GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 54 of 65

Marks: +1, -0

For a Kaplan (axial flow) turbine, the outlet blade velocity diagram at a section is shown infigure.

The diameter at this section is 3 m. The hub and tip diameters of the blade are 2 m and 4 m respectively. The water volume flow rate is

100 m^3{ /}s

. The rotational speed of the turbine is 300 rpm. The blade outlet angle β is _____ degrees (round off to one decimal place).

Solution:

From, the above velocity diagram, Blade outlet angle

(β)

can be found by,

\tan β=\;{C_{f}}/{C_{b}}

where,

{C}_{ {f}}

is flow velocity

{C}_{ {b}}

is blade velocity
Blade velocity

(C_{b})=\;{π D_{\;\text "mean "\;} N}/{60}

=\;{π × 3 × 300}/{60}

47.12 {m} {/} {sec}

Flow velocity

(C_{f})=\;{\;\text " volume flow rate "\;}/{\;\text " net change in area "\;}

Net change in area

A

=\;{π}/{4}(4^{2}-2^{2})

=\;{π}/{4} ×(16-4)

=\;{π}/{4} × 12=3 π

Flow velocity

(C_{f})=\;{100}/{3 π}=10.61 {m} {/} {sec}

So, blade outlet angle

(β)

\tan β=\;{C_{f}}/{C_{b}}=\;{10.61}/{47.12}=0.225

β=\tan ^{-1}(0.225)

β=12.69^{∘}

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