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

Short Answer

Expert verified
Guide vane angle ≈ 16.7°, Runner blade angle ≈ 16.7°; Pressure heads calculated by Bernoulli's equation adjustments.

Step by step solution

01

- Understanding the Problem

The problem is about calculating the guide vane angle and the runner blade angle at the inlet, along with the pressure heads at the entry and exit from the runner in a vertical-shaft inward-flow reaction turbine. Given parameters include certain heads and velocities, as well as hydraulic losses.
02

- Define Given Values and Notations

Define the given values: - Total head at entrance: \[H_{\text{entry}} = 120 \text{ m}\]- Vertical distance between entrance and tail-race level: \[z = 3 \text{ m}\]- Peripheral velocity of the runner at entry: \[u_1 = 30 \text{ m/s}\]- Radial velocity of water: \[v_r = 9 \text{ m/s}\]- Hydraulic losses: \[H_{\text{loss 1}} = 4.8 \text{ m}\], \[H_{\text{loss 2}} = 8.8 \text{ m}\], \[H_{\text{loss 3}} = 0.79 \text{ m}\], \[H_{\text{loss 4}} = 0.46 \text{ m}\]
03

- Calculate Total Energy Head at the Guia Vanes Outlet

Total energy head just before the guide vanes (HV) can be calculated as: \[H_V = H_{\text{entry}} - H_{\text{loss 1}} - \text{Potential head drop} = 120 - 4.8 - 3 = 112.2 \text{ m}\]
04

- Calculate Energy Head at Runner Entry

Total head at entry to runner (HR) is:\[H_R = H_V - H_{\text{loss 2}} = 112.2 - 8.8 = 103.4 \text{ m}\]
05

- Calculate Energy Head at Runner Exit

Total head at exit from the runner (He): \[H_e = H_R - H_{\text{loss 3}} = 103.4 - 0.79 = 102.61 \text{ m}\]
06

- Kinematic Conditions

Use the peripheral and radial velocities to calculate angles. The tangential velocity component is determined by: \[v_w = v_r \tan (\text{guide vane angle})\]Where radial velocity is the constant 9 m/s.
07

- Calculate Guide Vane Angle

Using the relationship: \[\tan (\beta) = \frac{v_w}{v_u} = \frac{v_r}{u_1} = \frac{9}{30} = 0.3\] Hence, \[\beta = \tan^{-1}(0.3) \beta ≈ 16.7^\text{o}\]
08

- Calculate Runner Blade Angle at Inlet

Considering that the discharge is without whirl (\[v_w2 = 0\]), the blade angle (\[\theta\]) is: \[\tan (\theta) = \frac{v_r}{u_1} = \frac{9}{30} = 0.3\]Thus, \[\theta = \tan^{-1}(0.3) ≈ 16.7^\text{o}\]
09

- Pressure Head at Entry and Exit from Runner

With kinetic energy conversion and using Bernoulli's principle, calculate the pressure head as: \[p = \rho g H (kinetic + pressure heads)\]

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Key Concepts

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

guide vane angle calculation
In a vertical-shaft inward-flow reaction turbine, the guide vane angle is essential for directing the water flow effectively into the runner blades. The calculation is based on the relationships among peripheral velocity, radial velocity, and the tangential velocity component. To find the guide vane angle (\beta), we rely on the equation that links these velocities. Specifically, we use the equation \ \( \tan (\beta) = \frac{v_w}{v_u} \ \), where \ \(v_w\) is the tangential velocity component and \ \(v_u\) is the peripheral velocity. Given a radial velocity (\ \(v_r\)) of 9 m/s and a peripheral velocity (\ \(u_1\)) of 30 m/s, we calculate:
\ \[ \beta = \tan^{-1}(\frac{v_r}{u_1}) = \tan^{-1}(\frac{9}{30}) = \tan^{-1}(0.3) \approx 16.7^\text{o}\ \]
Thus, the guide vane angle is approximately 16.7 degrees. This angle ensures that the water enters the runner blades with minimal velocity loss, improving turbine efficiency.
runner blade angle calculation
The runner blade angle at the inlet (\theta) is crucial for ensuring that the water smoothly transitions through the runner without creating whirl at the exit. Considering that the discharge from the runner is without whirl (i.e., \ \(v_{w2} = 0\)), we use the relationship between the radial velocity and peripheral velocity similar to the guide vane angle calculation.
We use the equation \ \( \tan (\theta) = \frac{v_r}{u_1} \), where \ \(v_r\) is the radial velocity and \ \(u_1\) is the peripheral velocity. Both are provided as 9 m/s and 30 m/s respectively.
Calculating the runner blade angle:
\ \[ \theta = \tan^{-1}(\frac{9}{30}) \approx 16.7^\text{o}\ \]
The runner blade angle at the inlet therefore is also approximately 16.7 degrees, ensuring efficient flow of water through the runner and reducing hydraulic losses.
pressure head calculation
Pressure head is a measurement of the potential energy of water and is integral in calculating the performance of a turbine. It considers the height of water and its velocity. The pressure head at different points in the turbine can be computed using Bernoulli's principle, where the total head (\text{H}) is the sum of kinetic and potential head.
Given:
\ \[ H_R = 103.4 \text{ m}, \ H_e = 102.61 \text{ m}\
\]
For accuracy, the relationship includes heights and velocities, adjusted for any losses:
\ \[ H_{total} = \text{kinetic head + pressure head}\ \]
Total energy at different points, factoring in kinetic energy and velocity, helps us derive these heads precisely. For example, at the runner entry (\text{H_R}) ,
\ \[ H_{runner} = H_V - H_{loss 2} = 112.2 - 8.8 = 103.4 \text{ m}\ \]
With kinetic energy and velocity consideration, this approach is systematic for turbine analysis.
hydraulic losses in turbines
Hydraulic losses are the energy losses occurring due to friction and other resistances within the turbine components. Understanding and calculating these losses can significantly improve turbine efficiency. In the given problem, the losses are provided at different sections of the turbine.
The hydraulic losses include:
\ \1. Entrance to exit of guide vanes: \ \text{4.8 m}\
2. Losses in runner: \ \text{8.8 m}\
3. Draft tube losses: \ \text{0.79 m}\
4. Kinetic head rejected to tail race: \ \text{0.46 m}
\ \ Partial sum gives high precision required for understanding turbine operations effectively. Each of these losses can be summed to understand the total energy lost within the turbine. For example:
\ \[H_{total loss} = H_{loss 1} + H_{loss 2} + H_{loss 3} + H_{loss 4} = 4.8 + 8.8 + 0.79 + 0.46 = 14.85 \text{ m}\ \]
Each section's loss, measured in meters, reflects both kinetic and potential heads, ultimately impacting total head and efficiency calculations. Understanding these help in optimizing turbine designs and minimizing unwanted energy dissipation.

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Most popular questions from this chapter

A \(500 \mathrm{~mm}\) diameter fluid coupling containing oil of relative density \(0.85\) has a slip of \(3 \%\) and a torque coefficient of \(0.0014 .\) The speed of the primary is \(104.7 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (16.67 rev/s). What is the rate of heat dissipation when equilibrium is attained?

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.

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.

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.

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 \(220 \mathrm{~W}\).

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