1 ‫موازنة الطاقة للعمليات بدون تفاعالت‬

2 Energy Balances on Nonreactive Processes

3 Fifth Lectures (10 Hours)

4 Outline Course
5 First Lecture ........................................................................................................................... 2
6 1. REVIEW FOR FIRST LAW OF THERMODYNAMICS...................................... 2

7 3. CHANGES IN PRESSURE AT CONSTANT TEMPERATURE .......................... 3

8 4. CHANGES IN TEMPERATURE ............................................................................. 3

9 4.1 Heat a Substance at Constant Volume ................................................................ 3
10 4.2 Heat a Substance at Constant Pressure ............................................................... 3

11 Second Lecture ....................................................................................................................... 7

12 Heat Capacity of a Mixture ........................................................................................ 7

13 EIGHT STEPS FOR EB CALCULATIONS ............................................................... 9

14 Third Lecture........................................................................................................................ 14
15 5. PHASE CHANGE OPERATIONS ......................................................................... 14
16 5.1 Latent Heats ....................................................................................................... 14
17 5.2 Estimation of Latent Heats ................................................................................. 19

18 Fourth Lecture...................................................................................................................... 22
19 6. HEAT OF MIXING AND SOLUTION (Ĥs) ........................................................ 22

20 Fifth Lecture ......................................................................................................................... 29

21 7. VAPOR–LIQUID EQUILIBRIUM CALCULATIONS ....................................... 29
22
23
24 Reference
25 Richard M. Felder. Elementary Principles of Chemical Processes, 3rd edition 2005.
 Energy Balance 2 / 13

1 Module Three
2 Energy Balances on Nonreactive Processes
3 First Lecture
4
5 Most of the problems given values of some process variables as temperatures, pressures,
6 phases, amounts or flow rates, and component mole fractions). You will also be asked to
7 calculate the heat transferred to or from the process system, which will require writing EB.
8
9 1. REVIEW FOR FIRST LAW OF THERMODYNAMICS
10 As an engineer designing a process, one of your principal jobs would be to account for the
11 energy that flows into and out of each unit. Based on first law of Thermodynamic:
12 For a closed system,
13 𝑄 − 𝑊 = ∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 (1)
14 For an open system at steady state,
15 𝑄̇ − 𝑊̇ 𝑠 = ∆𝐻̇ + ∆𝐸̇𝑘 + ∆𝐸̇𝑝 (2)

16 N If there is no appreciable vertical separation between the inlet and outlet ports, Drop Ėp

17 N If temperature changes of not more than a few degrees, and no phase changes, Drop Ėk

18 N If there are no moving parts (such as a pump or turbine), Drop Ws and W

19
20 The EB reduces to

21 𝑪𝒍𝒐𝒔𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 ̂ 𝒊 − ∑ 𝒏𝒊 𝑼
𝑸 = ∆𝑼 = ∑ 𝒏𝒊 𝑼 ̂ (𝟑𝒂)
𝑭𝒊𝒏𝒂𝒍 𝒊𝒏𝒊𝒕𝒊𝒂𝒍

22 𝑶𝒑𝒆𝒏 𝒔𝒚𝒔𝒕𝒆𝒎 𝑸̇ = ∆𝑯̇ = ∑ 𝒏̇ 𝒊 𝑯

̂ 𝒊 − ∑ 𝒏̇ 𝒊 𝑯
̂𝒊 (𝟑𝒃)
𝒐𝒖𝒕 𝒊𝒏

23
24 In adiabatic system (Q = 0), EB reduces to

25 𝑪𝒍𝒐𝒔𝒆 𝒔𝒚𝒔𝒕𝒆𝒎 ̂ 𝒊 − ∑ 𝒏𝒊 𝑼
∑ 𝒏𝒊 𝑼 ̂=𝟎 (𝟒)
𝑭𝒊𝒏𝒂𝒍 𝒊𝒏𝒊𝒕𝒊𝒂𝒍

26 𝑶𝒑𝒆𝒏 𝒔𝒚𝒔𝒕𝒆𝒎 ̂ 𝒊 − ∑ 𝒏̇ 𝒊 𝑯
∑ 𝒏̇ 𝒊 𝑯 ̂𝒊 = 𝟎 (𝟓)
𝒐𝒖𝒕 𝒊𝒏
27
 Energy Balance 1 / 13

1 3. CHANGES IN PRESSURE AT CONSTANT TEMPERATURE

2 If the pressure of a solid or liquid changes at constant temperature, you may write
3 ̂ = ∆𝑈
∆𝐻 ̂ + ∆(𝑃𝑉̂ ) = 𝑃 ∆𝑉̂

4
5 4. CHANGES IN TEMPERATURE
6 The quantity of heat required to produce a temperature change in a system can be
7 determined form of the first law of thermodynamics:

8
9 4.1 Heat a Substance at Constant Volume
10 The specific internal energy of a substance depends on temperature. If the temperature is
11 changed constant volume,
12
13 As T → 0, the ratio Û/T approaches a limiting value,
14 which is by definition the heat capacity at constant
15 volume (Cv).
16
17
18
19 Then

20

21 ̂ = ∫ 𝑪𝒗 𝒅𝒕
∆𝑼

22
23 4.2 Heat a Substance at Constant Pressure
24 Like internal energy, if Ĥ is the change in specific enthalpy resulting from a temperature
25 increase at constant pressure,
26 As T → 0, the ratio Ĥ/T approaches a limiting value, which is by definition the heat capacity
27 at constant pressure (Cp).
 Energy Balance 4 / 13

2 𝑇ℎ𝑒𝑛 ̂ = ∫ 𝑪𝒑 𝒅𝒕
∆𝑯

3 Heat capacities (Cv and Cp) are functions of temperature and are frequently expressed in
2 3
4 polynomial form (Cp = a + bT + cT + dT ). Values of the coefficients a, b, c, and d are given
5 in Table B.2.
6
7 Relationships between Heat Capacities are,

8
9
10 Example: Calculate the heat required to raise 200 kg of nitrous oxide from 20 °C to 150 °C in
11 a constant-volume vessel.
12 Where T is in °C.
13
14 Solution
15 The EB for this closed system is
150

16 𝑄 = 𝑈 = 𝑚 Û = 𝑚 ∫ 𝐶𝑣 (𝑇) 𝑑𝑇
20
150

17 𝑄 = 200 ∫ (0.855 + 9.42 × 10−4 𝑇)𝑑𝑇 = 24,200 𝑘𝐽

20
18
19 Example 8.1-2: Assuming ideal gas behavior, calculate the heat that must be transferred in
20 each of the following cases.
21 (a) A stream of nitrogen flowing at a rate of 100 mol/min is heated from 20 °C to 100 °C.
22 (b) Nitrogen contained in a 5-liter flask at an initial pressure of 3 bar is cooled from 90 °C to 30 °C.
23
24 Solution
 Energy Balance 5 / 13

1 (a) From Table B.2, the heat capacity of N2 at a constant pressure of 1 atm is

2
3 For ideal gas behavior,

4
5 Finally,
6 𝑸̇ = ∆𝐻̇ = 𝑛̇ ∆𝐻
̂ = 100 × 2.332 = 𝟐𝟑𝟑 𝒌𝑱/𝒎𝒊𝒏
7
8 (b) For the ideal gas equation of state,
9 Cv = Cp – R and R = 8.314 J/mol K
10 To evaluate Û,
11
12
13
14
15
16
17
18
19
20
21
22
23 Finally, ̂ = 0.497 × −1.25 = −𝟎. 𝟔𝟐𝟏 𝒌𝑱
𝑸 = ∆𝑼 = 𝑛 ∆𝑈
 Energy Balance 6 / 13

1 Example 8.1-1: Fifteen kmol/min of air is cooled from 430 °C to 100 °C. Calculate the required
2 heat removal rate using (a) heat capacity formulas and (b) specific enthalpies.
3 Solution:

4
5 Assume ideal gas behavior,
6 a. The hard way. Integrate the heat capacity formula in Table B.2.

7
8
9 b. The easy way. Use tabulated enthalpies from Table B.8.
10 Ĥ for air at 100 °C can be read directly from Table B.9 and Ĥ at 430 °C can be estimated by
11 linear interpolation from the values at 400 °C (11.24 kJ/mol) and 500 °C (14.37 kJ/mol).

12
 Energy Balance 7 / 13

1 Second Lecture
2 Heat Capacity of a Mixture
3 Method 1: Kopp’s rule is a simple empirical method for estimating the heat capacity of a solid
4 or liquid at or near 20 °C. Cp for a molecular compound is the sum of contributions (Table B.10)
5 for each element in the compound.
6 Table B.10: Atomic Heat Capacities for Kopp’s Rule
7
8
9
10
11
12
13
14
15
16 Example, the heat capacity of solid calcium hydroxide, Ca(OH)2 , would be estimated as

17
18 [The true value is 89.5 J/(mol °C).]
19
20 Method 2: A heat capacity for the mixture ((𝑪𝒑 ) 𝒎𝒊𝒙 ) in the following:
𝒏

21 (𝑪𝒑 ) 𝒎𝒊𝒙 = ∑ 𝒚𝒊 𝑪𝒑,𝒊 𝟖. 𝟑 − 𝟏𝟑

𝒊=𝟏

22 Where yi mass or mole fraction of the ith component

23 Cpi heat capacity of the ith component.
24 If Cpi is expressed in molar units (J/mol °C), then yi must be the mole fraction,
25 If Cpi is expressed in mass units (J/kg °C), then yi must be the mass fraction.
26 Then, Ĥ may be calculated as

27 ̂ = ∫(𝑪𝒑 ) 𝒎𝒊𝒙 𝒅𝑻
∆𝑯 𝟖. 𝟑 − 𝟏𝟒

28
 Energy Balance 8 / 13

1 Example 8.3-4: Calculate the heat required to bring 150 mol/h of a stream containing 60%
2 C2H6 and 40% C3H8 by volume from 0 °C to 400 °C.
3 Solution
4 The polynomial heat capacity formulas for ethane and propane given in Table B.2 are
5 substituted into Equation 8.3-13 to yield

6
7 By EB becomes

8
9
 Energy Balance 9 / 13

1 EIGHT STEPS FOR EB CALCULATIONS

2 Step 1. Perform all unknown variables by MB calculations.
3 Step 2. Write the EB and delete unnecessary terms.
4 Step 3. Choose a reference state (Phase, Temperature, and Pressure); otherwise, choose
5 one of the inlet or outlet states as the reference state for the species.
6 Step 4. Create an Enthalpy Table: ‫انشاء جدول االنثالبية للمدخالت واملخرجات‬
7  For a closed system, initial and final amounts of each species (mi or ni) and (Ûi).
8  For an open system, inlet and outlet stream component flow rates (ṁi or ṅi) and (Ĥi).

Substance ṅin Ĥin ṅin Ĥout

9
10 Step 5. Calculate all required values of Ûi (or Ĥi) and insert the values in the previous table.
11 Choose any convenient path from the reference state to the process state.
12
13 Step 6. Calculate U or Ḣ:
14
15 Step 7. Calculate any work, kinetic energy, or potential energy terms that you have not
16 dropped from the EB.
17 Step 8. Solve the EB for whichever variable is unknown [𝑄 (𝑘𝐽) 𝑜𝑟 𝑄̇ (𝑘𝐽/ℎ)] by eqs. 3a, 3b
 Energy Balance 30 / 13

1 Example 8.3-5 A stream containing 10% CH4 and 90% air by volume is to be heated from 20
2 °C to 300 °C. Calculate the required rate of heat input if the flow rate of the gas is 2000 L/min.
3 Molecular weight of mixture is 22
4
5 Solution:
6
7
8
9
10
11 Convert volumetric flow rate to molar flow rate.
𝐿 1 𝑚𝑜𝑙
12 𝒏̇ = 2000 × = 89.3 𝑚𝑜𝑙/𝑚𝑖𝑛
𝑚𝑖𝑛 22.4 𝐿
13
14 In other words, it is the specific enthalpy change for the process
15 CH4 (g, 20 °C, 1 atm) → CH4 (g, 300 °C, 1 atm)
16
17 References: CH4 (g, 20 °C, 1 atm), air (g, 25 °C, 1 atm)

18
19 We assume ideal gas behavior, and we always neglect heats of mixing of gases, so that the
20 enthalpy change is calculated for the heating of pure methane at 1 atm.
300

21 ̂1 = ∫ 𝐶𝑝, 𝐶𝐻 𝑑𝑡
𝐻 4
20

22 Substitute for Cp from Table B.2

23
 Energy Balance 33 / 13

1 The enthalpies of air at the inlet and outlet conditions relative to air at the reference state (Ĥ2
2 and Ĥ3, respectively) are determined from Table B.8 as:

3
4 The EB now yields

5 𝑸̇ = ∆𝑯̇ = ∑ 𝒏̇ 𝒊 𝑯
̂ 𝒊 − ∑ 𝒏̇ 𝒊 𝑯
̂𝒊
𝒐𝒖𝒕 𝒊𝒏

6
 Energy Balance 32 / 13

1 Example 8.3-6: A gas stream containing 8 mole% CO with CO2 at 500 °C is fed to a heat boiler.
2 The hot gas flows over the outside of the tubes. Liquid water at 25 °C is fed to the boiler in a
3 ratio 0.2 mol feed water/mol hot gas (inside the tubes). Heat is transferred from the hot gas
4 through the tube walls to the water, causing the gas to cool and the water to heat to its boiling
5 point and evaporate to form saturated steam at 5 bar. The boiler operates adiabatically (all
6 the heat transferred from the gas goes into the water). What is the temperature of the exiting
7 gas?
8 Solution: The flowchart for an assumed basis of 1 mol feed gas is shown below.

9
10 we chose the reference state for the steam tables (liquid water at the triple point) as our
11 reference for water, and knowing that the enthalpies in the steam tables are in kJ/kg, we list
12 the quantity of water in kg (m = 0.2 mol H2O × 0.018 kg/mol = 0.0036 kg).
13 We chose the reference states for CO and CO2 as the gas inlet temperature and 1 atm. We
14 assume ideal gas behavior so that deviations of the pressure from 1 atm have no effect on
15 enthalpies.
16 References: CO (g, 500 C, 1 atm), CO2 (g, 500 C, 1 atm), H2O (l, triple point)

17
18 The solution strategy will be to:
19 Step 1: Calculate Ĥ1 and Ĥ2 by integrating the heat capacity formulas of Table B.2 from
20 the reference temperature (500 °C) to the unknown T at the gas outlet,
21 Step 2: Look up Ĥ3 and Ĥ4 in the steam tables,
 Energy Balance 31 / 13

1 Step 3: Substitute for Ĥ1 through Ĥ4 in the EB for adiabatic system as shown in eq. (4)

2 ̂ 𝒊 − ∑ 𝒏̇ 𝒊 𝑯
𝟎 = ∑ 𝒏̇ 𝒊 𝑯 ̂𝒊
𝒐𝒖𝒕 𝒊𝒏

3 Step 4: Solve the resulting equation for T using a spreadsheet.

4 Step 5: Estimate the solution by neglecting all terms of second-order and higher in T.
5 Step 6: By trial-and-error manually. We insert the initial guess from Step 5 into fourth-
6 order polynomial to be find T value.

7
8 Integrating the expressions for Ĥ1 and Ĥ2 and substituting the resulting expressions and the
9 values of Ĥ3 and Ĥ4 into the EB eq. 4) yields the following equation:

10
11 Using MS Excel, neglecting all terms of second-order and higher in T:
12 0.03554 T - 12.16 = 0  T = 342 °C ‫لتحديد القيمة االبتدائية‬
13 Finally, We insert the initial guess for (T = 342 °C) into fourth-order polynomial to be solved.
14 After many iteration, T = 299 °C.
 Energy Balance 34 / 13

1 Third Lecture
2 5. PHASE CHANGE OPERATIONS
3 The molecules of a vapor, can move around relatively freely, are much more energetic than
4 the molecules of a liquid. Also, think about the fact that liquid molecules are held in close
5 proximity to each other by attractive forces between the molecules. The energy required to
6 overcome these forces when a liquid is vaporized is reflected in the higher internal energy of
7 the vapor molecules.
8 For example, for water at 100 °C and 1 atm, Ûliquid = 419 kJ/kg and Ûvapor = 2507 kJ/kg (Table B.5).
9
10 5.1 Latent Heats
11 Latent heat of the phase change is the specific enthalpy change associated with the transition
12 of a substance from one phase to another at constant T and P.
13 Latent heats are defined as follows:
14 1. Heat of fusion or melting. Ĥm (T, P) is the specific enthalpy difference between the solid
15 and liquid forms of a species. The heat of solidification is the negative of the heat of
16 fusion.
17 2. Heat of vaporization or condensation. Ĥv (T, P) is the specific enthalpy difference
18 between the liquid and vapor forms of a species. The heat of condensation is the
19 negative of the heat of vaporization.

20 ‫ تأثري درجة احلرارة على احلرارة الكامنة أكرب بكثري من الضغط‬:‫مالحظة للشطار‬
21
22 Example: The heat of vaporization of liquid water to steam at 100 °C and 1 atm is 40.6 kJ/mol.
23 Thus, the heat of condensation of water at 100 °C and 1 atm must be –40.6 kJ/mol.
24
25 Example: At what rate in kilowatts must heat be transferred to a liquid stream of methanol at
26 its normal boiling point (64.7 °C) to generate 1500 g/min of saturated methanol vapor?
27 Ĥv = 35.3 kJ/mol at Tbp from Table B.1.
28 Solution:
29 The EB after neglected unnecessary terms: 𝑸̇ = ∆𝑯̇ = 𝒏̇ ∆𝑯̇𝒗

30
 Energy Balance 35 / 13

1 Example: One hundred g-moles per hour of liquid n-hexane at 25 °C and 7 bar is vaporized
2 and heated to 300 °C at constant pressure. Neglecting the effect of pressure on enthalpy,
3 estimate the rate at which heat must be supplied.
4 For n-hexane, Ĥv = 28.85 kJ/mol at Tbp = 69 °C from Table B.1
5 Solution: An energy balance yields 𝑸̇ = ∆𝑯̇
6 The following illustrates possible paths from liquid hexane at 25 °C to vapor at 300 °C.
7 Actual Path
(Liq. 25 °C, 7 atm) 3 4 (Vap. 300 °C, 7 atm)
8
9 ĤA
ĤG
10
11 (Liq. 69 °C, 1 atm) 2 ĤD 1 (Vap. 69 °C, 1 atm)

12
13 Change hexane from a liquid to a vapor at 69 °C. Ĥv = 28.85 kJ/mol from Table B.1

14
15 To evaluate the overall Ĥ for the process,

16
17 Notice that the pressure change term in the first step (𝑉̂ ∆𝑃 = 0.0782 𝑘𝐽/𝑚𝑜𝑙) accounts for less
18 than 0.1% of the overall process enthalpy change. We will generally neglect the effects on ĤG.
 Energy Balance 36 / 13

1 Example: Acetone (Ac) is partially condensed out of a gas stream containing 66.9 mole% Ac
2 vapor and the balance nitrogen. Process specifications lead to the flowchart shown below. If
3 process operates at steady state, calculate the required cooling rate.
4 A heat of vaporization for Ac is Ĥv = 30.2 kJ/mol at Tbp = 56 °C and 1 atm.
5 .‫هنا تم إعطاء درجة الغليان وانثالبية التبخر التي هي عكس إشارة انتالبية التكثيف‬
6 From Table B8 (P 635)

7
𝑘𝐽
8 N2 (g): 𝐶𝑝 ( ) = 0.029 + 0.23 × 10−5 T + 0.5723 × 10−8 𝑇2 + 2.871 × 10−12 𝑇3
𝑚𝑜𝑙 ℃
9
10 The specific gravity (0.791) for liquid Acetone given in Table B.1.
11 𝑉̂ = 0.0734 L/mol
12
13
14
15
16
17
18 Solution:
19 3. Write the EB equation. (𝑄̇ = ∆𝐻̇)
20 2. Choose reference states for Acetone and Nitrogen.
21 The reference states may be chosen for computational convenience,‫الحسابات المالئمة‬
22 Table B.8 lists specific enthalpies of nitrogen relative to N2 (g, 25 °C, 1 atm), which makes
23 this state a convenient choice for nitrogen.
24 There are no tabulated enthalpy data for Ac in the text, so we will choose one of the
25 process stream conditions, Ac (liq, 20 °C, 5 atm), as the reference state for this species.
26 1. Construct an inlet–outlet table.
27 References: Ac (l, 20 °C, 5 atm), N2 (g, 25 °C, 1 atm)
 Energy Balance 37 / 13

1
2 Note: Since the liquid Ac leaving the condenser is at the reference state, we set its specific
3 enthalpy equal to zero.
4
5 5. Calculate all unknown specific enthalpies.
6 We construct hypothetical process paths from the reference states to the states of the
7 species and evaluate Ĥ for each path. The following process path from the reference state
8 [Ac (Liq, 20 °C, 5 atm)] to the state [Ac (Vap, 65 °C, 1 atm)].
9
10 (Liq. 20 °C, 5 atm) 1 (Vap. 65 °C, 1 atm)
5
11
12 (Liq. 65 °C, 1 atm) 2

13
(Liq. 56 °C, 1 atm) 3 4 (Vap. 56 °C, 1 atm)
14
15 ̂(1−2) = 𝑉̂ 𝑃 = 0.0297
𝐻 at constant 𝑇
𝑇2
16 ̂(2−3) = ∫ 𝐶𝑝 (𝑇) 𝑑𝑇 = 4.68
𝐻 at constant 𝑃
𝑇1

17 ̂(3−4) = ∆𝐻𝑣 = 30.2

𝐻 at constant 𝑃 𝑎𝑛𝑑 𝑇
𝑇2
18 ̂(4−5) = ∫ 𝐶𝑝 (𝑇)𝑑𝑇 = 0.753
𝐻 at constant 𝑃
𝑇1

19
20 Ĥ3 = (0.0297 + 4.68 + 30.2 + 0.753) kJ/mol = 35.7 kJ/mol.
21
22 In a similar manner, Ĥ1 from the reference state [Ac (Liq, 20 °C, 5 atm)]  [Ac (Vap, 20 °C, 5 atm)]
23 ̂ 𝟑 = ∆𝐻𝑣 = 30.2 𝑘𝐽/𝑚𝑜𝑙
𝑯
24 Ĥ2 from the reference state [N2 (g, 25 °C, 1 atm)]  [N2 (g, 65 °C, 1 atm)]
65
25 ̂2 = ∫ 𝐶𝑝 , 𝑁 𝑑𝑇 = 0.753
𝐻 at constant 𝑃
2
25
 Energy Balance 38 / 13

1 Ĥ4 from the reference state [N2 (g, 25 °C, 1 atm)]  [N2 (g, 20 °C, 5 atm)]
20
2 ̂4 = 𝑉∆𝑃 + ∫ 𝐶𝑝 , 𝑁 𝑑𝑇 = −0.1
𝐻 2
25

3
4 In a similar manner, we obtain the values for Ĥ2, Ĥ3, and Ĥ4 shown in the following table:

5
6 6. Calculate Ḣ:

7 𝑸̇ = ∆𝐻̇ = ∑ 𝑛̇ 𝑖 𝐻
̂𝑖 − ∑ 𝑛̇ 𝑖 𝐻
̂𝑖
𝑜𝑢𝑡 𝑖𝑛

8 Ḣ = (3.35 mol/s)(32.0 kJ/mol)+ [(63.55)(0)+ (33.1)( –0.10) – (66.9)(35.7) – (33.1)(1.16)] kJ/s

9 𝑸̇ = ∆Ḣ = – 𝟐𝟑𝟐𝟎 𝐤𝐉/𝐬
10 ‫ أخرى مثال‬Reference State ‫ اذا اختير‬:‫حل اخر‬

11 References State: Ac (Vap, 65 °C, 1 atm), N2 (g, 25 °C, 1 atm)

ṅin Ĥin ṅin Ĥout

Substance
(mol/s) (kJ/mol) (mol/s) (kJ/mol)
Ac (Vap) 66.9 0 3.35 Ĥ3

Ac (Liq) 0 – 63.55 Ĥ5

N2 33.1 Ĥ2 33.1 Ĥ4
12
References Ac (Liq)
13
State Bottom
14 (Vap. 65 °C, 1 atm) 1 5 (Vap. 20 °C, 5 atm)
15

16 4 (Liq. 20 °C, 1 atm)

17
(Vap. 56 °C, 1 atm) 2 3 (Liq. 56 °C, 1 atm)
18

19 Ĥ5 = Ĥ1-2 + Ĥ2-3 + Ĥ3-4 + Ĥ4-5

 Energy Balance 39 / 13

1 5.2 Estimation of Latent Heats

2 Several of the methods to estimate a standard Latent Heats are summarized below.
3 3. Chen’s equation provides roughly 2% accuracy:

4
5 Where Tb and Tc are the normal boiling point and critical temperature in Kelvin and Pc is the
6 critical pressure in atmospheres.
7
8 2. Clausius–Clapeyron equation estimated from vapor pressure (p*)

9
10 B and R are constant. It may be determined from a plot of ln p* versus 3/T
11
12 1. Watson’s correlation: A procedure for calculating the latent heat of vaporization at one
13 temperature from a known value at any other temperature was presented.

14
15 Where Tc is the critical temperature of the substance.
16
17 4. A formula for approximating a standard heat of fusion is

18
19
20 Example: The normal boiling point of methanol is 337.9 K, and the critical temperature is 513.2 K.
21 Estimate the heat of vaporization of methanol at 200 °C.
22 Solution:
23 By Watson’s Correlation

24

25
 Energy Balance 20 / 13

1 Example: An equimolar liquid mixture of benzene (B) and toluene (T) at 10 °C is fed
2 continuously to a vessel in which the mixture is heated to 50 °C. The liquid product is 40 mole%
3 B, and the vapor product is 68.4 mole% B. How much heat must be transferred to the mixture
4 per g-mole of feed?
5
6 Solution: Basis: 1 mol Feed
7
8
9
10
11
12
13 We start with a degree-of-freedom analysis:
14 3 Unknown variables (nV , nL , Q)
15 2 Material balances
16 1 Total energy balance
17 0 df
18
19 We next determine nV and nL from material balances, and then Q from an EB.

20
21 The EB for this process has the form Q = Ĥ. An enthalpy table appears as follows:

22
23 The process is not running at an unusually low temperature or high pressure, we neglect the
24 effects of pressure on enthalpy in the calculations of Ĥ1 through Ĥ4. The heat capacity and
25 latent heat data are calculated the outlet enthalpies are obtained from
26 Tables B.1 (Tbp = 80.1 °C for B and Tbp = 110.62 °C for T) (P. 628)
 Energy Balance 23 / 13

1 Table B.2. (P. 635)

4
5 The formulas represent Ĥ for the transitions from the reference states to the process states.

6
7 ̂𝑖 − ∑ 𝑛𝑖 𝐻
𝑸 = ∆𝐻 = ∑ 𝑛𝑖 𝐻 ̂𝑖 = 17.7 𝑘𝐽
𝑜𝑢𝑡 𝑖𝑛

8
 Energy Balance 22 / 13

1 Fourth Lecture
2 6. HEAT OF MIXING AND SOLUTION (Ĥs)
3 If you are mixed two liquids or dissolved a solid in a liquid, the mixture or solution became
4 quite hot. When two different liquids are mixed or when a gas or solid is dissolved in a liquid,
5 bonds are broken between neighboring molecules of the feed materials, and new bonds are
6 formed between neighboring molecules or ions in the product solution. A energy from broken
7 bonds is transferred from the solution to its surroundings as heat.
8 The heat of solution (Ĥs) is defined the change in enthalpy when a solute (gas or solid) is
9 dissolved in a liquid solvent.
10 The heat of mixing has the same meaning as the heat of solution when the process involves
11 mixing two fluids.
12 To calculate the enthalpy of the solution at a temperature,
𝑇𝑖
13 ̂𝑖 = ∆𝐻
𝐻 ̂𝑠 (𝑎𝑡 𝑇 = 25 °𝐶) + ∆𝐻
̂𝑖 = ∆𝐻
̂𝑠 + ∫ 𝐶𝑝,𝑖 𝑑𝑡
𝑇𝑓

14 r = mol of solute /mole of solvent )‫(نسبة املذاب اىل املذيب‬ ̂𝑠 ‫ جند‬:‫مثال‬

r ‫𝐻∆ من اجلدول ادناه بعد حساب قيمة‬
 Energy Balance 21 / 13

1
 Energy Balance 24 / 13

1 Example: Hydrochloric acid is produced by absorbing gaseous HCl in water. Calculate the heat
2 that must be transferred to or from an absorption unit if HCl (vap) at 100 °C and H2O (liq) at
3 25 °C are fed to produce 1000 kg/h of 20 wt% HCl (aq) at 40 °C.
4 MHCl = 36.5 mol/g and MH2O = 18 mol/g
5 Cp, HCl = 0.73 kcal/kg °C
6 1 kcal = 4.184 kJ ;
7 Solution: ‫نحول جميع وحدات الدخالت والمخرجات من وحدات الكتلة الى مول‬
8 In this case 1000 kg/h of 20 wt% HCl(aq).

9
10 Drawing and labeling the flowchart is.

11
12 The enthalpy table is:

13
14
15
16
17
18
 Energy Balance 25 / 13

1
2 r = mol of solute (gas or solid)/mole of solvent (liquid) = 44400/5480 = 8.1.
3 Form Table B.11: Ĥa = Ĥs (25 C, r = 8.1) = –67.4 kJ/mol HCl
4
5
6
7
8
9

10
11 By EB

12 𝑸̇ = ∑ 𝑛̇ 𝑖 𝐻
̂𝑖 − ∑ 𝑛̇ 𝑖 𝐻
̂𝑖
𝑜𝑢𝑡 𝑖𝑛

13 𝑄̇ = (5480 mol HCl/h)( –59 kJ/mol HCl) – (5480 mol HCl/h)(2.178 kJ/mol HCl)
14 𝑸̇ = – 𝟑. 𝟑𝟓 × 𝟏𝟎𝟓 𝒌𝑱/𝒉
 Energy Balance 26 / 13

1 Example: Sulfuric acid and water is liquid Mixture at 77 F. Bring a pure water from its reference
2 temperature of 32 °F to 77 °F while the sulfuric acid starts at 77 °F. calculate the specific
3 enthalpy (Btu/lbm) of a 40 wt.% sulfuric acid product solution at 120 °F?
4 Cp (H2O, liq) = 1 Btu/lbm °F Cp (H2SO4, liq) = 3.2 × 10-4 Btu/lbm °F
5 Solution: Ĥs
1 lbm (0.6 H2O +
6 0.4 H2SO4 77 °F)
7
Ĥ1 Ĥ3
3 1 4
8
(H2O 32 °F) 1 lbm (0.6 H2O + 0.4
9
H2SO4 120 °F)
10
2
11
(H2SO4 77 °F)
12

13

14
77
15 ̂1 = 0.6 𝑙𝑏𝑚 ∫ (𝐶𝑝 ) 𝐻2𝑂 𝑑𝑡 = 0.6 × 1 × (77 − 32) = 27 𝐵𝑡𝑢
∆𝐻
32

16

17
18 From Table B.11: Ĥs (r = 8.2, 77 °F) = –279 Btu/lbm
19 ̂2 = 𝑛 ∆𝐻
∆𝐻 ̂𝑠 = 0.4 × −279 = −111.6 𝐵𝑡𝑢
20

21
𝑛

22 (𝐶𝑝 ) 𝑚𝑖𝑥 = ∑ 𝑦𝑖 𝐶𝑝,𝑖 = 0.4 𝐶𝑝,𝐻2𝑆𝑂4 + 0.6 𝐶𝑝,𝐻2𝑂 = 0.67 𝐵𝑡𝑢/(𝑙𝑏𝑚 ℉)

𝑖=1
120
23 ̂3 = 1.0 𝑙𝑏𝑚 ∫
∆𝐻 (𝐶𝑝 ) 𝑚𝑖𝑥 𝑑𝑡 = 28.2 𝐵𝑡𝑢
77

24
 Energy Balance 27 / 13

1 Example : A 5 wt% H2SO4 solution at 60 F is to be concentrated to 40 wt% by evaporation of

2 water. The concentrated solution and water vapor emerge from the evaporator at 180 °F and
3 1 atm. Calculate the heat must be transferred to the evaporator to process 1000 lbm/h of the
4 feed solution.
5
6 Solution:

9
10 The enthalpy of water vapor at 180 F and 1 atm relative to liquid water at 32 F may be obtained
11 ̂𝑣 = 1138 Btu/lbm
from the steam tables as 𝐻

12 𝑸̇ = ∆𝑯̇ = ∑ 𝒎̇𝒊 𝑯
̂ 𝒊 − ∑ 𝒎̇𝒊 𝑯
̂𝒊
𝒐𝒖𝒕 𝒊𝒏

13
14
 Energy Balance 28 / 13

1
2 Figure 8.5-1: Enthalpy–concentration chart for H2SO4 –H2O.
3
 Energy Balance 29 / 13

1 Fifth Lecture
2 7. VAPOR–LIQUID EQUILIBRIUM CALCULATIONS
3 Enthalpy–concentration charts are particularly useful for two-component systems in which
4 vapor and liquid phases are in equilibrium (Figure 8.5-2).

5
6 Figure 8.5-2: Enthalpy–concentration diagram for the ammonia–water system at 1 atm.
7
8 For example, that a mixture of ammonia and water that is 40% NH3 by mass is contained in a
9 closed vessel at 140 °F and 1 atm (Point A on Figure 8.5-2). Since this point lies between the
10 vapor and liquid equilibrium curves, the mixture separates into two phases whose
11 compositions are found at points B and C.
12 In general, if F, L, and V are the total mass of the mixture, the mass of the liquid phase and
13 vapor phase, respectively, and xF, xL , and xV are the corresponding mass fractions of NH3, then
 Energy Balance 10 / 13

1
2 Substituting Equation 8.5-4 for F into Equation 8.5-5 yields

3
4
5 From the properties of similar triangles, the right side of
6 Equation 8.5-6 equals the ratio of distances AC/AB.
7 The mass fractions of the liquid and vapor phases are

8
9 It becomes a simple matter to determine the compositions, enthalpies, and relative
10 proportions of each phase.
11 ‫ هذه الطريقة باستخدام المخططات هي طريقة بدال الحسابات بالعادالت‬:‫باختصار‬
12
13 Example: An aqueous ammonia solution is in equilibrium with a vapor phase in a closed
14 system at 160 °F and 1 atm. The liquid phase accounts for 95% of the total mass of the system
15 contents. Use Figure 8.5-2 to determine the weight percent of NH3 in each phase and the
16 enthalpy of the system per unit mass of the system contents.
17
18 Solution: From Figure 8.5-2.
19
20
21
22
23
24
25
26
27
 Energy Balance 13 / 13

1 Example: A 30 wt% NH3 solution at 100 psia is fed at a rate of 100 lbm/h to a tank in which
2 the pressure is 1 atm. The enthalpy of the feed solution relative to the reference conditions
3 used to construct Figure 8.5-2 is 100 Btu/lbm. The vapor composition is to be 89 wt% NH3.
4 Determine:
5 (a) the temperature of the stream leaving the tank,
6 (b) the mass fraction of NH3 in the liquid product,
7 (c) the flow rates of the liquid and vapor product streams, and
8 (d) The rate at which heat must be transferred to the vaporizer.
9
10 Solution:
11
12 Basis: 100 lbm/h Feed
13
14

15
16
17
18
19
20
21
22
23
24
25
26