Impulse Turbine MCQ Quiz in தமிழ் - Objective Question with Answer for Impulse Turbine - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Design speed:

It is the net speed under which the turbine reaches its peak efficiency.

Specific speed: 

It is defined as the speed of a similar turbine working under a head of 1 m to produce a power output of 1 kW. The specific speed is useful to compare the performance of the various type of turbines. The specific speed differs for the different type of turbines and is the same for the model and actual turbine.

\({N_s} = \frac{{N\sqrt P }}{{{H^{\frac{5}{4}}}}}\)

Following are the range of specific speed of different turbines

• The specific speed of Pelton wheel turbine (single jet) is in the range of 10-35
• The specific speed of Pelton wheel turbine (multiple jets) is in the range of 35-60
• The specific speed of Francis turbine is in the range of 60-300.
• The specific speed of Kaplan/propeller turbine is greater than 300.

Concept:

Blade speed ratio is the ratio of blade velocity to the inlet jet velocity. It is denoted by ρ = \(\frac{u}{v}\)

For Pelton turbine (Impulse turbine) maximum blade efficiency is \({η _b} = {\cos ^2}α \) and Speed ratio (ρ) = \(\frac{{\cos α }}{2}\)

For maximum efficiency,

ηb = 1

\({1} = {\cos ^2}α \Rightarrow \alpha =0 \)

The Speed ratio

\(\frac uv =\frac{{\cos α }}{2}=\frac{cos 0}{2}=\frac 12\)

Therefore, v = 2u

where, u = blade velocity, v =  inlet absolute fluid flow velocity 

Concept:

In order to predict the behaviour of a turbine working under varying condition of head, speed output etc, the results are expressed in terms of quantities which may be obtained when the head on the turbine is reduced to unity.

The following are three important unit quantities.

Calculation:

Unit power is given as,

\({P_U} = \frac{P}{{{H^{3/2}}}}\)

∴ \( \frac{{{P_1}}}{{{{\left( {{H_1}} \right)}^{3/2}}}} = \frac{{{P_2}}}{{{{\left( {{H_2}} \right)}^{3/2}}}}\)

\(\frac{6}{{{{\left( {100} \right)}^{3/2}}}} = \frac{{{P_2}}}{{{{\left( {81} \right)}^{3/2}}}}\)

∴ P₂ = 4.4 MW

Concept:

Specific speed of turbine: The speed required to produce unit power working under a unit heat is called specific speed of turbine.

\({N_S} = \frac{{N\sqrt P }}{{{H^{5/4}}}}\)

Calculation:

N_S1 = N_S2, H₁ = H₂

\(\frac{{{N_1}}}{{{N_2}}} = 2\)

\(\therefore {N_1}\;\sqrt {{P_1}} \; = {N_2}\;\sqrt {{P_2}}\)

\(\frac{N_2}{N_1}=\sqrt\frac{P_1}{P_2}\)

\(\Rightarrow \frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{N_2}}}{{{N_1}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4} = 0.25\)

Concept:

Pelton wheel turbine:

• A Pelton wheel turbine is a tangential flow impulse turbine.
• In an impulse turbine total energy at the time of inlet of the turbine is only kinetic energy.
• A Pelton wheel turbine is used for the high head of water.
• The pressure of water both at entering and leaving the vanes is atmospheric.

Explanation:

In a Pelton turbine, energy conversion occurs in a specific sequence starting from the fluid's potential energy at the reservoir. The conversion steps are as follows:

Potential Energy:
Water stored at a high elevation has potential energy due to its height above the turbine.

Kinetic Energy:
Water is directed through nozzles where it is converted into high-velocity jets, thus transforming potential energy into kinetic energy.

Mechanical Energy:
The high-velocity jets impact the turbine buckets and cause them to rotate, thus converting kinetic energy into mechanical energy that drives the turbine shaft.

∴ The correct sequence is Potential energy - Kinetic Energy - Mechanical energy. 

Explanation:

Hydraulic turbines:

Hydraulic turbines may be considered as hydraulic motors or prime movers of a water–power development, which convert water energy (hydropower) into mechanical energy (shaft power).

• It is the device which transfer energy from the fluid to the rotor.
• The shaft power developed is used in running electricity generators directly coupled to the shaft of the turbine, thus producing electrical power.
• The hydraulic turbine is a rotary machine actuated by the impulse and/or reaction of a water current attacking the rotor (called the runner), which consists of a series of buckets or curved vanes or blades.

Classification of turbines on the basis of energy at inlet:

1. Impulse turbine: 

This has only kinetic energy at the inlet.

E.g- Pelton wheel.

2. Reaction turbine:

This has kinetic energy and pressure energy both at the inlet.

E.g- Francis turbine, Propeller turbine, Kaplan turbine, Thomson turbine

Nozzle: It is a pipe or tube of varying cross-sections. It is generally used to control the pressure or rate of flow.

Boiler: It is a closed vessel in which steam is produced from water by the combustion of fuel.

Concept:

Unit quantity

It is the parameter of the turbine which is defined when turbine operates under unit head. It is used to find out the speed, discharge, and power for same turbine.

Unit speed i.e. speed of the turbine under the unit head,

\({{\rm{N}}_{\rm{u}}} = \frac{{\rm{N}}}{{\sqrt {\rm{H}} }}\)

Calculation:

Given: 

P = 500 kW, H1 = 100 m, N1 = 200 rpm, and H2 = 81 m

\(\frac{{{{\rm{N}}_1}}}{{\sqrt {{{\rm{H}}_1}} }} = \frac{{{{\rm{N}}_2}}}{{\sqrt {{{\rm{H}}_2}} }} \Rightarrow \frac{{200}}{{\sqrt {100} }} = \frac{{{{\rm{N}}_2}}}{{\sqrt {81} }}\)

\(\Rightarrow {{\rm{N}}_2} = \frac{{200 \times 9}}{{10}} = 180{\rm{\;rpm}}\)

Unit discharge, \({{\rm{Q}}_{\rm{u}}} = \frac{{\rm{Q}}}{{\sqrt {\rm{H}} }}\)

Concept:

Hydraulic efficiency of the Pelton wheel is given by:

\({\eta _{h}} = \frac{{Runner\;power}}{{Kinetic\;energy\;per\;second}} = \frac{{2 \times \left( {{V_1} - u} \right) \times (1 + k\cos \phi )u}}{{V_1^2}}\)

For maximum efficiency,

When u = V₁/2

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

k = friction factor for blades

Calculation:

Given:

ϕ = 60°

As friction is neglected.

Maximum efficiency is:

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

\((\eta _h)_{max} = \left( {\frac{{1 + \cos 60^\circ }}{2}} \right) = \frac{{1 + \frac{1}{2}}}{2} = \frac{3}{4} \times 100 = 75\% \)

Concept:

The velocity diagram of a Pelton wheel is given below -

V₁, V₂ = Velocity of the jet at inlet and outlet.

u₁, u₂ = Velocity of the vane at inlet and outlet.

V_r1, V_r2 = Relative velocity of jet at inlet and outlet.

V_w1, V_w2 = Whirl velocity i.e. component of velocity parallel to the direction of motion.

V_f1, V_f2 = Flow velocity i.e component of velocity perpendicular to the direction of motion.

α = ∠ between the direction of jet and direction of motion.

θ = ∠ between V_r1 and the direction of motion.

β = ∠ between V₂ with the direction of motion.

ϕ = ∠ between V_r2 with the direction of motion.

Hydraulic efficiency is given by - 

\(η_h=\frac{R.P}{W.P}=\frac{\rho{Q}[V_{w1}\;\pm\;{V_{w2}}]u}{\rho{Q}gH}\)

Calculation:

Given:

For Pelton wheel 

V₁ = V_w1, u₁ = u₂ = u, V_r1 = V_r2.

If Nozzle is 100 % efficient -

\(⇒\rho{Q}gH=\frac{1}{2}\dot{m}{V_1}^{2}\)

Hydraulic efficiency is:

\(η_h=\frac{R.P}{W.P}=\frac{\rho{Q}[V_{w1}\;\pm\;{V_{w2}}]u}{\rho{Q}gH}\)

∴ \(η_h=\frac{\rho{Q}[V_{w1}\;\pm\;{V_{w2}}]u}{\frac{1}{2}\dot{m}{V_1}^{2}}\)

\(⇒ η_h=\frac{2[V_{w1}\;\pm\;{V_{w2}}]u}{{V_1}^{2}}\)   \([∵ \dot{m}=\rho{Q}]\)

V_w2 = V_r2cos ϕ - u

⇒ V_w2 = V_r1cos ϕ - u  (∵ V_r1 = V_r2)

⇒ V_w2 =  (V_w1 - u)cos ϕ -u (∵ V_r1 = V_w1 - u)

\(η_h=\frac{2[V_{w1}\;\pm\;{V_{w2}}]u}{{V_1}^{2}}\)

\( ⇒ {η _h} = \frac{{2\left[ {{V_1} + \left( {{V_1} - u} \right)\cos \phi - u} \right]u}}{{V_1^2}}\;\;\;[\because V_{w1}=V_{1}]\)

\( ⇒ \frac{{2u\left( {{V_1} - u} \right)\left[ {1 + \cos \phi } \right]}}{{V_1^2}}\)

For η_h to be maximum; \(\frac{d\eta_{h}}{du}=0\)

\(\frac{d\eta_{h}}{du}= 2(V_{1}-u)+2u(-1)\)

∴ 2u = 2V₁ - 2u

⇒ 4u = 2V₁

∴  \(\frac{u}{V_{1}} = \frac{{1}}{2}\Rightarrow 0.5\)

Explanation:

Hydraulic efficiency:

\( {\eta _h} = \frac{{W.D}}{{K.E}} = \frac{{\rho Q\left( {{V_1} - u} \right)\left( {1 + \cos \phi } \right)u}}{{\frac{1}{2}\rho aV_1^3}}\)

\( {\eta _h} = \frac{{2 {{(V_1} - u} )( {1 + \cos \phi } )u}}{{V_1^2}}\)

The efficiency will be maximum for a given value of V1 when

\(\frac{d}{{du}}\left( {{\eta _n}} \right) = 0 \Rightarrow u = \frac{{{V_1}}}{2}\)

i.e. Hydraulic efficiency of a Pelton wheel will be maximum when the velocity of the wheel is half the velocity of the jet of water at inlet.