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Chapter 13: Problem 3

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.

Short Answer

Expert verified
Calculate effective head, determine hydraulic power, find number of wheels, jet and wheel diameters, hydraulic efficiency, and kinetic energy percentage.

Step by step solution

01

Determine the Effective Head

Calculate the effective head by subtracting the head lost to friction from the gross head.
02

Calculate Hydraulic Power Available

Using the formula for hydraulic power, determine the power available from the effective head, taking into account the efficiency.
03

Determine Number of Wheels Required

Calculate the number of wheels needed by dividing the total power requirement by the power per wheel, considering the number of jets per wheel.
04

Calculate the Jet Velocity

Determine the velocity of the water jets using the continuity equation and the given discharge coefficient.
05

Calculate the Jet Diameter

Using the calculated jet velocity and the power equation, find the diameter of the jets.
06

Calculate the Wheel Diameter

Using the specific speed and the speed of the wheel, calculate the diameter of the Pelton wheels.
07

Determine Hydraulic Efficiency

Calculate the hydraulic efficiency using the given deflection angle and the reduction in relative velocity.
08

Calculate Kinetic Energy at Discharge

Determine the percentage of input power that remains as kinetic energy of the water at discharge using the kinetic energy equation.

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Key Concepts

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

Hydro-electric Power Calculation
Hydro-electric power generation harnesses the energy of falling or flowing water. The basic formula to calculate the hydraulic power is given by
\[ P_h = \rho \times g \times H \times Q \] where:
  • \(\rho\) is the density of water (approximately \( 1000 \; \text{kg/m}^3 \))
  • \( g \) is the acceleration due to gravity (\( 9.81 \; \text{m/s}^2 \))
  • \( H \) is the head or height of the water column (measured in meters)
  • \( Q \) is the flow rate (measured in cubic meters per second)
Effective hydraulic power considers system efficiency. If the efficiency is \(\eta\), the effective power is given as: \[ P_{eff} = P_h \times \eta \] This power is then converted into electrical power by the generator, considering further efficiencies. Understanding these calculations underpins designing efficient hydroelectric schemes.
Effective Head Determination
The effective head is the actual height of water used for power generation after accounting for energy losses in the system. Notably, friction losses in the pipeline affect the gross head. The formula is:
\[ H_{eff} = H_g - H_f \] where:
  • \( H_{eff} \) is the effective head.
  • \( H_g \) is the gross head (initial head).
  • \( H_f \) is the head lost due to friction.
For instance, with a gross head of 245 meters and a friction loss of 12 meters, the effective head calculates as:
\[ H_{eff} = 245 \; m - 12 \; m = 233 \; m \] It's critical to determine the effective head accurately for precise power calculations.
Hydraulic Efficiency Calculation
Hydraulic efficiency takes into account how effectively the energy in the water is converted into mechanical energy by the turbine. For Pelton wheels, this efficiency depends on blade shape and water jet interaction. The formula for hydraulic efficiency (\(\eta_h\)) can involve the deflection angle (\(\theta\)):
\[ \eta_h = \left( 1 - \frac{V_2}{V_1} \right) \cos( \theta) \] where:
  • \( V_2 \) is the reduced velocity of water after impact.
  • \( V_1 \) is the initial velocity of water.
  • \( \theta \) is the deflection angle of water.
Considering a reduction of 15% in velocity and a deflection of \(165^\text{o}\), hydraulic efficiency can be calculated as shown above. Such detailed calculations help in refining turbine designs for maximum power output.
Jet and Wheel Diameter Calculation
Diameter calculations are crucial for ensuring the proper functionality of Pelton wheels. To determine jet diameter, use the formula:
\[ d_j = \sqrt{ \frac{P}{ \pi \times \rho \times g \times H \times C_y \times V_j} } \] where:
  • \( P \) is power per wheel.
  • \( C_y \) is the discharge coefficient.
  • \( V_j \) is the jet velocity.
Further, the wheel diameter (\(D_w\)) is calculated considering the specific speed (\(n_{s}\)) and operational speed (\(n\)):
\[ n_{s} = \frac{n \times \sqrt{P}}{H^{5/4}} \] Adjusting these diameters ensures the Pelton wheel achieves the desired hydraulic performance, ultimately maximizing efficiency and power output.

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Most popular questions from this chapter

In a vertical-shaft inward-flow reaction turbine the sum of the pressure and kinetic heads at entrance to the spiral casing is \(120 \mathrm{~m}\) and the vertical distance between this section and the tail-race level is \(3 \mathrm{~m}\). The peripheral velocity of the runner at entry is \(30 \mathrm{~m} \cdot \mathrm{s}^{-1}\), the radial velocity of the water is constant at \(9 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and discharge from the runner is without whirl. The estimated hydraulic losses are: (1) between turbine entrance and exit from the guide vanes, \(4.8 \mathrm{~m}\) (2) in the runner, \(8.8 \mathrm{~m},(3)\) in the draft tube, \(790 \mathrm{~mm}\), (4) kinetic head rejected to the tail race, \(460 \mathrm{~mm}\). Calculate the guide vane angle and the runner blade angle at inlet and the pressure heads at entry to and exit from the runner

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?

The following duties are to be performed by rotodynamic pumps driven by electric synchronous motors, speed \(100 \pi / n \mathrm{rad} \cdot \mathrm{s}^{-1}(=50 / n \mathrm{rev} / \mathrm{s})\), where \(n\) is an integer: (a) \(14 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) of water against \(1.5 \mathrm{~m}\) head; \((b)\) oil (relative density \(0.80\) ) at \(11.3 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(70 \mathrm{kPa}\) pressure; \((c)\) water at \(5.25 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(5.5 \mathrm{MPa}\). Designs of pumps are available with specific speeds of \(0.20,0.60,1.20,2.83,4.0\) rad. Which design and speed should be used for each duty?

A vertical-shaft Francis turbine is to be installed in a situation where a much longer draft tube than usual must be used. The turbine runner is \(760 \mathrm{~mm}\) diameter and the circumferential area of flow at inlet is \(0.2 \mathrm{~m}^{2}\). The overall operating head is \(30 \mathrm{~m}\) and the speed \(39.27 \mathrm{rad} \cdot \mathrm{s}^{-1}(6.25 \mathrm{rev} / \mathrm{s}) .\) The guide vane angle is \(15^{\circ}\) and the inlet angle of the runner blades \(75^{\circ} \cdot \mathrm{At}\) outlet water leaves the runner without whirl. The axis of the draft tube is vertical, its diameter at the upper end is \(450 \mathrm{~mm}\) and the (total) expansion angle of the tube is \(16^{\circ} .\) For a flow rate of \(Q\left(\mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\right)\) the friction loss in the tube (of length \(l\) ) is given by \(h_{\mathrm{f}}=0.03 Q^{2} l .\) If the absolute pressure head at the top of the tube is not to fall below \(3.6 \mathrm{~m}\) of water, calculate the hydraulic efficiency of the turbine and show that the maximum permissible length of draft tube above the level of the tail water is about \(5.36 \mathrm{~m}\). (The length of the tube below tailwater level may be neglected. Atmospheric pressure \(\equiv 10.33 \mathrm{~m}\) water head.)

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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