Page 89 - Handbook of Energy Engineering Calculations

2. Sketch the T-S diagram for the cycle

To analyze any steam cycle, trace the flow of 1 lb (0.5 kg) of steam through
the system. Thus, in this cycle, 1 lb (0.5 kg) of steam leaves the steam
generator at point 2 and flows to the turbine. From state 2 to 3, 1 lb (0.5 kg)
of steam expands at constant entropy (assumed) through the turbine,
producing work output W = H − H , represented by area l-a-2-3 on the T-S
3
2
1
diagram, Fig. 16a. At point 3, some steam is bled from the turbine to heat the
feedwater passing through heater 1. The quantity of steam bled, m , lb is less
1
than the 1 lb (0.5 kg) flowing between points 2 and 3. Plot stages 2 and 3 on
the T-S diagram, Fig. 16a.
From point 3 to 4, the quantity of steam flowing through the turbine is 1 −

m lb. This steam produces work output W = H − H . Plot point 4 on the T-
1
4
3
2
S diagram. Then, area 1-3-4-12 represents the work output W , Fig. 16a.
2
At point 4, steam is bled to heater 2. The weight of this steam is m lb.
2
From point 4, the steam continues to flow through the turbine to point 5, Fig.
16a. The weight of the steam flowing between points 4 and 5 is 1 − m − m 2
1
lb. Plot point 5 on the T-S diagram, Fig. 16a.The work output between points
4 and 5, W = H − H , is represented by area 4-5-10-11 on the T-S diagram.
4
5
3
At point 5, steam is bled to heater 3. The weight of this bleed steam is m 3
lb. From point 5, steam continues to flow through the turbine to exhaust at
point 6, Fig. 16a. The weight of steam flowing between points 5 and 6 is 1 −

m − m − m lb. Plot point 6 on the T-S diagram, Fig. 16a.
1
2
3
The work output between points 5 and 6 is W = H − H , represented by
6
5
4
area 5-6-7-9 on the T-S diagram, Fig. 16a. Area Q represents the heat given
r
up by 1 lb (0.5 kg) of exhaust steam. Similarly, the area marked Q represents
a
the heat absorbed by 1 lb (0.5 kg) of water in the steam generator.