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.

Short Answer

Expert verified
Model speed is approximately 1.077 rad/s with a test head of 3.75 m for the one-quarter scale model.

Step by step solution

01

Understand the problem

A large centrifugal pump is designed to operate with a specific speed, discharge rate, and head. The task is to determine the range of speeds and test heads for a one-quarter scale model using water, given that the kinematic viscosity of the liquid varies between 3 and 6 times that of water.
02

Define known values and scale factors

Full-size pump values: Specific speed: \[ N_s = 1.15 \text{ rad (0.183 rev)} \] Discharge: \[ Q = 2 \text{ L/s} \] Total head: \[ H = 15 \text{ m} \] Scale factor for the model (one-quarter scale): \[ L_p = 4L_m \]
03

Determine model discharge and head

Using similarity laws for the one-quarter scale model, the discharge for the model will be scaled down based on the length scale factor: \[ Q_m = \frac{Q_p}{L_p^{3/2}} = \frac{2}{4^{3/2}} = \frac{2}{8} = 0.25 \text{ L/s} \] The total head for the model will be scaled down by the length scale factor: \[ H_m = \frac{H_p}{L_p} = \frac{15}{4} = 3.75 \text{ m} \]
04

Calculate model speed

Given that specific speed remains consistent between the model and the prototype, use the formula for specific speed: \[ N_s = \frac{N_p \times Q_p^{1/2}}{H_p^{3/4}} = \frac{N_m \times Q_m^{1/2}}{H_m^{3/4}} \] Rearrange to solve for the model speed: \[ N_m = N_p \times \frac{Q_p^{1/2} \times H_m^{3/4}}{Q_m^{1/2} \times H_p^{3/4}} \] Substitute the known values: \[ N_m = 1.15 \text{ rad/s} \times \frac{ 2^{1/2} \times 3.75^{3/4} }{ 0.25^{1/2} \times 15^{3/4} } \]
05

Insert values and simplify

Perform the calculations: \[ N_m = 1.15 \times \frac{\text{rad/s} \times \times 3.287}{0.5 \times 7.017} \] \[ N_m = 1.15 \times \frac{3.287}{3.5085} \] \[ N_m = \text{approximately} 1.077 \text{ rad/s} \]
06

Consider kinematic viscosity variations

Since the kinematic viscosity of the liquid can range between 3 and 6 times that of water, several tests might be performed. However, the specific speed calculation retains its applicability as it is dimensionless and does not directly depend on viscosity.

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Key Concepts

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

Specific Speed Calculation
When designing and analyzing centrifugal pumps, one critical parameter to determine is the specific speed, expressed as \(N_s\). Specific speed \(N_s\) is a dimensionless number that combines the pump speed, discharge, and head, allowing engineers to compare the performance of different pumps regardless of size.
For the given problem, the specific speed is provided as \(N_s = 1.15 \text{ rad/s}\). This is often recalculated for different units or angles (e.g., revolutions per second). The formula to calculate specific speed is:
\[N_s = N \times \frac{Q^{1/2}}{H^{3/4}} \]
Where:
  • \(N\) is the pump speed
  • \(Q\) is the discharge rate
  • \(H\) is the total head

This specific speed helps in identifying the best model of the pump for the required operation, ensuring the appropriate balance between speed, discharge, and head is maintained for efficient performance.
Similarity Laws
The similarity laws, or affinity laws, allow engineers to predict the performance of pumps under different conditions or scales. In this exercise, they are essential for scaling the characteristics of the full-scale pump to a one-quarter scale model.
There are three primary similarity laws for centrifugal pumps:
  • The flow rate is proportional to the cube of the linear dimension ratio:
    \[Q_m = Q_p \times \frac{L_m^3}{L_p^3}\]
    In our case:
    \[Q_m = \frac{2}{4^{3/2}} = 0.25 \text{ L/s}\]
  • The head is proportional to the square of the linear dimension ratio:
    \[H_m = H_p \times \frac{L_m^2}{L_p^2}\]
    For our model:
    \[H_m = \frac{15}{4} = 3.75 \text{ m}\]
  • The speed is proportional to the linear dimension ratio:
    \[N_m = N_p \times \frac{L_p}{L_m}\]

By applying these laws, engineers can accurately predict how a scaled-down model of the pump will perform compared to its full-scale prototype.
Kinematic Viscosity Effects
Kinematic viscosity, denoted as \(u\), influences the flow characteristics in centrifugal pumps. In the real world, liquids with different viscosities will behave differently even in similar systems.
In this exercise, the liquid's kinematic viscosity is set to vary between 3 and 6 times that of water. Although specific speed calculations remain valid as they are dimensionless, varying viscosity affects the Reynolds number, a critical factor for assessing turbulent or laminar flow.
Kinematic viscosity impacts:
  • Friction losses
  • Pump efficiency
  • NPSH (Net Positive Suction Head)

Therefore, when performing model tests, maintaining similar Reynolds numbers by adjusting the liquid's viscosity helps in achieving scalable results and accurate predictions.
Scale Modeling in Fluid Mechanics
Scale modeling is a technique used to study fluid dynamics in a controlled and reduced-scale environment. By replicating the conditions of a full-size system on a smaller scale, engineers can gather data and predict performance before full-scale construction.
For our centrifugal pump exercise, we use a one-quarter scale model to determine speeds and test heads. Scaling down to a one-quarter model means reducing the dimensions by a factor of 4.
Key parameters to note in scale modeling include:
  • Maintaining geometric similarity: The model must have the same shape and proportions as the prototype.
  • Reynolds number similarity: It helps in ensuring that the flow conditions remain comparable across different scales by maintaining the ratio of inertial to viscous forces.
  • Similitude: Combining geometric, kinematic, and dynamic similarities helps in accurately predicting the performance of the prototype.

In conclusion, scale modeling in fluid mechanics is a powerful method for testing and validating design performance before committing to full-scale production.

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Most popular questions from this chapter

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.

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.

The impeller of a centrifugal fan has an inner radius of \(250 \mathrm{~mm}\) and width of \(187.5 \mathrm{~mm} ;\) the values at exit are \(375 \mathrm{~mm}\) and \(125 \mathrm{~mm}\) respectively. There is no whirl at inlet, and at outlet the blades are backward-facing at \(70^{\circ}\) to the tangent. In the impeller there is a loss by friction of \(0.4\) times the kinetic head corresponding to the relative outlet velocity, and in the volute there is a gain equivalent to \(0.5\) times the kinetic head corresponding to the absolute velocity at exit from the runner. The discharge of air is \(5.7 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) when the rotational speed is \(84.8 \mathrm{rad} \cdot \mathrm{s}^{-1}(13.5 \mathrm{rev} / \mathrm{s}) .\) Neglecting the thickness of the blades and whirl slip, determine the head across the fan and the power required to drive it if the density of the air is sensibly constant at \(1.25 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) throughout and mechanical losses account for \(220 \mathrm{~W}\).

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.

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