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p0334

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Unformatted text preview: 378 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.34 Chapter 3, Problem 34 Problem: An inclined manometer is a large spherical container with an inclined tube of small diameter attached. The tube is inclined at an angle α to the horizontal and has a scale with markings separated by 1 cm. The manometer has been calibrated for a liquid of density ρ so that each marking on the scale corresponds to a change in pressure, ∆pb , of 100 Pa. If the manometer is filled with a fluid of density of 3 ρ, what 4 pressure change will each marking correspond to? ......... . . ....... . . . . . . . . . . . . . . . . . .. ... ... . 5 . .... . ... .... . .... .. ...... .... ...... .... 4 .. ........... . ......... .. .. . ... ... ... ... 3 .. ............ .. ... ...... . ... ... ... .... . .. ................ ......... ... ... 2 .. ..... .. . .. .. .. .. .. . ..... 1 .. ................. . ............ .... .. .. b .. . .. ................... .. .. . .... . ........... . . . 0. ................. . . ... ... . . ........ . . .......... . . . . .. . ................................................................ . ... ... ..... ....... ..... ........... . . . . . .. ......................................................................................................... ..................... .. .. ................................................................................................. ................. .. ........ . . . .... . . . ... ... ...... . .. . ... . ..... .. . .... ........ . . ... .... . .. . . . . . . . . . . . . .. ......... ...... . .. ........................................................................................................................... . ............................................................................................................................................................... . . .......................................................... ...... ....... ... ...................................................................................................................................... ...................... . ............................................................................................... .... .. . ...... .. ............................................................................... . . .. .................................................................................................. ...... ........................................ ................... . .............. .................................................... .. ....... ............................................................................................... ........................ . ................................................................................................. ....................................... .... .................................. ................................... ............. .... ............................... . .. . ... .... ......................................................................................................... ............. .................................................................................... . .. . .. . .................................................................................................. ... ................................................................................................... .. ............................................................................................. . ... ................. .... ............. .... .... ....... ... . ................................................................................. ..... ....... ............. .... .................... ..................................................................... ...... .................................................................. .. . ......................................................... ... ............................................. ..... ...................... ............. .... .. p α ρ Solution: From the hydrostatic relation, we know that p + ρgz = constant =⇒ pb = pa + ρgz Thus, since atmospheric pressure is a constant, changes in pb are related to changes in z according to ∆pb = ρg ∆z But, the distance measured in the direction tangent to the inclined tube, s, is related to z by ∆z = ∆s sin α =⇒ ρg ∆s sin α = ∆pb So, if we replace the fluid in the manometer with a fluid of density ρ, then the ratio of ¯ the pressure change, ∆pb , to the value corresponding to density ρ is ¯ ∆pb ¯ ρ ¯ = ∆pb ρ =⇒ ρ ¯ ∆pb = ∆pb ¯ ρ Therefore, if ρ = 3 ρ, the pressure change each marking corresponds to is ¯4 ∆pb = ¯ 3 ρ 4 ρ (100 Pa) = 75 Pa ...
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