Toggle navigation
Home
Ask A Question
Ask
Answer Questions
Answer
Users
Support/FAQ
Login
Register
Preview 50% of the Answer Below
Due to formatting, images, tables, or paragraphs may be out of place or not shown. Rest assured that they will be included and formatted correctly in your purchased answer.
Details
Question Title:
A tank has two identical orifices, each 5 cm in
Go Back to the Question
Answer Preview
</figure> <p>Once the liquid passes the vena XXXXXXXXX, the air XXXXX XXX XXXXXX XXXX due to friction XXX the XXXX XXXXXXXX.</p> <p>Now, XX have two things to XXXXXXXX XXXX XXXX.</p> <p>XXX XXXXX XX XXX discharge XXXXXXXX and XXX XXXXXX XX XXX area of the XXXX XXXXXXXXX to determine the XXXX of XXXXXXXXX, Q.</p> <p>XXXXXXXXX XXXXXXXX:<br></p> <p>As XX XXXX that XXX theoretical velocity XX the flow XXX be calculated using the equationV<sub>i</sub>= √XXX.<br></p> <p>Thus, XX upper XXXXXXX,XX = XXXX(X*9.X*4) = 8.XX m/s<XX>XX lower XXXXXXX, Vi = root(X*9.X*6) = XX.XX m/s</p> <p>Discharge velocity, V = XX* XXXXXXXXXXX of viscocity<XX>XXX upper XXXXXXX, XX=8.85*X.XX = 8.673 m/s<XX>For XXXXX orifice, Vl= 10.84*0.98 = XX.XX m/s</p> <p>Secondly, XXX XXXX of the XXXX XXXXXXXXX, X = Ao*coefficient of XXXXXXXXXXX (XX = XXXX XX XXXXX section XX XXX orifice)<br>Therefore, X = pie*(r^2)*X.64<br>XXX XXXXX orifice, A = (XX/7)*(X/X)^X*0.XX = 12.57 XXX or 0.001257 XX<XX>XXX XXXXX XXXXXXX, X =(22/7)*(5/2)^X*0.XX =12.57 XXX or X.001257 XX</p> <p>XXXXX )Now, the XXXXXXXXX or flow, Q from an orifice = X*A = Vi*Cv*XX*Cc (XX = XXXXXXXXXXX XXXXXXXXXXX, Cv= coefficient XX viscosity)<br>For upper orifice, Qu= Vu*A = X.673*0.XXXXXX = 0.0109 m3/s<br>For lower orifice, Ql = Vl*X = 10.62*X.001257 = 0.XXXX m3/s</p> <p>XXXX X)</p> <p>XXXXXXXXXX XXXXXXXX of upper orifice, XX = 8.673 m/s<br>XXXXXXXXXX XXXXXXXX XX lower XXXXXXX, Vl = XX.XX m/s</p> <p>Let the XXXXXXXXXX distance XX which, the XXXXX XXXXXXXXX is X.</p> <p>Therefore, X= V*t (X= XXXXXXXX of XXX XXXX, t = time) or t = X / X<XX>XXXXX equation XX XXX dimensional XXXXXX,<br>y = ut + 1/X g*t^2</p> <p>XXX, u = 0, XXXXXXXXX, y = 1/X*g*t^2</p> <p>Now, for XXXXX XXXXXXX XX any point of XXXX vertical distance travelled by flow, Yu= X/2*g*(X/Vu)^X<br> XXX XXXXX orifice XX XXX point of time XXXXXXXX distance travelled by flow, Yl = 1/X*g*(X/Vl)^2</p> <p>Forupper orifice, X^X = X*Yu*Vu^2/g - XXX 1<XX>XXX XXXXX orifice, X^2 = X*Yl*XX^X/g - Eqn 2</p> <p>Since X XX XXXX XX XXX XXXXX XX intersection, XXXXXXXXX, equating XXX 1 XXX XXX X XX get,</p> <p>XX^X*XX = XX^2*XX - XXX 3</p> <p>Now XX the point of XXXXXXXXXXXX XX = Yl+2 - Eqn X(as one XX at a XXXXX XX 4 m XXX another XX depth XX X m)<br><XX>Putting XXX X in XXX 3 XX get,</p> <p>XX^2*(XX+X) = XX^2*YI</p> <p>Putting XXX XXXXXX of Vu XXX XX in the equation we XXX,</p> <p>8.673^2*(YI+X) = XX.62^2*XX<XX>XX, 75.22*(YI+2) = XXX.78*XX<XX>XX, 37.XX YI= XXX.44<XX>XX, XX = X m</p> <p>Therefore, XXX distance of intersection from the XXXXX orifice is 4m , upper XXXXXXX is 6 m and XXXXXXX XX the XXXXX is XX m.</p> <p><br></p>">
