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PGE 421K Spring 2010 HWSolution3

# PGE 421K Spring 2010 HWSolution3 - PGE 421K Spring 2010...

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PGE 421K, Spring 2010 HOMEWORK 3 100 points Due M 22 Feb 1. Knowing the terms in the mass and energy balances can help you figure out appropriate (i.e. physically reasonable) forms of correlations, design equations, etc. The chart above appears in a Dowell Engineers Handbook from the 1950s and shows the volumetric flow rate that must be supplied to operate fluid jets through orifices of various sizes, as a function of the pressure drop across the jet. a. Start with the general form of the energy balance and derive the relationship between Q and Δ P from the energy balance. State the assumptions you make in the derivation. Assume steady state, isentropic, no shaft work, no change in potential energy. Mass balance: g1856g1865 g3004g3023 g1856g1872 = g1865 g4662 g3036g3041 −g1865 g4662 g3042g3048g3047 = 0 g1865 g4662 g3036g3041 = g1865 g4662 g3042g3048g3047 Steady state energy balance across the jet: 0 = g1843 g4662 +g1849 g4662 g3046 +g1865 g4662 g3036g3041 g4678g1834 g3553 g3036g3041 + g1873 g3036g3041 g2870 2 +g1859g1878 g3036g3041 g4679−g1865 g4662 g3042g3048g3047 g4678g1834 g3553 g3042g3048g3047 + g1873 g3042g3048g3047 g2870 2 +g1859g1878 g3042g3048g3047 g4679 20 pts

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Combine energy and mass balances: 0 = g1865 g4662 g3036g3041 g4678g1834 g3553 g3036g3041 −g1834 g3553 g3042g3048g3047 + g1873 g3036g3041 g2870 −g1873 g3042g3048g3047 g2870 2 g4679 Assume incompressible fluid. Then: g1834 g3553 g3036g3041 −g1834 g3553 g3042g3048g3047 = g1848 g3552 g4666g1842 g3036g3041 −g1842 g3042g3048g3047 g4667 and g1865 g4662 g3036g3041 = g1873 g3036g3041 g1827 g3036g3041 g1848 g3552 = g1873 g3042g3048g3047 g1827 g3042g3048g3047 g1848 g3552 where A in , A out are cross sectional areas of pipe and orifice. Hence: g1873 g3042g3048g3047 = g1873 g3036g3041 g1827 g3036g3041 g1827 g3042g3048g3047 Note: g1827 g3042g3048g3047 < g1827 g3036g3041 so g1873 g3042g3048g3047 > g1873 g3036g3041 . Substituting into the energy balance, we obtain: g1848 g3552 ∆g1842 = 1 2 g4666g1873 g3042g3048g3047 g2870 −g1873 g3036g3041 g2870 g4667 = 1 2 g4678g1873 g3042g3048g3047 g2870 −g1873 g3042g3048g3047 g2870 g3436 g1827 g3042g3048g3047 g1827 g3036g3041 g3440 g2870 g4679 Solving for g1873 g3042g3048g3047 : g1873 g3042g3048g3047 = g3497 2g1848 g3552 ∆g1842 1−g4672 g1827 g3042g3048g3047 g1827 g3036g3041 g4673 g2870 Flow through orifice is Q=u out A out so we have: g1843 = g1827 g3042g3048g3047 g3497 2g1848 g3552 ∆g1842 1−g4672 g1827 g3042g3048g3047 g1827 g3036g3041 g4673 g2870 orifice
b. Use the result of (a) to explain the shape of the curves in the chart. That is, what is the connection between the mathematical form of (a) and family of curves in the chart.

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