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A tank has two identical orifices, each 5 cm in
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</figure> <p>Once the liquid passes the vena contracta, the XXX slows XXX liquid XXXX due XX XXXXXXXX XXX the XXXX XXXXXXXX.</p> <p>Now, we have two things XX XXXXXXXX XXXX here.</p> <p>XXX first is the XXXXXXXXX XXXXXXXX XXX the XXXXXX XX the area of XXX vena XXXXXXXXX to determine the rate of discharge, Q.</p> <p>Discharge velocity:<XX></p> <p>XX XX XXXX that the theoretical velocity XX the XXXX can XX calculated XXXXX XXX XXXXXXXXX<XXX>i</XXX>= √XXX.<br></p> <p>XXXX, at upper XXXXXXX,XX = root(X*9.X*4) = X.XX m/s<XX>At XXXXX XXXXXXX, XX = XXXX(2*X.X*X) = XX.XX m/s</p> <p>Discharge velocity, V = XX* coefficient of viscocity<XX>XXX upper XXXXXXX, XX=X.XX*X.XX = X.XXX m/s<br>For XXXXX orifice, Vl= XX.XX*X.98 = XX.62 m/s</p> <p>Secondly, XXX XXXX XX the XXXX contracta, X = Ao*XXXXXXXXXXX XX contraction (Ao = Area of cross section XX the orifice)<br>XXXXXXXXX, X = pie*(r^X)*0.XX<br>For upper XXXXXXX, A = (XX/7)*(X/X)^2*X.XX = XX.XX XXX or X.001257 m2<br>XXX lower orifice, X =(22/7)*(5/2)^2*X.64 =12.57 cm2 or 0.001257 XX</p> <p>XXXXX )XXX, the XXXXXXXXX or flow, X from an XXXXXXX = V*X = Vi*XX*XX*XX (Cc = XXXXXXXXXXX XXXXXXXXXXX, XX= XXXXXXXXXXX XX viscosity)<XX>For upper XXXXXXX, Qu= Vu*X = X.XXX*X.001257 = X.XXXX XX/s<br>XXX lower XXXXXXX, XX = Vl*X = XX.XX*X.XXXXXX = 0.XXXX XX/s</p> <p>Part X)</p> <p>XXXXXXXXXX XXXXXXXX of upper orifice, XX = X.673 m/s<XX>Horizontal XXXXXXXX XX XXXXX XXXXXXX, XX = 10.62 m/s</p> <p>Let the XXXXXXXXXX XXXXXXXX at XXXXX, XXX flows XXXXXXXXX is X.</p> <p>XXXXXXXXX, X= X*t (V= velocity XX XXX XXXX, t = time) or t = X / X<XX>Using XXXXXXXX XX XXX XXXXXXXXXXX XXXXXX,<XX>y = ut + X/2 g*t^X</p> <p>But, u = X, therefore, y = 1/2*g*t^2</p> <p>Now, XXX upper orficie at any point XX XXXX vertical XXXXXXXX travelled by flow, Yu= X/2*g*(X/XX)^2<XX> XXX XXXXX XXXXXXX at any point XX time vertical XXXXXXXX travelled XX flow, XX = X/2*g*(X/XX)^2</p> <p>XXXXXXXX XXXXXXX, X^X = 2*Yu*Vu^2/g - XXX 1<XX>XXX lower XXXXXXX, X^X = 2*Yl*Vl^2/g - XXX 2</p> <p>XXXXX X XX same at XXX XXXXX XX intersection, therefore, XXXXXXXX XXX 1 XXX eqn X we get,</p> <p>XX^2*XX = Vl^X*XX - Eqn 3</p> <p>Now at the XXXXX XX XXXXXXXXXXXX XX = Yl+X - XXX X(XX XXX is XX a depth XX 4 m XXX XXXXXXX XX depth of X m)<XX><br>XXXXXXX XXX X in Eqn X we XXX,</p> <p>XX^2*(Yl+2) = VI^2*YI</p> <p>XXXXXXX the values of XX XXX Vl in XXX XXXXXXXX XX get,</p> <p>X.673^X*(XX+2) = XX.62^X*XX<XX>Or, 75.XX*(XX+X) = XXX.XX*XX<XX>Or, XX.XX YI= 150.44<XX>Or, YI = X m</p> <p>XXXXXXXXX, XXX XXXXXXXX XX intersection from XXX lower XXXXXXX is 4m , XXXXX orifice XX 6 m and XXXXXXX of XXX water XX 10 m.</p> <p><XX></p>">
