1. Energy Balances on Non-Reactive Processes
Forms of Energy
Energy Balances on Closed Systems
Energy Balances on Open Systems at Steady State
Elements of Energy Balance Calculations
Changes in Pressure and Temperature
Phase Change Operations

2. 1. Form of Energy
The First Law of Thermodynamics  Energy neither be created nor destroyed
3 components of total energy of a system
Kinetic energy (Ek)
Energy due to motion of an object or flowing stream relative to the surface of the earth.
Unit: Ek (J) or ĖK (J/s)
Potential energy (Ep)
Due to position in a gravitational or electromagnetic potential field or due to height of center of mass of object above a reference height.
Unit: Ep (J) or ĖP (J/s)
Internal energy (U)
Sum of rotational, vibrational, and electromagnetic energies of individual molecules, atoms & subatomic particles moving around in a body of material or a flowing stream.
Unit: U(J) or Ů (J/s)

3. Kinetic Energy Definition Energy of a moving object due to its motion
Formula Ek = ½ mu2
Where;
m = Mass (kg/g/Ibm)
u = Uniform velocity (m/s)
Units Joules (J) by an object / (J/s) fluid.

4. Ep = Mass x Free Fall Acceleration x Height
Potential Energy
Formula
Ep = Mass x Free Fall Acceleration x Height
Ep = mgz
Where;
m = Mass (kg/g/Ibm )
g = Acceleration of gravity(9.81 m/s2)
z = Height above a reference plane
Units Joules (J) by an object / (J/s) fluid.

5. Energy can be transferred as heat (Q)/work (W)
Heat & Work
2 forms of energy transfer between a system & its surroundings in a CLOSED system
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 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.
+Q (Heat added)
Work done positive and negative
+W (Work Done)
ETOT = EK + EP + U
Energy can be transferred as heat (Q)/work (W)

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

7. 2. Energy Balances on Closed Systems
A system is termed open / closed according to whether or not mass cross the system boundary.
Closed system: no mass crosses system boundaries.
Open system: mass crosses system boundaries.

8. 2. Energy Balances on Closed Systems
Since energy can neither be created nor destroyed.
The generation & consumption terms of the general balance drop out, leaving
Accumulation =
Input – Output + Generation – Consumption
Accumulation = Input – Output

9. 2. Energy Balances on Closed Systems
Accumulation = Input – Output
Mass balance closed system, we eliminated the input & output terms.
However, ENERGY can be transferred across the boundaries as HEAT/WORK.
Accumulation = Final value – initial value of the balanced quantity
Final system energy – Initial system energy = Net energy transferred to the system (in – out)

10. 2. Energy Balances on Closed Systems
Final system energy – Initial system energy = Net energy transferred to the system (in – out)
∆ is used to signify
(Final – Initial)

11. 2. Energy Balances on Closed Systems
First law thermodynamics for a closed system is
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)

12. 2. Energy Balances on Closed Systems
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/mole).

13. 2. Energy Balances on Closed Systems
In CLOSED system,
The energy balance reduce to

14. 3. Energy Balances on Open Systems at Steady State
General Energy balance for open system:
where
U = Internal Energy
Ek = Kinetic Energy
Ep = Potential Energy
Q = Heat
W = Work
∆ is used to signify
(Final-Initial).

15. 3. Energy Balances on Open Systems at Steady State
Work appearing in the equation is the combined flow work & shaft work.
Shaft work (Ws)
All other work transmitted across system boundary by moving parts (pistons, turbines, rotors, propellers,...), electrical currents, radiation.
Flow work (Wfl)
Also known as PV (pressure & 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.

16. 3. Energy Balances on Open Systems at Steady State
PROCESS UNIT
(m3/s)
(m3/s)
Pin (N/m2)
Pout (N/m2)
The net flow work is determined as:
The flow work is usually expressed in terms of pressure and volumetric flow rate:
If there is several input and output streams enter and leave the system, the PV products for each streams must be summed:

17. Definition of Enthalpy
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).

18. 3. Energy Balances on Open Systems at Steady State
It is known that the sum of Internal Energy, U and Wfl = PV is equal to enthalpy, H therefore we obtain;
The new energy balance equation now becomes

19. 3. Energy Balances on Open Systems at Steady State
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).

20. 3. Energy Balances on Open Systems at Steady State
Simplifying the first law for specific systems:
If there are no moving parts in the system and no energy is transferred by electricity/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:

21. 3. Energy Balances on Open Systems at Steady State
In open system,
The energy balance reduce to

22. Energy Balances Equation
In closed (batch) system,
In open (continuous), steady state system,
where;

23. Energy Equation Information
What they say…
What they mean…
Well insulated
Q = 0
Adiabatic
Rigid container
Volume doesn’t change
Wfl = 0
Isochoric
Constant volume
No mechanical parts, or no moving parts
Ws = 0
Isothermal
∆T = 0, but Q ≠ 0
Q not equal to zero?

24. 4. Elements of Energy Balance Calculations
What is a Reference State ?
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 a specified changes in state (temperature, pressure, phase) can be determined.
Specific internal energy at that state relative to the reference state.

25. 4. Elements of Energy Balance Calculations
H2O (liquid, 0.01°C, bar)  H2O (vapor, 400 °C, 10.0 bar), ΔÛ = 2958 kJ/kg
This does not mean that the absolute value of Û for water in specified state is 2958 kJ/kg; remember, we cannot know the absolute value of Û.
It means that Û of water vapor at 400 °C & bar is 2958 kJ/kg relative to water at the reference state [H2O (liquid, 0.01°C, bar)].

26. Example: Reference states and states properties
The enthalpy changes for CO going from a reference state of 0˚C and 1 atm to two other states are measured with following results.
Determine enthalpy changes of the process.
CO (g, 0oC, 1 atm)  CO (g, 100oC, 1 atm)
: Ĥ1= 2919 J/mol
CO (g, 0oC, 1 atm)  CO (g, 500oC, 1 atm)
: Ĥ2= J/mol

27. Example : Reference states and states properties
Ĥ for CO at (g, 100°C, 1 atm) and (g, 500°C, 1 atm) cannot be known absolutely.
CO (g, 0oC, 1 atm)  CO (g, 100oC, 1 atm): Ĥ1= 2919 J/mol
Assign value of specific enthalpy at a reference state, Ĥref = 0. Hence,
ΔĤ1 = Ĥ1 – Ĥref = Ĥ1 – 0 = Ĥ1
ΔĤ100 = Ĥ100 – 0 = 2919 J/mol
Similarly for Ĥ for CO at (g, 500°C, 1 atm)
CO (g, 0oC, 1 atm)  CO (g, 500oC, 1 atm): Ĥ2= J/mol
ΔĤ500 = Ĥ500 – 0 = J/mol
Since the reference state for both conditions is the same, the ΔĤ for CO from State 1 (g, 100°C, 1 atm) and State 2 (g, 500°C, 1 atm) can be calculated by
ΔĤ = Ĥ2 – Ĥ1 = – 2919 = J/mol
Reference State

28. Reference States
If two different physical property tables are used, make sure Ĥ1 and Ĥ2 are based on the same reference state.
If another reference state had been used to generate the specific enthalpy, Ĥ, they would have different values but the changes (Ĥ) still the same. Example :
4/8/2014 what is the two different tables.
Note that although the enthalpies values are different based on the reference state, but ΔĤ would still be the same, 12,141 J/mol.

29. Hypothetical Process Path
When a species passes from one state to another state, both ΔĤ and ΔÛ for the process are independent on the path taken from the first state to the second state.
A path, called hypothetical process path can be constructed consisting of several steps based on convenience, as long as the final state is reached starting from the initial/reference state.

30. Hypothetical Process Path
To evaluate changes in enthalpy or internal energy.
Û and Ĥ depends only on the state of the species & not depends on how the species reached its state.
Depend on T, P, & state of aggregation (solid, liquid, or gas). 3 2 1

31. Hypothetical Process Path
3 processes that can be used to calculate ΔĤ and ΔÛ associated with certain process are as following:
Changes in P at constant T & state of aggregation.
Changes in T, at constant P & state of aggregation.
Phase changes at constant T & P.
Since Ĥ is a state property, ΔĤ calculated for the hypothetical process path  which we constructed for convenience is the same as ΔĤ for the path actually followed by the process.
The same procedure can be followed to calculate ΔÛ for any process.

32. Hypothetical Process Path
Procedures to evaluate ΔĤ and ΔÛ when there is no table of Ĥ and Û are available for that particular process species.
How:
Need to construct hypothetical process path in order to simplify the calculations.
Where:
Starting point: Defined your reference state (Temperature, Pressure and Phase )
End point: The conditions of the stream of interest (inlets or outlet).

33. Hypothetical Process Path: Example 1
Example 1: Calculate ΔĤ for a process in which solid phenol at 25 oC and 1 atm is converted to a vapor phenol at 300 oC and 3 atm.
Cannot determine directly form enthalpy table. Hence must use hypothetical process path consist of several step.
ΔĤ = Ĥ (solid, 25oC, 1 atm) Ĥ (vapor, 300oC, 3 atm)
Data not available

34. Hypothetical Process Path: Example 1
ΔĤ = Ĥ (solid, 25oC, 1 atm) Ĥ (vapor, 300oC, 3 atm)
Check Table B.1: Normal melting points and boiling points, and standard heats of phase change (Reference: P = 1 atm; Tm = 42.5°C; and Tb = 181.4°C)
Changes in T
Changes in P
Changes in Phase
Changes in T
Changes in T
Changes in Phase

35. Procedure for Energy Balance Calculations
Draw and completely label a flow diagram. Do include T, P & phase (solid, liquid or gas phase).
Perform all the required material balance calculations FIRST.
Write the appropriate form of the energy balance (closed/open system) & delete any of the terms that are either zero/negligible for the given process system.

36. Procedure for Energy Balance Calculations
Choose a reference state (phase, T, & P) for each species involved in the process.
If using enthalpy table, choose the reference state as stated in the standard table (Table B.8)
If no table are available, choose either one inlet or one outlet condition as the reference state so that at least Ĥ or Û can be set to zero.)

37. Procedure for Energy Balance Calculations
Determine specific enthalpies of each stream component
Look up the information using tables (e.g Table B.1, B.2)
Other components - calculate them
Construct inlet-outlet enthalpy table of each species for close system/open system. Species with few phases are considered as separate.

38. Procedure for Energy Balance Calculations
For closed system, construct a table with columns for initial and final amounts of each species (mass/moles) and specific internal energies (Û) relative to the chosen reference states.
For an open system, construct a table with columns for inlet and outlet stream component flow rates (mass/moles) and specific enthalpies (Ĥ) relative to the chosen references states.

39. Procedure for Energy Balance Calculations
Calculate all required values of Ĥ or Û and insert the values into table.
Calculate the overall ΔĤ or ΔÛ for the system.

40. Procedure for Energy Balance Calculations
8. 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).
9. Solve the energy balance, (Find Q).

41. 5. Changes in Pressure (1st)
Changes in P at CONSTANT T & 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 is minimal (< 5 atm). Therefore we can assume ΔĤ = 0 and ΔÛ = 0.
If the pressure changes are large (> 5 atm), then use tables of thermodynamic properties (E.g. steam tables for water).
Solid & Liquid
Internal energy nearly independent of pressure, therefore ΔÛ ≈ 0. G

42. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
Sensible heat – Heat required/transferred to raise/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:
For close system Q = ΔU
For open system Q = ΔH
We neglected kinetic and potential energy changes and work.
Both the specific internal energy and specific enthalpy of substance is strongly dependent on temperature.

43. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase CLOSED SYSTEM
For close system, Q = ΔU.
It depends STRONGLY on temperature provided the system volume, V must remain CONSTANT.
The ΔU variation with temperature is shown in the following plot.
Curve slope = Heat capacity at
constant volume, Cv.
Since the plot is not a straight line,
Cv (slope of the curve) is a
function of temperature.

44. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase CLOSED SYSTEM
where Cv = Heat capacity at constant volume
Phase condition:
Ideal gas : Exact
Solid or Liquid: Good approximation
Nonideal gas : Valid only if V is constant

45. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase OPEN SYSTEM
For open system, Q = ΔH
ΔH like ΔU also depends STRONGLY on temperature provided the system pressure, P must remain CONSTANT.
The ΔH variation with temperature is shown in the following plot.
Curve slope = Heat capacity at
constant pressure, Cp.
Since the plot is not a straight
line, Cp (slope of the curve) is a
function of temperature. ΔH ΔT

46. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase OPEN SYSTEM
where Cp = Heat capacity at constant pressure
Phase condition:
Ideal gas : Exact
Nonideal gas : Exact only if Pressure is constant
Solid or Liquid :
* is usually negligible EXCEPT when there is large pressure changes and small temperature changes.

47. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
HEAT CAPACITY FORMULAS
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.
Heat capacities are functions of temperature and are expressed in polynomial form as following:
Cp = a + bT + cT2 + dT3 (Form “1”)
Values of coefficient a, b, c, and d are given in Table B.2 of Appendix B for number of species at 1 atm.
Relationship between Cp and Cv:
Cp = Cv for Liquids and Solids
Cp = Cv + R for Ideal gas
* Too complex for nonideal gases

48. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
HEAT CAPACITY FORMULAS

49. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
HEAT CAPACITY FORMULAS
How to use Table B.2
Be sure you use the correct functional form
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.

50. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
HEAT CAPACITY FORMULAS
For open system, we know previously
We know Cp as following (Table B.2): Cp = a + bT + cT2 + dT3
We combine both to obtain:
Be careful when you integrate! (T22 – T12) ≠ (T2 – T1)2

51. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase
HEAT CAPACITY FORMULAS
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

52. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase EXAMPLE
Calculate the heat, Q required to bring 150 mol/h of a stream containing 60% C2H6 (ethane) and 40% C3H8 (propane) by volume from 0˚C to 400˚C. Determine heat capacity for the mixture as part of the problem solution.
Solution:
For mixture:

53. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase EXAMPLE
From Table B:2
For Ethane: Cp = (49.37x10-3) x10-5T – 5.816x10-8T x10-12T3
For Propane: Cp = x x10-5T – 13.11x10-8T x10-12T3
Therefore:

54. 5. Changes in Temperature (2nd)
Changes in T at CONSTANT P & Phase EXAMPLE
To calculate heat, Q, we know for open system

55. 6. Phase Change Operations
Phase changes at CONSTANT T & P
Latent heat: Specific enthalpy change associated with the phase at constant temperature and pressure.
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).
Tabulated values of these latent heats for each species are in Table B.1

56. 6. Phase Change Operations
Phase changes at CONSTANT T & P
From Table B.1
ΔĤm (T, P):
Heat of fusion
(or heat of melting) at
Tm (melting temperature)
ΔĤv (T, P): Heat of vaporization at
Tb (boiling temperature)

57. Hypothetical Process Path
5/8/2014

58. Energy Balance Calculations (Example)
Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. The process operates at steady state. Calculate the required cooling rate.

59. Energy Balance Calculations (Example)
We will follow the procedure given preceding this example.
1. Perform required material balance calculations. None are required in this example.

60. Energy Balance Calculations (Example)
2. Write and simplify the energy balance.
For this open steady-state system, there are no moving parts in the system & no energy is transferred by electricity/radiation. No significant vertical distance separates the inlet & outlet ports. Phase changes & non negligible temperature changes occur. The energy balance reduces to

61. Energy Balance Calculations (Example)
3. Choose reference states for acetone and nitrogen.
Table B.8 lists specific enthalpies of nitrogen relative to N2(g, 25°C, 1 atm), which makes this state a convenient choice for nitrogen.
There are no tabulated enthalpy data for acetone in the text, so choose one of the process stream conditions, Ac(l, 20°C, 5 atm), as the reference state for this species, which will set the corresponding Ĥ value equal to zero rather than having to calculate it.

62. Energy Balance Calculations (Example)
Similar with reference state
4. Construct an inlet-outlet enthalpy table.

63. Energy Balance Calculations (Example)
5. Calculate all unknown specific enthalpies
Change in P
Change in T
Change in T
Change in phase
Tb = 56 oC
(Table B.1)

64. Energy Balance Calculations (Example)
Change in Pressure
Why 5-1?

65. Energy Balance Calculations (Example)
Change in Temperature

66. Energy Balance Calculations (Example)
Change in Temperature

67. Energy Balance Calculations (Example)
Change in Phase

68. Energy Balance Calculations (Example)
Change in Temperature

69. Energy Balance Calculations (Example)
5. Calculate all unknown specific enthalpies
Change in P
Change in T
Change in T
Change in phase

70. Energy Balance Calculations (Example)
Ĥ3 = ΔĤ for Acetone (l, 20 oC, 5 atm)  Acetone (v, 20 oC, 5 atm)
Tb = 56 oC
(Table B.1)
(l, 20 oC, 1 atm)
(l, 56 oC, 1 atm)
(v, 56 oC, 1 atm)
(v, 20 oC, 1 atm)
Try yourself!

71. Energy Balance Calculations (Example)
From Table B.8 Specific Enthalpies of Selected Gases
Ĥ2 = ΔĤ for N2 (25 oC, 1 atm)  N2 (65 oC, 1 atm) = 1.16 kJ/mol
Ĥ4 = ΔĤ for N2 (25 oC, 1 atm)  N2 (20 oC, 5 atm)
Try yourself!

72. Energy Balance Calculations (Example)
From Table B.8 Specific Enthalpies of Selected Gases
Ĥ2 = ΔĤ for N2 (25 oC, 1 atm)  N2 (65 oC, 1 atm) = 1.16 kJ/mol
Ĥ4 = ΔĤ for N2 (25 oC, 1 atm)  N2 (20 oC, 5 atm)
Try yourself!

73. Energy Balance Calculations (Example)
6. Calculate ΔḢ.

74. Energy Balance Calculations (Example)
6. Calculate kinetic and potential energy changes.
Since there is no shaft work and we are neglecting kinetic & potential energy changes, there is nothing to do in this step.
7. Solve the energy balance
Heat must be transferred from the condenser at a rate of 2320 kW to achieve the required cooling & condensation.

75. Procedure: Energy Balance Calculations
Summary:
Procedure: Energy Balance Calculations
Step
Notes
Perform material balance calculation
Write appropriate energy balance
Simplify negligible terms
Choose reference state (phase, T, P) for each species
Depends on available data
Construct process table with all information (with all phases)
Known or unknown for all phases
Calculate required values for energies (Û or Ĥ) and insert in table
Calculate ΔU or ΔH
Calculate other terms in energy balance
Terms not neglected
Find unknowns
15/8/14

76. Δ = final – initial = out – in = +/-
Summary:
Sensible heat Heat required/transferred to raise/lower the temperature of a substance without change in phase.
Latent heat Specific enthalpy change related with the phase at constant temperature and pressure.
Specific gravity  Ratio of the density () of a substance to the density of a reference (ref) substance at a specific condition
Specific volume  Volume per unit mass of a substance. Inverse of density. Unit: cm3/g; m3/kg; ft3/lbm
Cp = Cv for Liquids and Solids
Cp = Cv + R for Ideal gas
Δ = final – initial = out – in = +/-
15/8/14

77. Work Extra note: The opposite sign rule is sometimes used.
Work, W
Energy that flows in response to any driving force other 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.
The opposite sign rule is sometimes used.
The choice is subjective, as long as it is used consistently;
However, to avoid confusion when reading thermodynamic references, you should be sure which rule has been used.
+W (Work Done)
Work done positive and negative