Power Engineering
Exam Duration: 45 Mins Total Questions : 20
In an impulse turbine
- (a)
the entire kinetic energy is converted to pressure energy at the inlet to the turbine
- (b)
part of the kinetic energy is converted to pressure energy
- (c)
no part of pressure energy is converted to KE in the inlet
- (d)
the entire pressure energy is converted to velocity energy
In an inward flow reaction turbine, which of the following statements is correct?
- (a)
Water enters the outer periphery and moves radially inwards
- (b)
As speed increases, discharge increases
- (c)
Centrifugal head increases from inlet to outle
- (d)
All of the above
The function of guide vanes in a reaction turbine is
- (a)
to ensure shock less entry of water
- (b)
control flow rate of water on to the runner
- (c)
to guide the water at correct angle on to runner blades
- (d)
All of the above
Draft tube is fitted with the following purpose
- (a)
recover part of KE of water
- (b)
control the flow rate of water
- (c)
to guide water out of turbine
- (d)
avoid splashing of water
Cavitation occurs in centrifugal pumps due to
- (a)
very low suction pressure
- (b)
very high impeller speed
- (c)
Both (a) and (b)
- (d)
low discharge
Manometric head of a pump means
- (a)
total lift from pump centre line to the upper level of water
- (b)
head difference between two arms of a manometer connected between the suction and delivery
- (c)
pressure measured by a manometer at any point on the pump
- (d)
manometric. pressure of the pump casing
In order to avoid cavitation
- (a)
(NPSH)required< (NPSH)available
- (b)
(NPSH)required>(NPSH)available
- (c)
(NPSH)required = (NPSH)available
- (d)
None of the above
A jet of water with velocity of 10 m/s strikes to a plate as shown in figure. Diameter of jet is 25 mm.
Then force required to apply at lower edge of plate to keep it vertical is
- (a)
28.7 N
- (b)
49.09 N
- (c)
24.5 N
- (d)
zero
Force exerted by jet
F = \(\rho\)Av. (v-0)
= \(\rho\)Av2
\(=\rho \times \frac { \pi }{ 4 } { d }^{ 2 }.{ v }^{ 2 }\)
\(=1000\times \frac { 3.14 }{ 4 } \times { (0.025) }^{ 2 }\times { (10) }^{ 2 }\)
=49.09 N
As plate remains vertical, so taking moment about
O,
F x 10 = P x 20
\(\Rightarrow \quad P=\frac { 49.09 }{ 2 } =24.5N\)
A jet of water with velocity of 10 m/s strikes to a plate as shown in figure . Diameter of jet is 25 mm.
If P = 0, and Wplate= 10 g, then angle upto which plate swings freely
- (a)
60°
- (b)
90°
- (c)
30°
- (d)
45°
When P = 0
Then, let plate stops at angel '\(\theta\)' that is equilibrium condition
\(\therefore\)F x 10 = W x 10 sin\(\theta\)
\(\Rightarrow\) sin\(\theta\) = 0.5
\(\Rightarrow\) \(\theta\) = 30°
A Pelton turbine develops a power of 3600 kW at 400 rpm under a net head of 275 m. Its specific speed is
- (a)
18.29
- (b)
92.17
- (c)
89.43
- (d)
21.43
\({ N }_{ s }=\frac { N\sqrt { P } }{ { H }^{ 5/4 } } \)
\(=400\frac { \sqrt { 3600 } }{ { (275) }^{ 5/4 } } =21.43\)
A Pelton wheel produces power 4000 kW under a head of 250 m.
If coefficient of velocity is 0.98, then velocity of jet is
- (a)
70 m/s
- (b)
92.31 m/s
- (c)
68.6 m/s
- (d)
50.09 m/s
\(v={ C }_{ v }\sqrt { 2gH } \)
\(=0.98\sqrt { 2\times 9.81\times 250 } \)
= 68.3 m/s
A Pelton wheel produces power 4000 kW under a head of 250 m.
If nozzle area is 0.03 m2 then turbine efficiency is
- (a)
91%
- (b)
79%
- (c)
56%
- (d)
83%
\({ \eta }_{ turbine }=\frac { Shaft\quad power }{ Hydraulic\quad power } \)
\({ \eta }_{ \gamma }=\frac { { P }_{ s } }{ \rho QgH } =\frac { { P }_{ s } }{ \rho \times A\times v\times g\times H } \)
\(=\frac { 4000\times 1000 }{ 1000\times 0.03\times 68.6\times 9.8\times 250 } \)
= 0.79 or 79 %
In an inward flow reaction turbine discharge at outlet is radial. Velocity of flow at outlet is 2.5 m/s. If vane angle at outlet is 15 ° then speed of turbine is (given diameter at outlet is 0.6 m)
- (a)
301 rpm
- (b)
297 rpm
- (c)
310 rpm
- (d)
280 rpm
Outlet velocity diagram, as entry is radial
v2 =Vf2
= 25 m/s
\(tan\theta =\frac { { v }_{ 2 } }{ { u }_{ 2 } } \)
\(\Rightarrow { u }_{ 2 }=\frac { 2.5 }{ tan{ 15 }^{ o } } =9.3m/s\)
\(=\frac { \lambda { d }_{ o }N }{ 60 } \Rightarrow N=297rpm\)
In a Francis reaction turbine, discharge from the runner is radial at outlet . Blade speed at inlet is 10 m/s and whirl velocity at inlet is 12 m/s. If power developed is 100 kW then discharge is
- (a)
0.73 m3/s
- (b)
0.57 m3/s
- (c)
0.91 m3/s
- (d)
0.83 m3/s
We know P = pQ (vw1 , u1 vw2, u2)
As radial entry at outlet so
v2= vf2 and vw2
Power P = pQ ( Vw1) (u1)
1000 x 1000 = 1000 x Q x 12 x 10
\(\Rightarrow\) Q = 0.83 m3/s
Velocity at entrance and exit of a conical draft tube are 6 m/s and 1.3 m/s respectively. Tail water is 5 m below the entrance of draft tube. Assuming friction loss of 0.1 m, atmospheric pressure head as 10.3 m of water. Pressure head at entrance to draft tube
- (a)
4.27 m
- (b)
1.29 m
- (c)
3.65 m
- (d)
2.87 m
We know,
\(\frac { { P }_{ 2 } }{ W } =\frac { { P }_{ a } }{ W } -\left[ { H }_{ s }+\frac { { v }_{ 2 }^{ 2 }-{ v }_{ 3 }^{ 2 } }{ 2g } -{ h }_{ f } \right] \)
\(\frac { { P }_{ 2 } }{ W } =10.3\left[ 5+\frac { { \left( 6 \right) }^{ 2 }-{ \left( 1.3 \right) }^{ 2 } }{ 2\times 9.81 } -0.1 \right] \)
= 3.65 m
A reaction turbine with draft tube is shown in figure. Net head available is
- (a)
H
- (b)
H- Hs
- (c)
Hs
- (d)
-Hs
As draft tube increases the net head by permitting
the negati ve head to be established at outlet of runner.
\(\therefore\) Net head = (H - Hs) - (-Hs)
=H
The given figure belongs to centrifugal pump. Impeller is rotating anticlockwise and velocity diagram at outlet is shown. The vane is
- (a)
backward
- (b)
radial
- (c)
forward
- (d)
Cannot be determined
As \(\phi \) < 90°, backward vane
If \(\phi \) = 90°, radial vane
If \(\phi \) > 90°, forward vane
A centrifugal pump has speed 300 rpm, it consumes 17 kW power. It produces 30 m manometric head and delivers 1 m3/s discharge, then its specific speed is
- (a)
26.27
- (b)
29.21
- (c)
31.02
- (d)
23.41
For pump,
\({ N }_{ s }=\frac { N\sqrt { Q } }{ { H }_{ m }^{ 3/4 } } =\frac { 300\sqrt { 1 } }{ { \left( 30 \right) }^{ 3/4 } } \)
Ns = 23.41
A pump delivers a discharge to head of 30 m. Given, suction lift = 5.58 m, atmospheric head = 9.786, friction loss in inlet pipe = 0.3 rn, If vapour pressure is 3.0 kPa (absolute) then minimum value of NPSH is
- (a)
4.2 m
- (b)
3.6 m
- (c)
1.2 m
- (d)
7.1 m
\({ h }_{ v }=\frac { P_{ v } }{ \rho g } =\frac { 3\times { 10 }^{ 3 } }{ 1000\times 9.8 } \)
hv= 0.306 m
NPSH = ha - (hv + hs + hfs +\(\frac { { v }_{ s }^{ 2 } }{ 2g } \))
Vs is negligible for minimum NPSH.
\(\therefore\) NPSH =9.786 - (0.306 + 5.58 0.3)
=3.6
For a centrifugal pump, Thoma's cavitation factor is 0.12 and manometric head is 30 rn, then NPSH is
- (a)
2.1
- (b)
3.6
- (c)
4.2
- (d)
3.1
\({ \sigma }_{ c }\frac { NPSH }{ { H }_{ m } } \)
NPSH = 0.12 x 30
= 3.6