1

ChE 101.03

MASS & ENERGY BALANCES

Dr. Abdulhadi Al-Juhani

Department of Chemical Engineering, King Fahd University of Petroleum & Minerals

3-12-07

2

Terminal Objective:

upon completion of this module, you will be able to calculate:

- mass and energy balances- energy gains and losses for process equipment- utility requirements- refrigeration requirements

using appropriate drawings, diagrams, tables, equations, and process parameters

3

Enabling Objectives:

In order to achieve the terminal objective, you will be able to:

1. Interpret information contained in Process Flow Diagram (PFD) and Piping and Instrumentation Diagram (PID).

2. Calculate mass and energy balances, using PFD information.

3. Calculate boiler requirements and the effects of operating and design parameters on efficiency.

4. Calculate temperature changes due to energy gains and losses

5. Calculate utility requirements

6. Calculate refrigeration requirements.

4

Process Flow Diagram (PFD):

Figure 1: PFD of a Depropanizers/Debutanizer system at Ras Tanura.(Exercise: point out examples of the 10 items listed in the text)

5

6

Instrumentation and Piping Diagram (PID):

•The difference between PFD and PID, and the use of each

Figure 2: Typical PID diagram(Exercise: point out examples of the 15 items listed in the text)

7

8

9

Using mass and energy balances

Figure 3: PFD – NGL Absorption and Fractionation(Exercise: point out examples of 6 items listed in the text)

10

11

12

Using mass and energy balances (cont’d)

Figure 4: “PROCESS” Output (Saudi Aramco Format)

13

14

15

PFD Information

Figure 5: PFD – Atmospheric Section C.D.U.s No. 1 & 2 (Exercise: point out some missing information)

16

17

18

PID/PFD Data

B/CD = B/SD x 0.9

SpGr60 = 141.5 / (131.5 + API) ………………………..............(1)

to convert B/D to Ib/hr:Ib/hr = (B/D)60 x (350 Ib/B) x (SpGr60) x (D/24 hr)….....(2)

convert Ib/hr to Gpmgpm = (Ib/hr) / (SpGr) / 500………………………..(3)

where 1 B = 42 gal1 m3 = 264.17 gal1 Ib = 453.6 gdensity of water @ 60 oF = 8.33 Ib/gal = 350 Ib/B

19

Exercise 1

for each stream entering or leaving the attached figure, calculate the following:

- sp. gr. @ 60 oF- Ib/hr- Gpm @ actual conditions

then, check the overall material balance:- note: density of water @ 60 oF = 8.33 Ib/gal = 350 Ib/B

20

21

Exercise 1 (cont’d)Answer: Stream Feed OVHD Naphtha Lean Oil

API 77 118.5 68.8 68.8SpGr60

(eq 1)0.679 0.566 0.706 0.706

B/SD 32, 110 6, 308 9, 742 16, 060Ib/hr(eq 2)

317, 956 52, 067 100, 302 165, 351

Given SpGr

0.57 0.54 0.69 0.69

Gpm(eq 3)

1116 193 291 479

22

Calculating Energy Inputs, Outputs, and Losses

Steam Boiler Energy Balances:

Why we should have a blowdown here?

23

Example Problem 1:Calculate the amount of fuel required for the boiler in fig 6. Then, calculate the furnace efficiency

24

Example Problem 1 (cont’d):

Answer:

The following enthalpy data are available:

Stream T (oF) P (psia) HL(Btu/Ib)

HV(Btu/Ib)

Blowdown 370 174.7 343.5 1196.4

Steam 434 154.7 -- 1237.6

Feedwater 190 214.7 158.0 --

25

Example Problem 1 (cont’d):

Heat Balance: Water side/Process

Heat In:Feedwater = (910, 000 Ib/hr) (158 Btu/Ib) = 143.78 MBtu/hrheat absorbed = QATotal = 143.78 + QA

Heat Out: Steam = (805,000) (1,237.6) = 996.27 MBtu/hrBlowdown = (105, 000) (343.5) = 36.07 MBtu/hrTotal = 1, 032.34 MBtu/hr

Heat In = Heat Out143.78 + QA = 1, 032.34 QA = 888.56 MBtu/hr

26

ChE 101.03 V (19)

Example Problem 1, Cont’d

The furnace efficiency can be calculated as follows:

Process heat required = 888.56 x 106 Btu/hr (QA)Stack temperature = 650°F (Given)Percent excess air = 20 (Given)Furnace heat loss = 2% (Given)Fuel LHV = 19,400 Btu/lb (Given)Heat available at 650°F = 16,450 Btu/lb (Maxwell, p. 185)

= 54,016 lb/hr

Gross fuel = 1.02 x 54,016 = 55,096 lb/hr

Heat fired = 19,400 x 55,096 = 1,068.9 x 106 Btu/hr

= 83.1%LHV efficiency = 888.56 x 106

1,068.9 x 106 x 100

Net Fuel = 888.56 x 106

16,450

27ChE 101.03 V (20)

Example Problem 1, Cont’d

Shortcut Calculation:

What if the ambient temperature was 100°F instead of 80°F?

Increasing the ambient temperature by 20°F improved the efficiency by 0.5% assuming the same stack temperature.

Percent efficiency = 100 − 0.0237 + 0.000189( ) EA( )( )( ) Tst − TA( )[ ] 100100 + QL

⎛

⎝ ⎜

⎞

⎠ ⎟

= 100 - 0.0237 + 0.000189( )( )( ) 650 − 80( )[ ] 100/ 100 + 2( )[ ]

= 100 - 0.02748 570( )( )[ ]0.9804[ ] = 82.7%

Percent Efficiency = 100 - 0.02748( ) 650 − 100( )( )[ ] 0.9804[ ] = 83.2%

28

Exercise 2Calculate the fuel and Feedwater required for the boiler in fig 15. How can the furnace efficiency be improved. Use 2% for heat losses, and use 10% blowdown. The following enthalpy data are available:

Stream T (oF) P (psia) HL(Btu/Ib)

HV(Btu/Ib)

Blowdown 492 633 478.5 1203.1

Steam 700 600 -- 1351.8

Feedwater 180 -- 148.0 --

29

30

Exercise 2 (cont’d)

Answer:

Feedwater rate = FBlowdown = 0.1 FSteam product = 250, 000

water material balance: F = 250,000 + 0.1 F

F = 277, 788 Ib/hrBlowdown = 27, 778 Ib/hr

31

Exercise 2 (cont’d)

Heat Balance: Water side/Process

Heat In:Feedwater = (277, 778 Ib/hr) (148 Btu/Ib) = 41.11 MBtu/hrheat absorbed = QATotal = 41.11 + QA

Heat Out: Steam = (250,000) (1,351.8) = 337.95 MBtu/hrBlowdown = (27, 778) (478.5) = 13.29 MBtu/hrTotal = 351.24 MBtu/hr

Heat In = Heat Out41.11 + QA = 351.24 QA = 310.13 MBtu/hr

32ChE 101.03 V (26)

Exercise 2, Cont’dHeat in = Heat out 41.11 + QA = 351.24Heat absorbed QA = 351.24 - 41.11 = 310.13 MBtu/hrHeat loss (Given) = 2%Fuel LHV (Given) = 19,400

Heat available = 16,725 at 600°F Stack and 20% excess air Maxwell, p. 185

Net fuel = = 18,543

Gross fuel = 1.02 x 18,543 = 18,914Heat fired = 18,914 x 19,400 = 366.93 MBtu/hr

LVH efficiency = x 100 = 84.5%

Shortcut efficiency assuming the atmospheric air temperature is 100°F:

Efficiency= [100 -(0.0237 + 0.000189 EA)(TST - TA )][100/(100 + QL)]= [100 -(0.0237 + 0.000189)(20))(600 - 100)][100/(100 + 2)]= [(100 - 13.74)][0.9804] = 84.5%

310.13x106

16,725

310.13x106

366.93x106

33ChE 101.03 V (27)

Exercise 2, Cont’dTo Increase Efficiency:

• Lower stack temperature.- Add more surface to convection section and increase

boiler feedwater preheat.- Add more surface to convection section and preheat

another process stream. A 50°F reduction in stack temperature would increase efficiency from 84.5% to 85.9%.

• Reduce blowdown rate.- If boiler feedwater quality allows, the blowdown rate can be

reduced.- Reduction of blowdown from 10% to 2% would not increase

the efficiency, but would directly reduce fuel use by decreasing the process heat absorbed.

• Reduce percent excess air.- A reduction of excess air from 20% to 10% increases

efficiency from 84.5% to 85.4%.

34ChE 101.03 V (28)

Exercise 2, Cont’dFurnace Fuel Savings

Lower Reduce ReduceCase Base Stack Temp. Blowdown Excess Air

Percent blowdown 10 10 2 10

Heat absorbed, 310.13 310.13 302.63 310.13MBtu/hr

Stack temperature, °F 600 550 600 600

Excess air, percent 20 20 20 10

Furnace efficiency, 84.5 85.91 84.52 85.40percent

Fuel savings, percent Base 1.62 2.42 1.04

35

ChE 101.03 V (29)

Exercise 2, Cont’d

Case 1 2 3 4

BaseLower

Stack Temp.Reduce

BlowdownReduce %Excess Air

Heat in277,778 x 148 = 41.11 41.11 255,102 x 148 = 37.76 41.11

Heat out 27,778 x 478.5 = 13.29 5,102 x 478.5 = 2.44

250,000 x 1,351.8 = 337.95 250,000 x 1,351.8 = 337.95351.24 351.24 340.39 351.24

Heat absorbed 310.13 310.13 302.63 310.13Stack 600 550 600 600Percent excess air 20 20 20 10Heat loss 2% 2% 2% 2%Fuel LHV 19,400 19,400 19,400 19,400Heat avail.* 16,725 17,000 16,725 16,900Net fuel 18,543 18,243 18,095 18,351Gross fuel 18,914 18,608 18,457 18,718Heat fired 366.93 360.99 358.06 363.13LHV, percent eff. 84.52 85.91 84.52 85.40Fuel savings Base 1.62% 2.41% 1.04%

*Maxwell p. 185

Calculation for efficiency Improvement:

36

Heat Exchanger Energy Balances

The heat released by the hot stream must equals the heat absorbed by the cold stream

Hot

Cold

37ChE 101.03 V (28)

Example Problem 2

The hot stream is to be cooled to 300°F.

What is the cold stream outlet temperature?

Hot stream duty = (410 - 258) x 500,000 = 76 x 106 Btu/hr

Cold stream CP data (Maxwell, Pg. 93).

Temperature CP+B B CP200°F 0.677 0.08 0.597300°F 0.736 0.08 0.656400°F 0.794 0.08 0.714

Hot

500,000 lb/hr 500°F H = 410 Btu/lb

300°F H = 258 Btu/lb

Cold 10,000 lb/hr 200°F API = 70° VABP = 210°F

38ChE 101.03 V (31)

Example Problem 2, Cont’d

Assume for first trial cold stream out is same as inlet °F.

Avg CP = 0.597

Q = mCP²T ∆T = (TOUT - 200)

TOUT = QmCP

+ 200 = 76 x 10610,000 x 0.597 + 200 = 12, 930°F

39ChE 101.03 V (30)

Example Problem 2, Cont’d

It can be seen that this is an impossible solution. The hot stream cannot be cooled by the desired amount with the available cold stream. However, this problem illustrates both the enthalpy and heat capacity methods of heat exchanger heat balance.

To solve this problem in a practical manner, if the two flow rates are fixed, and only one heat exchanger shell is available, assume that the cold stream outlet equals the hot stream outlet, as shown below.

500°F

Hot

Cold

200°F

TX

TX

40

ChE 101.03 V (33)

Example Problem 2, Cont’d

1. (Average CP 300-500°F).

2.

3. Assume Cold CP = 0.597 (inlet condition)

4. Q = (m CP ²t) Hot = (m CP ²t) Cold

(500,000) (0.76) (500 – TX) = (10,000)(0.597) (TX – 200)

380 (500 – TX) = 5.97 (TX – 200)

190,000 – 380 TX = 5.97 TX – 1194

191,194 = 385.97 TX

TX = 495.4

Hot Cp =H1 – H2T1 – T2

= 410 – 258500 – 300 = 152

200 = 0.76

41ChE 101.03 V (34)

Example Problem 2, Cont’d

5. Recalculate average Cold CP.495° CP + B = 0.85 CP = 0.77

= 0.6835

Resolving 191,367 = 386.8 TX

TX = 494.7

This solution will be the best that can be achieved with the given conditions using one heat exchanger shell.

CP – 0.597 + 0.772

42

Exercise 3.1 (heat integration)We want to heat integrate a lean oil and feed stream in exercise 1, where it is proposed to use exchanger E6 to replace E1 in exercise 1 to provide preheat for the debutanizer feed

43ChE 101.03 V (36)

Exercise 3.1, Cont’d

Figure 16

Debutanizer Feed32,110 B/SD

77ÞAPI253ÞF

sp. gr. = 0.57

346ÞF

E-5

E-6 E-7B E-7A

E-2

E-4

DebutanizerOverhead

6,308 B/SD118.5ÞAPI

100ÞFsp. gr. = 0.54

Naphtha Product9,742 B/SD

130ÞF

Lean Oil16,060 B/SD

68.8ÞAPI100ÞF

sp. gr. = 0.69

E-6

115ÞF148 psig

143ÞF148 psig

154 psig322ÞF

P-2

D-1

E-3

P-1T-1

H=99.270x106 Btu/hr

258°FH=103.270

x106Btu/hr

LeanOil

Feed

44

Exercise 3.1 (cont’d)

Answer:

E6 Duty: 253 oF to 258 oF

Q = ∆H = 103.277 – 99.270 = 4.0 MBtu/hr

Q = mCp∆T

m = 165, 284 Ib/hr (from exercise 1)

1. Find the exit temperature for the lean oil leaving E6 after preheating the tower feed?

45

ChE 101.03 V (38)

Exercise 3.1, Cont’dAnswer , Cont’d:

Need average CP of the lean oil, so estimate exit temperature for first trial.

Use inlet CP to estimate exit temperature.

Ave CP = 0.681

∆T =

TOUT = 346 - 35.5 = 310.5°F

CP = 346 0.681 310.5 0.658

100 0.524

TOUT = 346 - 36 = 310°F

(346 – TOUT ) = QmCP

= 4 x 106165,284 x 0.681 = 35.5°F

Rate fromExercise 1

46

ChE 101.03 V (39)

Exercise 3.1, Cont’d

2. Calculate the heat duty of E7AB.

Answer:

Heat duty for E7AB

310°F → 100°F

CP = 0.66 0.524

Avg CP =

Q = mCP²T = 165,284 x 0.592 x (310 - 100) = 20.55 MBtu/hr

0.66 + 0.5242 = 0.592

47ChE 101.03 V (40)

Exercise 3.1, Cont’d

3. How much cooling water in lb/hr and 60°F gpm is required? Use 95°F to 120°F as the water temperature rise.

The heat capacity data for lean oil is:

Temperature,°F CP, Btu/lb°F

346 0.681

100 0.524

Use flow rates from Exercise 2.1. Feed = 317,829 lbs/hrL.O. = 165,284 lbs/hr

48ChE 101.03 V (41)

Exercise 3.1, Cont’d

Answer:

Cooling water rate.

Q = mCP²T

m = QCP ∆T = 20.55 x 10 6

1 x (120 - 95) = 822,000 lb/hr

gpm = 822,000 lbhr x hr

60 min xgal

8.33 lb = 1,645 gpm@60°F

310°F

120°F

95°F

100°F

49

Natural Heat Gain and Loss for Piping

Solar Heat Gain:

QS / L = 300 (Do L) / L ……………………………………………...(4)

Convection Heat Loss:

QC / L = 0.844(Do)0.75 (Tp – Ta)1.25 (Full pipe area) ………………..(5)

Radiation Heat Loss:

QR / L = 0.246 Do [(Tp/100)4 – (Ta/100)4] (1/2 pipe area) …………..(6)

50

Example Problem 3

Consider a 6 in. nominal size, Schedule 40, Grade B steel pipe filled with oil. The pipeline is unrestrained and above ground. The ambient temperature is 100 F. The line is blocked-in at 110 F and 150 psi. What is the maximum temperature the line will reach?

51

Example Problem 3 (cont’d)Answer:

Do = 6.625/12 = 0.552

assume a max. T of 160 F for the 1st trial:Tp = 160 + 460 = 620 RTa = 100 + 460 = 560 R

QC / L = 0.844(Do)0.75 (Tp – Ta)1.25

= 0.844(0.552)0.75 (620 – 560)1.25 = 90.25 Btu/hr/ft

QR / L = 0.246 Do [(Tp/100)4 – (Ta/100)4] = 0.246 (0.552) [(620/100)4 – (560/100)4] = 67.1 Btu/hr/ft

total heat loss: QT/L = QC/L + QR/L = 90.25 + 67.1 = 157.4 Btu/hr/ft

52

Example Problem 3 (cont’d)

Solar Heat Gain: QS / L = 300 (Do L) / L = 300 (0.552) = 165.6 Btu/hr/ft

total heat loss < total heat gain:therefore, the assumed max. T must be increased (e.g. to 163 F)

Trial Tp (oF) QC / L(Btu/hr/ft)

QR / L(Btu/hr/ft)

QT / L(Btu/hr/ft)

QS / L(Btu/hr/ft)

1 160 90.25 67.1 157.4

166.9

165.6

2 163 95.9 71.0 165.6

53

Heat Loss for Piping

Figure 7: Heat Loss of Horizontal Pipe at Night With Flow

ChE 101.03 V (46)

Pipe Heat Loss With Flow Diam. (ft ) Temperatures, °F Wind, mph Flow, lb/hr

1.0625 Air = 50 Sky = -50 5 50,000Pipe

Temp, °FQr /L

Btu/hr ftQc /L

Btu/hr ftQt /L

Btu/hr ft Cp

Btu/lb °FLength

ftCum.

Length, ftPipe

Temp, °R 1 200 741.2 1409.6 2150.8 0.5 121.2 121.2 660 2 190 682.4 1293.1 1975.5 0.5 132.3 253.4 650 3 180 626.3 1178.7 1805.0 0.5 145.2 398.6 640 4 170 572.8 1066.5 1639.3 0.5 160.4 559.0 630 5 160 521.7 956.6 1478.3 0.5 178.5 737.5 620 6 150 473.1 849.1 1322.2 0.5 200.5 938.1 610 7 140 426.8 744.4 1171.1 0.5 227.7 1165.7 600 8 130 382.7 642.4 1025.2 0.5 261.8 1427.5 590 9 120 340.9 543.7 884.6 0.5 306.0 1733.5 58010 110 301.1 448.4 749.5 0.5 365.0 2098.5 57011 100 263.4 357.0 620.4 0.5 447.2 2545.6 56012 90 227.7 270.1 497.8 0.5 568.1 3113.8 55013 80 193.8 188.5 382.3 0.5 760.2 3874.0 54014 70 161.8 113.6 275.4 0.5 1099.8 4973.8 53015 60 131.5 47.8 179.3 0.5 1771.6 6745.4 52016 50 103.0 0.0 103.0 0.5 3809.8 10555.1 51017 40 76.0 -47.8 28.3 0.5 500

Q = m Cp ∆t QT /L( )Ave =QT /L( )1 + Q T /L( )2

2 for line 1

L QT /L( )Ave = Q For line 1 Q T/L( )Ave = 2150.8 +1975.52 = 2063.2

L =m C p ∆t

(QT /L)Ave L =

50,000( ) 0.5( ) 10( )2063.2 =121.2

Figure 7

54

Exercise 3.2: calculating temperature change

1. A pipeline (Do = 1.0625 ft) is operating at night. The line contains 20o API hydrocarbon (VABP = 650 oF) flowing at a rate of 200, 000 Ib/hr. The initial temperature is 200 oF and the line length is 6300 ft. The average heat loss for the line has been calculated to be 898.4 Btu/hr/ft of line length with a wind velocity of 5 mph.

What is the exit temperature of the pipeline?

55ChE 101.03 V (48)

Exercise 3.2, Cont’dAnswer:

QLOSS = 898.4 x 6300 = 5.66 x 106 Btu/Hr

For initial guess at outlet temperature, use inlet temperature CP.

Temperature CP+B B CP

200 0.51 0.025 0.485

Inlet CP = Ave CP = 0.485

Q = QLOSS = m CP ∆Τ

∆T = 5.664 x 106200,000 x 0.485 = 58°F

TOUT = T - ∆T = 200 - 58 = 142°F

56

ChE 101.03 V (49)

Exercise 3.2, Cont’d

Calculate outlet CP - Maxwell (p. 93)

Temperature CP+B B CP

142 0.485 0.025 0.460

Ave CP = 0.460 + 0.4852 = 0.472

∆T = 5.664 x 106200,000 x 0.472 = 60°F

TOUT = 200 - 60 = 140°F

Check Q.

Q = 200,000 (0.472) (200 - 140) = 5.664 x 106 Btu/hr

57

ChE 101.03 V (50)

Exercise 3.2, Cont’dHEAT LOSS OF HORIZONTAL PIPE AT NIGHT

(Uses SADP-L-043 method with -50°F air temperature for radiation)

Diam. (ft ) Temperatures, °F Wind, mph Flow, lb/hr 1.0625 Air = 50 Sky = -50 5 200,000

PipeTemp, °F

Qr /LBtu/hr ft

Qc /LBtu/hr ft

Qt /LBtu/hr ft

Cp Btu/lb °F

Lengthft

Cum.Length, ft

PipeTemp, °R

1 200 741.2 1409.6 2150.8 0.485 467.7 467.7 660 2 190 682.4 1293.1 1975.5 0.480 759.7 1227.4 650 3 180 626.3 1178.7 1805.0 0.476 825.1 2052.5 640 4 170 572.8 1066.5 1639.3 0.469 901.4 2953.9 630 5 160 521.7 956.6 1478.3 0.467 997.0 3950.9 620 6 150 473.1 849.1 1322.2 0.462 1108.5 5059.4 610 7 140 426.8 744.4 1171.1 0.458 1247.5 6307.0 600 8 130 382.7 642.4 1025.2 0.454 1422.2 7729.1 590 9 120 340.9 543.7 884.6 0.450 1647.4 9376.6 58010 110 301.1 448.4 749.5 0.446 1947.5 11324.1 57011 100 263.4 357.0 620.4 0.442 560

58

ChE 101.03 V (51)

Exercise 3.2, Cont’d

2. An uninsulated horizontal drum 10 ft outside diameter and 40 ft long is exposed to full sunlight. The air temperature is 110°F. What is the maximum surface temperature that can be attained?

(A drum is the same geometry as a large pipe) Assume 187°F for the first trial and indicate if the next trial will require a hotter or cooler value.

59

ChE 101.03 V (52)

Exercise 3.2, Cont’dAnswer:

Diameter Do = 10 ft 2nd TrialTa = 110 + 460 = 570°R

Final temperature = 187°F (assume for first trial) 192°FTP = 187 + 460 = 647°R 652°RQC/L = 0.844Do

0.75 (TP - Ta)1.25

= 0.844 (10)0.75 (647 - 570)1.25 = 1082.6 Btu/hr/ft 1171.1 Btu/hr/ft

1848.8 Btu/hr/ft

QT/L = QR/L + QC/L = 1,082.6 + 1,714 = 2,796.6 Btu/hr/ft 3019.9 Btu/hr/ftQS/L = 300 Do = 300 x 10 = 3,000 Btu/hr/ftSince QT < QS, the assumption of 187°F is a little low.After second trial, QT ≅ QS, the assumption of 192°F is adequate.

QR /L = 0.246 D oTP100

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4-

Ta100

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

= 0.246(10) 647100

⎛ ⎝

⎞ ⎠

4- 570

100⎛ ⎝

⎞ ⎠

4⎡

⎣ ⎢

⎤

⎦ ⎥ = 1,714.0 Btu/hr/ft