Chapter 5
5.3 Macroscopic Energy Balance
The general energy balance equation has the form
Input   Output   Heat
 Accumulation  

 = 
  
 + 
 of Energy   of Energy   of Energy   to
added  Work

System   by
done 
System
Let Esys be the total energy (internal + kinetic + potential) of a system, m in be the mass flow
rate of the system input stream, and m out be the mass flow rates of the system output stream,
then
2
d
U sys  Ek ,sys  E p,sys  = m in  uin  Vin  gzin   m out
dt
2




V2
 uout  out  gzout  + Q  W (5.3-1)
2


where
Usys = system internal energy
Ek,sys = system kinetic energy
Ep,sys = system potential energy
uin , uout = internal energies per unit mass of the system inlet and outlet streams
Vin , 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 = 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 pump motor)
W f = 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 outlet
Pin
Pout
distance
= Force velocity
velocity
Rate of flow work done on the system fluid = PinAinVin = Pin Fin
Rate of work = Force
5-9
Rate of flow work done by the system fluid = PoutAoutVout = Pout Fout
Eq. (5.3-1) becomes

d
Vin2
U sys  Ek ,sys  E p,sys  = m in  uin   gzin   m out
dt
2





+ Q  W + Pin F  Pout F
s
in
2


Vout
 uout 
 gzout 
2


out
(5.3-2)
The internal energy can be combined with the flow work to give the enthalpy

P 
 in Fin uin + Pin Fin =  in Fin  uin  in  =  in Fin hin
 in 

In terms of enthalpies hin and hout
2
d
U sys  Ek ,sys  E p,sys  = m in  hin  Vin  gzin   m out
dt
2




V2
 hout  out  gzout  + Q  W s (5.3-3)
2


The internal energy and the enthalpy can be related to the heat capacities where
 h 
Cp = 
 , and Cv =
 T  p
 u 


 T  v
For constant values of Cp and Cv
h = Cp(T - Tref) and u = Cv(T - Tref)
For solid and liquid Cp  Cv
 = m in = m out , equation
If the system is at steady state with one inlet and one exit stream m
(5.3-3) is simplified to
hout  hin + g(zout  zin) +
W
1 2
Q

Vout  Vin2  =
 s
2
m
m
(5.3-4)
W
Q
, w = s be the heat added to the system and work done
m
m
by the system, respectively, per unit mass flow rate. Equation (5.3-4) becomes
Let  = (“out”)  (“in”), and q =
h + gz +
1
V2 = q  w
2
(5.3-5)
5-10
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 = q  w
(5.3-6)
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 ()
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 = q.
Example 5.3-1. ---------------------------------------------------------------------------------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.
Fo , To
Q
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.
5-11
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.
d
(VCpT) = FoCpTo - FCpT + Q
dt
d
Q
(VT) = FoTo - FT +
dt
C p
dV
= 5 - 3 = 2 => V = 10 + 2t
dt
Step #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
Q
dT
dV
+ T
= FoTo - FT +
C p
dt
dt
(10 + 2t)
dT
(800)( 7.481)
+ 2T = (5)(60) - 3T +
dt
(62.4)(1)
(2t + 10)
dT
= 395.91 - 5T
dt
dT
75 395.91  5T =
T
dt
t
 395.91  5T 
1
 2t  10  - 5 ln  395.91  5  75  =
0
 2t  10 
395.91 - 5T = 20.91 

 10 
2 . 5
1  2t  10 
ln 

2  10 
 2t  10 
=> T = 79.182 - 4.182 

 10 
2 . 5
at t = 20 min., T = 79.1oF
----------------------------------------------------------------------------------------------------Equation (5.3-5) and its differential form, equation (5.3-6), are not convenient for solving
engineering problems.
1
V2 = q  w
2
(5.3-5)
dh + gdz + VdV = q  w
(5.3-6)
h + gz +
5-12
We can use thermodynamics relations to convert the enthalpy term into a form that involves
temperature, pressure, and density changes across the system.
du = Tds  Pd(1/)
dh = du + d(P/) = Tds  Pd(1/) + d(P/)
dh = Tds  Pd(1/) + Pd(1/) +
dP

= Tds +
dP

(5.3-7)
For an idealized reversible process in which no energy dissipation occurs, the entropy change
arises from heat transfer across the system boundaries
Tds = q
In any real system, the process is irreversible and there is dissipation of energy, therefore
Tds = q + ef = du + Pd(1/)
dh = q + ef +
(5.3-8)
dP

In this equation ef represents the thermal energy generated due to the irreversibility of the
dP
system. Substituting dh = q + ef +
into the differential energy balance, Eq. (5.3-6),

gives
dP

+ gdz + VdV + ef =  w
This equation can be integrated along a streamline from the inlet to the outlet of the system to
give

Po
Pi
where ef =
 e
f
dP

+ g(zo  zi) +
,w=
1
(oVo2  iVi2) + ef + w = 0
2
(5.3-9)
 w , and  = kinetic energy correction factor,  = 2 for laminar
flow,  = 1 for turbulent flow.
The kinetic correction factor is due to the fact that the velocity profile is not uniform over the
cross-sectional area of flow. For uniform flow, the rate of kinetic energy entering a C.V. is
given as
1

E k =  V 2  VA
2

The kinetic energy per unit mass flow rate is then
5-13
1
E k
= V2
VA 2
For turbulent flow, the velocity profile is almost flat, therefore
1
E k
 V2   = 1 for turbulent flow
VA turbulent 2
Uniform
velocity profile
vz
V
Laminar
velocity profile
Figure 5.3-1 Laminar velocity profile in a pipe.
The velocity for laminar flow in a pipe is given as
  r 2 
r
vz = 2V 1     = 2V (1  2), where  =
R
  R  
The rate of kinetic energy entering a C.V. is
E k =
Therefore

R
0
1 2
 v z  2rdr
2

R
R
r r
E k =   v z3 rdr = R2  v z3   d  
0
0
R R
1
1
E k = R2  8 V3(1  2)3d = 8R2V3  (1   2 ) 3 d
0
0
1
E k = 8R2V3  8  = R2V3 = A V3
 
Therefore
E k
2
= V2, and  = 2 for laminar flow
VA la min ar 2
5-14