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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, XXX air XXXXX XXX liquid down due to XXXXXXXX XXX the XXXX diverges.</p> <p>Now, XX have XXX things to consider over XXXX.</p> <p>The first is XXX XXXXXXXXX XXXXXXXX XXX XXX second XX XXX XXXX of the vena contracta XX XXXXXXXXX the rate XX XXXXXXXXX, X.</p> <p>XXXXXXXXX XXXXXXXX:<XX></p> <p>As we XXXX XXXX the theoretical XXXXXXXX XX XXX flow XXX XX calculated using XXX equationV<sub>i</XXX>= √2gh.<br></p> <p>XXXX, at upper XXXXXXX,XX = root(2*X.8*X) = X.85 m/s<XX>XX lower orifice, Vi = XXXX(2*9.8*X) = 10.XX m/s</p> <p>XXXXXXXXX velocity, X = XX* XXXXXXXXXXX XX viscocity<XX>XXX XXXXX XXXXXXX, XX=X.XX*X.98 = 8.673 m/s<XX>XXX lower orifice, XX= XX.XX*0.98 = XX.XX m/s</p> <p>XXXXXXXX, the area of the XXXX XXXXXXXXX, A = Ao*coefficient of XXXXXXXXXXX (Ao = XXXX XX cross section of XXX XXXXXXX)<br>XXXXXXXXX, A = pie*(r^2)*0.64<br>For XXXXX orifice, X = (22/X)*(5/2)^2*X.64 = XX.XX XXX or 0.XXXXXX m2<br>For lower XXXXXXX, A =(22/7)*(5/X)^X*X.XX =XX.57 cm2 or X.001257 m2</p> <p>PartA )Now, the XXXXXXXXX or flow, Q from an orifice = X*A = XX*XX*XX*XX (Cc = XXXXXXXXXXX coefficient, Cv= XXXXXXXXXXX XX XXXXXXXXX)<br>XXX upper orifice, XX= Vu*X = X.XXX*0.001257 = 0.0109 m3/s<XX>For lower XXXXXXX, Ql = XX*X = 10.XX*0.001257 = 0.0133 XX/s</p> <p>XXXX B)</p> <p>Horizontal XXXXXXXX of XXXXX orifice, XX = 8.XXX m/s<XX>XXXXXXXXXX XXXXXXXX XX lower XXXXXXX, XX = XX.XX m/s</p> <p>Let the XXXXXXXXXX XXXXXXXX at XXXXX, the XXXXX XXXXXXXXX is X.</p> <p>XXXXXXXXX, X= V*t (X= velocity of XXX flow, t = XXXX) or t = X / V<br>Using XXXXXXXX XX XXX dimensional XXXXXX,<br>y = ut + X/2 g*t^2</p> <p>But, u = 0, XXXXXXXXX, y = 1/2*g*t^2</p> <p>Now, XXX upper orficie XX any point XX time vertical distance XXXXXXXXX XX flow, Yu= X/X*g*(X/XX)^X<br> XXX XXXXX orifice XX any XXXXX XX time vertical distance XXXXXXXXX by XXXX, Yl = X/2*g*(X/XX)^X</p> <p>Forupper orifice, X^X = X*XX*XX^X/g - Eqn X<XX>For XXXXX XXXXXXX, X^2 = 2*XX*XX^2/g - Eqn X</p> <p>Since X is same at XXX XXXXX XX XXXXXXXXXXXX, therefore, equating XXX X XXX XXX X XX XXX,</p> <p>XX^2*XX = XX^X*XX - Eqn 3</p> <p>XXX at the XXXXX XX XXXXXXXXXXXX Yu = Yl+2 - XXX X(XX one is XX a XXXXX of 4 m XXX another XX XXXXX XX 6 m)<XX><br>XXXXXXX XXX X in Eqn X we XXX,</p> <p>Vu^2*(XX+X) = VI^2*YI</p> <p>XXXXXXX the XXXXXX XX Vu XXX Vl in XXX XXXXXXXX we get,</p> <p>8.673^2*(XX+X) = 10.62^2*Yl<XX>XX, XX.22*(YI+X) = 112.78*XX<br>Or, 37.XX YI= XXX.XX<XX>XX, YI = X m</p> <p>XXXXXXXXX, the distance of intersection from the lower orifice is 4m , XXXXX XXXXXXX is 6 m and XXXXXXX of the water is 10 m.</p> <p><XX></p>">
