Steam Power Engineering

Vinayak N. Kulkarni
Department of Mechanical Engineering
Indian Institute of Technology – Guwahati

Lecture - 25
Examples: Steam Turbines

Welcome to the class. Today we are going to see the examples on steam turbines.
(Refer Slide Time: 00:36)

One example read that the velocity of steam entering a simple impulse turbine is 100 metre
per second and nozzle angle is 20 °. The main peripheral velocity of the plates is 400 metre
per second and the blades are symmetrical. If the steam enters blades without shock find the
blade angles; a. Neglecting the frictional effect on the blades, calculate tangential force on the
blades and diagram power for a mass flow of 0.75 kg per second.

Estimate also the axial thrust and the diagram efficiency. If relative velocity at the exit is
reduced by friction to 80% of it is inlet value estimate the axial thrust diagram power and
diagram efficiency. So we will first plot the velocity triangle and then we will note whatever
it is given to us. So let us say that this is V b, this is V 1, so this is V r 1 and then this is V r 2 and
this is V 2.

So what we have is, this is α, this is β 1, this is β 2 okay. So noting this we can proceed for a,
we are said that we have to neglect the friction factor. So K b =0. We have been told that the
turbine is impulse turbine okay. So since the turbine is impulse we have degree of reaction
equal to 0. So V r 1=V r 2 since we have K b =0. Further we are also told that blades are
symmetrical.

So β 2=β 1, so with this point we can start then we will note that directly V r 1 sin ⁡(β 1 ) which is
basically, V a, V a1 , parallel V a 2=V r 2 sin ⁡( β2 ) but what would happen V r 1=V r 2, β 1=β 2, so
V a1 =V a2 =V a, so we will have V a =V r1 sin ⁡(β 1 )=V 1 sin ⁡(α ). So this height, so for this we can
just equate this much part and then we can have one more thing that V r 1 cos ⁡( β1 )+V b .

So this is V b and this is V r 1 cos ⁡( β1 ) and this is V b, so this is totally plus. So for this plus we
have equal to V 1 cos ⁡(α). So we have V r 1 cos ⁡( β1 )=V 1 cos ⁡(α 1) basically, V 1 cos ⁡(α) – V b. We

V 1 sin ( α )
can divide these 2 equations, this and this, and then we have tan β1 = . So
V 1 cos ⁡(α ) – V b

V 1 sin ( α )
β 1=tan−1
[ V 1 cos ( α ) – V b ]
So now we have β=tan−1 V 1 is given to us, which is 1000 metre per second, nozzle angle is

1000 ×sin ( 20 )
told to us as β=tan
−1
[ ]
1000 × cos ( 20 ) – 400
is. So this β turns out to be 32.35 degree and

since blades are symmetrical β 2=β 1=32.35 °. So we found out the angle. Now we can move
ahead and then we can find out V r 1=¿ we know V r 1 sin β1 =V 1 sinα .

So we know V r 1 sin ( 32.35 ) =V 1 × 1000×sin ⁡(20). So we have V r 1=639.25 and this we know
since we have no friction V r 1=V r 2 and both are 639.25 meters per second. So now we know
V r 1 , V r2 , β1 , β 2 and then we can now calculate the desired thing.
(Refer Slide Time: 07:20)
So for that we can go ahead and calculate Δ V w =V r 1 cos ⁡( β 1)+V r 2 cos( β2 ), but we know
V r 1=V r 2 ,cos ⁡(β 1 )=cos ⁡( β2 ). So it is Δ V w =2V r1 ×cos ⁡( β1 ), so we have

Δ V w =2× 639.25 ×cos ⁡(32.35) so we have Δ V w =1080.07 meters per second. So we can find
out thrust, or force which is equal to change in momentum. So it is equal to
Thrust=force= ṁ Δ V w .

And we know we are given that ṁis 0.75 into change in velocity is 1080.07. So
Thrust=Force=810.05 N . So this is the tangential thrust. Now we can find out the axial
thrust which is equal to axial thrust=V r 1 sin ⁡( β1 ) – V r 2 sin ⁡( β2 ) but again V r 1=V r 2and β 1=β 2.
So this is equal to 0, so we have this is basically, axial change in axial velocity which is Δ V a .

So axial thrust=F a =ṁ ΔV a=0 . So power and since we are finding out that power from the
diagram, so which is diagram power is equal to the force, which is axial thrust into velocity
which is V b. So power diagram power P=810.05 × 400=324.02 kw. So this is the diagram

P
η D= .
power. So we can find out diagram efficiency which is equal to 1 2
ṁ V 1
2

3
324.02 ×10
So here we can put it as 1 , so we have diagram efficiency is equal to 0.864.
× 0.75 ×1000 2
2
Now I can move ahead for the second part of the same example where we are told that
 V r2
friction factor is 0.8. So we are told that =0.8. So we have V r 2=V r 1 ×0.8. Our
V r1
calculation of β, our calculation of V r 1 is intact, it will be same as what we did for part a.

So we will carry forward with the same V r 1, which is basically, 639.25 × 0.8, so
V r 2=511.4 m/ sec. So we have Δ V w =V r 1 cos ⁡( β 1)+V r 2 cos ⁡(β 2 ) and now this will be

639.25 × cos ⁡(32.35)+511.4 × cos ⁡(32.35). So we have Δ V w =972.06 m/sec. So we have

thrust =ṁ Δ V w=0.75 × 972.06 and this we can find out 729.05 Newton.

So power=thrust ×V b =729.05 × 400. So this gives us magnitude 291.62 kilowatt. Similarly,

we can find out change in axial velocity is equal to V r 1 sin ⁡(β 1 ) – V r 2 sin ⁡(β 2 )= Δ V a. So we
have axial thrust is equal to axial thrust maybe T a= ṁ(V r1 × sin β1 – V r2 ×sin β2 ) we can find
out T a gives you all numbers which are known to us.

So we can put V r 1 over here T a=639.25 × sin ( 32.35 ) – 511.4 ×sin ⁡(32.35), so axial thrust for
us is 51.3 Newton. So this is how we can solve the example, which is for the impulse turbine.
We will move ahead, in the next example and iterate that an impulse in turbine has number of
pressure stages.
(Refer Slide Time: 15:35)

Each having a row of nozzles and single ring of blade. The nozzle angle in the first stage is
20 ° and the blade exit angle is 30 ° with reference to the plane of rotation. The mean blade
speed is 130 meter per second and velocity of the steam leaving the nozzle is 330 meter per
second taking the blade friction factor as 0.8 and nozzle efficiency as 0.85, determine the
work done for stage per kg of steam and the stage efficiency.

If steam is applied in the first stage is at 20 bar 250 ° C and the condenser pressure is 0.04 bar
estimate the number of stages required, assume that the stage efficiency and the work done
are same for all stages and that the reheat factor is 1.08. So we start, so for that let us draw
again the velocity triangle. This is V 1 , this is V r 1 , this is V a1 , β1 α , β2 , Δ , this is V r 2 .

Now we are given that α =20 °, we are also told that β 2=30 ° , V b =130 m/ sec and
V 1=330m/ sec. So these are the things with us now we have also drawn the velocity triangle.
So let us move ahead with part a. So here as well, we can write down the expression as
V r 1 sin ⁡(β 1 )=V 1 sin ⁡(α ) and the also we can know V r 1 cos ( β1 ) +V b =V 1 sin ( α ) V 1 cos ⁡(α ).

We can put the V b or here and then we can divide and then what we got last time we will get

V 1 sin ( α ) 330 sin ( 20 )

the formula which is tan β1 = = and this gives us tan ⁡(β 1 ) as
V 1 cos ⁡(α ) – V b 330cos ⁡(20) – 130
0.627 and which gives us β 1=32.07 °. So we can have V r 1 equal to basically, what we know

V 1 sin ( α )
V r 1 from this equation we can write down V r 1= .
sin ⁡( β1 )

330 sin ( 20 )
So we can write down β as and then this gives us V r 1 as 212.5 m/sec. So
sin ⁡(32.07)
V r 2=V r 1 × frictionfactor =212.3× 0.8, so we have V r 2=170.25 m/ sec. So we have got this
all, so now we can move Δ V w =V r 1 cos ⁡( β 1)+V r 2 cos ⁡(β 2 ). So we have
Δ V w =212.5 ×cos(32.07)+170.25 ×cos ⁡(30) and this gives us Δ V w , β2 is given as 30 ° .

So now in this we can find out Δ V w and this turns out to be 372.3 m/sec and we will go ahead
and find out rest of the things till time we have found out Δ V w blades are not symmetrical in
this case that is why we do not have β 's as equal, so we have found out β 1 , β2 is already
given, now we have found on changing tangential velocity.
(Refer Slide Time: 21:40)
So tangential thrust is equal to basically, ṁ Δ V w and then power=ṁV b Δ V w , so power is
equal to ṁ, let us assume 1 kg per second, V b is told as 130 and Δ V w we have just find out
327.33 and then this power is 42.55 kilojoule per Kg. So we can have diagram efficiency

P
η D= .
1 2
ṁ V 1
2

So we know it is 2, these 2 will go in the numerator 2 into 42.55 divided by ṁ is

1× 330×330. So this is in kilowatt joule we can make it into joule 330 ×330 and this gives us
diagram efficiency as 78.15 %. Now we can find out stage efficiency and stage efficiency is
nozzle efficiency into diagram efficiency or blade efficiency. So nozzle efficiency is 0.85 and
diagram efficiency is 0.78.

So we get it as 0.66, so this is the stage efficiency. Now we will go to the part b, in part b
needs turbine efficiency in total that is equal to nozzle efficiency, blade efficiency into reheat
factor. So we have found out this answer, stage efficiency into reheat factor. So turbine
efficiency or internal efficiencyis ηT =ηstage × Reheat factor=0.66 × 1.06=0.7041 , we can find
out enthalpy that let to the nozzle from the given temperature and pressure condition.

That is equal to 2902.3 kilojoule per Kg, similarly, we will also note down entropy under
same equation which is 6.5466 and Sg corresponding the same saturation pressure we can
find out under steam level and then we can say that S2=S1 basically, dash and then from here
we actually need S1=S2 ' . So we have basically, S1=Sf + x2 Sfg and this gives us S since S '2 is
known and then we know Sf liquid saturation entropy for the condensation pressure.

Then this is the latent entropy for the condensation pressure and then we know x from here
which is x 'is 0.7757, knowing this we can find out ideal enthalpy h'2 at the exit of the nozzle
so rest of the enthalpy is converted into kinetic energy. So this is level enthalpy with the flow.
So this is equal to hf + x2 × hfg and then we can find out this as 2032.29 kilojoule per kg.

Δ h∨¿total h 2 – h1
So number of stages is equal to ¿ ¿ . So Δ h∨¿total= ¿ , here h2 – h1 is
Δ h∨¿stage ¿ Δ h∨¿ stage ¿

equal to turbine into h2 – h 2 ' . So h1 – h'2. So we know this efficiency is 0.7041 into turbine
efficiency is this into h1 and h1 is basically, the enthalpy at the entry to the nozzle and that is
2902.3. We are given that this is going to have boiler at 20 bar and 250 ° C.

So considering these 2 conditions we can find out h1, this is the condition for 2', this is the
condition for 1. Now in the state 1 we can find out h1 and we also found out h2. So it is minus
2032.29. So we have h2 – h1 =612.57 kj/ kg. So this is the h2 – h1 =Δ h∨¿total ¿. So we have

Δ h∨¿stage
n= ¿.
Δ h∨¿total ¿

So Δ h stage, Δ h total divided by Δ h stage and Δ h∨¿total ¿ is 612.57 and Δ h stage is 42.55
and basically, this is the power of diagram, so n = 14.39, which is almost equal to 15. So we
need 15 number of stages for this turbine. So here we have solved this example on impulse
turbine.