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First Law of Thermodynamics - h1012v1_85
Figure 19    T-s Diagram with Rankine Cycles

Thermodynamics Heat Transfer and Fluid Flow Volume 1 of 3
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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







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