Fluid Energy Equation and Applications of Bernoulli's Equation PDF

Energy Equation Expanded Bernoulli's equation More general flow equation than Bernoulli’s Includes energy loss or gain Recall - Restrictions on the use of the Bernoulli’s Equation.  Hydraulic head concept For an incompressible fluid flowing in a pipe, (horizontal, vertical...

Energy Equation Expanded Bernoulli's equation More general flow equation than Bernoulli’s Includes energy loss or gain Recall - Restrictions on the use of the Bernoulli’s Equation.  Hydraulic head concept For an incompressible fluid flowing in a pipe, (horizontal, vertical, or inclined) with smooth walls, the steady state Bernoulli equation is p v2 +z+ = constant = h g 2g p p = = piezometric head g  v2 = velocity head 2g z = geometric elevation Darcy’s Law Consider two points 1 and 2 separated by a distance s. Water levels in piezometers inserted at these points are h1 and h2 respectively from a datum. In groundwater flow, due to resistance against the flow within pores, Bernoulli equation takes the form, 2 2 p1 v1 p2 v2 + z1 + = + z2 + + h  2g  2g h - the head loss over the distance s  Motor hL hR 1 2 2 hA Valve Pump 1 hA = Energy added by mechanical device (Pump) hR = Energy removed (Motor/Turbine) hL = Energy losses (Friction + Minor losses)  Motor hL hR 1 2 2 hA Pump Valve 1 Total Energy per Unit Weight (from pts 1 and 2) h + hA − hR − hL = h ' 1 ' 2 But 2 P1 u h = + z1 + ' 1  1 2g 2 P2 u h = + z2 + ' 2  2 2g So 2 2 P1 u P2 u + z1 + 1 + hA − hR − hL = + z2 + 2  2g  2g  Energy equation is written in the direction of flow. Algebraic signs are very critical From above: Pump adds energy ( + hA ) Frictional losses remove energy ( − hL ) Motor extracts energy ( − hR ) Example 1 Large reservoir 4.5 m 10.0 m 2 10 cm Find hL Q = 0.03m3/s From point 1 →point 2 P1 u12 P2 u22 + z1 + + hA − hR − hL = + z2 +  2g  2g P1 and P2 Both Atmospheric pressure. Exposed to the atmosphere 2 u 01 Reservoir large. So valid for very large 2g surface area. hA = hR = 0 No mechanical devices Hence 2 2 u u z1 − hL = z2 + 2 or hL = z1 − z2 − 2 2g 2g  2 0.03m 3 /s  (Q A) 2    (0.1) / 4 2 hL = 14.5 − = 14.5m − 2g 2(9.81m/s 2 ) = 13.76 m These losses are frictional (pipe friction) and minor (valves, elbows) Geol Pump power Requirements Energy Transfer Pump power =  Weight flow rate Weight of fluid = Energy transfer rate PA = hAW = hAQ W = Q = Weight flow rate Mechanical Efficiency PA Power deliverd to fluid Em  = Em  1 PI Power put into pump Em = 0.7 − 0.9 Typical hydraulic pumps Fluid Motors (Turbines) PR  hRW = hRQ Po Power output of motor Em  =  1.0 PR Power delivered by fluid Example Given the following values B 296 kPa 0.053 m 20.0 m Check -28 kPa Valve Pump A 0.078 m  Determine the power of the pump, if the pump efficiency is 0.85. Solution Power delivered by pump to fluid: PA = hAW = hAQ 2 2 PA u PB u + zA + + hA − hL = A + zB + B  2g  2g  PB − PA 2 u −u 2 hA = + ( zB − z A ) + B A + hL  2g PA − PB 296 − (−28)kN / m 2 = = 38.3m  3 (0.86)(9.81kN / m ) Z B − Z A = 20.0 m u −u 2 2 (Q / AB ) − (Q / AA ) 2 B = A = 2g 2g hA = 38.4 m + 20.0 m + 1.7 m + 1.9 m = 62.0 m PA = hAW = hAQ PA Power deliverd to fluid Em  = PI Power put into pump If Em = 0.85, Calculate Power put into the pump PA PA PI = = Em 0.85 EXERCISE In a pipe conveying water, pressure gauges are inserted at points 1 and 2, where diameters are respectively 15 cm and 7.5 cm. When a discharge of 0.02 m3/s occurs and the flow is downward, the pressure at point 2 is 1.96 kN/m2greater 2than that at point 1. If losses are taken as h = K u1 , 2g where u1 is the velocity at point 1, determine the value of k. 2 2 5m 1 1 Venturi-Meter About 6 About 20 2 1 z2 z1 h Datum Civil  Venturi-meter - Device for measuring discharge. It consist of: Rapidly converging section, Throat and Diverging section (diffuser). ❖ The rapidly converging section increases velocity of flow and hence reduces the pressure. ❖ The gently diverging section (diffuser) returns the Venturi-meter to the original dimensions of pipe. Knowing the pressure difference between two points, the Discharge can be calculated. Energy loss in the Venturi-meter is assumed to be very small. Apply Bernoulli's equation along the streamline from point 1 to 2 About 6 About 20 2 1 z2 z1 h Datum 2 2 p1 u p2 u + + z1 = 1 + + z2 2 g 2 g g 2 g Using continuity equation, we can find velocity u2 Q = u1 A1 = u2 A2 u1 A1 u2 = A2  Substituting this into the B.E. and rearranging p1 − p2 u 2 A1   2 + z1 − z 2 =   − 1 1 g 2 g  A2      p1 − p2  2 g  + z1 − z 2  u1 =  g  2  A1    − 1  A2  Theoretical discharge is calculated as: Qideal = u1 A1 Actual discharge, taking into account losses due to friction: Qactual = Cd Qideal = Cd u1 A1  p1 − p2  2 g + z1 − z 2  Qactual = C d A1 A2  g  A1 − A2 2 2 Express discharge in terms of the manometer readings About 6 About 20 2 1 z2 z1 h Datum  Manometer Equation p1 + gz1 = p2 +  man gh + g ( z 2 − h) Simplify p1 − p2   man  + z1 − z 2 = h − 1 g    Hence actual discharge   man  2 gh − 1 Qactual = C d A1 A2    A1 − A2 2 2 Expression for discharge does not include any terms for elevation or orientation (z1 or z2) of the Venturi-meter. Hence the Venturi-meter can be at any convenient angle to function. The smoother the contraction, less the head loss C d → 1.0 Typically C d  0.96 − 0.98 The purpose of the diffuser in a Venturi-meter: Gradual and steady deceleration after the throat. It ensures that, pressure rises again to a value near the original value before the Venturi- meter. The angle of the diffuser is usually between 6 and 8 degrees. If the angle is wider than this limit, the flow might separate from the walls resulting in increased friction, energy and pressure loss.  If the angle is less than this limit, the meter becomes very long. Hence pressure losses again become significant. The efficiency of the diffuser, is rarely greater than 80%. Example Given the following : u B is f (p),  = 9.81kN / m 3 Find u A and Q B 100mm  0.60 m A Unknown “y” 400mm  1.50 m Specific gravity = 1.29 Known parameters z A − zB , D A , DB Consider Point A and B 2 2 pA uA pB uB + zA + = + zB +  2g  2g p A − pB u B − u A2 2 + ( z A − zB ) =  2g But z A − zB = 0.60m " y" is unknown Manometer equation p A +  ( y) +  (1.50) −  g (1.50) −  ( y) −  (0.60) = pB p A − pB =  (0.60 − 1.50) +  g (1.50) 𝛾𝑔 = (1.29)(𝛾𝑤 = 9.81𝑘𝑁/𝑚3)=12.65 𝑘𝑁/𝑚3 Now we need p A − pB , so divide by   𝑝𝐴 − 𝑝𝐵 𝛾𝑔 = −0.90 + (1.50) = 1.04 𝛾 𝛾 Using continuity Q=uA AA AAu A = AB u B uB = u A AB −3 125.6 10 uB = −3 u A = 15.9u A 7.9 10 u = 252.8u 2 B 2 A u − u = 252.8u − u = 251.8u 2 B 2 A 2 A 2 A 2 A  p A − pB u B − u A2 2 + ( z A − zB ) =  2g 251.8 1.04 − 0.60 = 𝑢𝐴2 (2)(9.81) (2)(9.81)(0.44) 𝑢𝐴 = = 0.185 m/s 251.8 𝑄 = 𝐴𝐴 𝑢𝐴 = 123.6 × 10−3 × 0.185 = 0.023m3 /s Two sides of a cylinder fitted with a piston are connected to the inlet and the throat of a horizontal venturi meter. If the diameters at the inlet and the throat are 10 cm and 6 cm respectively and the velocity the velocity of water in the pipe is 5 m/s, calculate the force on the piston. The diameter of the piston is 2 cm. Ignore losses and the area of the piston rod. 1 2 C Flow through a Small Orifice 1 Aactual orifice h 2 Vena-contracta Datum Consider flow from a tank through a hole (orifice) in the side close to the base. General arrangement of streamlines as above  Shape of holes edges is sharp – Hence Contact between hole and the fluid minimized. This means frictional losses decreased. Streamlines contract after the orifice to a minimum value when they become parallel. At this point, the velocity and pressure become uniform across the jet. Convergence is called vena-contracta. (‘Latin 'contracted vein').  To calculate the flow, it is important to know the amount of contraction Determine the velocity at the orifice using Bernoulli's equation. Apply Bernoulli's equation to the streamline joining: ❖ Point 1 at the surface of the reservoir to point 2 at the centre of the orifice. 2 2 p1 u1 p2 u2 + z1 + = + z2 +  2g  2g  Surface velocity head (u12/2g  0) is negligible and the pressure head (p1/ = 0) is zero. At the orifice, jet is open to the atmosphere so pressure head (p2/  = 0) zero. Datum line taken through the centre of the orifice, hence z1 = h and z2 =0 2 u2 0+h+0 = 0+0+ 2g u2 = 2 gh Theoretical velocity Discharge through the orifice Continuity Equation: Q = Au Hence: Qactual = Aactualu actual  Actual velocity u actual = Cv ut Cv - Coefficient of velocity (0.97 - 0.99) The Actual area of flow (jet) is area of the vena- contracta not the area of the orifice. Area at vena-contracta obtained by using the coefficient of contraction for the orifice Aactual = Cc Aorifice Cc - Coefficient of contraction  Discharge through the orifice Q = Au Qactual = Aactualuactual = (Cc Aorifice )(Cvut ) = CcCv Aorificeut = Cd Aorificeut = C d Aorifice 2 gh Cd - Coefficient of discharge Cd = CcCv Determination of Coefficient of Velocity Three methods used: Trajectory method (By measurement of coordinates) Momentum method Pitot-tube Method By measurement of coordinates (Trajectory method) H venacontracta x x y y Measure horizontal and vertical coordinates of the jet as it falls under gravity. Jet contracts and forms the vena-contracta at a distance of d/2 from plane of the orifice. Let u , be the velocity at the vena-contracta x and y , be the coordinates of a point on the jet after time t. Applying the equation of motion x = ut 1 2 y = 0 + gt 2  Eliminating t 2 2 yu x= g But u = Cv 2 gH So 2 2 yCv ( 2 gH ) x= = Cv 4 yH g 2 Knowing x and y, and the head x Cv = H, Coefficient of velocity is 4 yH obtained  A circular orifice, 2.0 cm diameter is made in the vertical wall of a tank. The jet fall vertically through 0.75 m while moving horizontally through a distance of 3.0 m. Calculate the coefficient of velocity if the head causing flow is 4.0 m. If the discharge is 1.80  10-3 m3/s, calculate C c and C d 2 2 x 3.0 Cv = = = 0.87 4 yH 4  0.75  4.0 Q 1.80  10−3 Cd = = = 0.64 A 2 gH   0.022  2  9.81  4.0 4 Cd 0.64 Cc = = = 0.74 Cv 0.87  EXERCISE A jet of water, issuing from a sharp edged vertical orifice under a constant head of 10.0cm, at a certain point, has the horizontal and vertical coordinates measured from the vena-contracta as 20.0cm and 10.5cm respectfully. Find the value of Cv. Also find the value of Cc if Cd=0.60. EXERCISE Water discharge at the rate of 98.2 litres/s through a 120mm diameter vertical sharp-edged orifice placed under a constant head of 10metres. A point, on the jet, measured from the vena-contracta of the jet has coordinates 4.5metres horizontal and 0.54 metres vertical. Find the co-efficient Cv, Cc and Cd of the orifice. Time taken for fluid level to fall through height, h dh h1 h2 OB In a small amount of time dt, Volume measured at the jet = Qdt = Aju jdt Volume Removed from tank VT = − AT dh The two are equal − AT dh = Aj u j dt Solving for time − AT dt = dh Aj u j So − AT dt = dh Cd Aorifice 2 gh − AT −1/ 2 dt = h dh Cd Aorifice 2 g Integrating from depth h1 to h2 t2 − AT h2   −1 / 2 dt = h dh t1 Cd Aorifice 2 g h1  − AT 1/ 2 (h2 − h1 ) 1/ 2 (t2 − t1 ) = Cd Aorifice 2 g 1/ 2 Time to drain a tank from h1 to h2 2𝐴 𝑇 1/2 1/2 t= (ℎ1 − ℎ2 ) 𝐶𝑑 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 Example What is the time taken to drain a tank from 4.5m to 0.7m. Tank diameter, D = 4.5m; Nozzle diameter = 700mm. Take Cd = 0.92 Solution: 4.52 0.07 m 2 AT =  = 15.91m 2 ; Aorifice =  = 0.004 m 2 4 4 AT = 15.91 / 0.004 = 3977.5 Aorifice 2(3977.5) 1/2 1/2 t= [(4.5) − (0.7) ] 0.92∗ 29.81m/s 2 t = 2940.5s = 49.0 min Time for a Tank to Empty 2𝐴 𝑇 1/2 1/2 t= (ℎ1 − ℎ2 ) h2 = 0 𝐶𝑑 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 Tank empty  EXERCISE A circular tank of diameter 1.5m contains water up to a height of 4m. An orifice of 40mm diameter is provided at its bottom. If Cd=0.62, find the height of water above the orifice after 10 minutes. An orifice of diameter 150mm is fitted at the bottom of a boiler drum of length 6m and a diameter of 2m. The drum is horizontal and contains water up to a height of 1.8m. Find the time required to empty the boiler. Take Cd=0.6. Notches and Weirs Depending on the shape Rectangular Triangular (V-Shape) Trapezoidal Others Depending on the form of the crest Sharp crested Broad crested Others GEOL SHARP CRESTED WEIR BROAD CRESTED WEIR b >0.66H. H v h b General Notch Equation b h H h Q = f {H and dimensions of notch (Weir)} Velocity t hrough strip h is u = 2 gh Discharge through the strip Q = Au = bh 2 gh Integrating from the free surface h=0 to the weir crest h= H H Theoretical discharge Q = Au =  2 gh (bh) 0 H 1 Qtheoretical = 2 g  bh 2 dh 0 Discharge varies with different shape of weir or notch Rectangular Notch Weir for irrigation project at Nobewam, Ghana Rectangular Notch B H For Rectangular Notch, the width (B) does not change with depth. b = constant = B Substitute (B) into the general weir equation H H 1 1 Qtheoretical = 2 g  Bh 2 dh = B 2 g  h 2 dh 0 0 2 = B 2g H 32 3 For actual discharge we introduce the coefficient of discharge C d 2 Qactual = Cd B 2 g H 3 2 3 2 Cd B 2 g = K Qactual = KH 3 2 3 K constant for the given notch (weir constant). Coefficient of discharge Cd depends on: Length of weir Head (Height of water above the crest) Degree of sharpness of crest Position of weir wall with respect to the side and bottom of the channel Example Find the discharge over a rectangular notch having width 4.5 m and a constant head of 1.5 m. Assume coefficient of discharge for the notch as 0.92 Soln. ▪ Width of notch L=B = 4.5 m ▪ Head over the notch, H = 1.5 m ▪ Coefficient of discharge Cd = 0.92 Discharge 2 Qactual = Cd B 2 g H 3 2 3 Exercise A rectangular notch is discharging 1000 l/s of water under a constant head of 1.5 m. Find the width of the notch, if Cd = 0.90 Calibration of a Rectangular Notch The Notch is usually used for determining the head, then the discharge of fluid (e.g. water). It must first be calibrated experimentally to find the constant K. 2 K = Cd B 2 g Qactual = KH 32 3 Two methods a. Plot a graph of Q against H3/2 Q , m 3 /s 32 3/2 H ,m Find the gradient of the straight line and hence determine K b. Using the relation Qactual = KH 32 Taking logs of both sides 3 log 𝑄𝑎𝑐𝑡𝑢𝑎𝑙 = log 𝐾 + log 𝐻 2 When H = 1, logH = 0, log Qactual = log K The value of K can be found Effect of error in the measurement of head on the discharge over a rectangular notch Qactual = KH 32 Differentiating Q, with respect to H, dQact 3 3 = KH 1/ 2  dQact = K H dH dH 2 2 Hence dividing through by Qact dQact 3 dH dQact 3 dH = Or  100 =  100 Qact 2 H Qact 2 H  Percentage error in  Percentage error in the   3  the discharge over  =  measuremen t of head  rectangula r notch  2 over the crest of the notch  Example The discharge over a rectangular notch 1.0 m wide was found to be 11.8 l/s. An error of 1 mm was made in the measurement of the head over the crest of the notch. Assuming the coefficient of discharge for the notch to be 0.82, calculate the percentage error in the discharge. Soln. Width of the notch B = 1.0 m. Discharge over the notch Q = 11.8*10-3m3/s Error in the head dH = 1 mm = 0.001m Coefficient of discharge Cd = 0.82  2 Qactual = Cd B 2 g H 32 3 2 11.8 10 =  0.82 1.0  2  9.81  ( H ) -3 3/ 2 3 H = 0.03 m Percentage error in discharge dQact 3 0.001  100 =   100 Qact 2 0.03 = 5.0 % Broad crested weirs. b >0.66H. H h V b H − Head of water over the weir on upstream side h − Head of water over the weir on down stream side L − Length of weir v − Velocity of water on the downstream side C d - Coefficient of discharge Applying the Bernoulli’s equation 2 u 0+0+ H = 0+ +h 2g u = 2 g ( H − h) Discharge Q = Cd  Area of flow  velocity = Cd  L  h  2 g ( H − h) = Cd  L  2 g  ( Hh 2 − h 3 )1/ 2 Determining Maximum Discharge d (Q) Differentiate w.r.t. variable h =0 dh d [ Hh 2 − h3 ] = 2 Hh − 3h 2 = 0 dh 2 h= H Qmax = 1.71C d LH 3/ 2 3 Hence Exercise A Broad-crested weirs is constructed across a channel in which the discharge is 30 m3/s. The cross-sectional area of the channel is 35 m2. If the crest of the weir is 2 m below the upstream water level, find the length of the crest. Take Cd = 0.97. Velocity of Approach Velocity of approach is assumed to be uniform over the weir Q = Discharge over weir ua = Velocity of approach to weir A = Cross sectional area of the channel on the upstream side of the weir Velocity of Approach Q = Area of channel  Velocity of approach bc is the channel width hc is the fluid height in the channel Q ua = Q = A  u a = hc bc u a hc bc Head over the weir (notch) on upstream side 2 u HT = H + H a = H + 2g 2 u Ha = Velocity Head 2g Exercise A 3 m long rectangular weir is discharging water under a constant head of 1 m. The channel approaching the weir is 5 m wide and 1.2 m deep. Determine the Total head over the weir. Take C d = 0.6 'V' or Triangular Notch (Weirs) For "V" notch (weir), relationship between width (b) and depth (h) is dependent on the angle of the "V". b Angle of the V is  and the width is, b, at depth h   b = 2( H − h) tan( ) 2 H 1 Qtheoretica l = 2 g  bh dh 2 0   H Qtheoretica l = 2 2 g tan( )  ( H − h)h dh 12 2 0  2 H 2 5 2 = 2 2 g tan( ) Hh 32 − h  2 3 5 0 8  52 = 2 g tan( ) H 15 2 Actual discharge 8  52 Qactual = Cd 2 g tan( ) H 15 2 Exercise Water is flowing over a right angled triangular notch for which the coefficient of discharge is 0.6. What will be the discharge if head is 50 cm. Exercise Find the depth and top width of a triangular notch capable of discharging a maximum quantity of 700 lit/sec and such that the head shall be 8 cm when the discharge is 6 lit/sec. The K for the notch is the same as for a right angled notch, that has a value of, K = 1.42.