Question

In a simple impulse turbine the nozzles are inclined at 20 deg to the direction of the moving blades. The steam leaves the nozzle at 375 m/s. The blade speed is 165m/s. Calculate suitable inlet and outlet angles for the blades in order that the axial thrust is zero. The relative velocity of the steam as it flows over the blades is reduced by 15% by friction. Also determine the power developed for a flow rate of 10 kg/sec​

Answer 1

Answer:

I dont know ok

Explanation:

✋✋

Answer 2

Answer:

The inlet blade angle is 55.60°

The outlet blade angle is -48.35°

Power developed is 378.13W

Explanation:

Given that,

Nozzle is inclined at an angle of 20° to direction of blade moving.

Hence,the angle made with axial direction is

$\alpha _{1} = 90 - 20 = 70 {}^{0}$

Also given that,

Blade speed(u)=165m/s

Nozzle speed

$v _{1} = 375ms {}^{ - 1}$

As axial thrust does not exist,the flow velocities remain same.

$= > v _{f1} = v _{f2} = v _{1}cos70 \\ = 375cos70 = 128.25ms {}^{ - 1}$

Let the inlet blade angle be

$\beta _{1}$

Therefore,from known inlet velocity diagram we have

$v _{r1}sin \beta _{1} = v _{1}sin \alpha _{1} - u \\ v _{r1}cos \beta _{1} = v _{1}cos\alpha _{1}$

from this we get

$tan \beta _{1} = \frac{375sin70 - 165}{375cos70} \\ = > \beta _{1} = 55.60 {}^{0}$

From this,we get

$v _{r1} = 227.07ms {}^{ - 1}$

Since,blade speed is reduced by 15%,

$v _{r2} = 0.85v _{r1} \\ = > v _{r2} = 193.01ms {}^{ - 1}$

From outlet velocity diagram,we have outlet blade angle

$\beta _{2} = - 48.35 {}^{0} \\ \alpha _{2} = - 9.19 {}^{0}$

Using relation,

$v _{2} cos \alpha _{2} = v _{r2}cos \beta _{2} \\ = > v _{2} = 129.93ms {}^{ - 1}$

Similarly inlet and exit tangential velocities are

$v _{t1} = v _{1}sin \alpha _{1} \\ = 375sin70 = 352.38ms {}^{ - 1} \\ v _{t2} = v _{2}sin \alpha _{2} \\ = 129.93sin( - 9.19)= - 20.75ms {}^{ - 1}$

Hence,power developed is

$p = m {}^{.} (v _{t1} - v _{t2}) \\ = 10(352.38 + 20.75) \\ = 378.13W$

#SPJ3