During a laboratory test on a water pump appreciable cavitation began when the pressure plus velocity head at inlet was reduced to \(3.26 \mathrm{~m}\) while the total head change across the pump was \(36.5 \mathrm{~m}\) and the discharge was \(48 \mathrm{~L} \cdot \mathrm{s}^{-1} .\) Barometric pressure was \(750 \mathrm{~mm} \mathrm{Hg}\) and the vapour pressure of water \(1.8 \mathrm{kPa}\). What is the value of \(\sigma_{\mathrm{c}}\) ? If the pump is to give the same total head and discharge in a location where the normal atmospheric pressure is \(622 \mathrm{~mm} \mathrm{Hg}\) and the vapour pressure of water \(830 \mathrm{~Pa}\), by how much must the height of the pump above the supply level be reduced?

Short Answer

Expert verified
The value of \(\sigma_{\mathrm{c}}\) is 0.364. The height of the pump must be reduced by 1.54 m.

Step by step solution

01

Convert barometric pressure to meters head (inlet condition)

Convert the given barometric pressure of 750 mm Hg to meters of water head using the relationship: 1 mm Hg = 0.0136 m of water head. Hence, \[ 750 \text{ mm Hg} \times 0.0136 \text{ m/mm Hg} = 10.2 \text{ m} \]
02

Calculate Net Positive Suction Head Available (NPSH)

NPSH is given by the equation: \[ \text{NPSH} = h_{atm} + h_{s} - h_{v} - h_f \] where \( h_{atm} \) = Ambient pressure head (10.2 m), \( h_{s} \) = Suction head (3.26 m), \( h_{v} \) = Vapor pressure head of water (1.8 kPa converted to head) \[ \text{1.8 kPa} = \frac{1.8 \times 10^3 \text{ Pa}}{9810 \text{ Pa/m}} = 0.18 \text{ m} \], and assuming friction head loss \( h_f \) is negligible, we get: \[ \text{NPSH} = 10.2 \text{ m} + 3.26 \text{ m} - 0.18 \text{ m} = 13.28 \text{ m} \]
03

Calculate the Value of Cavitation Number ( \( \sigma_c \))

Cavitation number \( \sigma_c\) is given by the equation: \[ \sigma_c = \frac{NPSH}{\text{Total head}} = \frac{13.28}{36.5} = 0.364 \]
04

Convert Atmospheric and Vapor Pressure for New Location to Meters Head

For the new conditions: 622 mm Hg is converted to meters head as: \[ 622 \times 0.0136 = 8.46 \text{ m} \] and vapor pressure of 830 Pa is converted to meters head as: \[ \frac{830}{9810} = 0.085 \text{ m} \]
05

Calculate New NPSH Available

Substitute values into the NPSH equation for new location: \[ \text{NPSH}_{new} = 8.46 + h_s - 0.085 \] Given \[ \sigma_c = 0.364 \], we rearrange to solve for suction head \( h_s \): \[ 0.364 = \frac{8.46 + h_s - 0.085}{36.5} \] Solving for \(h_s\), we get: \[0.364 \times 36.5 = 8.46 + h_s - 0.085 \Rightarrow h_s = 13.18 - 8.375 \Rightarrow h_s = 4.80 \text{ m}\]
06

Calculate the Change in Suction Head

The reduction in height is the difference between the initial and new suction head: \[ (3.26 - 4.80) = -1.54 \text{ m} \] Therefore, the pump must be reduced by 1.54 m.

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Key Concepts

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

Net Positive Suction Head (NPSH)
Net Positive Suction Head (NPSH) is a critical concept when considering pump operation, particularly to avoid cavitation. NPSH is a measure of the absolute pressure at the suction port of the pump and is given by the equation: \( \text{NPSH} = h_{atm} + h_{s} - h_{v} - h_{f} \). In this formula: - \( h_{atm} \): Ambient pressure head (converted from barometric pressure).
- \( h_{s} \): Suction head.
- \( h_{v} \): Vapor pressure head of the liquid being pumped.
- \( h_{f} \): Friction head loss, often considered negligible in many cases.
A higher NPSH ensures that the pump can avoid cavitation, which can cause serious damage. To calculate NPSH, each pressure must often be converted to consistent units such as meters of liquid, which can involve converting atmospheric pressures from units like mm Hg and vapor pressures from kPa or Pa.
Cavitation Number (Sigma)
The cavitation number, represented by \( \sigma \), is another key concept in understanding pump performance. It is a dimensionless number that describes the likelihood of cavitation occurring within the pump. Mathematically, it can be expressed as: \( \sigma = \frac{NPSH}{\text{Total head}} \). Here, NPSH is the Net Positive Suction Head discussed earlier, and the total head is the height increase provided by the pump. Cavitation number helps engineers assess whether a pump can operate safely without cavitation. Lower values of \( \sigma \) indicate a higher risk of cavitation, whereas higher values suggest safer operating conditions. Proper calculation and consideration of this number ensure the longevity and reliability of the pump system.
Barometric Pressure Conversion
A vital step in many pump and fluid dynamics calculations is converting barometric pressure into a compatible form, such as meters of water head. This conversion is essential when calculating NPSH. The barometric pressure is often provided in units like mm Hg (millimeters of mercury), which need to be converted using the relationship: \( 1 \text{ mm Hg} = 0.0136 \text{ m of water head} \). For example, a barometric pressure of 750 mm Hg can be converted as follows: \( 750 \times 0.0136 = 10.2 \text{ m} \). This value can then be incorporated into the NPSH formula. Appropriate conversion ensures accurate and meaningful results in your calculations, aiding in the proper design and operation of pump systems.
Vapor Pressure
Vapor pressure is the pressure at which a liquid begins to vaporize and is critical in avoidance of cavitation in pumps. When vapor pressure is reached, the liquid can form vapor bubbles, leading to cavitation. To include vapor pressure in NPSH calculations, it must often be converted to the same units as the other pressures, typically meters of liquid. This can be done using the conversion: \( \text{head} = \frac{ \text{Vapor Pressure (Pa)} }{ 9810 } \).
For instance, a water vapor pressure of 1.8 kPa (or 1800 Pa) can be converted as: \( \frac{1800}{9810} = 0.18 \text{ m} \). Such conversions ensure vapor pressure is accurately accounted for in NPSH calculations, helping to prevent cavitation by ensuring the liquid pressure does not drop below this point.
Pump Height Adjustment
Adjusting the height of a pump relative to the water supply level can directly influence cavitation occurrence. Proper height adjustment ensures that the NPSH is sufficient under varying operational conditions. For instance, in the example problem, when the atmospheric pressure and vapor pressure conditions at a new location are different, the required suction head (\( h_{s} \)) changes. Calculating the new NPSH: \( \text{NPSH}_{new} = h_{atm} + h_{s} - h_{v} \), and solving for the correct \( h_{s} \), helps determine by how much the pump’s height above the supply level must be adjusted. Proper calculations ensure operational efficiency and extend the pump’s lifespan by preventing cavitation under altered environmental conditions.

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Most popular questions from this chapter

An inward-flow reaction turbine has an inlet guide vane angle of \(30^{\circ}\) and the inlet edges of the runner blades are at \(120^{\circ}\) to the direction of whirl. The breadth of the runner at inlet is a quarter of the diameter at inlet and there is no velocity of whirl at outlet. The overall head is \(15 \mathrm{~m}\) and the rotational speed \(104.7 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (16.67 rev/s). The hydraulic and overall efficiencies may be assumed to be \(88 \%\) and \(85 \%\) respectively. Calculate the runner diameter at inlet and the power developed. (The thickness of the blades may be neglected.)

A centrifugal fan, for which a number of interchangeable impellers are available, is to supply air at \(4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) to a ventilating duct at a head of \(100 \mathrm{~mm}\) water gauge. For all the impellers the outer diameter is \(500 \mathrm{~mm}\), the breadth \(180 \mathrm{~mm}\) and the blade thickness negligible. The fan runs at \(188.5 \mathrm{rad} \cdot \mathrm{s}^{-1}\) ( \(\left.30 \mathrm{rev} / \mathrm{s}\right)\). Assuming that the conversion of velocity head to pressure head in the volute is counterbalanced by the friction losses there and in the impeller, that there is no whirl at inlet and that the air density is constant at \(1.23 \mathrm{~kg} \cdot \mathrm{m}^{-3}\), determine the most suitable outlet angle of the blades. (Neglect whirl slip.)

\(13.23\) 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.

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.

A quarter-scale turbine model is tested under a head of \(10.8 \mathrm{~m}\). The full-scale turbine is required to work under a head of \(30 \mathrm{~m}\) and to run at \(44.86 \mathrm{rad} \cdot \mathrm{s}^{-1}(7.14 \mathrm{rev} / \mathrm{s}) .\) At what speed must the model be run? If it develops \(100 \mathrm{~kW}\) and uses water at \(1.085 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) at this speed, what power will be obtained from the full-scale turbine, its efficiency being \(3 \%\) better than that of the model? What is the power specific speed of the full-scale, turbine?

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