Bourdon Gages and Pressure Calculation

February 5 Homework Solutions 2.5 Bourdon gages (see Video V2.2 and Fig. 2.13) are commonly used to measure pressure. When such a gage is attached to the closed water tank of Fig. P2.5 (copied from the text at the right), the gage reads 5 psi. What is the absolute pressure in the tank? Assume standard atmospheric pressure of 14.7 psia. The absolute pressure at the gage is is 5 psi + 14.7 psi = 19.7 psia. The fact that the gage is 6 in above the measuring line is irrelevant. The only difference in depth that contributes the pressure at the gage is the 12 in difference between the gage level and the air. Thus we can write that pair = 19.7 psia + γwater(12 in). Using γwater = 62.4 lbf/ft3 from Table 1-5 on the inside front cover gives the following result for the air pressure. p air = 19.7 lb f in 2 + 62.4 lb f ft 3 (12 in ) ft 3 1728 in 3 pair = 20.1 psia 2.11 In a certain liquid at rest measurements of the specific weight at various depths show the following variation: h(ft) 0 10 20 30 40 5 60 70 80 90 100 lb f 70 76 84 91 97 102 107 110 112 114 115 γ ft 3 The depth, h = 0, corresponds to a free surface at atmospheric pressure. Determine, through numerical integration of Eq. 2.4, the corresponding variation in pressure and show the results on a plot of pressure (in psf) versus depth (in feet). z We have to numerically integrate dp/dz = -γ or p = p0 – ∫ γdz , where p 0 = 0 (given). We can use z0 b a simple trapezoid rule ∫ a f ( x)dx ≈ N −1  ∆x  ( ) + 2 f a f (a + k∆x ) + f (b ) where N = (b – a)/∆x.  2  k =1  ∑ February 5 homework solutions Page 2 In this problem the variable of integration is z and the step size, ∆z = –10 ft. The values of f(z) are the specific weight taken from the table. Some example integrations give the following results: p10 ft = 0− lb f  lb f − 10 ft  lb f 70 3 + 76 3  = 730 2 2  ft ft  ft p 20 ft = 0−  lb f − 10 ft  lb f 70 3 + 2 76 3 2  ft ft  p30 ft = 0− lb f  lb f − 10 ft  lb f 70 3 + 2 76 3 + 84 3 2  ft ft ft  lb  lb   + 84 f  = 1540 f  ft 3  ft 2  lb  lb   + 91 f  = 2405 f  ft 3  ft 2  The complete results and graph, done in Excel, are shown below. h(ft) 0 10 20 30 40 50 60 70 80 90 100 γ (lbf/ft3) 70 76 84 91 97 102 107 110 112 114 115 p (psf) 0 730 1530 2405 3345 4340 5385 6470 7580 8710 9855 February 5 homework solutions 2.26 Page 3 A U-tube manometer contains oil, mercury, and water as shown in Fig. P2.26 (copied at the right). For the column heights indicated what is the pressure differential between pipes A and B? We see that there is a line of equal pressure near the bottom of the U-tube at the lower arrow for the dimension of the 12 in. Designating the two equal pressures on either side of the manometer at this point as pleft and pright allows us to write the following equations. pleft = p A + γ oil (7 in ) + γ Hg (12 in ) = p right = p B + γ H 2O (3 in + 12 in ) Take the data for specific weights from table 1-5 on the inside front cover (assuming the oil is SAE 30 weight oil and ignoring differences in temperature) are γwater = 62.4 lbf/ft3, γHg = 847 lbf/ft3, and γoil = 57.0 lbf/ft3. Rearranging the equation to solve for pB – pA and substituting the data gives. p B − p A = γ oil (7 in ) + γ Hg (12 in ) − γ H 2O (3 in + 12 in ) 847 lb f 62.4 lb f  57.0 lb f  ft 3 psia ⋅ in 2 ( ) ( ) ( ) = + − 7 in 12 in 15 in  ft 3 ft 3  ft 3  1728 in 3 1 lb f pB – pA = 5.57 psi 2.40 The differential mercury manometer of Fig. P2.40 (copied at the right) is connected to pipe A containing gasoline (SG = 0.65), and to pipe B containing water. Determine the differential reading, h, corresponding to a pressure in A of 20 kPa and a vacuum of 150 torr [use torr instead of mm Hg] in B. Call the pressure at the gasoline-mercury interface p1 and the pressure at the watermercury interface p2. The difference between these two pressures is due to the column of mercury with height h. p2 p1 p1 = p 2 + γ Hg h We can relate p1 and p2 to the pressures in A and B as follows. p1 = p A + γ gasol (h + 0.3 m ) p B = p 2 + γ water (h + 0.3 m ) Combining the two equations above gives the following expression for pA – pB: p A − p B = p1 − p 2 − γ water (h + 0.3 m ) − γ gasol (h + 0.3 m ) February 5 homework solutions Page 4 From the first equation we have p1 – p2 = γHgh. Substituting this result into the equation above for pA – pB gives. p A − p B = γ Hg h − γ water (h + 0.3 m ) − γ gasol (h + 0.3 m ) = γ Hg − γ water − γ gasol h − (0.3 m ) γ water + γ gasol ( ( ) ) Solving this equation for h gives h= ( p A − p B + (0.3 m ) γ water + γ gasol ) γ Hg − γ water − γ gasol The data for specific weights of water and mercury are taken from table 1-6 on the inside front cover (ignoring differences in temperature): γwater = 9.80 kN/m3 and γHg =133 kN/m3. The gasoline has a specific gravity of 0.65 and, assuming that its reference specific weight is that of water just found, we find the gasoline specific weight as (SG)(γwater) = (0.65)(9.80 kN/m3) = 6.37 kN/m3. The pressure in A is a gage pressure. The pressure in B is given as a vacuum of 150 mm Hg. A vacuum is the difference between the atmospheric pressure and the absolute pressure. By definition, pvacuum = patmospheric – pabsolute and pgage = –pvacuum. Thus the 150 torr vacuum is a gage pressure of –(150 torr)(101.325 kPa)/(760 torr) = –20.00 kPa. Substituting this pressure for pB, the given value of pA = 20 kPa, and the specific weight data from the previous paragraph into the equation for h gives h= ( p A − p B + (0.3 m ) γ water + γ gasol γ Hg − γ water − γ gasol ) =  9.81 kN 6.37 kN  ( ) + + 0 . 3 m   kPa ⋅ m 2 m3   m3 133 kN 9.81 kN 6.37 kN − − m3 m3 m3 [20 kPa − (− 20 kPa )] kN h = 0.384 m February 5 homework solutions 2.42 Page 5 The manometer fluid in the manometer of Fig. P2.42 (copied at the right) has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer. There is a common pressure on both sides of the manometer at the point of the lower water-gage-fluid interface. Writing the two equal pressures at this point in terms of the pressures in A and B gives the following equations. pleft = p A + γ water hwater ,left + γ gage hgage = p right = p B + γ water hwater ,rigfht For the configuration shown above, hwater,left = 2 ft, hgage = 2 ft, and hwater,right = 1 ft. We can solve this equation for pB – pA to obtain the following result. p B − p A = γ water hwater ,left + γ gage hgage − γ water hwater ,right Use the specific weight of water as 62.4 lbf/ft3 from Table 1-5 on the inside front cover for both the value of γwater and the reference specific weight for the gage fluid. This gives the specific weight of the gage fluid as (3.46)(62.4 lbf/ft3) = 215.9 lbf/ft3. Substituting these values and the original gage readings into the equation for pB – pA gives pB − p A = 62.4 lb f ft 3 (2 ft ) + 215.9 lb f ft 3 (2 ft ) − 62.4 lb f ft 3 (1 ft ) = 494.2 lb f ft 2 The new value of pB – pA is found as follows. + ∆p B − ( p A,old + ∆p A ) = ( p B − p A )old + ∆p B − ∆p A 494.2 lb f 0.9 lb f 144 in 2 − 1.3 lb f 144 in 2 811.0 lb f = + − = ft 2 in 2 ft 2 in 2 ft 2 ft 2 ( p B − p A )new = p B,old With the increased value of pB, the water-gage-fluid interface on the left will become lower by a value of ∆h so that the new value of hwater,left will become 1 ft + ∆h. This change will be reflected throughout the manometer system. The water-gage-fluid interface on the right will rise by the same value ∆h so that hwater,righy becomes 2 ft – ∆h. Both of these differences will change the height of the gage fluid by 2∆h to a new value of hgage = 2 ft + 2∆h. Substituting these new values of the gage heights and the new value of pB – pA into the original equation for pB – pA gives ( p B − p A )new = 811.0 lb f ft 2 = 62.4 lb f ft 3 (2 ft − ∆h ) + 215.9 lb f ft 3 (2 ft + 2∆h ) − 62.4 lb f ft 3 (1 ft + ∆h ) The products of specific weights and heights on the right-side of the equation are seen to be simply the original values for pB – pA = 494.2 lbf/ft2. This gives the following steps for obtaining ∆h. February 5 homework solutions 811.0 lb f ft 2 − 494.2 lb f ft 2 = Page 6 316.8 lb f ft 2  215.9 lb f 62.4 lb f 62.4 lb f = ∆h 2 − −  ft 3 ft 3 ft 3  215.9 lb f   = ∆h  ft 3  Solving this equation gives ∆h = 1.03 ft. Adding two times this value to the original value of hgage = 2 ft gives the required answer. hgage,new = 4.06 ft 2.46 Determine the change in elevation of the mercury in the left leg of the manometer of Fig. P2.46 (copied at the right) as a result of an increase in pressure of 5 psi in pipe A while the pressure in pipe B remains constant. The mercury pressures are equal at the mercury-water interface and the point in the incline that is at the same level as this interface. These two equal pressures can be written as follows. (In this equation we use a variable height for the following distances in the diagram: hwater= 18 in, hoil = 12 in, and ℓHg = 6 in = the difference between the water-mercury and the oilmrecury interfaces.) pleft = p A + γ water hwater = p right = p B + γ oil hoil + γ Hg  Hg sin 30 o Solving for the pressure difference gives p A − p B = γ oil hoil + γ Hg  Hg sin 30 o − γ water hwater When the pressure in A increases to pA,new, the water-mercury interface will drop by a value ∆h and the mercury-oil interface will rise a distance, ∆ℓ along the inclined length. These changes will cause hwater to increase by ∆h and hoil to decrease by ∆ℓ sin30o. The vertical difference in the mercury column will increase by the sum of ∆h and ∆ℓ sin30o. Even with all these changes we will still have the same relationship between the pressure difference and manometer measurements. This will give the equation below. ( ) [ ( ] ) p A,new − p B = γ oil hoil − ∆ sin 30 o + γ Hg ∆h +  Hg + ∆ sin 30 o − γ water (hwater + ∆h ) If we subtract the equation for the original pressure difference from this equation we obtain the following result for the increase in pressure in A. ( ) p A,new − p A = − γ oil ∆ sin 30 o + γ Hg ∆h + ∆ sin 30 o − γ water ∆h We appear to have two unknowns, ∆h and ∆ℓ. However these two are related because the volume of the mercury is constant. When the mercury drops by a height change of ∆h, a volume of mercury equal to πrwater2 is displaced from the vertical tube to the inclined tube. The volume, V, of fluid in an cylinder inclined at an angle θ with the horizontal plane is given by the equation V = February 5 homework solutions Page 7 πr2(ℓ + r/tanθ), where ℓ is the distance from the bottom of the cylinder to the closest location of the inclined plane. 1 If the value of ℓ is changed by ∆ℓ, keeping r and θ constant, the difference in volume is simply πr2∆ℓ, just as it would be for a vertical cylinder. Equating the two displaced volumes gives the following relationship between ∆h and ∆ℓ. 2 πrwater ∆h = 2 πrHg ∆  rHg ⇒ ∆h = ∆  rwater 2 2   0.125 in  ∆  = ∆  =   4  0.25 in   Before substituting this result into our equation for the pressure change, we have to get the necessary data. Table 1-5 in the inside front cover gives the values for γwater = 62.4 lbf/ft3 and γHg = 847 lbf/ft3. The oil has a specific gravity of 0.9 so we find its specific weight using the value of γwater = 62.4 lbf/ft3 just found: γoil = 0.9(62.4 lbf/ft3)r = 56.16 lbf/ft3. We can now substitute the equation ∆h = ∆ℓ/4 into the pressure increase equation, set sin30o to its value of 0.5, and substitute the specific weights just obtained to get the solution for ∆h. ( ) ( ) p A,new − p A = − γ oil 4∆h sin 30 o + γ Hg ∆h + 4∆h sin 30 o − γ water = − 2 γ oil + 3γ Hg − γ water ∆h ∆h = p A,new − p A 2 γ Hg − 2 γ oil − γ water (5 psi ) = 144 lb f psi ⋅ ft 2 = 0.304 ft 847 lb f 56.16 lb f 62.4 lb f −2 − 3 ft 3 ft 3 ft 3 The mercury level drops 0.304 ft.