( ) = k k ( ) +, where k = c c v =.4 for air = k γ + = = o R Model B: Ste #: Defe the system. System = air origally side the storage tank at any time. Assume no mixg between the origal air and the comg air. However the comressed origal air is at a uniform temerature at any time. Ste #: Fd equation that contas the temerature of the system. he temerature of the origal air side the tank can be obtaed from the energy balance on the closed system. Ste #3: Assume adiabatic oeration sce the air will be quickly ressurized from 4.7 sia to 000 sia. Durg this eriod the heat transfer from the air to the surroundgs (comg air, fuel, and tank) is negligible. Energy balance d(nu ) = dw = dv From the ideal gas law V = n R dv + Vd = n R d dv = Vd n R d d(nu ) = n c v d = Vd n R d N( c v + R )d = Vd n c d = Vd = n R d/ c d = R d/ 4-9

10 Ste #4: Secify the boundary conditions for the differential equation. At = = 4.7 sia, t = 0, = = 70 o F = = 530 o R At = = 000 sia, = Ste #5: Solve the resultg equation and verify the solution. he energy balance on the system becomes d R = c d ln = R c ln = R / c 000 = / 9.3 = 755 o R 4-0