Open System Control Volume

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: m_in(h_in pe_in ke_in) Q m_o_ut (h_out pe_out ke_out) W 1) Divide by m since, m_in m_out m. (h_in + pe_in + ke_in) + q = (h_out + pe_out + ke_out) + 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 pe_out + [(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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