Question: 16–133. Water leaves the impeller of the centrifugal pump with a velocity of 25 m/s and acceleration of 30 m/s², both measured relative to the impeller along the blade line AB. Determine the velocity and acceleration of a water particle at A as it leaves the impeller at the instant shown. The impeller rotates with a constant angular velocity of $$\omega = 15\ \text{rad/s}$$.

1. **Problem statement:** Water leaves the impeller of a centrifugal pump with velocity $$v_{rel} = 25\ \text{m/s}$$ and acceleration $$a_{rel} = 30\ \text{m/s}^2$$ relative to the impeller along blade line AB. The impeller rotates at angular velocity $$\omega = 15\ \text{rad/s}$$. We need to find the absolute velocity $$v$$ and absolute acceleration $$a$$ of the water particle at point A as it leaves the impeller. 2. **Given data:** - Relative velocity along AB: $$v_{rel} = 25\ \text{m/s}$$ - Relative acceleration along AB: $$a_{rel} = 30\ \text{m/s}^2$$ - Angular velocity of impeller: $$\omega = 15\ \text{rad/s}$$ - Radius from center to A: $$r = 0.3\ \text{m}$$ - Blade line AB is at an angle $$\theta = 30^\circ$$ above the leftward horizontal from A. 3. **Formulas and concepts:** - Absolute velocity $$\vec{v} = \vec{v}_{rel} + \vec{v}_{rot}$$ where $$\vec{v}_{rot} = \vec{\omega} \times \vec{r}$$ - Absolute acceleration $$\vec{a} = \vec{a}_{rel} + \vec{a}_{rot} + \vec{a}_{cor}$$ where - $$\vec{a}_{rot} = \vec{\alpha} \times \vec{r}$$ (angular acceleration term, zero here since $$\omega$$ is constant) - $$\vec{a}_{cor} = 2 \vec{\omega} \times \vec{v}_{rel}$$ (Coriolis acceleration) - $$\vec{a}_{rel}$$ is acceleration relative to impeller Since $$\omega$$ is constant, $$\alpha = 0$$, so $$\vec{a}_{rot} = 0$$. 4. **Step 1: Define coordinate system and vectors** - Let the impeller rotate counterclockwise about the z-axis. - Position vector $$\vec{r}$$ from center to A is vertical upward: $$\vec{r} = 0.3\hat{j}$$. - Angular velocity vector $$\vec{\omega} = 15\hat{k}$$ (out of plane). 5. **Step 2: Calculate rotational velocity $$\vec{v}_{rot}$$** $$\vec{v}_{rot} = \vec{\omega} \times \vec{r} = 15\hat{k} \times 0.3\hat{j} = 15 \times 0.3 (\hat{k} \times \hat{j}) = 4.5 (-\hat{i}) = -4.5\hat{i}$$ 6. **Step 3: Express relative velocity $$\vec{v}_{rel}$$ along AB** - AB is at $$30^\circ$$ above leftward horizontal from A. - Leftward horizontal is $$-\hat{i}$$, so direction of AB is: $$\hat{e}_{AB} = \cos 150^\circ \hat{i} + \sin 150^\circ \hat{j} = -\cos 30^\circ \hat{i} + \sin 30^\circ \hat{j} = -0.866\hat{i} + 0.5\hat{j}$$ - Relative velocity vector: $$\vec{v}_{rel} = 25(-0.866\hat{i} + 0.5\hat{j}) = -21.65\hat{i} + 12.5\hat{j}$$ 7. **Step 4: Calculate absolute velocity $$\vec{v}$$** $$\vec{v} = \vec{v}_{rel} + \vec{v}_{rot} = (-21.65 - 4.5)\hat{i} + 12.5\hat{j} = -26.15\hat{i} + 12.5\hat{j}$$ 8. **Step 5: Express relative acceleration $$\vec{a}_{rel}$$ along AB** $$\vec{a}_{rel} = 30(-0.866\hat{i} + 0.5\hat{j}) = -25.98\hat{i} + 15\hat{j}$$ 9. **Step 6: Calculate Coriolis acceleration $$\vec{a}_{cor} = 2 \vec{\omega} \times \vec{v}_{rel}$$** $$\vec{a}_{cor} = 2 \times 15\hat{k} \times (-21.65\hat{i} + 12.5\hat{j}) = 30 (\hat{k} \times (-21.65\hat{i} + 12.5\hat{j}))$$ Calculate cross products: $$\hat{k} \times \hat{i} = \hat{j}$$ $$\hat{k} \times \hat{j} = -\hat{i}$$ So: $$\vec{a}_{cor} = 30 (-21.65 \hat{j} - 12.5 \hat{i}) = -375\hat{i} - 649.5\hat{j}$$ 10. **Step 7: Calculate total absolute acceleration $$\vec{a}$$** Since $$\alpha = 0$$, $$\vec{a}_{rot} = 0$$, so $$\vec{a} = \vec{a}_{rel} + \vec{a}_{cor} = (-25.98 - 375)\hat{i} + (15 - 649.5)\hat{j} = -400.98\hat{i} - 634.5\hat{j}$$ 11. **Step 8: Calculate magnitudes and directions** - Velocity magnitude: $$|\vec{v}| = \sqrt{(-26.15)^2 + 12.5^2} = \sqrt{683.8 + 156.25} = \sqrt{840.05} \approx 28.98\ \text{m/s}$$ - Velocity direction angle $$\phi_v$$ from negative x-axis: $$\tan \phi_v = \frac{12.5}{-26.15} = -0.478$$ Since x is negative and y positive, angle is in second quadrant: $$\phi_v = 180^\circ - 25.6^\circ = 154.4^\circ$$ - Acceleration magnitude: $$|\vec{a}| = \sqrt{(-400.98)^2 + (-634.5)^2} = \sqrt{160784 + 402570} = \sqrt{563354} \approx 750.57\ \text{m/s}^2$$ - Acceleration direction angle $$\phi_a$$ from negative x-axis: $$\tan \phi_a = \frac{-634.5}{-400.98} = 1.582$$ Both components negative, angle in third quadrant: $$\phi_a = 180^\circ + 58.1^\circ = 238.1^\circ$$ **Final answers:** - Absolute velocity magnitude $$\approx 29.0\ \text{m/s}$$ at angle $$154^\circ$$ from positive x-axis (or 25.6° above left horizontal). - Absolute acceleration magnitude $$\approx 751\ \text{m/s}^2$$ at angle $$238^\circ$$ from positive x-axis.