chap09 - PROBLEM 9.1 FIND: SOLUTION Absolute pressure...

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PROBLEM 9.1 FIND: Convert to units of gauge pressure in N/m 2 SOLUTION Absolute pressure reference scale: 1 atm abs = 14.69 psia = 101.325 kPa abs = 101,325 N/m 2 abs 1 atm abs = 760 mm Hg abs = 406 in H 2 O abs Conversion factors: 14.69 psi = 101.325 kPa = 101,325 N/m 2 14.69 psi = 1 atm = 29.92 in Hg = 406 in H 2 O = 760 mm Hg 1 bar = 14.505 lb/in 2 = 100,000 N/m 2 p(gauge) = p(absolute) – p(reference) Using p(reference) = 101,325 N/m 2 : (a) 10.8 psia x 101325 N/m 2 /14.69 psi = 74 442 N/m 2 abs p = 74,442 – 101,325 N/m 2 = -26,883 N/m 2 = -26.883 kPa (b) 1.75 bars abs x 100,000 N/m 2 = 175,000 N/m 2 abs p = 175,000 – 101,325 N/m 2 = 73,675 N/m 2 = 73.675 kPa (c) 30.36 in H 2 O abs x 101,325 N/m 2 /406 in H 2 O = 7,577 N/m 2 abs p = 7,577 – 101,325 N/m 2 = -93,748 N/m 2 = - 93.748 kPa (d) 791 mm Hg abs x 101,325 N/m 2 /760mm Hg = 105,458 N/m 2 abs p = 105,458 – 101,325 N/m 2 = 4,133 N/m 2 = 4,133 kPa COMMENT Note that gauge pressure can be negative or positive relative to the reference pressure. Absolute pressure, which is measured relative to a perfect vacuum, is always a positive value.
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PROBLEM 9.2 FIND: Convert into absolute pressure SOLUTION p(absolute) = p(gauge) + p(reference) where here: p(reference) = 1 atm abs. Absolute pressure reference scale: 1 atm abs = 14.69 psia = 101.325 kPa abs = 101,325 N/m 2 abs 1 atm abs = 760 mm Hg abs = 406 in H 2 O abs Conversion factors: 14.69 psi = 101.325 kPa = 101,325 N/m 2 14.69 psi = 1 atm = 29.92 in Hg = 406 in H 2 O = 760 mm Hg (a) -0.55 psi + 14.69 psia = 14.14 psia = 0.963 atm abs (b) 100 mm Hg + 760 mm Hg abs = 860 mm Hg abs = 1.13 atm abs (c) 98.6 kPa + 101.325 kPa abs = 199.925 kPa abs = 1.97 atm abs (d) 7.62 cm H 2 O + 760 mm H 2 O abs = 836.2 mm H 2 O abs = 1.10 atm abs
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PROBLEM 9.3 KNOWN: H = 250 cm H 2 O p atm = 101.3 kPa abs FIND: The tank pressure p 1 . PROPERTIES: γ H2O = 997 kg/m 3 SOLUTION Referring to Figure 9.5 (manometer) p 1 – p 2 = p 1 - p atm p 1 = p atm + γ H2O H = 101,325 N/m 2 abs + (997 kg/m 3 )(250cm)(1m/100cm)(1N/kg-m/s 2 ) = 103,818 N/m 2 abs or = 0 N/m 2 + (997 kg/m 3 )(250cm)(1m/100cm)(1N/kg-m/s 2 ) = 2493 N/m 2
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PROBLEM 9.4 KNOWN: Deadweight tester provides a calibration pressure, p c Tester: W = 25.3 kg f = 2.58N (where g c = 9.8 kg-m/s 2 /kg f ) A p = 5.065 cm 2 W p = 5.38 kg f p amb = 770 mm Hg abs = 102.104 kPa abs z = 20 m (sea level datum) φ = 42 o FIND: p c PROPERTIES: γ air = 9.8 N/m 3 γ stainsteel = γ mass = 78.4 kN/m 3 SOLUTION Referring to the free-body diagram, Σ F y = 0 = p i A p + F B – W p - W - p amb A p W p amb A p amb (A – A p ) W p p i A p F B where F B is the buoyancy force. Neglecting F B , the indicated pressure is p i = 122,313 N/m 2 abs Now, the actual pressure provided by the deadweight tester is estimated by p c = p i (1 + e 1 + e 2 )
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where e 1 provides a correction for altitude effects and e 2 provides the correction for the neglected buoyancy effects. From (9.6a), e 1 = - 0.0003 and from (9.9), e 2 = - γ air / γ masses = -(9.8 N/m 3 /78,000N/m 3 ) = - 0.00012 Then, the actual deadweight pressure is estimated by p c = 122,313 N/m 2 (1 - 0.0003 - 0.00012) = 122,262 N/m 2 abs = 122.262 kPa abs COMMENT By including the correction factors e 1 and e 2 , we corrected for a known systematic error, thus reducing the uncertainty error sources of this measurement. The error corrections do not eliminate these systematic errors but do reduce them to the level of uncertainty in the corrections themselves.
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PROBLEM 9.5 KNOWN: Inclined tube manometer L = 5.6 cm H 2 O θ = 30 o FIND: H SOLUTION Referring to Figure 9.7, for an inclined manometer, sin HL ∆= Further, this deflection away from a null balance condition is referenced to the pressure on the open end of the tube. So the device measures a gauge (referenced to local atmosphere
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chap09 - PROBLEM 9.1 FIND: SOLUTION Absolute pressure...

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