Fluid Statics and Manometers

Solutions to Exercise One – Fluid Statics and Manometers 1. The drain plug in a bathtub is designed to seal properly when there is a 0.05 psi pressure applied to it. How deep must the water be in the tub for the plug to seal? (Problem 2.8 in text.) This is a simple application of the basic equation of fluid statics: p1 + γz1 = p2 + γz2 or p2 = p1 + γ(z2 – z1) = p1 + γh. In this case we want to find the depth, h = (p2 – p1)/γ. p1 = 0 psi(gage) at the free surface of the liquid and we are given that p2 = 0.05 psi(gage). Using γ = 62.4 lbf/ft3 for water from Table 1.5 on the inside front cover gives the following result.  0.05 lb f  144 in 2   0 − 2 ft 2 p 2 − p1  in  = = 0.115 ft = 1.38 in h= 62.4 lb f γ ft 3 2. The basic elements of a hydraulic press are shown in Figure P2.12 copied from the text at the right. The plunger has an area of 1 in2, and a force, F1, can be applied to the plunger through a lever mechanism (not shown) having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in2, what load, F2 can be raised by a force of 30 lbf applied to the lever? Neglect the variation of pressure with depth. (Problem 2.12 in text.) We first have to determine F1 from the information given about the applied force of 30 lbf and the mechanical advantage of 8 for the lever that is used to apply the force F1. From the definition of mechanical advantage as the ratio of the resultant force to the applied force we conclude that F1 = 8(30 lbf) = 240 lbf. We are told to neglect the variation of pressure with depth. This means that the pressure is constant throughout the hydraulic fluid. This pressure is found by dividing the value of F1 = 240 lbf by the 1 in2 area of the piston through which F1 is applied. This gives a pressure of 240 lbf/in2. Since the same pressure is exerted on the 150 in2 area of the piston that generates the force F2, we conclude that F2 = PA2 = (240 lbf/in2)(150 in2) so that F2 = 36,000 lbf . Page 2 3. A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system as shown in figure P2.25, copied from the text at the right. The liquid in the top part of the piping system has a specific gravity of 0.8, and the remaining parts of the system are filled with water. If the pressure gage reading at A is 50 kPa determine (a) the pressure in pipe B, and (b) the pressure head in torr (millimeters of mercury) at the top of the dome (point C). (Problem 2.25 from text.) pA (a) From the diagram we see that the fluid with a specific gravity of 0.8 fills the piping system on both sides at a point 4 m from the top branch. Thus the pressure on both sides of the piping system is the same at this point. We also see that this the level of the water on the left side of the piping system is the same as the level of the pressure gage in the tank which reads 50 kPa(gage). From these two observations we conclude that the manometer fluid at a distance 4 m from the top of the piping system has a pressure of 50 kPa(gage) on both sides of the system, which is pA. On the right side of the piping system, the difference in pressure between this level where the pressure = pA and the level of the pipe B, is given by 3 m of the fluid with the 0.8 specific gravity and 2 m of water. Thus we can write our basic equation for the variation of pressure with depth as follows: pA + γtop(3 m) + γwater(2 m) = pB. The specific weight of the top fluid is found by multiplying its specific gravity of 0.8 times the specific weight of water γwater= 9.80 kN/m3 from Table 1. on the inside front cover. This gives γtop = 7.84 kN/m3. Substituting the values for the specific weight of the top fluid and water and the value of pA = 50 kPa(gage) into the equation for pb gives the following result.  7.84 kN (3 m ) + 9.80 3kN (3 m ) kPa ⋅ m p B = p A + γ top (3 m ) + γ water (2 m ) = 50 kPa +  3 m  m  1 kN 2 PB = 93.1 kPa(gage) Because pB is measured relative to pA which is a gage pressure, pB is also a gage pressure. (b) We apply our usual relation between the levels C and A: pA = pC + γwater(3 m) Notice that the curvature of the tank has on effect on this relationship. We can solve thie equation for pC, and substitute the values of 50 kPa for pA and the specific weight of water used previously to give the desired result. pC = p A − γ water (3 m ) = 50 kPa − 9.80 kN kPa ⋅ m 2 ( ) 3 = 20.6 kPa m 1 kN m3 Page 3 We are asked to express this as a pressure head in mmHg. To do this we simply divide the pressure by the specific weight of mercury which is found from table 1.5 on the inside front cover to be 133 kN/m3. Thus the pressure head is h= pC γ Hg = 20.6 kPa = 0.155 m 133 kN kPa ⋅ m 2 1 kN m3 4. The cylindrical tank with hemispherical ends shown in figure P2.35, copied from the text at the right, contains a volatile liquid and its vapor. The liquid density is 800 kg/m3, and its vapor density is negligible. The vapor pressure is 120 kPa(abs) and the atmospheric pressure is 101 kPa(abs). Determine (a) the gage pressure reading on the pressure gage and (b) the height, h, of the mercury manometer. (Problem 2.35 from text.) The pressure at the gage will be the vapor pressure at the liquid vapor interface plus the γh term for the h = 1 m distance from the fluid-favor interface to the gage: pgage = pvapor = γliquidh. Since we are given the density of the liquid is 800 kg/m3 we have to compute the specific weight as the ρg product. We can assume a standard gravity giving γlquid= (800 kg/m3)(9.81 m/s2) = 7848 kg▪m/ m3▪s2 = 7848 N/m3. Thus the absolute pressure at the gage can be found as follows. p gage = p vapor + γ liquid (1 m ) = 120 kPa(abs ) + 7848 N kPa ⋅ m 2 ( ) = 127.848 kPa(abs ) m 1 1000 N m3 To get the gage pressure reading at the gage we have to subtract the atmospheric pressure of 101 kPa from this absolute pressure giving the result that pgage = 26.8 kPa(gage) . To find the height, h, of the mercury manometer, we note that the mercury-liquid interface on the left is at the same level as the pressure gage. So the mercury on both sides of the manometer at the level of the pressure gage has the same pressure as the gage, pgage = 26.8 kPa(gage). Since the mercury manometer is open to the atmosphere on the right we have the following equation for the height of the mercury column: pgage = patm + γHgh, where patm = 0 as a gage pressure. This gives the result that h = pgage/γHg = (26.8 kPa)(1 kN▪m2/kPa) / (133 kN/m3), where the specific weight of mercury is taken from Table 1.6 on the inside front cover. This gives the result that h = 0.202 m . Page 4 5. Pike’s Peak near Denver, Colorado has an elevation of 14,110 ft. (a) Determine the pressure at this elevation based on equation 2.12. (b) If the air is assumed to have a constant specific weight of 0.07647 lbf/ft3, what would the pressure be at this altitude? (c) If the air is assumed to have a constant temperature of 59oF, what would the pressure be at this altitude? For all three cases assume standard conditions at sea level as given in Table 2.1. (Problem 2.20 from text.) (a) This problem examines equations not covered in the lecture. Equation 2.12 is shown below followed by an explanation of the terms in this equation. g  βz  Rβ p ( z ) = p a 1 +   Ta  The equation accounts for the variation of pressure with distance for a gas, whose gas constant is R, in an atmosphere where the temperature changes linearly with elevation. From Table 2.1 on page 47 we find pa = 2116.2 lbf/ft2 and Ta = 518.67 R as the sea level pressure and temperature for the standard atmosphere. Both of these must be absolute values in this application because we are using the equation of state for an ideal gas. (R is the gas constant for air which is 1716 ft▪lbf/slug▪R. (See page 12 of the text for this value.) The temperature profile follows the equation T = Ta – βz, where z is the elevation. The value of β = 0.00650 K/m = 0.00357 R/ft for the standard atmosphere is given between equations 2.11 and 2.12 on page 48 of the text. Since this value is positive, the temperature will decrease with altitude. In deriving this equation the value of g is assumed constant and we will use the standard value of g = 32.174 ft/s2. With all the given data we can evaluate the dimensionless exponent g/Rβ in the pressure equation. 2 32.174 ft 1 lb f ⋅ s g slug ⋅ ft s2 = = 5.252 Rβ 1716 ft ⋅ lb f 0.00357 R slug ⋅ R ft We can now apply equation 2.12 to find the elevation at 14,110 ft.  βz  p ( z ) = p a 1 +   Ta  g Rβ 0.00357 R   ( 14110 ft )   2116.2 lb f  ft  1+ = 2   518.67 R ft     5.252 = 1237 psf(absolute) (b) If we assume a constant specific weight of 0.07647 lbf/ft3, we have our usual simple formula for constant specific weight. p = p a − γh = 2116.2 lb f ft 2 − 0.7647 lb f ft 3 (14,110 ft ) = 1037 psf(absolute) Page 5 Here the predicted pressure is less than in the previous case because the constant specific weight assumption keeps the value of γ too high when it is actually decreasing as accounted for in the previous calculations in case (a). (c) The case of an isothermal atmosphere is given by equation 2.10: dimensionless exponent in this equation is evaluated as follows: p( z ) = p a e − gh RTa . The 1 lb f ⋅ s 2 32.174 ft ( ) ft 14110 gh slug ⋅ ft s2 = = 0.51006 ft ⋅ lb 1716 RTa f (518.67 R ) slug ⋅ R We can now apply equation 2.10 to get the pressure as follows. p( z ) = p a e − gh RTa = 2116.2 lb f ft 2 e −0.51006 = 1271 psf(absolute) This answer is closer to the result of part (a) because it accounts for the effect of pressure on specific weight. From the temperature gradient of 0.00357 R/ft, we see that the temperature at 14,110 feet is 468.3 R. So the difference between this answer and part (a) is that we did not accounted for the decreasing temperature in the atmosphere in the calculation that we just did for part (c).