The impeller of a centrifugal pump has an outer diameter of \(250 \mathrm{~mm}\) and an effective outlet area of \(17000 \mathrm{~mm}^{2} .\) The outlet blade angle is \(32^{\circ} .\) The diameters of suction and discharge openings are \(150 \mathrm{~mm}\) and \(125 \mathrm{~mm}\) respectively. At \(152 \mathrm{rad} \cdot \mathrm{s}^{-1}(24.2 \mathrm{rev} / \mathrm{s})\) and discharge \(0.03 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the pressure heads at suction and discharge openings were respectively \(4.5 \mathrm{~m}\) below and \(13.3 \mathrm{~m}\) above atmospheric pressure, the measurement points being at the same level. The shaft power was \(7.76 \mathrm{~kW}\). Water enters the impeller without shock or whirl. Assuming that the true outlet whirl component is \(70 \%\) of the ideal, determine the overall efficiency and the manometric efficiency based on the true whirl component.

Short Answer

Expert verified
Overall efficiency and manometric efficiency can be computed using whirl components.

Step by step solution

01

Calculate the tangential velocity at the outlet

Using the equation for tangential velocity, we have: Equating: Calculating:
02

Determine the ideal outlet whirl component

Since water enters the impeller without shock or whirl, the ideal whirl component at the outlet is given by
03

Find the true outlet whirl component

Given:
04

Calculate the manometric head

Using Bernoulli's Equation
05

Determine the manometric efficiency

Given: true whirl component
06

Compute the overall efficiency

To find the overall efficiency

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Key Concepts

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

tangential velocity
In centrifugal pumps, tangential velocity refers to the speed at which fluid exits the impeller along the direction perpendicular to the radius.
For our problem, tangential velocity is calculated using the formula:
\[ u_2 = r_2 \times \text{angular velocity} \ \text{where } r_2 \text{ is the radius of the impeller's outer edge} \ \text{angular velocity} = 152 \text{ rad/s} \ r_2 = 250 \text{ mm} / 2 = 0.125 \text{ m}. \ \therefore u_2 = 0.125 \times 152 = 19 \text{ m/s} \]
This sets the stage for calculating other parameters like the ideal whirl component and efficiency characteristics.
ideal whirl component
The ideal whirl component represents the idealized scenario where there are no losses, and the fluid is ideal.
For our centrifugal pump, it's calculated based on the tangential velocity and outlet blade angle.
The formula used for this is:
\[ \text{Ideal Outlet Whirl Component } = u_2 \times \tan(\text{blade angle}) \ \text{Blade angle } = 32^\text{circ}. \ \text{Ideal Outlet Whirl Component } = 19 \times \tan(32) \ \tan(32) \text{ is } 0.625. \ \therefore \text{Ideal Outlet Whirl Component } = 19 \times 0.625 = 11.875 \text{ m/s} \]
Therefore, the ideal whirl component is 11.875 m/s.
manometric head
The manometric head reflects the energy imparted by the pump to the fluid relative to the atmospheric pressure.
Using Bernoulli's Equation, the manometric head can be calculated. Bernoulli's Equation helps in linking the velocity and pressure head for incompressible fluids.
Let's detail out the formula:
\[ H_m = \frac{\text{Pressure head (discharge)} - \text{Pressure head (suction)}}{g} +\frac{u_2^2}{2g} - \frac{u_1^2}{2g} \ \text{Given:} \ \text{ Pressure head (discharge)} = 13.3 \ \text{ and Pressure head (suction)} = -4.5 \ \ g \ \therefore H_m = \frac{13.3 - (-4.5)}{9.81} + \frac{19^2}{2 \times 9.81}\ = 1.8 + 18.4\ = 20.2 \ \therefore \text{ Manometric Head } (H_m) = 20.2 \text{ meters} \]
Thus, the manometric head for our problem is calculated to be 20.2 meters.
Bernoulli's Equation
Bernoulli's Equation describes the conservation of energy in fluid flow, particularly for incompressible, non-viscous fluids.
The equation is stated as:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \ \text{where:} \ P_1 \text{ and } P_2 \text{ are pressure energies at two points.} \ \rho \text{ is fluid density.} \ v_1 \text{ and } v_2 \text{ are fluid velocities.} \ h_1 \text{ and } h_2 \text{ are height energies.} \]
In the context of the centrifugal pump, Bernoulli's Equation helps us calculate the manometric head by equating the pressure differences and velocities at the suction and discharge points.
manometric efficiency
Manometric efficiency helps to measure how efficiently the pump converts the mechanical energy of the fluid into pressure energy.
It is given by the formula:
\[ \text{Manometric Efficiency} (\text{η}_m) = \frac{g \times H_m}{u_2 \times \text{True Outlet Whirl Component}} \ \text{Given } H_m = 20.2. \ \text{True Outlet Whirl Component } = 0.7 \times 11.875 = 8.3125 \ \therefore \text{η}_m = \frac{9.81 \times 20.2}{19 \times 8.3125} = 1.07 \times \frac{20.2}{8.3125} = 0.353 \text{ or } 35.3\text{%} \]
In essence, the manometric efficiency for our given conditions is 35.3%.
overall efficiency
Overall efficiency gauges the pump's complete performance, indicating how well it converts the input mechanical energy into useful hydraulic energy.
This is calculated using both the shaft power and the power imparted to the fluid.
The formula is:
\[ \text{Overall Efficiency} (\text{η}_o) = \frac{\rho \times g \times Q \times H_m}{\text{Power Input} (\text{shaft power})} \ \text{Given:} \rho = 1000 \text{ kg/m}^3, g = 9.81 \text{ m/s}^2, Q = 0.03 \text{ m}^3/\text{s}, H_m = 20.2 \text{ meters, Shaft Power = 7.76 kW}.\ \ \therefore \text{power input} = 7760 \text{ W}. \ \text{η}_o = \frac{1000 \times 9.81 \times 0.03 \times 20.2}{7760} = \frac{5700}{7760}= 0.735 \text{ or } 73.5\text{%} \]
Thus, the overall efficiency of the pump is 73.5%.

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Most popular questions from this chapter

In a hydro-electric scheme a number of Pelton wheels are to be used under the following conditions: total output required \(30 \mathrm{MW} ;\) gross head \(245 \mathrm{~m} ;\) speed \(39.27 \mathrm{rad} \cdot \mathrm{s}^{-1}\) \(\left(6.25\right.\) rev/s); 2 jets per wheel; \(C_{\mathrm{Y}}\) of nozzles \(0.97 ;\) maximum overall efficiency (based on conditions immediately before the nozzles) \(81.5 \% ;\) power specific speed for one jet not to exceed \(0.138\) rad \((0.022\) rev); head lost to friction in pipe-line not to exceed \(12 \mathrm{~m} .\) Calculate \((a)\) the number of wheels required, (b) the diameters of the jets and wheels, (c) the hydraulic efficiency, if the blades deflect the water through \(165^{\circ}\) and reduce its relative velocity by \(15 \%,(d)\) the percentage of the input power that remains as kinetic energy of the water at discharge.

A vertical-shaft Francis turbine, with an overall efficiency of \(90 \%\), runs at \(44.86 \mathrm{rad} \cdot \mathrm{s}^{-1}(7.14 \mathrm{rev} / \mathrm{s})\) with a water discharge of \(15.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1} .\) The velocity at the inlet of the spiral casing is \(8.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and the pressure head at this point is \(240 \mathrm{~m}\), the centre-line of the casing inlet being \(3 \mathrm{~m}\) above the tail-water level. The diameter of the runner at inlet is \(2.23 \mathrm{~m}\) and the width at inlet is \(300 \mathrm{~mm}\). The hydraulic efficiency is \(93 \%\). Determine \((a)\) the output power, \((b)\) the power specific speed, ( \(c\) ) the guide vane angle, \((d)\) the runner blade angle at inlet, (e) the percentage of the net head which is kinetic at entry to the runner. Assume that there is no whirl at outlet from the runner and neglect the thickness of the blades.

The following data refer to a Pelton wheel. Maximum overall efficiency \(79 \%\), occurring at a speed ratio of \(0.46 ; \mathrm{C}_{\mathrm{v}}\) for nozzle \(=0.97\); jet turned through \(165^{\circ}\). Assuming that the optimum speed ratio differs from \(0.5\) solely as a result of losses to windage and bearing friction which are proportional to the square of the rotational speed, obtain a formula for the optimum speed ratio and hence estimate the ratio of the relative velocity at outlet from the buckets to the relative velocity at inlet.

A large centrifugal pump is to have a specific speed of \(1.15 \mathrm{rad}\) \((0.183 \mathrm{rev})\) and is to discharge liquid at \(2 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against a total head of \(15 \mathrm{~m}\). The kinematic viscosity of the liquid may vary between 3 and 6 times that of water. Determine the range of speeds and test heads for a one-quarter scale model investigation of the full-size pump, the model using water.

\(13.23\) A fluid coupling is to be used to transmit \(150 \mathrm{~kW}\) between an engine and a gear-box when the engine speed is \(251.3 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (40 rev/s). The mean diameter at the outlet of the primary member is \(380 \mathrm{~mm}\) and the cross-sectional area of the flow passage is constant at \(0.026 \mathrm{~m}^{2}\). The relative density of the oil is \(0.85\) and the efficiency of the coupling \(96.5 \%\). Assuming that the shock losses under steady conditions are negligible and that the friction loss round the fluid circuit is four times the mean velocity head, calculate the mean diameter at inlet to the primary member.

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