Fluid Mechanics Past Paper PDF

2500 l/s of water flows through a 50 cm diameter pipe. Calculate the mean velocity. If the diameter is reduced to 25 cm, what would be the mean velocity? 2500 l/s of water flows through a 50 cm diameter pipe. Calculate the mean velocity. If the diameter is reduced to 25 cm, what would be the mean v...

2500 l/s of water flows through a 50 cm diameter pipe. Calculate the mean velocity. If the diameter is reduced to 25 cm, what would be the mean velocity? 2500 l/s of water flows through a 50 cm diameter pipe. Calculate the mean velocity. If the diameter is reduced to 25 cm, what would be the mean velocity? Solution Q = Au 𝜋 u = Q/A = 2.50/( ∗ 0.50 2) = 12.75 m/s 4 In the second case: 𝜋 u = Q/A = 2.50/( ∗ 0.25 2) = 51 m/s 4  Conservation of Energy/Bernoulli’s Equation Energy can neither be created nor destroyed. It can only be transferred from one form to another. Conservation of Energy (applied to fluid) In fluid dynamics, total energy on a fluid is made up of three components. Consider fluid flow through a pipe. Element of fluid L Direction u of flow z Reference level 1. Potential energy: Relative to some reference level. PE = wz 2. Kinetic Energy: Due to velocity wu 2 KE = 2g 3. Pressure Energy: Also known as flow energy or flow work. wp WD =  Work necessary to move the fluid element against Pressure Proof of work done Consider that a fluid element moves a distance L Force on fluid = pA Element of fluid L p u Work done = Force  distance = pAL A z Volume V = AL Reference level w weight of fluid, w = V ; Hence V =  Therefore w Work done = pAL = pV = p  Total energy E possessed by the moving fluid element E = WD + PE + KE = 1 + 2 + 3 So 2 wp wu E= + wz + , Nm  2g Consider Conservation of Energy 2 wp1 wu1 E1 = + wz1 +  2g z2 2 z1 wp 2 wu 2 E2 = + wz 2 +  2g Reference level E1 = E2 If no energy is removed or added 2 2 wp1 wu1 wp 2 wu 2 E1 = + wz 1 + = E2 = + wz 2 +  2g  2g or 2 2 wp1 wu1 wp 2 wu 2 + wz 1 + = + wz 2 +  2g  2g Dividing by , w 2 2 2 p1 u1 p2 u2 pn un + z1 + = + z2 + =... = + zn + =H  2g  2g  2g Each term represents a form of Energy per unit weight or “head” p   Pressure head    z  Potential (Elevation ) head Sum is Total Head, H u2   Velocity head  2g  The Equation 2 2 2 p1 u1 p2 u2 pn un + z1 + = + z2 + =... = + zn + = H = const  2g  2g  2g 2 p u +z+ = H = const  2g Bernoulli’s Equation Restrictions in the application of the Bernoulli’s equation Flow is steady. Density is constant (which also means the fluid is incompressible). Friction losses are negligible. The equation relates the states at points along a single streamline, (not on two different streamlines).  Consider a horizontal uniform pipe 2 2 u1 u2 2g 2g p1 p2   H1 H2 z1 z2 Reference level  Consider a horizontal uniform pipe 2 2 u1 u2 2g 2g p1 p2   H1 H2 Reference z1 = 0 z2 = 0 level  Consider an inclined non-uniform pipe 2 u2 2 u1 2g 2g H2 H1 Exercise 14 Example Oil of specific gravity 0.75 flows through a 15 cm diameter pipe under pressure of 98.1 kN/m2. If the datum is 3 m below the centre of the pipe and the total head with to the datum is 20 m. Calculate the discharge. Example Oil of specific gravity 0.75 flows through a 15 cm diameter pipe under pressure of 98.1 kN/m2. If the datum is 3 m below the centre of the pipe and the total head with to the datum is 20 m. Calculate the discharge. Solution 𝑃 𝑉2 H= + + z  2𝑔 98.1 𝑣2 𝑣2 20 = + + 3 = 13.33 + +3 9.81 𝑥 0.75 2𝑔 2𝑔 V = 8.48 m/sec Discharge Q = VA = 8.50 X Τ4 𝑋 0.15 𝑋 0.15 = 0.15 𝑐𝑢𝑚𝑒𝑐𝑠 Exercise Consider the water flow in a converging duct without losses due to friction in the figure below. The parameters at sections 1 and 2 of the pipe are given. Determine the value of Z2. P2 = 15Pa 𝑢2 = 5 m/s 𝜌2 = 1000 kg/m3 17  Exercise In the pipe shown, 0.5 cm3/s of water flows from A to B. The diameter of the pipe at A and B are respectively 30 cm and 60 cm. If the pressure head at A is 7 m of water, find the pressure head at B. Neglect all losses B A 8m 3m Datum 18 Exercise Calculate Q and pressure ( p) at B, C, D, E Soln. Assumptions for solving 1. Exposed to air p  =0  2. Large Volume 2 u  0 2g Find knowns ▪ At Point A p A is Atmospheri c pressure 2 uA ▪ Area is very large so assume 0 2g  ▪ At point F the pressure pF is atmospheric From diagram, F is 1.2+1.8 = 3.0 m below point A Apply Bernoulli’s equation to points A and F 2 2 pA uA pF uF + zA + = + zF +  2g  2g 2 uF z A = zF + 2g  u F = ( z A − z F )2 g z A − z F = 3m vF = 3 * 2 * 9.81m/s = 7.67m/s 2 D 2  (25mm) 2  1m 2  Q = Au = uF =    6 (7.67 m/s ) 2  4  4  10 mm  = 3.77 10 −3 m 3 /s Pressures at B, C , D, E Write Bernoulli’s equation for points A and B 2 2 pA uA pB uB + zA + = + zB +  2g  2g  pA 2 =0 uA  =0 2g pB uB 2  uB  2 zA = + zB + ; PB =  ( z A − z B ) −   2g  2g  z A = zB Q 3.77 10 −3 m 3 /s Q = AB u B uB = = −3 = 3m/s A 1.26 10 m 2  3 2 m 2 2 /s  pB = 9.81kN/m 0 − 3 2  = − 4.5kN/m 2  ( 2)(9.81m/s )  Note: ▪ Unlike fluid statics, Pressures at the same elevation may NOT be the same when fluid is moving. ▪ It is possible to have a negative pressure lower that the vapor pressure which leads to gas Bubbles formation (Cavitations) Exercise Determine the velocity of efflux from the nozzle in the wall of the reservoir as shown below. Also find the discharge through the nozzle. The water level in the reservoir is 4 m above the centre line of the nozzle.  = 100mm Nozzle Exercise U = 3 m/s 2m 1m U = 10 m/s Water is flowing in a channel as shown at a depth of 2 m and velocity of 3 m/s. It then flows down a chute into another channel where the depth is 1 m and the velocity is 10 m/s. Assuming frictionless flow, determine the difference in elevation of the channel beds. The velocities are assumed to be uniform over the cross sections. Exercise A jet of water vertically issuing from a 25 mm diameter nozzle is directed upward. Calculate the diameter of the jet at a point 5 m above the nozzle if the velocity with which the jet leaves the nozzle is 15 m/s. Apply Bernoulli’s equation. Exercise A pipe of varying section has the cross-sectional area of 25 cm2, 50 cm2 and 12 cm2 at points A, B and C respectively (Refer to diagram). Find the discharge and the absolute pressure at B if the atmospheric pressure is 10,.3 m of water. Depth of water in the reservoir is 10 m. Neglect losses.  D 10 m A B 15 m 10 m C 2m  p1 u12 p2 u22 + + z1 = + + z2 = H 1 g 2 g g 2 g At the reservoir, p1=0 U1