Chemical engineering balancing of mass and energy

1. Energy and
Energy Balances
Ch E 201
Material and Energy Balances

2. Energy Balances are used to answer…
• How much power is required to pump 1250 m3/h of
water from a storage vessel to a process unit?
• How much energy is required to convert 2000 kg of
water at 30°C to steam at 180°C?
• How rapidly must steam be supplied to a distillation
column separating a 2000 mol/h feed?
• How much coal must be burned to produce enough
energy each day to generate enough steam to run
the turbines that supply electricity to a city of
500,000 people?

3. Forms of Energy
• Kinetic energy is due to translational motion of the
system as a whole relative to some frame of
reference (surface of the earth) or to rotation of the
system about some axis.
• Potential energy is due to position of the system in a
potential field (e.g., gravitation, electromagnetic).
• Internal energy is all energy possessed by a system
other than kinetic or potential energy.
– such as energy due to the motion of molecules relative to the center of mass
of the system, rotational or vibrational motion and the electromagnetic
interactions of the molecules, and to the motion and interactions of the
atomic and subatomic constituents of the molecules.

4. First Law of Thermodynamics
• The First Law of Thermodynamics states that
energy can neither be created nor destroyed.
• This principle underlies all energy balances and is
also referred to as the law of conservation of energy.

5. Energy Transfer in a Closed System
• Consider a closed system (i.e., one that no mass is
transferred across it’s boundaries), energy may be
transferred as heat or work. Heat & work are flows.
• Heat is energy that flows as a result of a temperature
gradient between system and surroundings. Flow
direction of is always from high to low temperature.
• Work is energy flow resulting from any driving force
other than a temperature gradient. Work is positive
if done by the system on the surroundings.
– e.g., if a gas in a cylinder expands and moves a piston against a
restraining force, the gas does work on the piston and energy is
transferred as work from the gas to the surroundings.

6. Kinetic and Potential Energy
• Kinetic Energy (Ek)
– where overdots represent per unit time (flow rates)
– u ≡ velocity
– m ≡ mass
• Potential Energy (Ep)
– g ≡ gravitational constant
– z ≡ height above a reference point
2
2
1
k mu
E  2
2
1
k u
m
E 
 
mgz
Ep  gz
m
Ep

 
 
1
2
p z
z
g
m
E 

 

 
2
1
2
2
2
1
k u
u
m
E 

 


7. Closed System Energy Balances
• An integral energy balance may be derived for a
closed system between two instants of time.
• Energy may be transferred across system boundaries
in a closed system, though mass is not.
    W
Q
E
E
U
E
E
U pi
ki
i
pf
kf
f 






      W
Q
E
E
+
E
E
U
U pi
pf
ki
kf
i
f 





W
Q
E
+
E
U p
k 






final system
energy






initial system
energy






net energy
transfer to system







8. Closed System Energy Balances
• When applying a closed system energy balance:
– If no temperature changes, phase changes, or chemical reactions
occur in a closed system, and if pressure changes are less than a few
atmospheres, then ΔU=0.
– If a system is not accelerating, then ΔEk =0.
– If a system is not rising or falling, then ΔEp =0.
– If a system and surroundings are at the same temperature or the
system is perfectly insulated, then Q=0.
– If there are no moving parts or electric currents or radiation at the
system boundaries, then W=0.
U Ek + Ep  Q W

9. Closed System Energy Balance
• A 25°C gas is in a cylinder with a movable piston.
• The cylinder is placed in boiling water with the piston
clamped in place, and 2.00 kcal of heat is transferred
to the gas, raising the temperature to 100°C, also
increasing the pressure.
U Ek + Ep  Q W

Ek = 0 system is stationary
Ep  0 no vertical displacement
W  0 no moving boundaries
0 0 0
UQ
2 kcal 1000 J
0.23901cal
 8370 J

10. Closed System Energy Balance
• The piston is released, and the gas does 100J
of work in moving the piston to its new equilibrium
position.
U Ek + Ep  Q W

Ek = 0 system is stationary at initial and final state
Ep  0 assumed negligible
U  0 T of ideal gas does not change
0 0
0
0 Q W
0 Q 100 J
Q  100 J

11. Open System Steady-State Energy Balance
• In an open system, mass crosses system boundaries.
• Work is done on the system by mass entering;
work is done on the surroundings by mass leaving.
• Work is a rate of energy transfer, and therefore must
be quantified in the energy balance.

12. Work – Flow and Shaft
• Shaft work is the rate of work done by the process
fluid on a moving part within the system (such as a
pump rotor).
• Flow work is the rate of work done by the fluid at the
system outlet minus the rate of work doen on the
fluid at the system inlet.
• The net rate of work done by an open system on its
surroundings is the sum of shaft and flow work:
fl
s W
W
W 

 


13. Flow Work
• In a 1-inlet/outlet system, fluid at pressure Pin enters
at volumetric flow rate Vin and exits at Pout and Vout.
• Fluid entering has work done on it by the fluid
behind it at the rate of Win = PinVin; fluid leaving has
work done on it by fluid behind it at Wout = PoutVout.
• Net rate of work done is thus Wfl = PoutVout – PinVin.
• For several streams, Wfl = Σ(PoutVout) – Σ(PinVin)

14. Specific Properties
• A specific property is an intensive quantity divided by
an extensive property of the process material.
• Specific properties are denoted with the hat (^) over
the variable representation.
– Total internal energy of a mass of material
– Rate of internal energy tranfer for mass flow rate
Û
m
U 

Û
m
U 
 


15. Specific Enthalpy
• Specific Enthalpy is a property that occurs in the
open system energy balance:
   
   
mol
J
mol
J
mol
J
atm
L
J
mol
L
mol
J
K
mol
atm
L
K
mol
J
mol
L
mol
J
6295
2495
3800
Ĥ
3
.
101
63
.
24
atm
1
3800
Ĥ
08206
.
0
314
.
8
63
.
24
atm
1
3800
Ĥ
V̂
p
Û
Ĥ














16. Steady-state Open-System Energy Balance
• input = output (no reactions, steady-state operation)
• input
– total Ek, Ep, and U entering system with process streams
– total energy transferred to the system as heat
• output
– total Ek, Ep, and U leaving system with process streams
– total energy transferred from system as work
W
E
E
Q
outputs
j
inputs
j



 

 
 W
Q
E
E
inputs
j
outputs
j



 

 


17. Steady-state Open-System Energy Balance
• Evaluating the energy summations:
W
Q
E
E
inputs
j
outputs
j



 

 

 
j
2
j
j
j
j
j
j
2
j
j
j
j
j
pj
kj
j
j
gz
2
u
Û
m
E
gz
m
2
u
m
Û
m
E
E
E
U
E




















18. Steady-state Open-System Energy Balance
• Evaluating the work term:
W
Q
E
E
inputs
j
outputs
j



 

 


 




inputs
j
j
j
outputs
j
j
j
s
s
fl
V̂
P
m
V̂
P
m
W
W
W
W
W







 
j
2
j
j
j
j gz
2
u
Û
m
E 

 


19. Steady-state Open-System Energy Balance
• substitute:
W
Q
E
E
inputs
j
outputs
j



 

 


 


inputs
j
j
j
outputs
j
j
j
s V̂
P
m
V̂
P
m
W
W 



 
j
2
j
j
j
j gz
2
u
Û
m
E 

 


20. Steady-state Open-System Energy Balance
• collect terms, substitute for enthalpy, define
differences between in and out flows with Δ:
s
p
k
s
inputs
j
2
j
j
j
outputs
j
2
j
j
j
s
inputs
j
2
j
j
j
j
j
j
outputs
j
2
j
j
j
j
j
j
inputs
j
j
j
outputs
j
j
j
s
inputs
j
2
j
j
j
outputs
j
2
j
j
j
W
Q
E
E
H
W
Q
gz
2
u
Ĥ
m
gz
2
u
Ĥ
m
W
Q
gz
2
u
V̂
P
Û
m
gz
2
u
V̂
P
Û
m
V̂
P
m
V̂
P
m
W
Q
gz
2
u
Û
m
gz
2
u
Û
m




















































































































21. Turbine Energy Balance
• 500 kg/h of steam drives a turbine. Steam enters at 44 atm,
450°C at 60 m/s, and leaves a point 5 m below the turbine
inlet at atmospheric pressure with a velocity of 360 m/s.
• The turbine delivers 70 kW of shaft work and heat loss is
estimated at 10,000 kcal/h.
• Calculate the specific enthalpy change.

22. Turbine Energy Balance
   
    
     
kg
kJ
s
m
hr
kg
2
s
m
2
s
m
hr
kg
2
1
hr
kcal
4
1
2
2
1
2
2
2
m
s
p
k
s
650
H
m
5
81
.
9
500
360
60
500
kW
70
10
H
z
z
g
m
u
u
W
Q
E
E
W
Q
H
2

































 
unit conversions left for the student

23. Reference State and State Properties
• It is not possible to know the absolute value of the
specific internal energy or the specific enthalpy.
• Change (Δ) in a specific property can be determined.
• To permit quantification of the difference between
various states, it is convenient to define a reference
state, and to list changes from this state for a series
of other states.
• State properties only depend on
the state of the system, not how
that state was reached.
CO
1atm

24. Tabulated Enthalpy Data
• What reference state used to generate these values?
• Calculate ΔĤ and ΔÛ for the transition of saturated
methyl chloride vapor from 50°F to 0°F. What
assumption is necessary?

25. Tabulated Enthalpy Data
   
 
     
  m
3
m
m
lb
BTU
psia
ft
73
.
10
BTU
987
.
1
lb
BTU
1
1
2
2
lb
BTU
96
.
4
920
.
1
99
.
51
969
.
4
90
.
18
05
.
6
Û
V̂
P
V̂
P
Ĥ
V̂
P
Ĥ
Û
05
.
6
28
.
202
23
.
196
F
50
Ĥ
F
0
Ĥ
Ĥ

























assume enthalpy
independent of pressure

26. Steam Tables
• Steam tables tabulate H2O specific quantites.
• Separate tables exist for saturated liquid/vapor
(Appendix B.5 and B.6) and superheated vapor (B.7).
• The saturated table gives specific quantities on the
equilibrium lines.
V̂
,
Û
,
Ĥ
• The superheated
tables provide
specific quantities
in the 1-phase
regions.

27. Use of the Steam Tables
• Determine p*, of sat’d steam at 133.5°C.
Table B.6 at 133.5°C
p* = 3.0 bar
Û = 2543.0 kJ/kg
Ĥ = 2724.7 kJ/kg
V̂
,
Û
,
Ĥ
kg
/
m
606
.
0
V̂ 3


28. Use of the Steam Tables
• Show that water at 400°C and 10 bar is
superheated steam and determine its specific
properties relative to the triple point of liquid water.
• Also determine Tdp.
Table B.7 at 400°C, 10.0 bar
Û = 2958 kJ/kg
Ĥ = 3264 kJ/kg
Tdp = 179.7°C
kg
/
m
307
.
0
V̂ 3


29. Steam Turbine Energy Balance
• Steam at 10 bar absolute with 190°C of superheat
is fed to a turbine at 2000 kg/h. The turbine
operation is adiabatic, and the effluent is saturated
steam at 1 bar. Calculate the work output of the
turbine in kW
(neglect ΔEk and ΔEp).
s
p
k W
Q
E
E
H 



 






0 0 0
 
   
kW
292
W
292
2675
3201
2000
W
H
H
m
W
H
s
s
kJ
s
3600
h
1
kg
kJ
h
kg
s
out
in
s













 
  kJ/kg
2675
d
’
sat
,
bar
1
Ĥ
kJ/kg
3201
C
370
,
bar
10
Ĥ
out
in



from steam tables
to surroundings

30. Energy Balance Procedures
• Draw and label flowchart properly.
• Include all necessary information to determine
specific enthalpy of each stream component,
including all known T and P values.
• Show “states of aggregation” of process materials
(i.e., phase) when not obvious (H2O(s), H2O(l), H2O(g)).

31. Energy Balance, 1-component process
• Determine Q(dot)
• Note that mass
balance is trivial
(inlet flows summed to obtain outlet flow)
• Determine specific enthalpies of all streams.
– can use steam tables for system of water
– asssume liquid water enthalpies are independent of P
– sat’d effluent condition also provides T
• simplify energy balance s
p
k W
Q
E
E
H 



 






0 0

32. Energy Balance, 1-component process
• evaluate
enthalpy change
0 0
     
 
kJ/min
10
61
.
7
H
9
.
271
175
7
.
125
120
2793
295
H
Ĥ
m
Ĥ
m
H
5
kg
kJ
min
kg
kg
kJ
min
kg
kg
kJ
min
kg
in
i
i
out
i
i









 






s
p
k W
Q
E
E
H 



 







33. Energy Balance, 1-component process
• evaluate kinetic
energy change
– assume Ek,in=0
0 0
kJ/min
10
61
.
7
H 5



    
   
min
kJ
3
k
3
2
2
2
cm
100
m
1
2
kg
m
s
60
min
1
min
kg
min
kg
2
1
k
2
2
2
2
k
10
02
.
6
E
m
N
10
kJ
s
/
m
kg
N
cm
3
1166
.
0
295
295
E
R
V̂
m
2
m
A
V
2
m
2
mu
E
3



































s
p
k W
Q
E
E
H 



 







34. Energy Balance, 1-component process
• substitute
into energy
balance
0 0
kJ/min
10
61
.
7
H 5



min
kJ
3
k 10
02
.
6
E 


min
kJ
5
min
kJ
3
min
kJ
5
k
10
67
.
7
Q
10
02
.
6
10
61
.
7
Q
Q
E
H















Note relative magnitude of enthalpy
term versus kinetic energy
contribution.
It is often safe to neglect Ek and Ep
contributions as a first approximation.
s
p
k W
Q
E
E
H 



 







35. Energy Balance, 2-component process
• Liquid containing
ethane and n-butane
is heated. Determine heat flow.
• No need for a material balance (1 stream in/out).
0 0
0

 



in
i
i
out
i
i Ĥ
m
Ĥ
m
H
Q 



s
p
k W
Q
E
E
H 



 







36. Energy Balance, 2-component process
• Obtain enthalpies
from Perry’s Handbook.
   
 
   
 
kJ/kg
112
kg/s
1.0
kJ/s
112
kJ/s
112
Q
0
.
30
400
.
0
3
.
314
600
.
0
2
.
130
400
.
0
5
.
434
600
.
0
Q
Ĥ
m
Ĥ
m
H
Q
kg
kJ
s
H
C
kg
kg
kJ
s
H
C
kg
kg
kJ
s
H
C
kg
kg
kJ
s
H
C
kg
in
i
i
out
i
i
10
4
6
2
10
4
6
2










 








37. Simultaneous Material/Energy Balances
• Sat’d steam at
1 atm discharged
from a turbine at
1150 kg/h.
• Superheated steam
at 300°C is needed. To produce it, the turbine discharge is
mixed with superheated steam available from another source
at 400°C. The mixer operates adiabatically.
• Calculate amount of superheated steam at 300°C produced
and the required volumetric flowrate of the 400°C steam.
• Specific enthalpies can be retrieved from the steam tables (as
shown in the flowchart).

38. Simultaneous Material/Energy Balances
• 2 unknowns,
1 independent
balance.
• Material/energy
balances must therefore be solved simultaneously.
0 0
0
2
1
h
kg
m
m
1150 
 


 



in
i
i
out
i
i Ĥ
m
Ĥ
m
0
H 


0
s
p
k W
Q
E
E
H 



 







39. Simultaneous Material/Energy Balances
• 2 unknowns,
1 independent
balance.
• Material/energy
balances must therefore be solved simultaneously.
2
1
h
kg
m
m
1150 
 

     
kg
kJ
1
kg
kJ
h
kg
kg
kJ
2
in
i
i
out
i
i
3278
m
2676
1150
3074
m
Ĥ
m
Ĥ
m






 

h
kg
2
h
kg
1 3390
m
,
2240
m 
 


40. Simultaneous Material/Energy Balances
• 2 unknowns,
1 independent
balance.
• Specific volume
of steam at 400°C, 1 atm ≈ 3.11 m3/kg.
– In the absence of data, the ideal gas EOS could be used to
approximate the specific volume.
   h
m
kg
m
h
kg
1
1
1
3
3
6890
11
.
3
2240
V̂
m
V 

 

h
kg
2
h
kg
1 3390
m
,
2240
m 
 


41. Mechanical Energy Balances
• To this point, focus has been on heat flows and internal
energy and enthalpy changes (kinetic and potential energy
and work terms negligible).
– typical of chemical processes such as reactors, distillation columns,
heat exchangers, and evaporators
• Another important class of operations involves an absence of
heat flows, and instead the kinetic and potential energy and
work terms are important.
– flow of fluids through valves and pumps between tanks, reservoirs,
wells and process units
• Analysis of energy flows in such systems is conveniently
performed by a mechanical energy balance.

42. Mechanical Energy Balances
• Combine the open-system energy balance with an
expression of the law of conservation of momentum
(Ch E 305) to obtain:
– where ρ is the liquid density
m
W
m
Q
Û
z
g
2
u
P s
2
























43. Mechanical Energy Balances
• In many cases, only slight amounts of heat are
transferred to/from the surroundings, there is little
temperature change from inlet to outlet.
• In such cases, the term in parentheses has a positive
component called the friction loss.
m
W
m
Q
Û
z
g
2
u
P s
2























m
W
F̂
z
g
2
u
P s
2












44. Mechanical Energy Balances
• This is the mechanical energy balance, valid for
steady-state flow of an incompressible fluid.
– Methods for estimating friction losses for flows through
pipes, orifices, nozzles, elbows, etc. will be learned in 305.
• For a frictionless process in which no shaft work is
performed, mechanical energy balance reduces to:
m
W
F̂
z
g
2
u
P s
2











0
z
g
2
u
P 2







The Bernoulli Equation

45. Use of The Bernoulli Equation
• Water flows through system shown. Estimate the
pressure required at pt 1, neglecting friction losses.
0
z
g
2
u
P 2







  
   
s
m
2
s
m
2
cm
10
m
2
2
cm
5
.
0
L
10
m
s
60
min
min
L
1
24
.
4
u
0
.
17
20
A
V
u
3
3






46. Use of The Bernoulli Equation
• Water flows through system shown. Estimate the
pressure required at pt 1, neglecting friction losses.
0
z
g
2
u
P 2







      
Pa
10
56
.
4
P
m
50
81
.
9
2
24
.
4
0
.
17
1000
atm
1
P
5
2
s
m
2
s
m
2
s
m
m
kg
2
2
3








47. Siphoning
• Gasoline is to be siphoned from
a tank. The friction loss in the
is 0.8 ft⋅lbf/lbm. How long
will it take to siphon 5.00 gal,
neglecting the change in liquid level in
the tank during the process and assuming
both points 1 and 2 are at 1 atm.
m
W
F̂
z
g
2
u
P s
2










 ① 1 atm, z = 2.5 ft, u = 0 ft/s
② 1 atm, z = 0.0 ft, u = ?
0 0 V
V
t



48. Siphoning
    
     s
ft
2
lb
lb
ft
s
lb
174
.
32
ft
lb
1
s
ft
2
2
2
1
5
.
10
u
0
8
.
0
ft
5
.
2
174
.
32
u m
f
2
m
f
2 



 


     
 
s
ft
3
2
in
12
ft
1
2
s
ft
2
3
10
58
.
3
V
in
125
.
0
5
.
10
A
u
V








① 1 atm, z = 2.5 ft, u = 0 ft/s
② 1 atm, z = 0.0 ft, u = ?
0 0
min
1
.
3
10
58
.
3
gal
5
V
V
t
min
s
60
s
ft
3
gal
ft
1337
.
0
3
3













 

m
W
F̂
z
g
2
u
P s
2












49. Hydraulic Power Generation
• Water flows from an elevated reservoir through a
conduit to a turbine at a lower level and out of the
turbine through a similar conduit. At ① the pressure
is 207 kPa, and at ② is 124 kPa. What must be the
water flow rate if turbine output is 1.00 MW?
m
W
F̂
z
g
2
u
P s
2











0
0

50. Hydraulic Power Generation
0
0
     
 
        s
kg
m
kg
s
N
s
m
kg
10
m
kPa
m
/
N
10
s
MW
m
N
10
s
m
3
3
s
915
m
103
81
.
9
kPa
83
MW
00
.
1
m
m
103
81
.
9
m
kg
10
kPa
207
124
MW
00
.
1
z
g
P
W
m
2
2
3
2
3
5
2






















m
W
F̂
z
g
2
u
P s
2










