ChE 101.04ChE 101.04
MASS & ENERGY BALANCESMASS & ENERGY BALANCESPREPARATION & USEPREPARATION & USE
Dr. Abdulhadi Al-Juhani
Department of Chemical Engineering, King Fahd University of Petroleum & Minerals
4-12-07
Recycle & Purge Streams
ChE 101.04 V (36)
Boiler Blowdown Method
where: FBFW = Feedwater flow, lb/hr.FBD = Blowdown flow, lb/hr.FSTM = Steam flow, lb/hr.CBFW = Solids concentration in boiler feedwater, ppm.CBD = Solids concentration of blow down (circulating boiler water), ppm.X = Percent blowdown, percent of boiler feed water.
FSTM
FBFW
CBFW
FBD
C BD
BoilerDrum
FBFW
FBD
CBFW
X = (100)
FBFW = FBD C BD
CBFW FBD
FBFWC BD
X100= =
Recycle Streams
Hydrotreating Purge
ChE 101.04 V (37)
Treat Gas, TG
FEED F
Recycle, RPurge, P
Liquid Product L
Seperation Drum
Reactor
Figure 4
Recycle Streams
• Heat of Reaction
Chemical Reactions
· Overall Mass Balance Not Changed
· Mole Balance and Composition Change
· Reversible Reactions
· Irreversible Reactions
· Predict Reaction Products
Exercise 3
Directions:
Water is fed into a boiler from a demineralizer plant. There arespecifications on solids and chlorides in the boiler water. If thechloride specification is met, the solids specification isautomatically met, since demineralized water is good qualityboiler feedwater and the chloride specification is the harder ofthe two specifications to achieve.
The chloride content of the boiler water must be kept equal to orlower than 8 wppm (weight parts per million). The feedwater hasa chloride content of 0.2 wppm. The boiler raises 200,000 Ib/hr ofsteam. Calculate the boiler feedwater and blowdown rates.
Exercise 3, Cont’d
Answer:
( )100FFXBFW
BD
FBFW CBFW = FBD CBD
BFW
BD
BD
BFWFF025.0
82.0
CC
==⎟⎠⎞
⎜⎝⎛=
FBD = 0.025 (FBFW)FBFW = FStm + FBD
FBFW = FStm + .025 FBFW
0.975 FBFW = FStm = 200,000FBFW = 205,128 FBD = 5128
FBD
CBD
FBFW
CBFW
FSTM
X =
2. How would you fix them?
Exercise 4
Introduction:
This Exercise will be worked on as a class discussion. In
Exercise 3 you prepared a mass and energy balance
around a C3/C4 splitter system. The owner of this
1. What would you do to find the bottlenecks?
Answer:
Exercise 4, Cont'd
1. Unless there are known limitations, a unit testwould be the first step. This would determine themajor limitations and uncover other areas that areclose to limiting. The class should suggestpossible areas at which to observe bottlenecks.
2. The instructor should propose bottlenecks and askthe class how they would find and solve them. A listof proposed bottlenecks should include:
First Step Unit Test• Tower flooding upper section• Tower flooding lower section• Overhead exchanger• Pumps
• Reboiler• Feed/effluent exchanger
tube leak • Control valve
Evaluation, Cont'd
The refinery has 12,629 B/D of naphtha to hydrotreat. Theprocess licenser has the following process requirementsand data:• Treat gas rate must provide a minimum of 241 SCF
H2/bbl feed.
• The hydrogen partial pressure in the reactor must be aminimum of 56.6 psia.
• The hydrogen consumption is 5.83 SCF/bbl. Theconsumed hydrogen appears as H2S.
Evaluation, Cont'd
The refinery Information are as follows:
• Treat gas is 69.23 mole percent hydrogen. Theremainder is methane.
• The feed is totally vaporized in the reactor
• Feed information:moles per hour (mph)
iC5C5C6/310310/360360/395395/430430 +
MWsp. Gr.60
0.190.19
182.74223.52202.23166.67175.23950.76
152 0.785
Note that since there is insufficient time to run acomputer flash, assume a perfect split ofcomponents and that all the feed components leavein the product.
Evaluation, Cont'd
Calculate
1. The material balance around this unit.
2. The reactor pressure in psia.
Evaluation, Cont’d
Answer:
Feed rate:
( )( ) orhrlbs144518
hr24Dx
gallbs785.833x
Bgal42x
DB1629 =
950.76 mph x 152 mw = 144,515 lbs/hr
Hydrogen treat gas rate : 241 SCFH2/bbl feed.
22 Hmph6.334
SCF379molex
hrscf2.126816
hr24Dx
BHSCF241x
DB1629 ==
Evaluation, Cont'd
Total treat gas: Treat gas is 69.23 mole percent H2
0.6923 X = 334.6 mphX = 483.3 mph
Treat Gastotal treat gas ratehydrogen
methane
mph483.3-334.6
148.7 mph
Evaluation, Cont’d
Hydrogen consumed:
Consumption = 5.83 SCF/bbl feed
mph1.8SCF379
molexhr
SCF79.3067bbl
SCF83.5xhr24
DxDB1629 ==
• H2S created = H2 consumed =8.1 mph.
• Offgas composition (mph):
Treat Gas mph Consumed mph Offgas mph148.7 CH4 0 CH4 148.7 CH4
334.6 H2 8.1 H2 326.5 H2
483.3 8.1 8.1 H2S 483.3
E v a lu a tio n , C o n t’d
• M in im u m H 2 p a rt ia l p re s s u re = 5 6 .6 p s ia .T h is w ill o c c u r a t th e o u tle t w h e re th ep re s s u re a n d H 2 c o n te n t a re a t th em in im u m .
molestotalHmoles 2 x p re s s u re = 5 6 .5
76.9503.4835.326
+ x p re s s u re = 5 6 .5 =
0 .2 2 7 7 x p re s s u re
R e a c to r o u tle t p re s s u re = 2 4 8 .6 p s ia =2 3 3 .9 p s ig
ChE 101.04 V (57)
Evaluation, Cont’d
Feed
950 mph144,518 lbs/hr
Treat Gas
248.6 psia
Reactor
Offgas
Product950 mph144,258 lbs/hr
MW162
2379669
3048
lbs/hrmphCH4 148.7
H2 334.6483.3
CH4 148.7 H2 326.5
H2S 8.1483.3
mph16
234.1
MW2379653276
3308
lbs/hr
Evaluation; Cont'dEvaluation; Cont'd
Mass BalanceMass Balance
n
FeedTreat Gas
Ibs/hr
144,5183,048147,566
Out Ibs/hr Off gas 3,308Product 144.25(By difference)
147,566