/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A manufacturer tests a 1: 10 sca... [FREE SOLUTION] | 91Ó°ÊÓ

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A manufacturer tests a 1: 10 scale model of a pump in the laboratory. The model pump has an impeller diameter of \(200 \mathrm{~mm}\) and a rotational speed of \(3450 \mathrm{rpm},\) and when the head across the pump is \(40 \mathrm{~m}\), the pump delivers a flow rate of \(10 \mathrm{~L} / \mathrm{s}\) at an efficiency of \(84 \%\). The prototype pump is to develop the same head as the scale model; however, because of its increased size, the prototype pump is expected to have an efficiency of \(90 \%\). (a) What is the power supplied to the model pump? (b) What is the rotational speed, flow rate, and power supplied to the prototype pump under homologous conditions? Assume water at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) 466.43 W (b) 1091.29 rpm, 316.2 L/s, 13.76 kW

Step by step solution

01

Calculate the Power Supplied to the Model Pump

First, we need to calculate the power supplied to the model pump using the formula for hydraulic power:\[ P = \frac{\rho \cdot g \cdot H \cdot Q}{\eta} \]where \( \rho = 1000 \, \mathrm{kg/m^3} \) (density of water), \( g = 9.81 \, \mathrm{m/s^2} \) (acceleration due to gravity), \( H = 40 \, \mathrm{m} \) (head), \( Q = 10 \, \mathrm{L/s} = 0.01 \, \mathrm{m^3/s} \) (flow rate), and \( \eta = 0.84 \) (efficiency of the model pump). Substituting these values in, we find:\[ P = \frac{1000 \times 9.81 \times 40 \times 0.01}{0.84} = 466.43 \, \mathrm{W} \]Thus, the power supplied to the model pump is approximately 466.43 watts.
02

Calculate the Scale Factor for Rotational Speed and Diameters

The model is a 1:10 scale model, which means each linear dimension of the prototype is 10 times that of the model. Therefore, the diameter of the prototype is ten times that of the model pump's impeller diameter. For rotational speed, because of scale similarity laws, the rotational speed of the prototype \( n_p \) relates to that of the model via:\[ n_p = n_m \times \sqrt{\frac{D_m}{D_p}} \]where \( n_m = 3450 \, \mathrm{rpm} \) and \( D_m = 200 \, \mathrm{mm} \) and \( D_p = 10 \times D_m = 2000 \, \mathrm{mm} \). Substituting the values, we get:\[ n_p = 3450 \times \sqrt{\frac{200}{2000}} = 3450 \times 0.3162 \approx 1091.29 \, \mathrm{rpm} \]Thus, the rotational speed of the prototype is approximately 1091.29 rpm.
03

Calculate the Flow Rate of the Prototype

For flow rate, we apply the rule of similarity that states flow rate is determined by the cube of the scale factor of diameters, i.e.,\[ Q_p = Q_m \times (\frac{D_p}{D_m})^{3/2} \]Substitute the respective values into the equation:\[ Q_p = 0.01 \times (\frac{2000}{200})^{3/2} = 0.01 \times 10^{3/2} = 0.01 \times 31.62 \approx 0.3162 \, \mathrm{m^3/s} \]Hence, the flow rate of the prototype is approximately 0.3162 cubic meters per second or 316.2 liters per second.
04

Calculate the Power Supplied to the Prototype Pump

Given that the prototype's efficiency is 90%, we use the same power equation:\[ P_p = \frac{\rho \cdot g \cdot H \cdot Q_p}{\eta_p} \]Where \( Q_p = 0.3162 \, \mathrm{m^3/s} \) and the others remain the same, substitute the known values:\[ P_p = \frac{1000 \times 9.81 \times 40 \times 0.3162}{0.90} = 13760.12 \, \mathrm{W} \]Therefore, the power supplied to the prototype pump is approximately 13,760 watts or 13.76 kilowatts.

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Key Concepts

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

Pump scale modeling
Pump scale modeling involves creating smaller versions of pumps to study and predict the behavior of prototype pumps under similar conditions. This technique simplifies experimentation, allowing manufacturers to make improvements and adjustments before full-scale production.

Scale models follow geometric scaling laws, meaning that linear dimensions like length, diameter, and height, are reduced by a specific ratio, known as the scale factor. In the provided example, the model is a 1:10 scaled-down version of the prototype, which means the model's dimensions are 10 times smaller.

This scaling helps engineers understand how changes in size affect the pump's performance, flow rate, efficiency, and power requirements. By testing and analyzing the scale model, insights can be gained into the larger prototype’s behavior, reducing risks and costs associated with full-scale testing.
Rotational speed scaling
Rotational speed scaling is crucial in model testing to ensure that dynamics mirror the larger prototype. The scaling of the rotational speed involves a mathematical relationship between the speeds of the model and prototype pumps. The rotational speed of the prototype is adjusted by the square root of the ratio of the impeller diameters.

In the case study, the model's rotational speed is 3450 rpm, and given the prototype’s impeller diameter is ten times that of the model, the rotational speed is recalculated as follows:
  • Apply the formula: \( n_p = n_m \times \sqrt{\frac{D_m}{D_p}} \)
  • Using the known values: \( n_p = 3450 \times 0.3162 \approx 1091.29 \) rpm
This ensures that the physical forces experienced by the scaled model align closely with those anticipated in the larger pump.
Flow rate scaling
Flow rate scaling utilizes the cube of the scale factor for diameters. This means the expected flow rate in the prototype is higher due to its larger size. The formula for this scaling is expressed as:

\[ Q_p = Q_m \times \left(\frac{D_p}{D_m}\right)^{3/2} \]
In our analysis, with a model flow rate of 10 L/s, and a scale factor of 10 for the diameters, the prototype's flow rate turns out to be:
  • \( Q_p = 0.01 \times 31.62 \approx 0.3162 \) m³/s
  • Equivalent to 316.2 L/s
Flow rate scaling is pivotal to predicting how changes in pump size affect the volume of fluid processed over time.
Efficiency in pumps
Efficiency is a measure of how well a pump converts input energy into useful hydraulic power. The efficiency of both a model and prototype pump is a critical parameter because it directly affects their operational cost and effectiveness.

In the exercise, the model pump’s efficiency is 84%, while the prototype's anticipated efficiency is 90%. By knowing the efficiency, engineers can compute the power required to achieve desired output conditions.
  • For the model: \( P = \frac{1000 \times 9.81 \times 40 \times 0.01}{0.84} = 466.43 \) W
The same formula is employed for the prototype given its higher efficiency:
  • \( P_p = \frac{1000 \times 9.81 \times 40 \times 0.3162}{0.90} \approx 13760.12 \) W or 13.76 kW
Understanding and optimizing efficiency ensure that the pump operates effectively while minimizing energy consumption.

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Most popular questions from this chapter

If the performance curve of a certain pump model is given by $$ h_{\mathrm{p}}=30-0.05 Q^{2} $$ where \(h_{\mathrm{p}}\) is the head added in \(\mathrm{m}\) and \(Q\) is the flow rate in \(\mathrm{L} / \mathrm{s}\), what is the performance curve of a pump system containing \(n\) of these pumps in series? What is the performance curve of a pump system containing \(n\) of these pumps in parallel?

A pump has an impeller diameter of \(300 \mathrm{~mm}\) and a rotational speed of \(1500 \mathrm{rpm}\). At the best efficiency operating point, the pump adds a head of \(9 \mathrm{~m}\) at a flow rate of \(25 \mathrm{~L} / \mathrm{s} .\) What is the specific speed of the pump? What type of pump is this likely to be?

Tests on a pump under standard atmospheric conditions show that when water at \(20^{\circ} \mathrm{C}\) is pumped at \(60 \mathrm{~L} / \mathrm{s}\) and the head added by the pump is \(40 \mathrm{~m},\) cavitation occurs when the pressure head plus velocity head on the suction side of the pump is \(3.9 \mathrm{~m}\). (a) Determine the required net positive suction head and the cavitation number of the pump. (b) If this same pump is operated on a mountain under the same flow rate and added head condition but the temperature of the water is \(5^{\circ} \mathrm{C}\) and the atmospheric pressure is \(90 \mathrm{kPa}\), by how much must the elevation of the pump above the sump reservoir be reduced compared with the test condition? Assume that the friction loss in the suction pipe remains approximately the same and that the sump reservoir is open to the atmosphere in both cases.

A pump is to be used to withdraw water from a reservoir at a rate of \(1500 \mathrm{~L} / \mathrm{s}\). When operating at this flow rate, the head loss in the suction pipe is estimated to be \(2.3 \mathrm{~m}\), and the pump specifications give the required net positive suction head as \(2.9 \mathrm{~m}\). Standard sea-level atmospheric conditions are expected at the site, and under worstcase conditions, the temperature of the water in the reservoir is \(25^{\circ} \mathrm{C}\). What is the maximum allowable elevation of the suction side of the pump above the reservoir water surface?

A single-jet Pelton wheel hydropower plant is to be operated such that the shaft power developed by the Pelton wheel is equal to \(15 \mathrm{MW}\) when the head just upstream of the nozzle is equal to \(1600 \mathrm{~m}\). The Pelton wheel is constructed such that is has a diameter of \(4 \mathrm{~m}\), has a deflection angle of \(170^{\circ},\) and rotates at a controlled speed of 600 rpm. Determine the appropriate nozzle diameter to be used in the project. Assume water at \(20^{\circ} \mathrm{C}\) and a nozzle loss coefficient of 0.03 .

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