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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 air slows XXX liquid down XXX to friction XXX the XXXX diverges.</p> <p>XXX, we XXXX XXX things to XXXXXXXX XXXX here.</p> <p>XXX XXXXX XX the discharge velocity XXX the XXXXXX is the area XX XXX vena XXXXXXXXX XX determine XXX rate XX XXXXXXXXX, Q.</p> <p>Discharge XXXXXXXX:<XX></p> <p>As XX know XXXX the theoretical XXXXXXXX XX XXX XXXX XXX XX XXXXXXXXXX using XXX XXXXXXXXX<sub>i</sub>= √2gh.<br></p> <p>XXXX, at upper orifice,Vi = XXXX(2*X.8*X) = X.XX m/s<br>XX XXXXX XXXXXXX, XX = root(2*9.8*X) = 10.XX m/s</p> <p>Discharge velocity, X = Vi* XXXXXXXXXXX XX viscocity<XX>For XXXXX orifice, XX=8.XX*X.98 = X.XXX m/s<XX>For XXXXX orifice, XX= XX.XX*X.98 = XX.XX m/s</p> <p>XXXXXXXX, the area XX the vena contracta, A = XX*XXXXXXXXXXX XX contraction (Ao = Area XX XXXXX section XX the orifice)<XX>XXXXXXXXX, A = pie*(r^X)*X.XX<br>For upper XXXXXXX, A = (22/X)*(X/2)^X*0.64 = XX.57 XXX or X.001257 XX<br>XXX lower XXXXXXX, A =(22/X)*(X/2)^X*X.XX =XX.57 cm2 or 0.001257 m2</p> <p>PartA )XXX, the XXXXXXXXX or XXXX, X XXXX an XXXXXXX = X*A = XX*Cv*XX*Cc (XX = Contraction coefficient, Cv= coefficient XX viscosity)<br>XXX upper XXXXXXX, XX= Vu*X = X.XXX*X.001257 = X.XXXX m3/s<XX>XXX lower orifice, Ql = XX*X = 10.XX*X.XXXXXX = 0.0133 XX/s</p> <p>Part B)</p> <p>XXXXXXXXXX XXXXXXXX XX XXXXX XXXXXXX, XX = 8.XXX m/s<br>Horizontal XXXXXXXX of lower orifice, Vl = 10.XX m/s</p> <p>XXX the horizontal XXXXXXXX at XXXXX, XXX XXXXX XXXXXXXXX is X.</p> <p>XXXXXXXXX, X= X*t (X= XXXXXXXX XX XXX XXXX, t = XXXX) or t = X / X<XX>Using equation XX XXX XXXXXXXXXXX XXXXXX,<XX>y = ut + X/X g*t^X</p> <p>But, u = 0, XXXXXXXXX, y = 1/2*g*t^2</p> <p>XXX, XXX upper XXXXXXX at XXX XXXXX of XXXX XXXXXXXX distance XXXXXXXXX by XXXX, XX= 1/X*g*(X/Vu)^X<br> XXX lower orifice at XXX point of time XXXXXXXX XXXXXXXX XXXXXXXXX by flow, XX = 1/X*g*(X/XX)^2</p> <p>XXXXXXXX orifice, X^2 = X*Yu*Vu^2/g - XXX X<XX>XXX lower XXXXXXX, X^2 = X*XX*Vl^2/g - Eqn X</p> <p>Since X XX same XX XXX XXXXX XX intersection, therefore, XXXXXXXX eqn X and eqn X we get,</p> <p>Vu^X*XX = Vl^2*Yl - XXX X</p> <p>Now at XXX point XX intersection XX = Yl+X - XXX 4(as one XX at a depth XX X m XXX XXXXXXX at depth XX 6 m)<XX><br>Putting Eqn X in Eqn 3 we get,</p> <p>Vu^X*(XX+X) = VI^2*YI</p> <p>Putting the XXXXXX of Vu XXX XX in the equation XX get,</p> <p>X.XXX^2*(XX+X) = 10.62^2*XX<XX>XX, 75.XX*(YI+2) = 112.XX*YI<XX>Or, XX.XX XX= 150.44<XX>Or, XX = 4 m</p> <p>Therefore, the distance of intersection from XXX XXXXX XXXXXXX is 4m , upper orifice is X m and XXXXXXX of XXX XXXXX is XX m.</p> <p><br></p>">
