Unit 3 of ME 3003 (Mechanical/Automobile Engineering)

 Unit 3: Impact of Jets from your ME 3003 (Mechanical/Automobile Engineering). These are short notes for revision purpose. please refer you Reference book & College study materials for complete study.

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3. IMPACT OF JETS

The impact of jets refers to the force exerted by a high-velocity fluid jet (such as water or steam) when it strikes a surface. This is an important concept in fluid machinery, especially turbines and nozzles, where energy is transferred from the jet to the surface to do useful work.

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3.1 Impact of Jet on Fixed and Vertical Flat Plates

When a jet strikes a fixed, vertical flat plate, the following occurs:

• Fixed Plate (Normal Impact):

  • When the jet strikes the plate at a right angle (normal), the fluid velocity is completely absorbed by the plate. The jet’s momentum is transferred to the plate.
  • The force exerted by the jet on the plate is given by the rate of change of momentum.

  Formula for Force:

  $$F = \dot{m} \cdot V = \rho A V^2$$

  • Where:

    • $F$ = Force exerted by the jet on the plate
    • $\dot{m}$ = Mass flow rate of the jet (kg/s)
    • $V$ = Velocity of the jet (m/s)
    • $\rho$ = Density of the fluid (kg/m³)
    • $A$ = Cross-sectional area of the jet (m²)

  • In the case of normal impact, the velocity change is maximal (from $V$ to zero), so the force is calculated as above.

• Vertical Flat Plate (Oblique Impact):

  • If the jet strikes at an oblique angle, the velocity components are split into two:
    • One component parallel to the surface (which does not change and is not involved in the impact).
    • The other component normal to the surface (which is absorbed and generates the force).
  • The force exerted by the jet can be calculated similarly, but only the normal component of velocity contributes to the force.

  Force Calculation for Oblique Impact:

  $$F = \rho A V_n V$$
  • Where $V_n$ is the normal component of the velocity and is calculated as $V \cos \theta$, where $\theta$ is the angle of impact.

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3.2 Impact of Jet on Curved Vanes

The impact of a jet on a curved vane (like in turbines) is a more complex case. The curve of the vane alters the direction of the jet as it strikes the surface, and the velocity is changed both in magnitude and direction.

• Curved Vane (Turbine Blade):

  • The curved vane redirects the jet, causing a change in the direction of the fluid's velocity. In turbines, this is used to extract energy from the fluid.

  • The work done by the jet on the vane is related to the change in velocity of the jet as it strikes and leaves the vane.

  • If the vane is curved and moving, the relative velocity between the jet and the vane needs to be considered. This involves both the velocity of the jet and the velocity of the vane itself.

  • Work Done on the Vane: The work done by the jet on the vane is related to the change in momentum of the fluid.

  Formula for Work Done: $$W = \dot{m} \cdot (V_{\text{inlet}} - V_{\text{outlet}}) \cdot R$$

  • Where $R$ is the radius of the vane and $V_{\text{inlet}}$ and $V_{\text{outlet}}$ are the velocities of the jet at the points of entry and exit, respectively.

  • Efficiency:

    • Efficiency is the ratio of the useful energy (work done by the vane) to the total energy supplied by the jet.

    Efficiency Formula:

    $$\eta = \frac{\text{Useful Work}}{\text{Total Energy Supplied by Jet}} = \frac{\text{Work Done}}{\text{Energy of Jet}} = \frac{W}{\frac{1}{2} \rho A V^2}$$

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3.3 Simple Numericals on Work Done and Efficiency

Let's go through a couple of simple examples involving work done and efficiency of jets.

Example 1: Impact of Jet on Fixed Plate

A jet of water with a velocity of 10 m/s strikes a fixed flat plate normally. The cross-sectional area of the jet is $0.01 \, \text{m}^2$. Calculate the force exerted on the plate.

Solution:

• Given:

  • $V = 10 \, \text{m/s}$
  • $A = 0.01 \, \text{m}^2$
  • $\rho = 1000 \, \text{kg/m}^3$ (density of water)

• The mass flow rate $\dot{m} = \rho A V$.

  $$\dot{m} = 1000 \times 0.01 \times 10 = 100 \, \text{kg/s}$$

• The force $F$ exerted on the plate is:

  $$F = \dot{m} \times V = 100 \times 10 = 1000 \, \text{N}$$

Thus, the force exerted on the fixed plate is 1000 N.

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Example 2: Impact of Jet on Curved Vane

A jet of water strikes a curved vane at a velocity of 12 m/s. The angle of impact is 30°, and the vane is moving with a velocity of 4 m/s. The mass flow rate of the jet is $0.2 \, \text{kg/s}$. Find the work done by the jet on the vane.

Solution:

• Given:

  • $V = 12 \, \text{m/s}$
  • $V_{\text{vane}} = 4 \, \text{m/s}$
  • $\dot{m} = 0.2 \, \text{kg/s}$
  • The angle of impact $\theta = 30^\circ$.

• First, calculate the relative velocity of the jet with respect to the vane. This is the velocity of the jet minus the velocity of the vane (in the direction normal to the vane).

  The normal component of the velocity $V_n = V \cos \theta$:

  $$V_n = 12 \times \cos 30^\circ = 12 \times 0.866 = 10.4 \, \text{m/s}$$

• The work done on the vane is:

  $$W = \dot{m} \cdot (V_{\text{inlet}} - V_{\text{outlet}}) \cdot R$$

  Assuming the exit velocity is zero (complete absorption of kinetic energy):

  $$W = 0.2 \cdot 10.4 \cdot R$$

  You would need the radius $R$ of the vane to calculate the exact work done.

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Summary of Key Concepts

• Impact of Jet on Fixed Plates: The force is proportional to the velocity squared and the cross-sectional area of the jet.
• Impact of Jet on Curved Vanes: The work done by the jet on the vane depends on the change in velocity, and efficiency depends on how effectively energy is transferred from the fluid to the vane.
• Work Done: Calculated using the change in momentum of the jet as it strikes the vane.
• Efficiency: The ratio of useful work done by the jet to the total energy provided by the jet.