Chapter 14, Problem 68P

To determine

The net head produced by the pump.

The required brake horsepower.

The angle at which the fluid impinges on the impeller blades a critical parameter in the design of the centrifugal pump.

Answer to Problem 68P

The net head produced by the pump in $\text{cm}$ of water column height is $0.294\text{cm}$.

The required brake horsepower is $\text{13,699}\text{W}$.

When there is a swirl at the inlet, the net head produced by the pump and the required brake horsepower will decrease. Thus, the angle at which the fluid impinges has an important role in the design of the centrifugal pumps.

Explanation of Solution

Given information:

The speed of the pump is $750\text{rpm}$, the entry angle of the water in the impeller is $7°$, the exit angle of the water in the impeller is $35°$, the inlet radius is $12\text{cm}$ at which the blade width is $18\text{cm}$, the outlet radius is $24\text{cm}$ at which the blade width is $16.2\text{cm}$, the volume flow rate is $0.573{\text{m}}^{\text{3}}/\text{s}$ and the efficiency is $100\%$.

The density of water is $998\text{kg}/{\text{m}}^{\text{3}}$, the density of air is $1.2\text{kg}/{\text{m}}^{\text{3}}$ and the acceleration due to gravity is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the angular velocity of the centrifugal pump in $\text{rad}/\text{s}$.

   $\omega =\frac{2\pi \dot{n}}{60}$    ...... (I)

Here, the speed of the pump is $\dot{n}$.

Write the expression for the normal velocity component at outlet of pump.

   $V_{2,n}=\frac{\dot{V}}{2\pi r_2{b}_2}$    ...... (II)

Here, the volume flow rate is $\dot{V}$, the outlet radius is $r_2$ and the blade width at exit is $b_2$.

Write the expression for the tangential velocity component at outlet of pump.

   $V_{2,t}=V_{2,n}\tan {\alpha }_2$    ...... (III)

Here, the normal velocity component at outlet of pump is $V_{2,n}$ and the exit angle of the water in the impeller is ${\alpha }_2$.

Write the expression for the normal velocity component at inlet of pump.

   $V_{1,n}=\frac{\dot{V}}{2\pi r_1{b}_1}$    ...... (IV)

Here, the inlet radius is $r_1$ and the blade width at inlet is $b_1$.

Write the expression for the tangential velocity component at inlet of pump.

   $V_{1,t}=V_{1,n}\tan {\alpha }_1$    ...... (V)

Here, the normal velocity component at inlet of pump is $V_{1,n}$ and the inlet angle of the water in the impeller is ${\alpha }_1$.

Write the expression for the linear velocity.

   $v=r_1.\omega$    ...... (VI)

Write the expression for the total head.

   $H=\left(\frac{v}{g}\right)\left(r_2{V}_{2,t}-r_1{V}_{1,t}\right)$    ...... (VII)

Here, the linear velocity of the centrifugal pump in is $v$ and the acceleration due to gravity is $g$.

Write the expression for the head in $\text{cm}$ of water column.

   $H_{water}=H\left(\frac{{\rho }_{air}}{{\rho }_{water}}\right)$    ...... (VIII)

Here, the density of air is ${\rho }_{air}$, the density of water is ${\rho }_{water}$, and the total head is $H$.

Write the expression for the brake horse power.

   $\text{bhp}={\rho }_{water}gH\dot{V}$    ...... (IX)

Calculation:

Calculating the parameters when inlet angle is $7°$.

Substitute $750\text{rpm}$ for $\dot{n}$ in the Equation (I).

   $\begin{array}{c}\omega =\frac{2\pi \left(750\text{rpm}\right)}{60}\\ =\frac{4712.38898\text{rpm}}{60}\\ =78.54\text{rad}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $24\text{cm}$ for $r_2$ and $16.2\text{cm}$ for $b_2$ in the Equation (II).

   $\begin{array}{c}{V}_{2,n}=\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(24\text{cm}\right)\left(16.2\text{cm}\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(388.8{\text{cm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^{\text{2}}}\right)\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{0.244290{\text{m}}^{\text{2}}}\\ =2.346\text{m}/\text{s}\end{array}$

Substitute $2.346\text{m}/\text{s}$ for $V_{2,n}$ and $35°$ for ${\alpha }_2$ in the Equation (III).

   $\begin{array}{c}{V}_{2,t}=\left(2.346\text{m}/\text{s}\right)\tan 35°\\ =2.346\text{m}/\text{s}\times 0.70020\\ =1.643\text{m}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $12\text{cm}$ for $r_1$ and $18\text{cm}$ for $b_1$ in the Equation (IV).

   $\begin{array}{c}{V}_{1,n}=\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(12\text{cm}\right)\left(18\text{cm}\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(216{\text{cm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^{\text{2}}}\right)\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{1357.168{\text{m}}^{\text{2}}}\\ =4.22\text{m}/\text{s}\end{array}$

Substitute $4.22\text{m}/\text{s}$ for $V_{1,n}$ and $0°$ for ${\alpha }_1$ in the Equation (V).

   $\begin{array}{c}{V}_{1,t}=\left(4.22\text{m}/\text{s}\right)\tan 7°\\ =4.22\text{m}/\text{s}\times 0.12278\\ =0.51813\text{m}/\text{s}\end{array}$

Substitute $78.54\text{rad}/\text{s}$ for $\omega$ and $12\text{cm}$ for $r_1$ in the Equation (VI).

   $\begin{array}{c}v=12\text{cm}\times 78.54\text{rad}/\text{s}\\ =12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\times 78.54\text{rad}/\text{s}\\ =0.12\text{m}\times 78.54\text{rad}/\text{s}\\ =9.4248\text{m}/\text{s}\end{array}$

Substitute $12\text{cm}$ for $r_1$, $24\text{cm}$ for $r_2$, $0.51813\text{m}/\text{s}$ for $V_{1,t}$, $1.643\text{m}/\text{s}$ for $V_{2,t}$, $9.4248\text{m}/\text{s}$ for $v$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (VII).

   $\begin{array}{c}H=\left(\frac{9.4248\text{m}/\text{s}}{9.81\text{m}/{\text{s}}^{\text{2}}}\right)\left\{\left(24\text{cm}\right)\left(1.643\text{m}/\text{s}\right)-\left(12\text{cm}\right)\left(0.51813\text{m}/\text{s}\right)\right\}\\ =\left(0.96073\text{s}\right)\left\{\begin{array}{l}\left(24\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(1.643\text{m}/\text{s}\right)-\\ \left(12\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(0.51813\text{m}/\text{s}\right)\end{array}\right\}\\ =\left(0.96073\text{s}\right)\left\{\left(0.24\text{m}\right)\left(1.643\text{m}/\text{s}\right)-\left(0.12\text{m}\right)\left(0.51813\text{m}/\text{s}\right)\right\}\\ =2.442\text{m}\end{array}$

Substitute $2.442\text{m}$ for $H$

   $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ and $1.2\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{air}$ in the Equation (VIII).

   $\begin{array}{c}{H}_{water}=2.442\text{m}\left(\frac{1.2\text{kg}/{\text{m}}^{\text{3}}}{998\text{kg}/{\text{m}}^{\text{3}}}\right)\\ =2.442\text{m}\times \left(1.202404\times {10}^{-3}\right)\\ =2.93627\times {10}^{-3}\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =0.294\text{cm}\end{array}$

Substitute $2.442\text{m}$ for $H$

   $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$

   $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in the Equation (IX).

   $\begin{array}{c}\text{bhp}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.442\text{m}\right)\left(0.573{\text{m}}^{\text{3}}/\text{s}\right)\\ =13,699\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}\right)\\ =13,699\text{W}\end{array}$

Calculating the parameters when inlet angle is $0°$.

Substitute $750\text{rpm}$ for $\dot{n}$ in the Equation (I).

   $\begin{array}{c}\omega =\frac{2\pi \left(750\text{rpm}\right)}{60}\\ =\frac{4712.38898\text{rpm}}{60}\\ =78.54\text{rad}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $24\text{cm}$ for $r_2$ and $16.2\text{cm}$ for $b_2$ in the Equation (II).

   $\begin{array}{c}{V}_{2,n}=\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(24\text{cm}\right)\left(16.2\text{cm}\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(388.8{\text{cm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^{\text{2}}}\right)\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{0.244290{\text{m}}^{\text{2}}}\\ =2.346\text{m}/\text{s}\end{array}$

Substitute $2.346\text{m}/\text{s}$ for $V_{2,n}$ and $35°$ for ${\alpha }_2$ in the Equation (III).

   $\begin{array}{c}{V}_{2,t}=\left(2.346\text{m}/\text{s}\right)\tan 35°\\ =2.346\text{m}/\text{s}\times 0.70020\\ =1.643\text{m}/\text{s}\end{array}$

Substitute $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $12\text{cm}$ for $r_1$ and $18\text{cm}$ for $b_1$ in the Equation (IV).

   $\begin{array}{c}{V}_{1,n}=\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(12\text{cm}\right)\left(18\text{cm}\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{2\pi \left(216{\text{cm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^{\text{2}}}\right)\right)}\\ =\frac{0.573{\text{m}}^{\text{3}}/\text{s}}{1357.168{\text{m}}^{\text{2}}}\\ =4.22\text{m}/\text{s}\end{array}$

Substitute $4.22\text{m}/\text{s}$ for $V_{1,n}$ and $0°$ for ${\alpha }_1$ in the Equation (V).

   $\begin{array}{c}{V}_{1,t}=\left(4.22\text{m}/\text{s}\right)\tan 0°\\ =4.22\text{m}/\text{s}\times 0\\ =0\text{m}/\text{s}\end{array}$

Substitute $78.54\text{rad}/\text{s}$ for $\omega$ and $12\text{cm}$ for $r_1$ in the Equation (VI).

   $\begin{array}{c}v=12\text{cm}\times 78.54\text{rad}/\text{s}\\ =12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\times 78.54\text{rad}/\text{s}\\ =0.12\text{m}\times 78.54\text{rad}/\text{s}\\ =9.4248\text{m}/\text{s}\end{array}$

Substitute $12\text{cm}$ for $r_1$, $24\text{cm}$ for $r_2$, $0\text{m}/\text{s}$ for $V_{1,t}$, $1.643\text{m}/\text{s}$ for $V_{2,t}$, $9.4248\text{m}/\text{s}$ for $v$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (VII).

   $\begin{array}{c}H=\left(\frac{9.4248\text{m}/\text{s}}{9.81\text{m}/{\text{s}}^{\text{2}}}\right)\left\{\left(24\text{cm}\right)\left(1.643\text{m}/\text{s}\right)-\left(12\text{cm}\right)\left(0\text{m}/\text{s}\right)\right\}\\ =\left(0.96073\text{s}\right)\left\{\left(24\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(1.643\text{m}/\text{s}\right)\right\}\\ =\left(0.96073\text{s}\right)\left\{\left(0.24\text{m}\right)\left(1.643\text{m}/\text{s}\right)\right\}\\ =3.157\text{m}\end{array}$

Substitute $3.157\text{m}$ for $H$

   $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ and $1.2\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{air}$ in the Equation (VIII).

   $\begin{array}{c}{H}_{water}=3.157\text{m}\left(\frac{1.2\text{kg}/{\text{m}}^{\text{3}}}{998\text{kg}/{\text{m}}^{\text{3}}}\right)\\ =3.157\text{m}\times \left(1.202404\times {10}^{-3}\right)\\ =3.8\times {10}^{-3}\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =0.38\text{cm}\end{array}$

Substitute $3.157\text{m}$ for $H$

   $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$

   $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.573{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in the Equation (IX).

   $\begin{array}{c}\text{bhp}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(3.157\text{m}\right)\left(0.573{\text{m}}^{\text{3}}/\text{s}\right)\\ =17710.4082\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}\right)\\ =17710.4082\text{W}\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ =17.710\text{kW}\end{array}$

Conclusion:

The net head produced by the pump in $\text{cm}$ of water column height is $0.294\text{cm}$.

The required brake horsepower is $\text{13,699}\text{W}$.

When there is a swirl at the inlet, the net head produced by the pump and the required brake horsepower will decrease. Thus, the angle at which the fluid impinges has an important role in the design of the centrifugal pumps.