EGH463 Plant and Process Design Report: Energy Balance Calculations

This report presents an analysis of plant and process design, focusing on energy balance calculations and water evaporation. It includes detailed calculations for steam conditions, enthalpy, and mass flow rates at various stages. The report provides insights into the quantity of water evaporated per hour and the minimum cooling water required. Furthermore, it calculates the quantity of total fresh water produced. The report also references relevant standards for process design and control, and includes a P&ID diagram for an integrated gasification combined cycle (IGCC) schematic, demonstrating the application of thermodynamics and heat transfer principles in plant design. The assignment solution contributes to the understanding of power cycles and industrial processes.

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.