plik

��Chapter 5 Control Volume Approach and Continuity Principle Problem 5.1 A 10-cm-diameter pipe contains sea water that ows with a mean velocity of 5 m/s. Find the volume ow rate (discharge) and the mass ow rate. Solution The discharge is = where is the mean velocity. Thus = 5 � � 0 12 4 = 0 0393 m3/s From Table A.4, the density of sea water is 1026 kg/m3 The mass ow rate is = = 1026 � 0 0393 = 40 3 kg/s 29 30CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE Problem 5.2 The velocity pro le of a non-Newtonian uid in a circular conduit is given by 1 2 � �2� = 1 max where max is the velocity at the centerline and is the radius of the conduit. Find the discharge (volume ow rate) in terms of max and Solution The volume ow rate is Z = For an axisymmetric duct, this integral can be written as Z = 2 0 Substituting in the equation for the velocity distribution Z 1 2 � �2� = 2 max 1 0 Recognizing that 2 = 2 we can rewrite the integral as Z 1 2 � �2� = max 1 2 0 Z 1 2 � �2� � �2 = max 2 1 0 or Z 1 = max 2 [1 ]1 2 0 2 = max 2 [1 ]3 2 |1 0 3 2 = max 2 3 31 Problem 5.3 A jet pump injects water at 120 ft/s through a 2-in. pipe into a secondary ow in an 8-in. pipe where the velocity is 10 ft/s. Downstream the ows become fully mixed with a uniform velocity pro le. What is the magnitude of the velocity where the ows are fully mixed? Solution Draw a control volume as shown in the sketch below. Because the ow is steady X V � A = 0 Assuming the water is incompressible, the continuity equation becomes X V � A = 0 The volume ow rate across station is � �2 X 8 V � A = 10 � 4 12 where the minus sign occurs because the velocity and area vectors have the opposite sense. The volume ow rate across station b is � �2 X 2 V � A = 120 � 4 12 32CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE and the volume ow rate across station c is � �2 X 8 V � A = � 4 12 where is the velocity. Substituting into the continuity equation � �2 � �2 � �2 2 8 8 120 � 10 � + � = 0 4 12 4 12 4 12 � � 120 � 22 + 10 � 82 = 82 = 17 5 ft/s Problem 5.4 Water ows into a cylindrical tank at the rate of 1 m3 min and out at the rate of 1.2 m3 min. The cross-sectional area of the tank is 2 m2 Find the rate at which the water level in the tank changes. The tank is open to the atmosphere. Solution Draw a control volume around the uid in the tank. Assume the control surface moves with the free surface of the water. 33 The continuity equation is Z X + V � A = 0 The density inside the control volume is constant so Z X + V � A = 0 X + V � A = 0 The volume of the uid in the tank is = Mass crosses the control surface at two locations. At the inlet V � A = and at the outlet V � A = Substituting into the continuity equation + = 0 or = 1 1 2 = 2 = 0 1 m/min 34CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE Problem 5.5 Water ows steadily through a nozzle. The nozzle diameter at the inlet is 2 in., and the diameter at the exit is 1.5 in. The average velocity at the inlet is 5 ft/s. What is the average velocity at the exit? Solution Because the ow is steady, the continuity equation is X V � A = 0 Also, because the uid is incompressible, the continuity equation reduces to X V � A = 0 Draw a control surface that includes the inlet and outlet sections of the nozzle as shown. At the inlet, station 1, � �2 2 (V � A)1 = 5 � � 4 12 At the exit, station 2, � �2 1 5 (V � A)2 = 2 � � 4 12 35 Substituting into the continuity equation � �2 � �2 X 2 1 5 V � A = 5 � � + 2 � � = 0 4 12 4 12 or � �2 2 0 2 = 5 � = 8 89 ft/s 1 5 Problem 5.6 Air ows steadily through a 10 cm-diameter conduit. The velocity, pressure, and temperature of the air at station 1 are 30 m/s, 100 kPa absolute, and 300 K. At station 2, the pressure has decreased to 95 kPa absolute, and the temperature re- mains constant between the two stations (isothermal ow). Find the mass ow rate and the velocity at station 2. Solution The mass ow rate is = The density is obtained from the equation of state for an ideal gas. = At station 1 100 � 103 N/m2 1 = = 1 16 kg/m3 287 J/kgK � 300 K The ow rate is = 1 16 � 30 � � 0 12 = 0 273 kg/s 4 Because the ow is steady, the continuity equation reduces to X V � A = 0 36CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE which states that the rate of mass ow through station 1 will be equal to that through station 2. The air density at station 2 is 95 � 103 N/m2 2 = = 1 10 kg/m3 287 J/kgK � 300 K The mass ow is the same at each station. Thus ( )1 = ( )2 So 1 1 16 2 = 1 = 30 � = 31 6 m/s 2 1 10 Problem 5.7 Water ows steadily through a 4-cm diameter pipe that is 10-m long. The pipe wall is porous, leading to a small ow through the pipe wall. The inlet velocity is 10 m/s, and the exit velocity is 9 m/s. Find the average velocity of the water that is passing through the porous surface. Solution The ow rate is steady, and the uid is incompressible so the continuity equation reduces to X V � A = 0 Draw a control surface around the pipe. The entrance is station 1, the exit is station 2, and the surface of the porous pipe is station 3. For station 1 (V � A)1 = 10 � � 0 042 4 37 For station 2 (V � A)2 = 9 � � 0 042 4 For the porous surface (V � A)3 = 3 � � 0 04 � 10 The continuity equation is 10 � � 0 042 + 9 � � 0 042 + 3 � � 0 04 � 10 = 0 4 4 or 3 = 0 001 m/s Problem 5.8 Water is forced out of a 2-cm diameter nozzle by a 6-cm-diameter piston mov- ing at a speed of 5 m/s. Determine the force required to move the piston and the speed of the uid jet ( 2). Neglect friction on the piston and assume irrotational ow. The exit pressure ( 2) is atmospheric. Solution When ow is irrotational, the Bernoulli equation applies. Applying this equation along the nozzle centerline between locations 1 and 2 gives 2 1 12 2 2 + = + (1) 2 2 The continuity principle is 1 1 = 2 2 So 0 062 2 = (5 m/s) 0 022 = 45 m/s (2) Letting 2 = 0 kPa gage and combining Eqs. (1) and (2) 2 22 1 1 = 2 452 52 = (1000) 2 = 1 MPa 38CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE Since the piston is moving at a constant speed, the applied force is balanced by the pressure force. = 1 1 � � = (1 MPa) � 0 062 4 m2 = 2 83 kN Problem 5.9 The sketch shows a fertilizer sprayer that uses a Venturi nozzle. Water moving through this nozzle reaches a low pressure at section 1. This low pressure draws liquid fertilizer (assume fertilizer has the properties of water) up the suction tube, and the mixture is jetted to ambient at section 2. Nozzle dimensions are 1 = 3 mm, 2 = 9 mm, and = 150 mm. Determine the minimum possible water speed ( 2) at the exit of the nozzle so that uid will be drawn up the suction tube. Solution Since we are looking for the lower limit of operation, assume inviscid ow so that the Bernoulli equation applies. Also assume that the pressure at 1 is just low enough to draw uid up the suction tube, meaning there is no ow in the suction tube. Identify locations 1 to 3 as shown by the points in the sketch below. Applying the hydrostatic principle (constant piezometric pressure in a uid of con- stant density) between 1 and 3 gives 3 = 1 + ( 1 3) 39 Let 3 = 0 kPa gage, and let ( 1 3) = ( + 1 2). 1 = ( + 1 2) (1) Applying the Bernoulli equation between 1 and 2 gives 2 1 12 2 2 + = + (2) 2 2 The continuity principle is 1 1 = 2 2 (3) Let 2 = 0 kPa gage, and combine Eqs. (2) and (3). � � 22 4 2 1 = 1 (4) 2 4 1 Combine Eqs. (1) and (4). � � 2 2 4 2 ( + 1 2) = 1 2 4 1 � � �4! 2 1000 � 2 9 9800(0 15 + 0 003 2) = 1 2 3 So 2 = 0.193 m/s Problem 5.10 Show that the velocity eld � � 2 2 3 V = 2i j + k 2 2 3 satis es the continuity equation for an incompressible ow and nd the vorticity at the point (1,1,1). Solution The continuity equation for the ow of an incompressible uid is � V = + + = 0 Substituting in the velocity derivatives 2 + 2 0 so continuity equation is satis ed. 40CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE The equation for vorticity is = � V � � � � � � = i + j + k Substituting in the velocity derivatives � � � � � 2 3 2 = i + + j (2 0) + k 0 2 2 3 2 Substituting values at point (1,1,1) 2 = i+2j k 3

Wyszukiwarka

Podobne podstrony:
9 Practice Problems
11 Practice Problems
1 Practice Problems
12 Practice Problems
3 Practice Problems
3 Practice Problems
7 Practice Problems
14 Practice Problems
15 Practice Problems
4 Practice Problems
13 Practice Problems
8 Practice Problems
6 Practice Problems
10 Practice Problems
2 Practice Problems
iesol b1 achiever practice paper 2
Zespoły posturalne problem cywilizacyjny(1)
A Balaban Polskie problemy ustrojowe 2003

więcej podobnych podstron