A centrifugal pump which runs at \(104.3 \mathrm{rad} \cdot \mathrm{s}^{-1}(16.6 \mathrm{rev} / \mathrm{s})\) is mounted so that its centre is \(2.4 \mathrm{~m}\) above the water level in the suction sump. It delivers water to a point \(19 \mathrm{~m}\) above its centre. For a flow rate of \(Q\left(\mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\right)\) the friction loss in the suction pipe is \(68 Q^{2} \mathrm{~m}\) and that in the delivery pipe is \(650 Q^{2} \mathrm{~m}\). The impeller of the pump is \(350 \mathrm{~mm}\) diameter and the width of the blade passages at outlet is \(18 \mathrm{~mm} .\) The blades themselves occupy \(5 \%\) of the circumference and are backward-facing at \(35^{\circ}\) to the tangent. At inlet the flow is radial and the radial component of velocity remains unchanged through the impeller. Assuming that \(50 \%\) of the velocity head of the water leaving the impeller is converted to pressure head in the volute, and that friction and other losses in the pump, the velocity heads in the suction and delivery pipes and whirl slip are all negligible, calculate the rate of flow and the manometric efficiency of the pump.

Step by step solution

01

Calculate the total head

The total head (\(h_t\)) consists of the static head (the vertical distance the water is lifted) and the friction losses in both the suction and delivery pipes. The static head can be found by adding the distances above and below the pump center: \[ h_{static} = 2.4 m + 19 m = 21.4 m \] The friction losses can be represented as: \[ h_{friction} = 68Q^2 (suction) + 650Q^2 (delivery) = 718Q^2 \] So, the total head is: \[ h_t = h_{static} + h_{friction} = 21.4 + 718Q^2 \]

02

Compute the radial component of velocity

Since the flow is radial at the inlet and remains unchanged through the impeller, the radial component of velocity (\(v_r\)) remains the same. The volumetric flow rate \(Q\) is related to the impeller dimensions by: \[ Q = A_r \cdot v_r \] where \(A_r\) is the cross-sectional area at the impeller outlet. The area is: \[ A_r = \frac{{\pi \left(D_{\text{impeller}} - t\right)^2}}{4} - \frac{{n \cdot b \cdot t}}{360} \] Given that the impeller diameter \(D_{\text{impeller}} = 0.35 m\), blade passage width \(b = 0.018 m\), blade angular coverage \(n = 0.05\), and blade angle \(t = 35^\circ\): \[ A_r = \frac{{\pi \left(0.35 - 0.018\right)^2}}{4} - \frac{{0.05 \cdot 0.018 \cdot 35}}{360} = 0.0093 \ m^2 \] If we assume the blades are backward-facing, then the radial component of velocity is: \[ v_r = \frac{Q}{A_r} = \frac{Q}{0.0093} \]

03

Find Blade Speed at outlet

We need the blade speed at the outlet. The formula for blade speed (\(u_2\)) at the impeller outer radius (\(R_2\)) is: \[ u_2 = \omega \cdot R_2 \] Given: \(\omega = 104.3 \ rad/s\) and \(R_2 = 0.35m / 2 = 0.175m\). Therefore, \[ u_2 = 104.3 \cdot 0.175 = 18.252 \ m/s\]

04

Calculate the theoretical head

The theoretical head (\(h_{theoretical}\)) can be found using the Euler equation for turbomachinery: \[ h_{theoretical} = \frac{{u_2 \cdot v_u2}}{g} \] However, we need the tangential component of the velocity \(v_{u2}\). For backward-facing blades: \[ v_{u2} = u_2 - v_r \cdot \tan(\beta_2) \] Given: \(\beta_2 = 35^\circ\): \[ v_{u2} = 18.252 - \left( \frac{Q}{0.0093}\right) \cdot \tan(35) \]

05

Compute the Rate of Flow \(Q\)

Equate the total head to the theoretical head to solve for \(Q\): \[ h_t = h_{static} + 718Q^2 = \frac{{u_2^2}}{2g} \] Substitute the values: \[ 21.4 + 718Q^2 = \frac{{(18.252 - \left( \frac{Q}{0.0093}\right) \cdot \tan(35))^2}}{2x9.81} \]

06

Find Manometric Efficiency

Manometric Efficiency (\(\eta_m\)) is given by: \[ \eta_m = \frac{h_{static}}{h_t} = \frac{21.4}{21.4 + 718Q^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

total head calculation

The total head is a crucial concept in centrifugal pump analysis, representing the energy imparted to the fluid by the pump. It comprises two main components: static head and friction head. The static head is the vertical distance the water is lifted, either above or below the pump center. In our case:
\[ h_{static} = 2.4 \text{ m} + 19 \text{ m} = 21.4 \text{ m} \]
Next, the friction head accounts for energy losses due to friction in the suction and delivery pipes. It's given by:

• Suction Pipe loss: 68\( Q^2 \)
• Delivery Pipe loss: 650\( Q^2 \)

Thus, the friction head is: \[ h_{friction} = 68Q^2 + 650Q^2 = 718Q^2 \]
Combining these, the total head becomes: \[ h_t = h_{static} + h_{friction} = 21.4 \text{ m} + 718Q^2 \text { m} \]
Understanding total head is essential for determining pump performance and efficiency.

radial component of velocity

The radial component of velocity (\( v_r \)) is associated with how fluid particles move as they exit the impeller. In this problem, we assume that water flow remains radial and unchanged through the impeller. The radial velocity can be derived using the volumetric flow rate (\( Q \)) and the cross-sectional area (\( A_r \)).
First, calculate \( A_r \) using impeller dimensions:
Given:

• Impeller diameter (\( D_{impeller} \)): 0.35 m
• Blade width (\( b \)): 0.018 m
• Blade angular coverage (\( n = 0.05\))
• Blade angle (\( t = 35^\text{circ} \))

The cross-sectional area is: \[ A_r = \frac{\pi \times (D_{impeller} - t)^2}{4} - \frac{n \times b \times t}{360} \= \frac{\pi \times (0.332)^2}{4} - \frac{0.05 \times 0.018 \times 35}{360} \approx 0.0093 \text{ m}^2 \]
Next, the radial velocity (\( v_r \)) is: \[ v_r = \frac{Q}{A_r} \]
This component of velocity helps us understand fluid flow at the impeller outlet.

blade speed

Blade speed (\( u_2 \)) at the impeller outlet is another critical factor. It describes the speed at which impeller blades move, directly affecting the energy imparted to the fluid. Blade speed is calculated using:
\[ u_2 = \frac{\text{impeller diameter}}{2} \times \text{angular velocity} \]
Given angular velocity (\( \text{\omega} \)): 104.3 rad/s and impeller outer radius (\( R_2 = \frac{0.35}{2} = 0.175 \)). The blade speed is:
\[ u_2 = 104.3 \times 0.175 = 18.252 \text{ m/s} \]
Blade speed plays a pivotal role in determining the centrifugal force exerted on the fluid. This force, in turn, influences the head and pressure developed by the pump.

theoretical head

The theoretical head (\( h_{theoretical} \)) is a measure of the ideal energy transfer from the pump to the fluid, without considering losses. It's determined by the Euler equation for turbomachinery:
\[ h_{theoretical} = \frac{u_2 \times v_{u2}}{g} \]
To calculate this, we need the tangential component of the velocity (\( v_{u2} \)). For backward-facing blades: \[ v_{u2} = u_2 - v_r \times \tan(\beta_2) \]
Given \( \beta_2 = 35^\text{circ} \) and substituting known values: \[ v_{u2} = 18.252 - \frac{Q}{0.0093} \times \tan(35) \]
Finally, substituting in the head equation: \[ h_{theoretical} = \frac{18.252 \times (18.252 - \frac{Q}{0.0093} \times \tan(35))}{9.81} \]
The theoretical head helps estimate how effectively a pump can convert mechanical energy into fluid kinetic and potential energy.

manometric efficiency

Manometric efficiency (\( \text{\eta_m} \)) is a key performance metric of centrifugal pumps, indicating how effectively the pump converts input energy into useful work. It's defined as the ratio of static head to total head:
\[ \text{\eta_m} = \frac{h_{static}}{h_t} \]
Given static head (\( h_{static} \)) = 21.4 m and total head (\( h_t \)) = 21.4 + 718\( Q^2 \):
\[ \text{\eta_m} = \frac{21.4}{21.4 + 718Q^2} \]
Manometric efficiency indicates how much energy is successfully used to lift the water, with higher values representing more efficient pumps. Understanding manometric efficiency helps in selecting pumps for desired applications and achieving optimal performance.