FIRST LAW OF THERMODYNAMICS
Thermodynamics
Example 1 illustrates the use of the control volume concept while solving a first law problem
involving most of the energy terms mentioned previously.
Example 1: Open System Control Volume
The enthalpies of steam entering and leaving a steam turbine are 1349 Btu/lbm and 1100
Btu/lbm, respectively. The estimated heat loss is 5 Btu/lbm of steam. The flow enters
the turbine at 164 ft/sec at a point 6.5 ft above the discharge and leaves the turbine at 262
ft/sec. Determine the work of the turbine.
Solution:
min(hin
pein
kein)
Q
m out (hout
peout
keout)
W
1) Divide by
m since,
min
mout
m.
(hin + pein + kein) + q = (hout + peout + keout) + w
where:
q
= heat added to the system per pound (Btu/lbm)
w
= work done by the system per pound (ft-lbf/lbm)
2) Use Joule’s constant J = 778 ft-lbf/Btu for conversions and substitute known values.
1349 Btu/lbm + (6.5/778) Btu/lbm + [(164)2/2(32.17)(778)] Btu/lbm +
(-5 Btu/lbm) = 1100 Btu/lbm + 0 peout + [(262)2/2(32.17)(778)] Btu/lbm + w
Note: The minus sign indicates heat out of the turbine.
3) Solve for work, w.
1349 Btu/lbm + 8.3548 x 10-3 Btu/lbm + 0.5368 Btu/lbm - 5 Btu/lbm = 1100
Btu/lbm + 1.37 Btu/lbm + w
1344.54 Btu/lbm = 1101.37 Btu/lbm + w
w = 1344.54 Btu/lbm - 1101.37 Btu/lbm
w = 243.17 Btu/lbm
HT-01
Page 60
Rev. 0
