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Unformatted text preview: Set 03 2.58* The pressure at the bottom of the dip tube is h (a) That is the depth at the end of the dip tube. The total depth is 4.807 + 0.5 =5.3074 ft. (b) If we use 2.000 psig, the calculated depth is 4.800 ft. The difference is 0.0074 ft or 0.15% of the total. (c) The pressure scale is often converted to the level measurement. In this case it would be a linear function converting the Pressure on the gauge at the top to level in feet: 2.59 If we ignore the gas density, then P = (gh )fluid = (gh)manometer fluid = hmanometer 1.5m kg kg lbm manometer = 998.2 3 = 1497.3 3 = 93.45 3 hfluid 1.0 m m m ft If we account for the gas density, then P = (gh)fluid (fluid - gas )= (gh)manometer( manometer - gas ) h h kg fluid = m m - m - 1 gas = 1497.3 - (1.5 - 1) 1.20 = 1496.7 3 h h m f f which is a difference of 0.04%. Set 03 2.62* (a) Here we will use the pressure-depth relationship for an ideal gas (equation 2.16, page 42) Note the molecular weight for CH4 is 16 lbm/lbmole and P1=1014.7 psia =0.1954 (b) For constant density: we use equation 2.9: Note we use the ideal gas law to determine the density: . or 1.7% error 2.63. Let's examine one of the 200 ft hills: 2 1 Adding the above equations yields: 3 Thus, for each of these "rises", there is a pressure drop of 17.3 psig. Ten such rises would account for a static pressure drop of 173 psig > 150 psig, thus, no flow... 3.7* OR V2 = V1 A1 ft 500 ft 2 ft m =1 = 0.145 2 = 0.476 A2 s 7150 ft s s Set 03 3.12 ; The velocity increases as the pressure falls. This has the paradoxical consequence that friction, which causes the pressure to fall, causes the velocity to increase. 3.14* ; 3.22 Let represent system volume, which is constant, and let represent velocity. . d (Vc )system dt
Or, = (cQ)in - (cQ)out where c is the concentration Here Vsys is constant and cout = csys, and cin = 0 so that V Also, . Separating the variables, we have: dc = -Qc dt which is the same as Eq 3.16 (after taking the exponential of both sides) with concentration replacing density, and the Q / V one tenth as large as in Ex. 3.7. One can look up the answers on Fig 3.5, if one takes the vertical scale as c / c0 and multiplies the time values on the horizontal scale by 10. 3.23 ...
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This homework help was uploaded on 04/18/2008 for the course CHE 218 taught by Professor Smith during the Spring '08 term at SDSMT.
- Spring '08