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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 XXXXXXXXX, XXX XXX slows XXX liquid XXXX due XX friction XXX the flow diverges.</p> <p>Now, we have XXX XXXXXX XX consider over here.</p> <p>The XXXXX XX XXX discharge velocity and XXX second is the area XX XXX vena XXXXXXXXX XX XXXXXXXXX the XXXX of XXXXXXXXX, X.</p> <p>XXXXXXXXX velocity:<br></p> <p>XX we know XXXX XXX theoretical XXXXXXXX of XXX XXXX XXX be calculated XXXXX the XXXXXXXXX<sub>i</sub>= √XXX.<br></p> <p>Thus, at XXXXX orifice,Vi = XXXX(2*9.X*4) = X.85 m/s<br>XX XXXXX XXXXXXX, XX = root(X*X.X*X) = 10.84 m/s</p> <p>XXXXXXXXX XXXXXXXX, X = Vi* XXXXXXXXXXX XX XXXXXXXXX<XX>XXX upper XXXXXXX, Vu=8.XX*0.98 = 8.XXX m/s<XX>For lower orifice, XX= 10.84*0.98 = 10.62 m/s</p> <p>Secondly, XXX area of XXX vena contracta, A = Ao*XXXXXXXXXXX of contraction (XX = Area of cross XXXXXXX of the XXXXXXX)<XX>XXXXXXXXX, X = XXX*(r^X)*X.XX<br>For XXXXX XXXXXXX, X = (XX/7)*(5/X)^2*X.64 = 12.XX XXX or 0.001257 m2<br>XXX XXXXX orifice, X =(XX/7)*(X/X)^2*0.64 =12.57 XXX or 0.XXXXXX XX</p> <p>XXXXX )XXX, XXX discharge or flow, X from an XXXXXXX = V*A = Vi*Cv*XX*Cc (XX = XXXXXXXXXXX XXXXXXXXXXX, XX= coefficient of viscosity)<br>XXX upper orifice, Qu= Vu*X = 8.XXX*0.001257 = 0.0109 m3/s<br>XXX lower orifice, Ql = XX*X = XX.XX*X.XXXXXX = X.XXXX m3/s</p> <p>XXXX X)</p> <p>Horizontal velocity XX XXXXX XXXXXXX, XX = 8.XXX m/s<XX>XXXXXXXXXX XXXXXXXX XX lower orifice, XX = 10.XX m/s</p> <p>Let XXX XXXXXXXXXX XXXXXXXX XX which, the XXXXX intersect is X.</p> <p>Therefore, X= V*t (V= XXXXXXXX of the flow, t = XXXX) or t = X / X<XX>Using XXXXXXXX of XXX dimensional motion,<br>y = ut + X/2 g*t^X</p> <p>XXX, u = X, therefore, y = 1/X*g*t^2</p> <p>Now, for XXXXX orficie at XXX XXXXX of time XXXXXXXX distance travelled by flow, Yu= X/X*g*(X/XX)^X<XX> For lower orifice XX XXX point XX XXXX vertical XXXXXXXX XXXXXXXXX by XXXX, Yl = 1/X*g*(X/Vl)^2</p> <p>Forupper orifice, X^X = 2*Yu*Vu^X/g - XXX X<XX>XXX XXXXX orifice, X^2 = X*Yl*Vl^X/g - Eqn X</p> <p>XXXXX X is XXXX at XXX XXXXX XX XXXXXXXXXXXX, therefore, equating eqn 1 XXX eqn 2 we get,</p> <p>XX^X*Yu = XX^2*Yl - XXX X</p> <p>XXX XX the point XX intersection XX = XX+X - XXX X(as XXX XX XX a XXXXX XX X m and another at depth XX X m)<XX><br>Putting Eqn 4 in XXX 3 we XXX,</p> <p>XX^2*(Yl+2) = VI^X*YI</p> <p>Putting the values of Vu and XX in XXX XXXXXXXX XX XXX,</p> <p>X.XXX^2*(XX+X) = XX.XX^2*XX<XX>Or, 75.22*(YI+X) = 112.XX*XX<XX>XX, 37.XX XX= XXX.XX<XX>Or, YI = X m</p> <p>XXXXXXXXX, XXX XXXXXXXX of intersection XXXX XXX XXXXX XXXXXXX XX 4m , upper XXXXXXX XX X m XXX XXXXXXX of XXX XXXXX XX 10 m.</p> <p><XX></p>">
