UNIT-2: PRESSURE AND ITS MEASUREMENTS Q 1 State and prove Pascal’s Law (July 2013, 2014, jan 2015, July 2015) Ans:Consider a small vertical cylinder of fluid in equilibrium, where positive z is pointing vertically upward. Suppose the origin z = 0 is set at the free surface of the fluid. Then the pressure variation at a depth z = -h below the free surface is governed by ( p  p) A  W  pA pA  gA z  0 p  - gz dp dz  - g or - Eq .(1) (as z → Therefore, the hydrostatic pressure increases Linearly with depth at the rate of the specific weight  = ρg of the fluid. Homogeneous fluid: ρ is constant By simply integrating the above equation-1: ∫ dp  - ∫  g dz p  -gz  C C is constant of integration When z = 0 (on the free surface), p= C = p0= (the atmospheric pressure). Hence, p  - g z  p0 Pressure given by this equation is called ABSOLUTE PRESSURE, i.e., measured above perfect vacuum. However, for engineering purposes, it is more convenient to measure the pressure a
A pipe contains petrol of specific gravity 0.8. The flow is upwards from the point ‘A’ to ‘B’ as shown in Fig. Q.(2c). A differential mercury manometer is connected at the two points ‘A’ and ‘B’ of the pipe which are 0.3m apart. Find the deflection in mercury column if the pressure difference between two points is 18kN/m2 (July 2013,July 2014, jan 2015) Ans: Given (pA- pB) = 18 kN/m2, (ZB - ZA) = 0.3m, Sp.Gr Petrol = 0.8, Sp.Gr. Mercury = 13.6 Equating the pressure about X-X on both limbs of U-tube manometer (p = w×Sp.Gr×H) pA  (x  y)  0.8  9.81  pB  0.3 0.8  9.81 y  0.8  9.81 x 13.6  9.81 (pA – pB) = 18 = 15.6456 + 125.86 x x = 0.125m
FR = Where  γAyc sin θ = (γ h c )A hc : Pressure at the centroid = (Pressure at the centroid)  Area Magnitude of a force on an INCLINED plane Dependent on  Area, and Depth of centroid Perpendicular to the surface (Direction) i) Position of FR on y-axis ‘yR’ : y coordinate of the point of action of FR Moment about x axis: FR yR  (Ayc sin ) yR  A ydF  A sin y dA   sin A y dA  h R h 2 dA = A h c A I I = x where h c A  A y2dA : 2nd moment of areaor, by using the parallel-axis theorem, Ix  Ixc  Ayc2 (The center of pressure below the centroid)
trapezoidal channel 2m wide at the bottom and 1m deep has side slopes 1:1. Determine: i) Total pressure ii) Centre of pressure, when it is full of water (July 2014, Jan 2015) Ans: Given B = 2m Area of flow A = (B+sy)y = 3m2 The combined centroid will be located based on two triangular areas and one rectangle (shown as G1, G2, G2) y  A1  h1  A2  h2  A3  h3 A1  A2  A3 The total area A = 3m2 Area of rectangle = 2 ×1 = 2m2 Area of Triangle = 1 ×1×1=0.5m2 2       1 1 (21)0.5   (11)0.333  (11)0.333 y   2  2  3   4  0.444m 9 The total pressure P = w×A× y = 9810×3×0.444 = 13080N Centre of pressure The centroidal moment of Inertia of Rectangle and Triangle is IG1  2 13 12  0.1667m 4 at 0.5m from water  surface IG1  113 36  0.028m 4 at 0.333m from water  surface h  y  I Eq.1 A y The moment of Inertia about combined centroid can be obtained by using parallel axis theorem
I  0.1667  0.00618   A d 2  G G 2 2 2 I  0.1667  0.00618  20.028  0.0062   0.2408m 4 Substituting in Eq.1, The centre of pressure from free surface of water h  y  I g A y h  0.444  0.2408  0.6252m 3 0.444 Q 4Derive the expression for total pressure and centre of pressure on a vertical plate submerged in a static liquid (Dec 2013,July 2015) Ans:Vertical Plane surface submerged in liquid Consider a vertical plane surface of some arbitrary shape immersed in a liquid of mass density  as shown in Fig. y dy Let A = Total area of the surface y = Depth of Centroid of the surface from the free surface G = Centroid of the immersed surface C = Centre of pressure h = Depth of centre of pressure
Q 6 A circular plate 2.5m diameter is immersed in water, its greatest and least depth below the free surface being 3m and 1m respectively. Find (i) The total pressure on one face of the plate and(ii) Position of centre of pressure (Dec 2013, July 2015) Ans: d = 2.5m,   Sin 1  2   2.5    53.13 h  1  1  2m A   d 2    2.52  4.909m2 4  IG  64 4 d 4    2.54  1.917m4 64 F = wA h = 9.81×4.909×2 = 96.31 kN Q 7 Prove that for a plate submerged in horizontal position in water the center of pressure is same as centroid of the plate. (July 2013, July 2015) Ans: Consider a plane horizontal surface immersed in a static fluid. As every point of the horizontal surface is at the Water same depth from the free surface of the liquid, the pressure intensity will be P equal to p =  g h, where ‘h is the h depth of surface (Refer Fig.). Horizontal Plane Surface Let ‘dA’ be the Elemental area and ‘A’ be the surface area Then total force dP = p × dA c.p Area ‘A’ P  gAh The centre of pressure Fig. Total Pressure on a Horizontal Plane Surface I Sin 2 hcp  h  Eq.(1) For Horizontal surface  0 , substituting in Eq.(1) Ah
Q.6 A stream function is given by  = 2x2-2y2. Determine the velocity and velocity potential function at (1, 2) (Dec 2013, jan 2015) Ans: Given:  = 2x2-2y2   4 x  v; v  4 x  Velocity at (1,2), x    4 y  u; u  4 y  Velocity at (1,2), y v  4 u  8 Units Units Resultant velocity V(1,2) =  8.94 Units   u x   v y   x   x  (4 y)  4 y    4  x  y  C ( f (  (4 x)  4 x    4  x  y  C ( f (  C C  C    C  (4 x  )   4 x    4 x       4 x  4 x  0 From Eq.1 y y C  0 Integrating y y y y C  0  x  y   4  x  y    4 1 2  8 Units Q. 7 In a flow the velocity vector is by V = 3xi + 4yj -7zk. Determine the equation of the stream line passing through a point M (1, 4, 5). (July2013, july 2015) Ans: Given the Velocity vector V = 3xi+4yj -7zk  u = 3x ; v = 4y; w = -7z To determine the equation of the stream line passing through a point M (1, 4, 5) The 3-D equation of streamline is given by, dx  dy  dz u v w dx  dy  dz Eq.1 3x 4 y  7z The streamline equation at M (1, 4, 5), x =1, y = 4, z = 5 Substituting the values of x, y, and z in Eq.1 dx  dy  dz
Q 8The velocity potential  for a two dimensional flow is by (x2 – y2) + 3xy. Calculate: i) the stream function  and ii) the flow rate passing between the stream lines through (1,1) and (1, 2). (July 2013, 2014) Ans:  = (x2 – y2) + 3xy To determine the  function d   x dx   y d  v dx  Eq.(2) As per definition of velocity potential () and stream function ();     u and      v x y  y x    u    (2 x  3 y)   and  (2 y  3x)     v x y y  x  Substituting the value of u and v in terms of x and y in equation 2, we obtain d  v dx  udy  (2 y  3x)dx  (2 x  3 y)dy d  2 y  3xdx  2 x  3 Eq.3 Integrating the equation-3 (partially w.r.t ‘x’ the ‘dx-term’ and w.r.t ‘y’ the ‘dy-term’)  3   3  3 2 2    2 xy  x2    2 xy  y2  = 4 xy  x  y   2   2  2 The flow rate passing between the stream lines through (1, 1) and (1, 2). The equation of stream function is given by The value of Point streamline at (1, 1) is obtained by substituting x = 1, y = 1   4 xy  3 x2  y2   4 11 3 (12 12 )  7Units (1,1) 2 2 The value of Point streamline at (1, 2) is obtained by substituting x = 1, y = 2   4 xy  3 x2  y2   4 1 2  3 (12  22 )  15.5Units (1,2) 2 2 The flow rate passing between the stream through (1, 1) and (1, 2) (1,2) - (1,1) = (15.5-7)
Ans: Given the components of velocity  y 3 2   2 x 3  u    3 2x  x  y ,  v   x  y   2 y    The condition for Irrorational flow  v   u        x   y  v   2 x3  u   y 3  LHS   x  y x x   2 y   and RHS    y y  3 2x  x  y   i.e. LHS = (y2 – x2) and RHS = (y2 – x2) Hence the flow is Irrorational The corresponding stream function ‘’can be obtained by using following relationship   x  v   x  y 2  2 y  3 Eq.1 x     y3  u     2x  x2  y Eq.2 y   Integrating Eq.1 with respect to ‘x’   x y 2  2  x  y  x  C  f ( Eq.3 2 12 1 Differentiating Eq.3 with respect to ‘y’   x 2  y  2 x  C 1 y y C y 3 1   y 3 y 4 Integratin g, C 1   12 y 4 C 1   C ; (assu min g C  0)
Where ‘as’is the acceleration in the direction of s. as= dV/dt, ‘V’isa function of ‘s’and t. cos  dz a  V dV If the flow is steady, = 0, ds and ds .On substituting and dividing the equation by ρgdA, we can obtain Euler's equation: dp  gdz  vdv  0  Eq.(1)  The above equation is known as Euler’s equation of motion. Bernoulli’s Equation: Integrating the Eq(1) the ‘total energy head’ H of the fluid is found by adding the three types of mechanical energy possessed by the fluid at that point. H  p  v  z g 2g As water flows between two points, or sections of a pipeline or channel, no energy can be created or destroyed (fundamental law of physics concerning conservation of energy). If
Ans: The pipe is of uniform area. Therefore the velocities at section 1-1 and section 2-2 are same. V1  V2  Q  0.25    3.537 m / s  A   0.32   4  The pressure intensity is also same p1 = p2 = 400 kPa = 400 ×1000 N/m2 Force along x-axis ‘Fx’: (Dynamic + Static) Force F   W Q  V V  Cos135  p A  p A cos135 x g 1 2 1 1 2 2 F  9810  0.25  3.54  3.54  (0.707  400 1000    0.32  400 1000    0.32  (0.707) x 9.81 4 4 Fx = 49750.51 N ( ) Force along Y-axis ‘Fx’: (Dynamic + Static) Force F   W Q  0  V  Sin135  p A Sin135 x g 2 2 2 F  9810 0.25 0  3.54(0.707 4001000   0.32 (0.707) x 9.81 4 Fy = -20605.51 N () R  R  R = 53848.9 N tan   20605.5 49750.5   tan1  22.5 
Q.8 A vertical pipeline carrying water changes in diameter from 200mm at a position ‘A’ to 500mm at another position ‘B’ which is 3m at a higher level. If the pressure at ‘A’ and ‘B’ are 80kN/m2 and 60kN/m2 respectively and discharge is 200 litres/sec, determine the loss of head and direction of flow (Dec2013, July 2014 ) Ans: Given Q = 200 lit/sec = 0.2m3/s, DA = 200mm = 0.2m, DB = 500mm = 0.5m (ZB – ZA)= 3m, pA = 80kN/m2, pB = 60kN/m2, The velocity at ‘A’ and ‘B’ can be obtained by continuity equation Q VA   0.2    6.366 m / s  AA Q VB  AB   0.22   4   0.2    0.52   4   1.018 m / s Energy head at ‘A’ EA  Z A p A   w 2 A 2  g EA  0  80  9.81 2  g  10.21m Total Energy head at ‘B’ EB  ZB pB  w 2 B 2  g EB  3  60  9.81 2  g  9.169m Since EA> EB, The direction of flow will be from “A” to “B” i.e. Upward  The loss of head = (EA – EB) = (10.21-9.169) = 1.041m
UNIT – VI PIPE FLOW Q.1 Define (i) Hydraulic gradient (ii) Energy gradient (July 2013,July 2014,Jan 2015) Ans:(i) Hydraulic gradient: A Line joining the peizometric heads at various points in a flow is known as Hydraulic Grade Line (HGL) (ii) Energy gradient: It is a line joining the elevation of total energy of a flow measured above a datum, i.e. 2 EGL Line lies above HGL by an amount V /2g. Q. 2Distinguish between compound pipe and equivalent pipe (July 2014, Jan 2015) Ans: Compound pipe Equivalent pipe lent pipe is such that the entire system is replaced by a single pipe of uniform diameter D, but of the same length L=L1+ L2+ L3 such that the head loss due to friction for both the pipes, viz equivalent pipe & the compound pipe are the same. For a compound pipe or pipes in series. hf  hf1  hf2  hf3 h  8 fL1Q 8 fL2Q 8 fL3Q    (1) f g 2 D5 g 2 D5 g 2 D5 For an equivalent pipe h  8 fLQ    (2) f g 2 D5 Equating (1) & (2) and simplifying L L1 L2 L3 5 5 5 5 1 2 3   D   L 1 5 L  L L   1  2  3   D5 D5 D5   1 2 3 
Q. 3 At a sudden enlargement of water main from 240mm to 480mm diameter, the hydraulic gradient rises by 10mm. Estimate the rate of flow. (July 2014, july 2015) Ans: Rise of hydraulic gradient line = 10mm From Continuity equation Q   4  0.242 V1    0.482 4 V2 V1 = 4× V2 Loss of head in sudden expansion H   V1  V2   exp ansion  2   2  g  Applying Bernoulli’s equation between upstream and downstream of sudden expansion p V 2 p V 2 1  Z  1  2  Z  2  H Eq.1 1 2  g  2 2  g exp ansion  p  p  Given rise of hydraulic gradient line = 10mm = 0.01m = 2 1  w The elevation is same for horizontal enlargement Z1 = Z2 and also V1 = 4× V2 Substituting in eq.1 p1  Z 4V   2  p2  Z  V2   V1  V2   2 1 2  g  2  1 2  g  2  2  g   p  p  15 V 2 9 V 2 2 1  0.01  2  2  w V2  0.181m / s 2  g 2  g Q    0.482  0.181  0.03275m3 / s 4 The rate of flow Q = 0.03275 m3/s
Consider RHS of eq(1) Mass x acceleration = × Vol × change in velocity /time   Q  V1  V2     (iii)  =volume/time × change in velocity a2  p1  p2   pQV1 V2  Substitution (ii) & (iii) in eq(i)  p1  p2   V2 V1 V2  Both by (Specific weight ‘’)  p  p  V V V   1 2   2 1 2    (iv)    g Applying Bernoulli’s equation between (1) and (2) with the centre line of the pipe as  p  p  V 2 V 2 1 2   1 2 2g  hL 2V V V  V 2 V 2  and hL  2g Considering head loss due to sudden expansion hLonly. 2 2 2 2 2 2 Z1  Z 2 2isV1Vh2 orViz1 ontVa2l hL  h  2V1V2  2V2  V1 V2 Z  p1 1  V 2  1  2g 2g Z2  2g p V 2 2  2  2g
difference of water level as 20m. Determine the discharge through the pipe. If an additional pipe of diameter 200mm and length 1200m is attached to the last 1200m of the existing pipe, find the increase in discharge. Take f = 0.015 and neglect minor losses. (Dec2013,July 2014) Ans; Case-1: When single pipe connects two reservoirs H = 20m, f = 0.015, L1 = 2000m D1 = 0.2m V  Q  Q A1   2    D1   4  4 fL V 2 H  1 1  2gD 4 fL1Q    32 fL1Q   2 gD5 1 2 1 2gD1   D1  4   H  20  32  0.015  2000  Q  2  g  (0.2)5 Q  0.0254 m3 / sec Case-2: When Pipeline is branched midway for connecting two reservoirs H = 20m, f = 0.015, L1 = 800m, D1 = 0.2m L2 = 1200m, D2 = 0.2m L3 = 1200m, D3= 0.2m Q1 = Q2 + Q3 = 2Q2  Q  Q2 =  1   2  4 fL V 2 4 fL V 2 H  1 1  2 2 2gD1 32 fL Q 2 2gD2 32 fL Q 2 32  0.015  800  Q 2 32  0.015  1200  0.5Q 2 20  1 1  2 2  1  1  2 gD 5  2 gD 5  2  g  0.25  2  g  0.25 Q  0.0342 m 3 / sec
Case- 2 Parallel Pipe f Length L Discharge Q1 D1 = D2 = d Q = Q1 + Q2 = 2Q1 (as both the diameter and length are same) 2 f  L  Q1  Q hf 1  hf 2     Q1   0.5  Q 2  g  D  A  2 2     2 hf 1  hf 2  f  L   0.5  Q   0.0207  f  L  Q 2  9.81 d    d 2  d  4  Equating the values of hfand hf1 from Eq.1 and Eq.2 Q.8 Define Pipes in Series (Dec 2013, Jan2015) Ans: Pipes in Series or Compound Pipe D1, D2, D3, D4 are diameters. L1, L2,L3, L4 are lengths of a number of Pipes connected in series (hf)1, (hf)2, (hf)3 &(hf)4 are the head loss due to friction for each pipe. The total head loss due to friction hf for the entire pipe system is given by hf  hf1  hf2  hf3  hf4 h  8 fL1Q 8 fL2Q 8 fL3Q 8 fL4Q f g 2 D5 g 2 D5 g 2 D5 g 2 D5
Leq  1800 1200 600  0.45  0.55  0.45  0.35    Leq= 4318.2 m Given Leq= 3600 m 3600  1800 1200 600  D5  0.55  0.45  0.35  eq   Deq= 0.39 m
x CV  CV  Q. 5 A 25mm diameter nozzle discharges 0.76m3/minute, when the head is 60m. The diameter of the jet is 22.5mm. Determine the values of Cc, Cv, Cd and loss of head due to fluid resistance (July 2014) Ans: Given Nozzle Diameter D= 25mm, Q = 0.76 m3/min, H = 60m, djet= 22.5mm Values of coefficients  2 Area of jet d j 4 2 22.52 CC  Area of  Nozzle  4  D 2 D  0.81 C  Qact  Qth  0.01267   0.0252  4  0.7522 From the relationship Cd = CC × CV Cv  Cd Cc  0.7523  0.928 0.81 Loss of head (HL) due to fluid resistance can be obtained by applying Bernoull’s Equation between the outlet of the nozzle and jet of water p V 2 p V 2 1  Z  1  2  Z  2  H Eq.1 1 2  g  2 2  g L
4 fL V 2 H  1 1  2gD 4 fL1Q    32 fL1Q   2 gD 5 1 2 1 2gD1   D1  4   H  0.3  32  0.04 1500 Q  2  g  (0.2)5 Q  0.0343 m3 / sec Case-2: When Pipeline is branched midway for connecting two reservoirs H = 20m, f = 0.015, L1 = 800m, D1 = 0.2m L2 = D2 = 0.2m L3 = 1200m, D3= 0.2m Q1 = Q2 + Q3 = 2Q2  Q  Q2 =  1   2  4 fL V 2 4 fL V 2 H  1 1  2 2 2gD1 32 fL Q 2 2gD2 32 fL Q 2 32  0.04  750  Q 2 32  0.04  750  0.5Q 2 20  1 1  2 2  1  1  2 gD 5  2 gD 5  2  g  0.65  2  g  0.65 Q  0.04338 m 3 / sec Increase in Discharge = (Q1 – Q) = (0.04338-0.0343) = 0.00908m3/sec The velocity of water in a 60cm diameter and 15mm thick cast iron pipe (E=1.04x1011 Pa) is changed from 3 m/s to zero in 1.25 s by closure of a valve i) if the pipe length is 800m what will be the water hammer pressure at the valve? What will be the corresponding pressure rise if the closure takes place in; ii) 2s and iii) 0.8s respectively? Bulk module of elasticity of water is 2.11 x 109 N/m2. (July2013, July 2015) Ans:Given: D = 60cm = 0.6m, L = 800m, t = 15mm = 0.015m E = 1.04 × 1011 Pa, K = 2.11 × 109 N/m2 , V = 3m/s Case-1 Time of Closure Tc = 1.25 sec
and Venturimeter: Orifice meter is a small insertion while venturimeter is long insertion. The typical Cd values for orifice meter is 0.68 while for venturimeter it is around 0.9. Orifice meter is cheaper as compared to venturimeter. A Pitot tube was used to measure the velocity of water at the centre of a 25cm diameter pipe. The stagnation and static pressure heads are indicated as 6m and 5m of water head. Given the coefficient of velocity Cv= 0.98. Determine the velocity at the centre of the pipe Ans: Given Cv = 0.98, hstagnation = 6m, hstatic = 5m (Dec 2013, July 2014) h= (6-5) = 1m The velocity through tube V  CV  V  0.98  4.34m / s Derive the equation for the discharge through Venturimeter 2015) (July 2014, July Ans:Venturimeter: The venturimeter consists essentially of a convergence in a pipeline followed by a short parallel sided ‘throat’ and then a divergence.
On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid surface. Discharge through the dq=area x velocity. dQ  Lxdhx Therefore discharge is given by 1 Q H 2  dq  L   H  ha  dh 0 0 3 3  L 2  H  ha 2  ha 2    3 3 Q  2 C  L  H  h 2  ha 2  3 d  a    Empirical Formula 3 3 (i)Francis Formula Q  1.84  L  0.1nH   H  ha  2  ha 2    Q3.Water flows over a rectangular weir 1m wide at a depth of 15cm and afterwards passes through a right angled weir. Taking Cd for rectangular weir 0.62 and for triangular 0.59 calculate the depth over the triangular weir (July2014,July 2015) Ans: Given For Rectangular Weir Cd = 0.62, H = 15cm, L = 1m For Triangular Weir Cd = 0.59,  = 90 Let the depth over triangular weir be ‘H1’. As the same discharge passes through both the weirs equating the discharge for Rectangular and triangular weir    Q  2  C  3  H 2  8  C   tan  H 52 3 d Re c tan gular 15 d Triangular  2  1    Q  2  0.62 1  0.1532  8  0.59  tan 90  H 52 3 H1  0.3573m 15  2 
ADVANTAGES OF TRIANGULAR NOTCH OR WEIR OVER RECTANGULAR NOTCH OR WEIR A triangular notch or weir is preferred to a rectangular weir or notch due to following reasons The expression for discharge for a right-angled V-notch or weir is very simple. For measuring low discharge, a triangular notch gives more accurate results than rectangular weir or notch. In case of triangular notch, only one reading, i.e., (H) is required for the computation of discharge. Ventilation of a triangular notch is not necessary. Q10. Define velocity approach. How does the velocity of approach affect the discharge over aweir? (July 2014 , Jan 2015) Ans: It is the velocity of water which approaches just upstream of weir or notch. It is affected by the nappe drawdown and correction ‘ha’need to be applied based on velocity of head calculated with velocity of approach ‘Va’ Velocity of approach (Va) The total head over the weir will be the sum of static head (H) and velocity head (ha), V2 velocity head h  a is due to the Velocity of the liquid approaching the notch. a 2a On similar lines, considering a strip of uniform thickness ‘dh’ at a depth h below the liquid surface. Discharge through the strip dq=area x velocity. dQ  Lxdhx Therefore Total discharge is given by 1 Q H 2  dq  L   H  ha  dh 0 0 3 3  L 2  H  ha 2  ha 2    3 3 Q  2 C  L  H  h 2  ha 2  3 d  a    Empirical Formula 3 3 Formula Q  1.84  L  0.1nH   H  ha  2  ha 2   
formula Q   0.405  0.003  L H 2  H    (iii)) Rehbock formula  Q  0.403  0.053 H  0.011 L 3  H  0.00112  Z  Considering velocity of approach and End contraction, we have 3 3 Q  2 C  L  0.1nH   H  h  2  ha 2  3 d  a    Considering velocity of approach and End contraction, we have 3 3 Q  2 C  L  0.1nH   H  h  2  ha 2  3 d  a    Q 11A rectangular channel 2m wide has a discharge of 250 LPS is measured by a right-angled V-notch. Find the position of the notch from the bed of the channel if maximum depthof water is not to exceed 1.3m.Take Cd = 0.62. (July 2013 , july 2015) Ans: Given: Q = 250 LPS = 0.25m3/s, Cd = 0.62,  = 90, Maximum depth in the channel not to exceed 1.3m The discharge through a right angled triangular notch is given by, The position of apex of the = Maximum Depth – Height of water over from the bed of channel of water in channel V-notch notch The position of apex of the from the bed of channel = (1.3 – 0.5) = 0.8m
Without considering velocity of approach Q  Cd  2  L  3 3 H 2 (Without end contraction) Q  0.6  2 10  3 3  2 2  50.11 m3 / s Q 50.11 Velocity of approach Va   =1 m/s A 20  2.5 2 2 a Approach Head Ha    0.051m 2  g 2  9.81 Discharge equation considering end contraction Q  Cd  2  L  3  (H  Ha 32  H 32 Q  0.6  2 10  3  (2  0.05132  0.05132  = 51.81m3/s