Chapter 14, Problem 1P

To determine

The initial blade angle and the relative velocity of the water flows onto the blades of the impeller.

Explanation of Solution

Given:

Mean radius $\left(r_m\right)$ is $\text{75}\text{mm}$.

Discharge $\left(Q\right)$ is $0.095{\text{m}}^{\text{3}}/\text{s}$.

Angular speed of the impeller $\left(\omega \right)$ is $80\text{rad}/\text{s}$.

Calculation:

Calculate the velocity at the midpoint of the impeller $\left(U\right)$.

  $\begin{array}{c}U=r_m\omega \\ =\left(75\text{mm}\right)\left(80\text{rad}/\text{s}\right)\\ =\left(75\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\left(80\text{rad}/\text{s}\right)\\ =6\text{m}/\text{s}\end{array}$

Calculate the axial velocity of the water through the impeller $\left(V_a\right)$.

  $\begin{array}{c}{V}_a=\frac{Q}{A}\\ =\frac{0.095{\text{m}}^{\text{3}}/\text{s}}{\pi \left[{\left(0.1\text{m}\right)}^2-{\left(0.05\text{m}\right)}^2\right]}\\ =4.0319\text{m}/\text{s}\end{array}$

Draw the kinematic diagram of the water at the lead of the blade as shown in Figure (I).

Calculate the initial blade angle $\left({\beta }_1\right)$.

  $\begin{array}{c}\tan {\beta }_1=\frac{V_a}{U}\\ =\frac{4.0319\text{m}/\text{s}}{6\text{m}/\text{s}}\end{array}$

  ${\beta }_1=33.9°$

Thus, the initial blade angle is $33.9°$.

Calculate the relative velocity of the water ${\left(V_{\text{rel}}\right)}_1$.

  $\begin{array}{l}\cos {\beta }_1=\frac{U}{{\left(V_{\text{rel}}\right)}_1}\\ \cos \left(22.6°\right)=\frac{6\text{m}/\text{s}}{{\left(V_{\text{rel}}\right)}_1}\\ {\left(V_{\text{rel}}\right)}_1=7.23\text{m}/\text{s}\end{array}$

Thus, the relative velocity of the water flows on to the blades of the impeller is $7.23\text{m}/\text{s}$.