Plant and Process Design

This study material provides information about plant and process design. It covers topics such as temperature and enthalpy of steam and condensate, quantity of water evaporated per hour, and the minimum quantity of cooling sea water required. The material also discusses process plant design and project management. References are provided for further reading.

PLANT AND PROCESS DESIGN
Name of Student
Institution Affiliation
SECTION D
The temperature of the sea water in the two vessels
From the data inflow of salt water is at 800C
The temperature and enthalpy of the steam/vapour and the condensate at each step
Temperature and enthalpy of the steam/vapour and the condensate at each step 600t/h and 800C

Lp saturated steam flow is 100t/h
Table 1: Showing pressure at different conditions
Steam conditions Pressure (kPa abs)
Lp Steam 220
No1 vapour 180
No2 vapour 140
Table 2: Showing Temperature and enthalpy of the steam and condensate at every step
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
From the energy balance equations to obtain the mass
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)
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)
2211w1+ 600-w1(2211)= 2231.8 w2

2211w1 + 1326600-600w1= 2231.8 w2
1611w1-2231.8w2=1326600 ………………………………………………………………..1
And from
W2(2231.8)+(600-w1-w2)(22231.8)= 2257.7 w2
13339080-22231.8w1-22231.8w2=25.8w2
-22231.8w1-22206w2=13339080 …………………………………………………………….2
W1= 1326600−2231.8 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
Here w1 is the energy used in evaporating water in stage 1 w2 is the energy used in evaporating
water in stage 2.
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
X = 3502.5kg/h
Hence,
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 the below equation;
From Log (140) m =7.90298 ( 373.16
20 −1) +5.02808 log 373.16
20
Where ∆ T the temperature change, p is is the pressure while m is the water mass
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;
Fresh water mass = p (temp of preheat)
hp saturated
Fresh water mass = 220(80)
100

Fresh water mass = 17600
100
Fresh water mass = 176kg
References
Helmus, F. P. (2011). Process Plant Design: Project Management from Inquiry to Acceptance. Liverpool:
John Wiley & Sons.
Seferlis, P. (2014). The Integration of Process Design and Control. Florida: Elsevier.