/*! 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 68 In the figure is shown a detail ... [FREE SOLUTION] | 91Ó°ÊÓ

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In the figure is shown a detail of the stationary nozzle diaphragm \(A\) and the rotating blades \(B\) of a gas turbine. The products of combustion pass through the fixed diaphragm blades at the \(27^{\circ}\) angle and impinge on the moving rotor blades. The angles shown are selected so that the velocity of the gas relative to the moving blade at entrance is at the \(20^{\circ}\) angle for minimum turbulence, corresponding to a mean blade velocity of \(315 \mathrm{m} / \mathrm{s}\) at a radius of \(375 \mathrm{mm}\). If gas flows past the blades at the rate of \(15 \mathrm{kg} / \mathrm{s}\), determine the theoretical power output \(P\) of the turbine. Neglect fluid and mechanical friction with the resulting heat-energy loss and assume that all the gases are deflected along the surfaces of the blades with a velocity of constant magnitude relative to the blade.

Short Answer

Expert verified
Power output depends on velocity differences, calculated by trigonometric principles. Calculate using formula: \(P = \dot{m} \times u \times \Delta V_t\).

Step by step solution

01

Identify Given Data

The problem gives the following information: - Velocity angle at stationary nozzle diaphragm: \(\alpha = 27^{\circ}\)- Relative velocity angle at entrance to moving blade: \(\beta = 20^{\circ}\)- Mean blade velocity: \(u = 315 \, \text{m/s}\)- Radius: \(r = 375 \, \text{mm} = 0.375 \, \text{m}\)- Mass flow rate: \(\dot{m} = 15 \, \text{kg/s}\)
02

Understand Velocity Relationships

The gas velocity at the fixed nozzle can be decomposed into two components, axial velocity \(V_x\) and tangential velocity \(V_{f,t}\). Given the angle, \(V_x = V_f \cos \alpha\) and \(V_{f,t} = V_f \sin \alpha\). When entering the moving blade, velocity is \(20^{\circ}\) relative to the blade.
03

Calculate Velocity Components

Using the law of cosines, the tangential components at the nozzle and at the blade should match. Thus, initial velocity components:\[ V_f \sin(\alpha) = u + V_r \sin(\beta) \] with \(V_r = V_f\) because velocity is constant relative to blades.
04

Calculate Gas Velocities

From step 3, solve for \(V_f\)\[ V_f \sin(27^{\circ}) = 315 + V_f \sin(20^{\circ}) \]Thus: \[ V_f = \frac{315}{\sin(27^{\circ}) - \sin(20^{\circ})} \]
05

Calculate Change in Tangential Velocity

Once we have \(V_f\), calculate the change in tangential kinetic energy. The change relates to power:\[ \Delta V_t = V_{f,t} - V_{r,t} = V_f \sin(27^{\circ}) - (315 + V_f \sin(20^{\circ})) \]
06

Calculate Theoretical Power Output

Using the kinetic energy principle, power output is \(P = \dot{m} \times u \times \Delta V_t\). Utilize the calculated \(\Delta V_t\) from step 5 for the final power output value.

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Key Concepts

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

Velocity Components in Fluid Mechanics
When dealing with fluid mechanics, breaking down the velocity of a gas or liquid into components is crucial for understanding the flow dynamics. In the exercise, we initially observe how the gas exits the stationary nozzle diaphragm at an angle of \(27^{\circ}\). To analyze this motion fully, the velocity can be decomposed into two main components: axial and tangential.
The axial component refers to the straight-line motion parallel to the axis of the turbine, and it is calculated using \( V_x = V_f \cos \alpha \). Meanwhile, the tangential component, which affects the motion parallel to the circumference, is found using \( V_{f,t} = V_f \sin \alpha \).
These components are important because they determine how the gas interacts with the moving turbine blades. Consistently understanding these two components helps predict the gas path as it transfers energy to the turbine.
Turbine Blade Dynamics
Turbine blade dynamics examines how the angles and speeds of incoming gases affect energy transfer and overall performance of a turbine. In the provided exercise, the critical aspect is the angle of the gas relative to the blade at entry, ideally set at \(20^{\circ}\).
This specific angle is critical as it minimizes turbulence—unpredictable swirls and eddies that can cause energy loss. By maintaining a constant velocity magnitude relative to the blade as the gas moves, energy losses are minimized, allowing for the design of highly efficient turbine systems.
The dynamics also depend on the relationship between the blade speed \(u = 315 \text{ m/s}\) and the radial position \(r = 0.375 \text{ m}\). Understanding these relationships is vital to optimize the energy capture and translate it into mechanical work.
Energy Conversion in Turbines
The primary function of a turbine is the conversion of kinetic energy from the gas into mechanical energy which can then be used to generate electricity. In this exercise, the theoretical power output is the focus, calculated from the change in tangential velocity.
Initially, one calculates the gas velocity \(V_f\) at the nozzle using the relationship: \( V_f \sin(27^{\circ}) = 315 + V_f \sin(20^{\circ}) \). This equation results from equating the tangential components, a step necessary to solve for \(V_f\).
Once \(V_f\) is known, the change in tangential kinetic energy, \(\Delta V_t\), is computed as \[ \Delta V_t = V_f \sin(27^{\circ}) - (315 + V_f \sin(20^{\circ})) \]. The power output is then derived as:
  • \( P = \dot{m} \times u \times \Delta V_t \)
  • where \(\dot{m} = 15 \text{ kg/s}\) is the mass flow rate.
These calculations underscore the elegant transformation of kinetic energy into usable mechanical energy, demonstrating the efficiency and potential of turbine systems.

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