Chapter 4
4.2 Energy Balances on Nonreactive Processes
The general energy balance equation has the formEnergy of on Accumulati = Energy of Input − Energy of Output + System to added Heat − System by done Work
Let Esys be the total energy (internal + kinetic + potential) of a system, minbe the mass flow rate of the system input stream, and moutbe the mass flow rates of the system output stream, then dt d
(
)
sys p sys k sys E E U + , + , = min in + in +gzin V u 2 2−mout + out + out
out gz V u 2 2 + Q − W (3.1-1) where
Usys = system internal energy Ek,sys= system kinetic energy Ep,sys = system potential energy
in
u , uout = internal energies per unit mass of the system inlet and outlet streams in
V , Vout = average velocity of the system inlet and outlet streams Q = rate of heat added to the system
W = rate of work done by the system
The net rate of work done by the system can be written as W= Ws + Wf
where s
W = rate of shaft work = rate of work done by the system through a mechanical device (e.g., a pump motor)
f
W = rate of flow work = rate of work done by the system fluid at the outlet minus rate of work done on the system fluid at the intlet
Pin Pout
Rate of work = Force×distance
time = Force×velocity
Rate of flow work done by the system fluid = PoutAoutVout = PoutFout Eq. (3.1-1) becomes dt d
(
)
sys p sys k sys E E U + , + , = min in + in +gzin V u 2 2−mout + out + out
out gz V u 2 2 + Q − Ws + PinFin−PoutFout (3.1-2) The internal energy can be combined with the flow work to give the enthalpy
in inF
ρ
uin + PinFin =ρ
inFin + in in in P uρ
=ρ
inFin hinIn terms of enthalpies hin and hout
dt d
(
)
sys p sys k sys E E U + , + , = min + in + in in gz V h 2 2−mout + out + out
out gz V h 2 2 + Q − Ws (3.1-3)
The internal energy and the enthalpy can be related to the heat capacities where cp = p T h ∂ ∂ , and cv = v T u ∂ ∂
For constant values of cp and cv
h = cp(T - Tref) and u = cv(T - Tref) For solid and liquid cp≈cv
If the system is at steady state with one inlet and one exit stream m = min = mout, equation (4.2-3) is simplified to hout−hin + g(zout−zin) +
(
2 2)
2 1 in out V V − = m Q − m Ws (4.2-4)Let ∆ = (“out”) − (“in”), and q = m Q
, w = m Ws
be the heat added to the system and work done by the system, respectively, per unit mass flow rate. Equation (4.2-4) becomes
∆h + g∆z +
2
This equation also applies to a system comprising the fluid between any two points along a streamline within a flow field. If these two points are only infinitesimal distance apart, the differential form of the energy equation is obtained
dh + gdz + VdV =
δ
q −δ
w (4.2-6)The d() notation represents a total or “exact” differential and applies to the change in state properties that are determined only by the initial and final states of the properties. The
δ
() notation represents an “inexact” differential and applies to the change in properties that depend upon the path taken from the initial to the final point of the properties. The forms of energy can be classified as either mechanical energy, associated with motion or position, or thermal energy, associated with temperature. Mechanical energy is considered to be an energy form of higher quality than thermal energy since it can be converted directly into useful work. Mechanical energy includes potential energy, kinetic energy, flow work, and shaft work. Internal energy and heat are thermal energy forms that cannot be converted directly into useful work. For systems that involve significant temperature changes, the mechanical energy terms are usually negligible compared with the thermal terms. In such cases the energy balance equation reduces to a “heat or enthalpy balance”, i.e. dh =δ
q.Example 4.2-1.---
In a residential water heater, water at 60oF flows at a constant 5 GPM into the 100 gallons tank and leaves at 3 GPM. Initially the tank has 10 gallons of 75oF water in it. The tank gas heater heats the tank contents at a constant rate of 800 Btu/min. Assume perfect mixing, determine the temperature of the discharge water after 20 min. of operation.
Water: cp = cv = 1 Btu/(lb.oF), density = 62.4 lb/ft3. Unit conversion 1 ft3 = 7.481 gal.
Q
Fo, To F, T
Solution ---
Step #1: Define the system.
Step #2: Find an equation that contains the temperature of the discharge water. The energy balance for the system gives the desired equation.
Step #3: Apply the energy balance on the system with the reference temperature Tref = 0oF. Neglect the changes in kinetic and potential energies compared with the changes in thermal energies. dt d (
ρ
Vc pT) =ρ
FocpTo -ρ
FcpT + Q dt d (VT) = F oTo - FT + p Q cρ
dt dV = 5 - 3 = 2 => V = 10 + 2tStep #4: Specify the initial condition for the differential equation. At t = 0, T = 75oF
Step #5: Solve the resulting equation and verify the solution. V dt dT + T dt dV = FoTo - FT + p Q c
ρ
(10 + 2t) dt dT + 2T = (5)(60) - 3T + ) 1 )( 4 . 62 ( ) 481 . 7 )( 800 ( (2t + 10) dt dT = 395.91 - 5T − T T dT 75395.91 5 = + t t dt 02 10 -5 1 ln × − − 75 5 91 . 395 5 91 . 395 T = 2 1 ln + 10 10 2t 395.91 - 5T = 20.91 5 . 2 10 10 2t+ − => T = 79.182 - 4.182 5 . 2 10 10 2t+ − at t = 20 min., T = 79.1oF ---Example 4.2-2
As part of a space experiment, a small instrumentation package is release from a space vehicle. It can be approximate by a solid aluminum sphere, 10 cm in diameter. If we take the surrounding space to be 0oK, how long will it take for the temperature of the package to change from 300oK to 30oK. Physical properties of aluminum: thermal conductivity k = 211
W/m⋅oK, emissivity
ε
= 0.05, densityρ
= 2702 kg/m3, and heat capacity cp = 798 J/kg⋅oK. Solution 2R T = 0 K sur o Ts qr
Figure E4.2-2 Heat leaving the sphere by radiation. The heat flow from the sphere by radiation is given by
qr= A
εσ
(Ts4 – Tsur4) Vρ
cp dt dT = q in – qout + qgen = – qr= – Aεσ
Ts4 = – Aεσ
T4 30 300T4 dT = – V c A pρ
εσ
t dt 0 – 30 300 3 3 − T = – V c A pρ
εσ
t = –ρ
cpRεσ
3 t t =εσ
ρ
cp(R/3) − 3 3 300 1 30 1 t = 8 10 67 . 5 05 . 0 ) 3 / 1 . 0 ( 798 2702 − × × × × − 3 3 300 1 30 1 = 1.56×108 s = 1809 daysHuman Body Temperature Regulation2
Humans are homeotherms, or warm-blood animals, and can regulate body temperature rather than have it adjusted by the external environment. The average internal temperature of the human body (core temperature) is maintained to within ±0.5oC around the average temperature of 37oC. The surface skin temperature is lower and under normal conditions is around 31oC. The body exchanges heat with its ambient by surface radiation Qr (W), by
conduction through direct contact with solid surfaces Qk (W), by surface convection heat
transfer Qku (W), and by surface evaporation energy conversion Slg= − Mlg∆hlg where Mlg
(kg/s) is the evaporation rate and ∆hlg (J/kg) is the heat of evaporation for water.
When ambient is at a temperature T∞ lower than the skin, heat flows out of the body into the ambient. In addition to the surface heat losses, heat is loss through gas, liquid, and solid discharges. This heat loss is sustained by the generation of heat by conversion of chemical bonds to thermal energy in metabolic reactions, i.e. chemical reactions, and temporarily by the reducing the body temperature. When ambient is at a higher temperature, heat flows out of the body by surface evaporation or sweating since the body must use its energy to break the physical bonds of the liquid.
The fat tissues have a thermal conductivity of one third of the other tissues, and therefore, act as an insulator. The blood flow to the surface is controlled by increasing it for heating purposes (vasodilatation) or by decreasing it for reducing heat losses (vasoconstriction). The conversion of energy, blood flow rate, and sweating are controlled by the nervous system feedback mechanisms and this control originates from the hypothalamus in the brain. There are temperature sensors throughout the body that are heat-sensitive neurons, which send higher frequency signals to the brain as the temperature increases.
Example 4.2-3. ---
Consider a person under a condition of hypothermia that is generating a maximum heat under severe shivering of 400 W. However, the total heat loss due to convection and radiation is 800 W. The body energy content (
ρ
cvV) is assumed to be 5×105 J/K. Determine how long itwill take for the body temperature to drop by 10oC.
Solution --- Neglect any other energy loss, the heat balance for the body is
dt dE
= accumulated energy change (W) =
ρ
cvVdt dT = – qout + qgen 5×105 dt dT = – 800 + 400 = – 400 t = 1,250×10 = 12,500 s = 3.47 hr.
Example 4.2-4.---
During an emergency launch operation, to fill a missile with RP-4 (a kerosene-based fuel), the ullage volume of the fuel storage tank is first pressurized with air from atmospheric pressure to a pressure of 150 psia. The air is available from large external storage tanks at high pressure (1000 psia, 70oF). This operation is to be completed as rapidly as possible. After the 150 psia pressure level is reached, the main transfer valve is opened and fuel flows at a steady rate until the missile is loaded. It is necessary to maintain a constant gas pressure of 150 psia inside the fuel tank during transfer.
The fuel storage tank can be approximated as a right circular cylinder 40 ft tall and 10 ft in diameter and is originally filled to 90% of capacity. Transfer of fuel to a residual volume of 10% must be completed in 18 minutes.
What problems would you anticipate if the inlet gas control valve were to malfunction and the gas space above the fuel were to reach full storage pressure (1000 psia)? (The fuel tank has been hydrostatically tested to 4000 psia.)
(Ref.11)
Air
1000 psia Air 150 psia
to Missile
Solution --- a. Identify any problem if the air above the fuel reachs 1000 psia
Model A:
Step #1: Define the system.
System = air inside the storage tank at any time. Assume perfect mixing of the air so that the air temperature is uniform at any time.
The temperature of air inside the tank can be obtained from the energy balance on the system. The temperature of air will increase with pressure and it might be high enough to ignite the fuel.
Step #3: Assume adiabatic operation since the air will be quickly pressurized from 14.7 psia to 1000 psia. During this period the heat transfer from the air to the surroundings (fuel and tank) is negligible.
Energy balance
dt
d (nu ) = in
n hin
where n = moles of air in the tank at any time and nin = inlet molar flow rate of air dn
dt = nin , u = cv(T−Tref), hin = cp (Tin−Tref)
Step #4: Specify the boundary conditions for the differential equation. At p = 14.7 psia, t = 0, T = T1 = 70oF = 70 + 460 = 530oR
At p = 1000 psia, T = T2
Tin = 70oF = 530oR
Step #5: Solve the resulting equation and verify the solution. The energy balance on the system becomes
dt
d (nu ) = in h dn
dt
Only the highest temperature when P = 1000 psia is important, the time derivative can be cancelled out.
d(nu) = hindn => d nu( ) = hin dn => n2u2 −n1u1 = hin (n2−n1) Let Tref = 0oR then
n2cvT2−n1cvT1 = (n2−n1)cpTin
Note cp = cv + R for ideal gas where R = gas constant
2 1 p V p V R − R cv = 2 1 2 1 p V p V RT −RT cpTin=> (p2 - p1) cv = 2 1 2 1 p p T −T cpTin
T2 is the only unknown in the above equation
2 2 p T = 11 p T +
(
2 1)
v p in p p c c T − = 1(
2 1)
1 p in v p in c p T p p c c T T − T2 = 1 2 1 ( 2 1) 1 in in kT T p kT p + p −p T = 1 2 1 1 2 2 (in ) in kT kT p p p T p p − + , where k = p v c c = 1.4 for air T2 = 1 2 1 1 1 in in kT T p p Tγ
+ − = − × + × 1 530 530 4 . 1 1000 7 . 14 1 530 4 . 1 = 737.7oR Model B:Step #1: Define the system.
System = air originally inside the storage tank at any time. Assume no mixing
between the original air and the incoming air. However the compressed original air is at a uniform temperature at any time.
Step #2: Find equation that contains the temperature of the system.
The temperature of the original air inside the tank can be obtained from the energy balance on the closed system.
Step #3: Assume adiabatic operation since the air will be quickly pressurized from 14.7 psia to 1000 psia. During this period the heat transfer from the air to the surroundings (incoming air, fuel, and tank) is negligible.
Energy balance d(nu) = dW = −pdV
From the ideal gas law pV = n R T
pdV + Vdp = n R dT −pdV = Vdp – n R dT
d(nu) = nc dT = Vdp – n R dTv N(cv + R)dT = Vdp
Step #4: Specify the boundary conditions for the differential equation. At p = p1 = 14.7 psia, t = 0, T = T1 = 70oF = 70 + 460 = 530oR
At p = p2 = 1000 psia, T = T2
Step #5: Solve the resulting equation and verify the solution. The energy balance on the system becomes
T dT = p R c dp p ln 1 2 T T = p R c ln 2 1 p p T2 = T1 / 2 1 p R c p p = 530 3 . 29 / 314 . 8 7 . 14 1000 = 1755 oR
