Question 14.6: Preliminary Design of a Centrifugal Pump A centrifugal pump ...

For a given flow rate, net head, and dimensions of a centrifugal pump, we are to design the blade shape (leading and trailing edge angles). We are also to estimate the horsepower required by the pump.
Assumptions   1  The flow is steady. 2  The liquid is incompressible. 3  There are no irreversible losses through the impeller. 4  This is only a preliminary design.
Properties   For refrigerant R-134a at T = 20°C, v_f = 0.0008157  m^3/kg. Thus \rho = 1/v_f = 1226  kg/m^3.
Analysis   We calculate the required water horsepower from Eq. 14–3,

\dot{W}_{\text {water horsepower }}=\rho g \dot{V} H             (14.3)

= 43,600 W

The required brake horsepower will be greater than this in a real pump. However, in keeping with the approximations for this preliminary design, we assume 100 percent efficiency such that bhp is approximately equal to \dot{W}_{\text {water  horsepower }} ,

\text { bhp } \cong \dot{W}_{\text {water  horsepower }}=43,600  W \left(\frac{ hp }{745.7  W }\right)=58.5  hp

We report the final result to two significant digits in keeping with the precision of the given quantities; thus, bhp ≈ 59 horsepower.
In all calculations with rotation, we need to convert the rotational speed from \dot{n} (rpm) to 𝜔 (rad/s), as illustrated in Fig. 14–43,

\omega=1720 \frac{\operatorname{rot}}{\min }\left(\frac{2 \pi  rad }{\operatorname{rot}}\right)\left(\frac{1  min }{60  s }\right)=180.1  rad / s            (1)

We calculate the blade inlet angle using Eq. 14–25,

\dot{V}=2 \pi b_1 \omega r_1^2 \tan \beta_1             (14.25)

\beta_1=\arctan \left(\frac{\dot{V}}{2 \pi b_1 \omega r_1^2}\right)=\arctan \left(\frac{0.25  m ^3 / s }{2 \pi(0.050  m )(180.1  rad / s )(0.10  m )^2}\right)=23.8^{\circ}

We find \beta_2 by utilizing the equations derived earlier for our elementary analysis.
First, for the design condition in which V_{1, t} = 0, Eq. 14–17 reduces to

H=\frac{1}{g}\left(\omega r_2 V_{2, t}  –  \omega r_1 V_{1, t}\right)                   (14.17)

Net head:                        H=\frac{1}{g}(\omega r_2 V_{2, t}  –  \omega r_1 \underbrace{\cancel{V _{1, t}}}_0)=\frac{\omega r_2 V_{2, t}}{g}

from which we calculate the tangential velocity component,

V_{2, t}=\frac{g H}{\omega r_2}                     (2)

Using Eq. 14–12, we calculate the normal velocity component,

\dot{V}=2 \pi r_1 b_1 V_{1, n}=2 \pi r_2 b_2 V_{2, n}        (14.12)

V_{2, n}=\frac{\dot{V}}{2 \pi r_2 b_2}                 (3)

Next, we perform the same trigonometry used to derive Eq. 14–23, but on the trailing edge of the blade rather than the leading edge. The result is

V_{1, t}=\omega r_1  –  \frac{V_{1, n}}{\tan \beta_1}          (14.23)

V_{2, t}=\omega r_2  –  \frac{V_{2, n}}{\tan \beta_2}

from which we finally solve for \beta_2,

\beta_2=\arctan \left(\frac{V_{2,  n}}{\omega r_2  –  V_{2, t}}\right)                 (4)

After substitution of Eqs. 2 and 3 into Eq. 4, and insertion of the numerical values, we obtain

\beta_2 = 14.7°

We report the final results to only two significant digits. Thus our preliminary design requires backward-inclined impeller blades with \beta_1 ≅ 24° and \beta_2 ≅ 15°.
Once we know the leading and trailing edge blade angles, we design the detailed shape of the impeller blade by smoothly varying blade angle 𝛽 from \beta_1 to \beta_2 as radius increases from r_1 to r_2. As sketched in Fig. 14–44, the blade can be of various shapes while still keeping \beta_1 ≅ 24° and \beta_2 ≅ 15°, depending on how we vary 𝛽 with the radius. In the figure, all three blades begin at the same location (zero absolute angle) at radius r_1; the leading edge angle for all three blades is \beta_1 = 24°. The medium length blade (the brown one in Fig. 14–44) is constructed by varying 𝛽 linearly with r. Its trailing edge intercepts radius r_2 at an absolute angle of approximately 93°. The longer blade (the black one in the figure) is constructed by varying 𝛽 more rapidly near r_1 than near r_2. In other words, the blade curvature is more pronounced near its leading edge than near its trailing edge. It intercepts the outer radius at an absolute angle of about 114°. Finally, the shortest blade (the blue blade in Fig. 14–44) has less blade curvature near its leading edge, but more pronounced curvature near its trailing edge. It intercepts r_2 at an absolute angle of approximately 77°. It is not immediately obvious which blade shape is best.
Discussion   Keep in mind that this is a preliminary design in which irreversible losses are ignored. A real pump would have losses, and the required brake horsepower would be higher (perhaps 20 to 30 percent higher) than the value estimated here. In a real pump with losses, a shorter blade has less skin friction drag, but the normal stresses on the blade are larger because the flow is turned more sharply near the trailing edge where the velocities are largest; this may lead to structural problems if the blades are not very thick, especially when pumping dense liquids. A longer blade has higher skin friction drag, but lower normal stresses. In addition, you can see from a simple blade volume estimate in Fig. 14–44 that for the same number of blades, the longer the blades, the more flow blockage, since the blades are of finite thickness. In addition, the displacement thickness effect of boundary layers growing along the blade surfaces (Chap. 10) leads to even more pronounced blockage for the long blades. Obviously some engineering optimization is required to determine the exact shape of the blade.