1. Unit 4: Energy Balance

2. Subunit for unit 4.0
4.1 Forms of energy: The First Law of Thermodynamics
4.2 Various form of energy
4.3 Definitions of system (closed or open system)
4.4 Balance on various type of unit process (single/multiple unit operation, reactive and non-reactive)
4.5 Mechanical Energy Balances

3. Importance of ENERGY BALANCE
Considerable amounts of energy are required to run most chemical processes.
Energy balance is used to
account for the energy that flows into or out of each unit of a process
determine the net energy requirements for the process
design ways to reduce the energy requirements so as to improve process profitability

4. THE FIRST LAW OF THERMODYNAMICS
4.1: FORMS OF ENERGY
THE FIRST LAW OF THERMODYNAMICS

5. LEARNING OUTCOME At the end of Unit 4.1, students should be able to :
Describe first law of thermodynamics
Define the terms; kinetic energy, potential energy and internal energy.

6. THERMODYNAMICS
Thermodynamics is the study of the patterns of energy change.
Thermo means energy
Dynamics means patterns of change

7. THERMODYNAMICS Thermodynamics deals with: a) Energy
b) The stability of molecules and direction of change.

8. The first law of thermodynamics
Energy can be exchanged between the system and its surroundings but the total energy of the system and the surrounding is constant.
The energy of the universe remains constant.
1st law of thermodynamics stated that
“ energy is conserved” or
“ energy can neither created nor destroyed”

9. THERMODYNAMICS In thermodynamics, you need to know this terms!!
1- System : a collection of matter within defined boundaries.
2- Surroundings
3- Boundary

10. EXAMPLE

11. THE FIRST LAW OF THERMODYNAMICS
The total energy of a system has three components:
Kinetic Energy
Potential Energy
Internal Energy

12. Ep = Mass x Free Fall Acceleration x Height = mgh
Kinetic Energy
Energy due to the translational motion of the system as a whole relative to some frame of reference (usually the earth’s surface) or to rotation of the system about some axis.
Ek = ½ mv2
Where; m = Mass (kg/g/Ibm); v = Uniform velocity (m/s)
Potential Energy
Energy due to the position of the system in a potential field (such as a gravitational or electromagnetic field).
Ep = Mass x Free Fall Acceleration x Height = mgh
Where; m = Mass (kg/g/Ibm ); g = Acceleration of gravity (9.81 m/s2); h = Height above a reference plane

13. EXAMPLE
Imagine a brick resting on window ledge 3 meters high. As the brick rests on the ledge it has potential energy.
If you knock the brick out of the ledge, the energy is converted to kinetic energy as the brick accelerated towards the ground.

14. Internal Energy
All energy possessed by a system other than kinetic and potential energy, such as energy due to the motion of molecules relative to the center of mass of the system, to the rotational and vibrational motion and the electromagnetic interactions of the molecules, and to the motion and interactions of the atomic and subatomic constituents of the molecules.

15. THERMODYNAMICS
Two types of exchange can occur between the system and surroundings.
1-Energy exchange (heat, radiation….etc)
2-Matter/Mass exchange (movement of molecules across the boundary between the system and the surroundings)

16. THERMODYNAMICS
The total energy of a system before the process has taken place, called the initial state.
While the total energy of a system after the process has taken place, called final state.
The equilibrium state is that condition in which no further change is occurring within the system or between the system and its surroundings.

17. Heat & Work
Two forms of energy transfer between a system & its surroundings
Heat, Q
Energy that flows as a result of temperature difference between a system and its surroundings.
Q as positive if heat flows to system from surroundings.
Work, W
Energy that flows in response to any driving force other THAN a temperature difference, such as a force, a torque or a voltage.
Work is defined as positive when it is done by the system on the surroundings.

18. Heat and work changes
The energy of a system will change if heat is transferred to or from the system or work is done by the system.
1st law stated that, if some amount of heat (Q) is added into the system, it must either do work or increase the total energy of the system.

19. 4.2 Various form of energy

20. LEARNING OUTCOME At the end of Unit 4.2, students should be able to :
Describe various forms of energy and the example

21. Various form of energy
Energy has a number of different forms, all of which measure the ability of an object or system to do work on another object or system. 
In other words, there are different ways that an object or a system can possess energy.

22. Energy Energy exists in many forms.
Energy can be moved from one object to another.
Energy can be changed from one form to another.
Energy cannot be created or destroyed.

24. 4.3 Definitions of system
CLOSED OR OPEN SYSTEM

25. LEARNING OUTCOME At the end of Unit 4.3, students should be able to :
Explain the differences of closed and open system
Write energy balance on closed and open system

26. Types of system:

27. Energy transfer occur in:
Closed system
No mass is transferred across the system boundaries while the process is taking place.
Example: Batch process
Open system
Mass crosses system boundaries.
Example: Semi batch and continuous system

28. Energy Balances on Closed Systems
no mass crosses system boundaries

29. First law thermodynamics for a closed system is
U = Internal energy
Ek = Kinetic energy
Ep = potential energy
Q = Heat
W = Work
∆ is used to signify Final-Initial
Simplifying the first law for specific systems:
If no temperature change, no phase changes, no chemical reaction occur and pressure changes < a few atmospheres, then ∆U ≈ 0
If system is not accelerating, Ek= 0
If system is not rising/falling, Ep = 0
If Tsystem = Tsurroundings or the system is perfectly insulated, then Q = 0 (adiabatic)
If no moving parts /electrical currents at system boundary, W=0

30. If system is not accelerating If system is not rising/falling
EXAMPLE
For a system;
If system is not accelerating
If system is not rising/falling
If no moving parts /electrical currents at system boundary
First law thermodynamics for a closed system is
U = Internal energy
Ek = Kinetic energy
Ep = potential energy
Q = Heat
W = Work

31. One important property for energy balance on closed system is specific internal energy, Û (kJ/kmol).
^ symbol is used to denote the specific property (divided by mass or by mole).

32. Energy Balances on Open System
General Energy balance for open system:
𝑄 − 𝑊 =∆ 𝑈 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝
where;
U = Internal Energy
Ek = Kinetic energy
Ep = Potential Energy
Q = Heat
W = Work
∆ is used to signify Final-Initial

33. 𝑊 = 𝑊 𝑆 + 𝑊 𝐹 Shaft work (Ws)
𝑊 = 𝑊 𝑆 + 𝑊 𝐹
Shaft work (Ws)
All other work transmitted across system boundary by moving parts (pistons, turbines, rotors, propellers,...), electrical currents, radiation.
Flow work (WF)
Also known as PV (pressure and volumetric flow rate) work.
Rate of work done by the fluid at the system outlet minus the rate of work done on the fluid at the system inlet.

34. ENTHALPY H = U + PV Thermodynamic Definition of Enthalpy (H):
U = Internal energy of the system
P = Pressure of the system
V = Volume of the system
Changes in enthalpy mainly when:
Heating or cooling a solid, liquid or gas.
Phase changes (evaporation, condensation, freezing, and melting).

35. Energy Balances on Open System
General Energy balance for open system:
𝑄 − 𝑊 =∆ 𝑈 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝
𝑄 − 𝑊 𝑆 =∆ 𝑈 +𝑃 𝑉 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝
𝑄 − 𝑊 𝑆 =∆ 𝐻 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝
𝑊 = 𝑊 𝑆 + 𝑊 𝐹
P 𝑉

36. ∆ 𝐻 = 𝑛 𝐻 𝑂𝑢𝑡 − 𝐻 𝐼𝑛 ∆ 𝐻 = 𝑚 𝐻 𝑂𝑢𝑡 − 𝐻 𝐼𝑛
One important property for energy balance on open system is specific enthalpy, Ĥ (kJ/kmol).
^ symbol is used to denote the specific property (divided by mass or by mole flow rate).
∆ 𝐻 = 𝑛 𝐻 𝑂𝑢𝑡 − 𝐻 𝐼𝑛
∆ 𝐻 = 𝑚 𝐻 𝑂𝑢𝑡 − 𝐻 𝐼𝑛

37. Simplifying the first law for specific systems:
If there are no moving parts in the system and no energy is transferred by electricity or radiation, Ws=0;
If no significant vertical distance separates the inlet and outlet ports, ΔEp=0;
If the system is not accelerating, ΔEk=0;
Then energy balance equation become:
𝑄 − 𝑊 𝑆 =∆ 𝐻 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝

38. Reference states
It is not possible to know the absolute value of Û and Ĥ for pure species at a given state.
However, the change in ΔÛ and ΔĤ corresponding to specified changes in state (temperature, pressure, phase) could be determined

39. THANK YOU!

40. 4.4 Balance on various type of unit process
Single/Multiple unit operation, Reactive and Non-reactive

41. LEARNING OUTCOME At the end of Unit 4.4, students should be able to :
Construct hypothetical path from reference state to the process state
Calculate internal energy and enthalpy changes from initial to final states
Demonstrate the energy balance for various types of processes

42. Hypothetical Process Path
Hypothetical process path : A path constructed consisting of several steps based on convenience, as long as the final state is reached starting from the initial/reference state.
Divide any processes into few steps based on common processes to construct a hypothetical path and calculate ΔĤ and ΔÛ :
Change in Pressure at constant Temperature & constant Phase.
Change in Temperature at constant Pressure & constant Phase.
Phase changes at constant Temperature & constant Pressure.
Mixing at constant Temperature & constant Pressure.
Chemical reaction at constant Temperature & constant Pressure.

43. Hypothetical Process Path (Condition 1) Changes in Pressure at CONSTANT Temperature and Phase
Ideal Gases
By definition, ΔĤ ≈ 0 and ΔÛ ≈ 0, (molecules do not interact, so changing pressure does not change the internal energy and enthalpy).
Non-Ideal Gases
ΔÛ and ΔĤ are considerably small when pressure changes are minimal (< 5atm). Therefore we can assume ΔĤ = 0 and ΔÛ = 0.
If the pressure changes are large (> 5atm), then use Tables of Thermodynamic Properties (E.g. steam tables for water).
Solid & Liquid
Internal energy nearly independent of pressure, therefore ΔÛ ≈ 0.

44. Hypothetical Process Path (Condition 2) Changes in Temperature at CONSTANT Pressure and Phase
Sensible heat – Heat required or transferred to raise or lower the temperature of a substance without change in phase.
The quantity of heat required to produce a temperature change in a system can be determined from First Law of Thermodynamics. By neglecting kinetic and potential energy changes and work;
For close system Q = ΔU
For open system Q = ΔH
Both the specific internal energy and specific enthalpy of substance is strongly dependent on temperature.

45. For close system, Q = ΔU.
Internal energy depends STRONGLY on temperature provided the system volume, V must remain CONSTANT. The ΔU variation with temperature is shown in the following plot.
where Cv = Heat capacity at constant volume
Heat capacity–The amount of heat required to raise the temperature of one mole or one gram of a substance by one degree Celsius without change in phase.
Unit: J/mol.K or cal/g.oC.
Phase condition:
Ideal gas: Exact
Solid or Liquid: Good approximation
Nonideal gas: Valid only if V is constant

46. For open system, Q = ΔH
ΔH like ΔU also depends STRONGLY on temperature provided the system pressure, P must remain CONSTANT.
where Cp = Heat capacity at constant pressure
Phase condition:
Ideal gas: Exact
Nonideal gas: Exact only if Pressure is constant
Solid or Liquid:
usually negligible EXCEPT when there is large pressure changes and small temperature changes.

47. Example Cp = a + bT + cT2 + dT3 (Form “1”)
HEAT CAPACITY
HOW TO USE TABLE B.2 (HEAT CAPACITIES)
Be sure you use the correct functional form
Example Cp = a + bT + cT2 + dT3 (Form “1”)
Temperature units are sometimes K and sometimes °C
Positive exponent in table (for a, b, c, and d) heading means you use NEGATIVE exponent in the expression.
E.g., if given in the table heading, a x 103 = Therefore use a = x 10-3.

48. HOW TO OBTAIN internal energy or enthalpy change???
For example:
For open system
We know Cp as following (Table B.2): Cp = a + bT + cT2 + dT3
(T24 – T14) ≠ (T2 – T1)4
Be careful when you integrate! (T22 – T12) ≠ (T2 – T1)2

49. To determine the heat capacity, Cp for a mixture of gases or liquid, calculate the total enthalpy change as the sum of the enthalpy changes for pure components.
where (Cp)mix = Heat capacity of the mixture
yi = Mass or moles fraction of each component
Cpi = Heat capacity of each component

50. Hypothetical Process Path (Condition 3) Phase changes at CONSTANT Temperature and Pressure
Phase changes occur from solid to liquid phase, and from liquid to gas phase, and the reverse. Phase change such as melting and evaporation are usually accompanied by large changes in internal energy, U and enthalpy, H
LATENT HEAT: Specific enthalpy change associated with the phase at constant temperature and pressure. (Refer Table B.1)
ΔĤm (T, P): Heat of fusion (or heat of melting) is the specific enthalpy difference from SOLID TO LIQUID forms of a species at T and P.
ΔĤv (T, P): Heat of vaporization is the specific enthalpy difference from LIQUID to VAPOR forms of a species at T and P.
From LIQUID TO SOLID, Heat of solidification is NEGATIVE value of heat of fusion, ΔĤm (T, P).
From VAPOR TO LIQUID, Heat of condensation is NEGATIVE value of heat of vaporization, ΔĤv (T, P).

51. Energy Balance Procedures
Draw and completely label a flow diagram. Do include temperature, pressure & phase (solid, liquid or gas phase).
Perform all the required mass balance calculations FIRST.
Write the appropriate form of the energy balance (closed or open system) and delete any of the terms that are either zero or negligible for the given process system.
𝑄 − 𝑊 𝑆 =∆ 𝐻 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝

52. Choose a reference state (phase, temperature, and pressure) for each species involved in the process.
Construct inlet-outlet enthalpy table of each species for close system or open system. Species with few phases are considered as separately.
For example:

53. Calculate all required values of Ĥ or Û and insert the values into table. Determine specific enthalpies of each stream component (Table B.1, Table B.2, steam table, etc.)
Calculate the overall ΔĤ or ΔÛ for the system.
Calculate any other term in the energy balance equation (e.g work, kinetic energy, or potential energy) (if its applicable to the particular energy balance).
Solve the energy balance, (Find Q).

54. 4.4 Balance on various type of unit process
EXERCISE
Single/Multiple unit operation, Reactive and Non-reactive

55. EXERCISE
1. Calculate ΔĤ for Acetone (l, 15 °C) → acetone (l, 55 °C)

56. EXERCISE
2. 20 mole/hr of benzene vapor at 580°C is cooled and converted to a liquid at 25°C in a continuous condenser.
Construct a hypothetical path from initial and final states of the process
Calculate the specific enthalpy change for each step
Determine the specific enthalpy changes from initial and final states of the process
Calculate the heat, 𝑄 is transferred from the benzene in the condenser.

57. EXERCISE
3. Determine the specific enthalpy (kJ/mol) of n-hexane vapor at 200°C and 2.0 atm relative to n-hexane liquid at 20°C and 1.0 atm, assuming ideal gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step.
Construct a hypothetical path from initial and final states of the process
Calculate the specific enthalpy change for each step
Determine the specific enthalpy changes from initial and final states of the process

58. EXERCISE
4. Calculate the heat, Q required to bring 150 mol/h of a stream containing 60% C2H6 (ethane) and 40% C3H8 (propane) from 0˚C to 400˚C. Determine heat capacity for the mixture as part of the problem solution.

59. EXERCISE
Evaporation process of a mixture is shown in flow chart below. Neglect the effects of pressure changes on enthalpies; calculate the required heat input rate in kilowatts by taking [benzene (l, 90˚C), toluene (l, 90˚C)] as reference for enthalpy calculations.
35 mol/s
0.714 mol B/mol
0.286 mol T/mol
(55oC, v)
𝑛 𝐹 mol/s
0.67 mol B/mol
0.33 mol T/mol
(90oC, l)
𝑛 𝐿 mol/s
0.345 mol T/mol
0.655 mol B/mol
(55oC, l)

60. steps
completely label a flow diagram
Mass balance for each species

61. appropriate form of the energy balance (open system)
reference state (phase, temperature, and pressure) for each species involved in the process – refer question [benzene (l, 90˚C), toluene (l, 90˚C)]
Construct inlet-outlet enthalpy table of each species involved
𝑄 − 𝑊 𝑆 =∆ 𝐻 +∆ 𝐸 𝑘 +∆ 𝐸 𝑝
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25 H1
T (l)
45.43
35.42 H2
B (v) -
24.99 H3
T (v)
10.01 H4
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
0 (Ref) H1
T (l) H2
B (v) - H3
T (v) H4

62. Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25 H1
T (l)
45.43
35.42 H2
B (v) -
24.99 H3
T (v)
10.01 H4
Calculate all required values of Ĥ and insert the values into table. Determine specific enthalpies of each stream component using Table B.1 and Table B.2

63. Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25
H1 =
T (l)
45.43
35.42
H2 =
B (v) -
24.99 H3
T (v)
10.01 H4

64. Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25
H1 =
T (l)
45.43
35.42
H2 =
B (v) -
24.99 H3
T (v)
10.01 H4

65. Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25
H1 =
T (l)
45.43
35.42
H2 =
B (v) -
24.99
H3 =
T (v)
10.01 H4

66. Calculate the overall ΔĤ for the system.
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
B (l)
92.24
0 (Ref)
67.25
H1 =
T (l)
45.43
35.42
H2 =
B (v) -
24.99
H3 =
T (v)
10.01
H4 =
Calculate the overall ΔĤ for the system.
Solve the Q for the system

67. EXERCISE
Condensation process of a mixture is shown in flow chart below. Neglect the effects of pressure changes on enthalpies; calculate the required heat input rate in kilowatts by taking [acetone (v, 20˚C), toluene (v, 20˚C)] as reference for enthalpy calculations.
𝑛 𝑉 mol/s
0.092 mol A/mol
0.908 mol T/mol
(20oC, v)
200 mol/s
0.669 mol A/mol
0.331 mol T/mol
(65oC, l)
𝑛 𝐿 mol A/s
(20oC, l)

68. TRY YOURSELF! ANSWER Substance INPUT OUTPUT nin (mol/s) Hin (kJ/mol)
Reference: acetone (v, 20˚C), toluene (v, 20˚C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
A (l)
133.8
H1 = –26.134
H3 = –
T (l)
66.2
H2 = –30.896 -
A (v)
6.7074
0 (Ref)
T (v)
66.200

69. Methods for reactive process
Heat of Reaction (ΔĤr) Method
Heat of Formation (ΔĤof) Method

70. Heat of Reaction/Enthalpy of Reaction, ΔĤr
Enthalpy change for a process in which stoichiometric quantities of reactants at T & P react completely in single reaction to form products at the same temperature and pressure.
If ΔĤr is negative, the reaction releases heat (exothermic reaction).
If ΔĤr is positive, reaction requires heat (endothermic reaction).
The standard heat of reaction, ΔĤor is the heat of reaction when both reactants and products are at specific reference T and P of 25˚C and 1 atm.

71. Heat of Reaction/Enthalpy of Reaction, ΔĤr
The relationships between enthalpy change, ΔH and heat of reaction/enthalpy of reaction, ΔĤr is given as following
Extent of reaction, ξ (measure of how far a reaction has proceed)
νA= Stoichiometric coefficient of a reactant or reaction product A
nA,r = Number of moles of A consumed or generated at T0 and P0
modulus sign – values must be positive

72. 1) Heat of Reaction (ΔĤr) Method
This method is preferable when there is a SINGLE reaction for which ΔĤor is known or given.
Procedures:
Complete the material balance calculations on the reactor as much as possible.
Choose reference state for specific enthalpy, ΔĤ calculations. The best choice is generally 25oC and 1 atm.
Prepare the inlet-outlet enthalpy table, inserting the known values.
Calculate the extent of reaction, ξ.
Calculate ΔĤ for the reactor. For a single reaction in a continuous process we can use:
∆ 𝐻 =ξ∆ 𝐻 𝒓 𝑶 + 𝑜𝑢𝑡 𝑛 𝑖 𝐻 𝑖 - 𝑖𝑛 𝑛 𝑖 𝐻 𝑖
Substitute the calculated values into the energy balance equation for open system.

73. example
The standard heat of reaction for the propane combustion process is given below:
Calculate the rate at which heat, Q must be transfer to or from the reactor.

74. Material balance calculations
Reference state 25 oC and 1 atm because the specific enthalpy for most of the compounds in this example (O2(g), N2(g), CO2(g), and H2O(g)) is available from the Physical Property Table (Table B.8).
Inlet-outlet enthalpy table, inserting the known amounts for all components. Remember if the components is at its reference state (E.g. C3H8(g) inlet condition is 25 oC and 1 atm = Reference state), therefore the enthalpy value for this stream is 0.
Reference: C3H8 (g), N2 (g), O2 (g), CO2 (g), H2O (l) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100
0 (Ref) -
O2 (g)
600 H1 H3
N2 (g)
2256 H2 H4
CO2 (g)
300 H5
H2O (v)
400 H6

75. Calculate the extent of reaction, ξ.
Reference: C3H8 (g), N2 (g), O2 (g), NO (g), H2O (l) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100
0 (Ref) -
O2 (g)
600 H2 H4
N2 (g)
2256 H3 H5
CO2 (g)
300 H6
H2O (v)
400 H7
Calculate the extent of reaction, ξ.
ξ= ( 𝑛 𝐶3𝐻8 ) 𝑜𝑢𝑡 − ( 𝑛 𝐶3𝐻8 ) 𝑖𝑛 𝑣 𝐶3𝐻8
When we want to calculate extent of reaction, choose ANY species A (can be any product or reactant) in which the feed and product flow rates are known. In this example, we choose C3H8 as species A. Therefore ξ can be calculated as

76. From Table B.8 (Specific Enthalpies of Selected Gases) and Table B.1
Reference: C3H8 (g), N2 (g), O2 (g), NO (g), H2O (l) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100
0 (Ref) -
O2 (g)
600 H2 H4
N2 (g)
2256 H3 H5
CO2 (g)
300 H6
H2O (v)
400 H7
Calculate each unknown stream component enthalpy, Ĥi, as ΔĤ for the species going from its reference state to the process state, and insert the enthalpies in the table.
From Table B.8 (Specific Enthalpies of Selected Gases) and Table B.1
Ĥ2 = ΔĤ for O2 (25 oC) —> O2 (300 oC) = 8.47 kJ/mol
Ĥ3 = ΔĤ for N2 (25 oC) —> N2 (300 oC) = 8.12 kJ/mol
Ĥ4 = ΔĤ for O2 (25 oC) —> O2 (1000 oC) = kJ/mol
Ĥ5 = ΔĤ for N2 (25 oC) —> n2 (1000 oC) = kJ/mol
Ĥ6 = ΔĤ for CO2 (25 oC) —> CO2 (1000 oC) = kJ/mol
Ĥ7 = ΔĤ for H2O (25 oC, l) —> H2O (1000 oC, g) =??
(At reference state, 25 oC, water is in liquid phase but at 1000 oC, it will be in gas phase). Therefore;

77. Reference: C3H8 (g), N2 (g), O2 (g), NO (g), H2O (l) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100
0 (Ref) -
O2 (g)
600
8.47
32.47
N2 (g)
2256
8.12
30.56
CO2 (g)
300
48.60
H2O (v)
400
81.71
Calculate ΔH
𝑄=∆ 𝐻
∆ 𝐻 =ξ∆ 𝐻 𝒓 𝑶 + 𝑜𝑢𝑡 𝑛 𝑖 𝐻 𝑖 - 𝑖𝑛 𝑛 𝑖 𝐻 𝑖
𝑜𝑢𝑡 𝑛 𝑖 𝐻 𝑖
𝑖𝑛 𝑛 𝑖 𝐻 𝑖
Q = ∆ 𝐻 =ξ∆ 𝐻 𝒓 𝑶 + 𝑜𝑢𝑡 𝑛 𝑖 𝐻 𝑖 - 𝑖𝑛 𝑛 𝑖 𝐻 𝑖

78. Heat of Formation, ΔĤof
A formation reaction of a compound is the reaction in which the compound is formed from its elemental constituents as they normally occur in nature.
The enthalpy change associated with the formation of 1 mole of the compound at a reference temperature and pressure (usually 25 oC and 1 atm) is the standard heat of formation of the compound, Δ 𝐻 of
Standard heats of formation for many compounds are listed in Table B.1

79. 2) Heat of Formation (ΔĤof) Method
This method is preferable for MULTIPLE reactions and single reaction for which ΔĤor is not known.
Procedures:
Complete the material balance calculations on the reactor as much as possible.
Choose reference state for specific enthalpy, ΔĤ calculations.
- This is the step that distinguishes the Heat of Reaction Method with this one.
- Choose the elemental species that constitute the reactants and products in the state in which the elements are found at 25ºC and 1 atm [C(s), H2(g),etc].
Prepare the inlet-outlet enthalpy table, inserting the known values.

80. Calculate each unknown ΔĤ for the species.
- For a reactant or product, START with the ELEMENTAL SPECIES at 25ºC and 1 atm (references) and form 1 mol of the process species at 25ºC and 1 atm (Ĥ=ΔĤof from Table B1).
- Then bring the species from 25ºC and 1 atm to its process state (Can use heat capacities from Table B.2, specific enthalpies from Table B.8 and B.9, and latent heats from Table B.1).
Calculate for the reactor. For both single and multiple reactions, the formula is
Substitute the calculated values into the energy balance equation for open system.

81. example The propane combustion process is given below:
Calculate the rate at which heat, Q must be transfer to or from the reactor. Given
Heat of formation (kJ/mol) for Propane is – (l), (g)
Heat of formation (kJ/mol) for Carbon dioxide is (l), (g)

82. Material balance calculations
Taking elemental species [ C (s), H2 (g), O2 (g), N2 (g)] at 25 oC and 1 atm
Inlet-outlet enthalpy table
Reference: C (s), H2 (g), O2 (g), N2 (g) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100 H1 -
O2 (g)
600 H2 H4
N2 (g)
2256 H3 H5
CO2 (g)
300 H6
H2O (v)
400 H7

83. Calculate each unknown stream component enthalpy, Ĥi
Reference: C (s), H2 (g), O2 (g), N2 (g) (1 atm, 25°C)
Calculate each unknown stream component enthalpy, Ĥi
Form (∆ 𝐻 𝑓 𝑜 ) and bring to the process state using Table B.1, Table B.2 or Table B.8
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100 H1 -
O2 (g)
600 H2 H4
N2 (g)
2256 H3 H5
CO2 (g)
300 H6
H2O (v)
400 H7

84. Calculate ΔH 𝑄 Reference: C (s), H2 (g), O2 (g), N2 (g) (1 atm, 25°C)
Substance
INPUT
OUTPUT
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Hout (kJ/mol)
C3H8 (g)
100
-103.8 -
O2 (g)
600
8.47
32.47
N2 (g)
2256
8.12
30.56
CO2 (g)
300
-344.9
H2O (v)
400 𝑄

85. 4.5 Mechanical Energy Balances

86. LEARNING OUTCOME At the end of Unit 4.5, students should be able to :
Solve mechanical energy balance for unknown variable

87. Mechanical energy balance
Mechanical energy balance accounts for energy flows in process operation involve the flow of fluids to, from, and between tanks, reservoir, wells, and process units.
For steady state flow of an incompressible fluid; the mechanical energy balance is
∆𝑃 𝜌 + ∆ 𝑢 𝑔∆𝑧+ 𝐹 = − 𝑊 𝑠 𝑚
where shaft work, WS is work by fluid on moving elements in process line
𝐹 is frictional energy loss in system, 𝑚 is mass flow rate
ΔP = P2 – P1; Δ u2 = u22 –u12; Δz = z2 –z1

88. Mechanical energy balance
Simplified form of mechanical energy balance is Bernoulli equation
where no shaft work is performed (WS =0) and frictionless processes ( 𝐹 =0)

89. EXAMPLE
Water flows between two points 1,2. The volumetric flow rate is 20 litres/min. Point 2 is 50 m higher than point 1. The pipe internal diameters are 0.5 cm at point 1 and 1 cm at point 2. The pressure at point 2 is 1 atm. Estimate the pressure at point 1 if friction losses are negligible.
P2 – P1
= P1
= u22 –u12
1000 kg/m3
9.81 m/s2
z2 –z1
= 50 m

90. 𝑉 = 20 𝐿 𝑚𝑖𝑛 × 1 𝑚 𝐿 1 𝑚𝑖𝑛 60 𝑠
= 𝑚 3 /s
𝑢= 𝑉 𝐴
𝜋 𝑟 2
u1 = /(π(0.0025)2) = m/s
u2 = / (π(0.005)2) = 4.24 m/s
( P1)/ [(4.24)2 – (16.97)2]/ (50) = 0
P1 = Pa (4.6 bar)

91. EXERCISE
An aqueous solution with SG = 1.12 flows through a channel with a variable cross section. Data taken at two axial positions in the channel are shown below.
Point 2 is 6 meters higher than point 1.
a) Calculate the velocity at point 2, neglecting the friction.
b) If the pipe diameter at point 2 is 0.06 m, what is the diameter at point 1?
Point 1
Point 2
Pressure
1.5 x 105 Pa
9.77 x 104 Pa u
5.00 m/s ?

92. QUIZ 4 (30 minutes) GOOD LUCK!

93. THANK YOU You’re done with all units!!
GOOD LUCK WITH QUIZ, TEST 2 and FINAL