University Plant and Process Design Report - EGH463

Verified

Added on  2022/12/30

|7
|687
|43
Report
AI Summary
This report analyzes plant and process design, focusing on Section C's heat transfer coefficient calculation, and Section D's multi-stage evaporation process. It includes calculations for the overall heat transfer coefficient, mass flow rate, and energy balance equations for steam conditions. The report further addresses the quantity of water evaporated, and the minimum cooling water required for condensation, referencing relevant equations and data. The assignment also includes the P&ID diagram for IGCC schematic. The report concludes with a list of references, demonstrating a comprehensive understanding of plant design principles and their application in industrial processes. This assignment is a great example of how engineering principles are applied to solve real world problems.
Document Page
Running head: PLANT AND PROCESS DESIGN
PLANT AND PROCESS DESIGN
Name of Student
Institution Affiliation
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
PLANT AND PROCESS DESIGN 2
SECTION C
The following factor for River water below is 500C is 0.0001 m2k/W
We have to find overall heat transfer coefficient
1
U = 1
h 0 +
´ ln Do
Oi
2 πKL
× πDL + Rfouling
1
U = 1
10000 +
´ ln 1.35
1.25
2 ×16.3
×2.54 ×2+104
Uoval = Overal heat transfer coef
= 5406.7W/m2k
Document Page
PLANT AND PROCESS DESIGN 3
Data not give find the mass flow
Rate let us assume m= 5kg/s
For condensing Cmax=
Cmin=4.18w/kgh
(mcp ) Min =20900
εMax= 1-e NT
(C)( D)
Cmin( Tmax ) = 3
25= 1- e NT =0.3
NTU=1.2= A
mcp
1.2× 20.9× 103
5406.7 =400π × × L
L=1.16 meters
SECTION D
Given that inflow of salt water =600t/h at 800C
Lp saturated steam flow = 100t/h
Given the table
Steam conditions Pressure (kPa abs)
Lp Steam 220
No1 vapour 180
No2 vapour 140
Document Page
PLANT AND PROCESS DESIGN 4
Temperature and enthalpy of the steam and condensate at each step is
Steam condition Pressure (kPa) Saturation
temperature
Enthalpy of
satrurated steam
(kJ/kg)
Enthalpy of
condensate
(kj/kg)
4 steam 220 13.3 2710.9 517.6
No 1 vapour 180 116.9 2701.7 490.9
No2 Vapour 140 104.3 2690.2 458.4
Use the energy balance equations to obtain the mass
517.6us+600(2410.9-517.6)=w1(2701.7-490.7)
W1(2701.7-490.7)+(600-w1)(2701.7-490-7)=w2(2690.2-458.4)
W2(2690.2-458.4)+(600-w1-w2)(2690.2-458.4)= w2(2675.1-417.4)
Here w1 is the energy used in evaporating water in stage 1 w2 is the energy used in evaporating
water in stage 2.
W2(2231.8)+(600-w1-w2)(22231.8)= 2257.7 w2
13339080-22231.8w1-22231.8w2=25.8w2
-22231.8w1-22206w2=13339080 …………………………………………………………….1
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
PLANT AND PROCESS DESIGN 5
From
W1(2701.7-490.7)+(600-w1)(2701.7-490-7)=w2(2690.2-458.4)
2211w1+ 600-w1(2211)= 2231.8 w2
2211w1 + 1326600-600w1= 2231.8 w2
1611w1-2231.8w2=1326600 …………………………………………………………….2
W1= 13266002231.8 w 2
1611
W1= 823.463-1.3853w2
823.463-1.3853 [ -22231.8] -22206w2=13339080
W2 = 3683.33
1611w1-(2231.8*3683,33)=1326600
1611w1-822038.894= 1326600
W1 = 1333.73
Quantity of water evaporated per hour
517.6X+600(2410.9-517.6)=w1(2701.7-490.7)
517.6 X+600(1893.3) = 2211(1333.73)
517.6X+1135980=2948877.03
517.6 X = 1812897.03
Document Page
PLANT AND PROCESS DESIGN 6
X = 3502.5kg/h
Therefore;
Quantity of water evaporated per hour = 3502.5kg
The minimum quantity of cooling sea water required to fully condense the No. 2 vapour
if the initial water temperature is 25 C and a temperature rise of 5 C is allowed for the
cooling water.
From Goff Gratch equation;
From Log (140) m =7.90298 ( 373.16
20 1) +5.02808 log 373.16
20
Where P is the pressure, T isthe changetemperature, m is the mass of water
2.146m =7.90298*17.658+ 5.02808 *log 18.658
2.146m =7.90298*17.6586.39
2.146m=139.555
M= 65.03 kg
The quantity of total fresh water produced (do not count the LP steam condensate)
Mass of fresh water can be obtained by the following formula;
Mwaterfresh = p(temp of preheat)/ (hp saturated)
Mwaterfresh = 220*80 / 100
Mwaterfresh = 17600/100
Mwaterfresh = 176kg
Document Page
PLANT AND PROCESS DESIGN 7
References
Backhurst, H. (2013). Process Plant Design: Heinemann Chemical Engineering Series. Florida:
Elsevier Science.
Moran, S. (2015). An Applied Guide to Process and Plant Design. Manchester: Elsevier Science.
Sinnott, R. (2017). Chemical Engineering Design: Principles, Practice and Economics of Plant
and Process Design. Liverpool: Butterworth-Heinemann.
Towler, G. (2012). Chemical Engineering Design: Principles, Practice and Economics of Plant
and Process Design. Hull: Elsevier.
chevron_up_icon
1 out of 7
circle_padding
hide_on_mobile
zoom_out_icon
[object Object]