Problem 3.4 - 0.571 14640 The change in altitude is then...

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Problem 3.4 [Difficulty: 3] Given: Boiling points of water at different elevations Find: Change in elevation Solution: From the steam tables, we have the following data for the boiling point (saturation temperature) of water T sat ( o F) p (psia) 195 10.39 185 8.39 The sea level pressure, from Table A.3, is p SL = 14.696 psia Hence T sat ( o F) p/p SL 195 0.707 185 0.571 From Table A.3 p/p SL Altitude (m) Altitude (ft) 0.7372 2500 8203 0.6920 3000 9843 0.6492 3500 11484 0.6085 4000 13124 0.5700 4500 14765 Then, any one of a number of Excel functions can be used to interpolate (Here we use Excel 's Trendline analysis) p/p SL Altitude (ft) 0.707 9303 Current altitude is approximately 9303 ft
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Unformatted text preview: 0.571 14640 The change in altitude is then 5337 ft Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points p/p SL Altitude (m) Altitude (ft) p/p SL Altitude (m) Altitude (ft) For 0.7372 2500 8203 0.6085 4000 13124 0.6920 3000 9843 0.5700 4500 14765 Then 0.7070 2834 9299 0.5730 4461 14637 The change in altitude is then 5338 ft Altitude vs Atmospheric Pressure z = -39217(p/p SL ) + 37029 R 2 = 0.999 2500 5000 7500 10000 12500 15000 0.55 0.60 0.65 0.70 0.75 p/p SL Altitude (ft) Data Linear Trendline...
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