--- title: "Rotating tube with quadrant, speed and head" subject: "Fluid Mechanics" chapter: 2 key: "semester_3/fluid_mechanics:rotating-tube-abcd-speed" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rotating-tube-abcd-speed site: BCT Engineering --- # Rotating tube with quadrant, speed and head Fluid Mechanics · Chapter 2: Pressure and Head - Asked in: 2076 Asoj, 5+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rotating-tube-abcd-speed - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2 ## Question A tube ABCD has the end A open to atmosphere and the end D closed as shown in figure below. The portion ABC is vertical while the portion CD is a quadrant of radius 250 mm with its centre is B, the whole being arranged to rotate about its vertical axis ABC. If the tube is completely filled with water to a height in the vertical limb of 300 mm above C find (a) the speed of rotation which will make the pressure head at D equal to pressure head at C, (b) the value and position of the maximum pressure head in the curved portion CD when running at the speed. [Figure: vertical limb A-B-C with 50 mm and 250 mm marked, quadrant CD of radius 250 mm centred at B] ## Short answer (a) Equating pressure heads at D and C gives ω = √(2g/R) = 8.86 rad/s (84.6 rpm). (b) Maximum pressure head in the curved part is 0.3625 m of water, at 60° from C along the arc. ## Answer **Reading of the figure:** $B$ is on the axis, $C$ is on the axis 250 mm below $B$, and the quadrant $CD$ (radius $R=0.25$ m, centre $B$) curves from $C$ up to $D$, which is level with $B$ and 0.25 m from the axis. The water surface in $A$ is 0.3 m above $C$ (on the axis), and $A$ is open. ```text A o <- water surface, 0.3 m above C | | B +------- D (r = 0.25 m) | / | / quadrant, centre B C +-- ``` For rigid-body rotation, the pressure head at a point at radius $r$ and height $z$ above $C$ is $$ h_p=h_C+\frac{\omega^{2}r^{2}}{2g}-z,\qquad h_C=0.3\ \text{m} $$ #### (a) Speed for $h_D=h_C$ At $D$: $r=0.25$ m and $z=0.25$ m. $$ \frac{\omega^{2}(0.25)^{2}}{2g}=0.25\ \Rightarrow\ \omega^{2}=\frac{2(9.81)}{0.25}=78.48 $$ $$ \omega=8.859\ \text{rad/s},\qquad N=\frac{60\omega}{2\pi}=84.6\ \text{rpm} $$ #### (b) Maximum head in the curved part Let a point on the quadrant make angle $\theta$ at $B$ measured from $BC$. Then $r=R\sin\theta$, $z=R(1-\cos\theta)$. With $\dfrac{\omega^{2}R^{2}}{2g}=R$: $$ h_p=0.3+R\sin^{2}\theta-R(1-\cos\theta)=0.3+R(\cos\theta-\cos^{2}\theta) $$ $$ \frac{dh_p}{d\theta}=0\ \Rightarrow\ \sin\theta(2\cos\theta-1)=0\ \Rightarrow\ \cos\theta=0.5,\ \theta=60^\circ $$ $$ h_{p,max}=0.3+0.25(0.5-0.25)=0.3625\ \text{m of water} $$ Position: at $60^\circ$ from $C$ along the arc, i.e. 0.2165 m from the axis and 0.125 m above $C$. **Answer: (a) 8.86 rad/s (84.6 rpm); (b) maximum head 0.3625 m of water, at $60^\circ$ from C on the arc.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the pressure head at radius r and height z above C? A. h_C + omega^2 r^2/2g - z B. h_C - omega^2 r^2/2g + z C. h_C + omega r/g + z D. h_C + omega^2 r^2/g Answer: A. Rotation adds omega^2 r^2/2g and rising by z subtracts z. (Key point: Pressure head in rotation) 2. At D (r = z = 0.25 m), what condition gives h_D = h_C and what omega results? A. omega^2 r^2/2g = 0 so omega = 0 B. omega^2 (0.25)^2/2g = 0.25 so omega = 8.86 rad/s C. omega^2 (0.25)^2/2g = 0.3 so omega = 9.7 rad/s D. omega = 84.6 rad/s Answer: B. Rotation head must equal the rise of 0.25 m, giving omega^2 = 78.48, omega = 8.86 rad/s (84.6 rpm). (Key point: Speed for equal heads) 3. Setting dh_p/d(theta) = 0 gives which angle of maximum pressure head on the arc? A. 30 degrees B. 45 degrees C. 60 degrees D. 90 degrees Answer: C. sin(theta)(2 cos(theta) - 1) = 0 gives cos(theta) = 0.5, theta = 60 degrees. (Key point: Maximum head location) 4. What is the maximum pressure head in CD? A. 0.3625 m B. 0.3 m C. 0.55 m D. 0.25 m Answer: A. h = 0.3 + 0.25(0.5 - 0.25) = 0.3625 m of water. 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