Environmental Engineering Reference
In-Depth Information
Example 3:
Interval # 1 in the cold section:
The interval start temperature is 550 K
The interval end temperature is 520 K
The flow specific heat of this cold stream is F1Cp1= 10 kW/K
Then,
Q stream #1(cooling capacity) in interval #1= 10*(560-520) = 400 kW
Upon the completion of this step,
2.2
We can now obtain the collective loads (capacities) of the hot (cold) process streams.
These collective loads (capacities) are calculated by summing up the individual
loads of the hot process streams that pass through that interval and the
collective cooling capacity of the cold streams within the same interval
These calculations for the above problem is shown in the following tables
Table 3. Exchangeable Loads for Process Hot Streams Intervals
Interval
Load of H1, kW
Load of H2, kW
Total Load, kW
0.0*(560-520)= 0.0
0.0*(560-520)= 0.0
0.0+0.0= 0.0
1
10*(520-390)= 1300
0.0*(520-390)= 0.0
1300+0.0= 1300
2
10*(390-380)= 100
0.0*(390-380)= 0.0
100+0.0= 100
3
5*(380-330)= 250
10*(380-350)= 500
500+250= 750
4
5*(330-310)= 100
5
0.0*(330-310)= 0.0
0.0+ 100= 100
6
0.0*(310-300)= 0.0
5*(310-300)= 50
0.0+50= 50
Table 4.Cooling Capacities for Process Cold Stream Intervals
Interval
Capacity of C1, kW
Capacity of C2, kW
Total Load, kW
10*(550-510)= 400
0.0*(550-510)= 0.0
400+0.0= 400
1
10*(510-380)= 1300
10*(380-370)= 100
0.0*(510-380)= 0.0
5*(380-370)= 50
1300+0= 1300
100+50= 150
2
3
4
5*(370-320)= 250
10*(370-320)=500
500+250= 750
0.0*(320-300)= 0.0
5
6
10*(320-300)= 200
0.0*(300-290)= 0.0
200+ 0.0= 200
0.0*(300-290)= 0.0
0.0+0.0= 0
