Chapter 10, Problem 1P

The geometry of a centrifugal water pump is r₁ = 10 cm, r₂ = 20 cm, b₁ = b₂ = 4 cm, β₁ = 30°, β₂ = 15°, and it runs at speed 1600 rpm. Estimate the discharge required for axial entry, the power generated in the water in watts, and the head produced.

To determine

The discharge required for axial entry, the power generated in the water in watts, and the head produced.

Explanation of Solution

Given:

Inlet impeller radius is $r_1=10\text{cm}$.

Outlet impeller radius is $r_2=20\text{cm}$.

Inlet blade angle is ${\beta }_1=30°$.

Outlet blade angle is ${\beta }_2=15°$.

Speed of the centrifugal water pump is $N=1600\text{rpm}$.

Inlet tangential velocity is $V_{t_1}=0$

Impeller inlet width $b_1=4\text{cm}$.

Impeller outlet width $b_2=4\text{cm}$.

Calculation:

Calculate the angular speed of centrifugal water pump $\left(\omega \right)$.

  $\begin{array}{c}\omega =\frac{2\pi N}{60}\\ =1600\frac{\text{rev}}{\min }\times 2\pi \frac{\text{rad}}{\text{rev}}\times \frac{\text{rev}}{60\text{s}}\\ =167.46\text{rad}/\text{s}\end{array}$

Calculate the runner speed at inlet $\left(U_1\right)$.

  $\begin{array}{c}{U}_1=\omega r_1\\ =167.46\frac{\text{rad}}{\text{sec}}\times 10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ \text{=}16.746\text{m}/\text{sec}\end{array}$

Calculate the runner speed at outlet $\left(U_2\right)$.

  $\begin{array}{c}{U}_2=\omega r_2\\ =167.46\times 20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =33.492\text{m}/\text{s}\end{array}$

Draw inlet and outlet velocity diagram as shown in Figure (1).

From Figure (1).

  $\begin{array}{l}\cos \beta =\frac{U-V_t}{V_{r_b}}\\ \sin \beta =\frac{V_n}{V_{r_b}}\\ \tan \beta =\frac{V_n}{U-V_t}\end{array}$

At inlet, $V_{t_1}=0$.

Calculate $V_{n_1}$.

  $\begin{array}{l}\tan \beta =\frac{V_n}{U-V_t}\\ V_{n_1}=U_1\tan {\beta }_1\\ V_{n_1}=16.74\text{}\text{}\text{m}/\text{s}\times \tan 30°\\ V_{n_1}=9.66\text{m/s}\end{array}$

Using continuity equation calculate the discharge required at the entry $\left(Q\right)$.

  $\begin{array}{c}Q=2\pi r_1{b}_1{V}_n\\ =2\times \pi \times \left[10\text{}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right]\times \left[4\text{}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right]\times \left(9.66\text{m}/\text{s}\right)\\ =0.2427{\text{m}}^3/\text{s}\end{array}$

Thus, the discharge required at the entry $\left(Q\right)$ is $\underset{\_}{0.2427{\text{m}}^3/\text{s}}$.

At Outlet,

Calculate tangential velocity at outlet $\left(V_{t_2}\right)$.

  $\begin{array}{c}{V}_{t_2}=U_2-\frac{Q}{2\cdot \pi \cdot r_2\cdot b_2}\cdot \cot ({\beta }_2)\\ =33.492\text{m}/\text{s}-\frac{0.2427{\text{m}}^3/\text{s}}{2\times 3.14\times \left[20\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right]\times 4\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\cdot \cot (15°)\\ =15.466\text{m}/\text{s}\end{array}$

Calculate the theoretical power generated in the water $\left({\dot{W}}_m\right)$.

  $\begin{array}{c}{\dot{W}}_m=\left(U_2{V}_{t_2}-U_1{V}_{t_1}\right)\dot{m}\\ =\left(U_2{V}_{t_2}\right)\rho Q\\ =\left[\begin{array}{l}(33.492\text{m}/\text{s}\times 15.466\text{m}/\text{s})\\ \times 1000\text{kg}/{\text{m}}^3\times 0.2427{\text{m}}^3/\text{s}\end{array}\right]\\ =\text{125715}\text{.51}\text{W}\end{array}$

Thus, the power generated in the water $\left({\dot{W}}_m\right)$ is $\underset{\_}{\text{125715}\text{.51}\text{W}}$.

Calculate the theoretical head $\left(H\right)$.

  $\begin{array}{c}H=\frac{1}{g}\left(U_2{V}_{t_2}-U_1{V}_{t_1}\right)\\ H=\frac{1}{g}\left(U_2{V}_{t_2}\right)\\ =\frac{1}{9.81{\text{m}/\text{s}}^2}\times (33.492\text{m}/\text{s}\times 15.466\text{m}/\text{s})\\ =52.08\text{m}\end{array}$

Thus, the theoretical head $\left(H\right)$ is $\underset{\_}{52.08\text{m}}$.