Post on 16-May-2018
Fifth Semester B.E. III IA Test, 2014
PES INSTITUTE OF TECHNOLOGY, BANGALORE
(Bangalore South Campus)
Hosur Road, 1KM before Electronic City, Bangalore-560 100
Department of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
Date: 12-11-2014 Max Marks: 50
Subject & code: Turbomacnines (10ME56) Time: 90 Mins.
Name of faculty: P. Dandapani Semester & Section: V ‘B’
Scheme and Solutions 1. A double jet Pelton wheel is required to generate 7500KW. When the available head at the base of
the nozzle is 400m. The jet is deflected to 1650 and the relative velocity of the jet is reduced by 15%
in passing over the buckets. Determine the diameter of the jet, total flow rate and force exerted by
the jet in tangential direction. Assume generator efficiency=95%, overall efficiency=80%, speed
ratio = 0.47 and Cv=0.97.
Solution: -
Diameter = 0.1344m--------------3M
Discharge = 2.515m3/s------------3m
tangential Force = 215.08kN-------4m
2. A penstock supplies water from the reservoir to the Pelton wheel with a gross head of 500m. 1/3rd
of
the gross head is lost due to friction in the penstock. The rate of flow of water through the nozzle
fitted at the end of the penstock is 2m3/s. the angle of deflection of jet is 165
0. Determine the power
given by the water to the runner and the hydraulic efficiency of the Pelton wheel. Take speed ratio =
0.45 and Cv = 1.
Solution: -
u = 36.39 m/s
V1 = 80.87m/s
Vu1 = 6.57m/s---------------6m
p = 5.4MW----------------------2m
efficiency = 50%---------------2m
Fifth Semester B.E. III IA Test, 2014
PES INSTITUTE OF TECHNOLOGY, BANGALORE
(Bangalore South Campus)
Hosur Road, 1KM before Electronic City, Bangalore-560 100
Department of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
3. The inward flow reaction turbine is to develop 300KW at 200rpm. the active head at the turbine is
18m. Determine the outlet and inner diameters, the inlet and exit angle for the moving vane and the
exit angle of the guide vane. Assume the inlet
diameter is twice the outlet diameter. Hydraulic
efficiency = 80%, a constant radial velocity of
flow as 3.6m/s through the runner, mechanical
efficiency = 95%, width ratio = 0.1, water
leaves the runner radially.
Solution: -d1= 1.4m
d2= 0.7m
nozzle angle = 6.7
blade angles = 26.5, 12.9-------------10m
4. A Kaplan turbine produces 44000KW under a head of 25m with an overall efficiency of 90%.
Taking the value of speed ratio =1.6, flow ratio = 0.5 and hub diameter as 0.35 times the outer
diameter. Find i) Diameter of runner ii) speed of the runner iii) specific speed.
Solution: -discharge =199.3m3/s
N= 132rpm
Ns=496.88rpm-------------10m
5. For a single stage impulse steam turbine, prove that the maximum blade efficiency is given by
where K=Vr2/Vr1 and C = cosβ2/cosβ1, α1= nozzle angle and β1, β2 is blade
angles at inlet and outlet, and Vr2,Vr1 are outlet and inlet relative velocity.
Solution: -
Energy transfer as a work per Kg of steam is
E=W=U(Vu1+Vu2)/gc
From velocity triangle,
∆Vu = Vu1+Vu2=Vr1Cosβ1+ Vr2Cosβ2
Vr1Cosβ1= V1Cosα1-U
∆Vu = V1Cosα1-U [1+(Vr2Cosβ2/ Vr1Cosβ1)]
∆Vu = V1Cosα1-U[1+KC]
where K=Vr2/Vr1, C = cosβ2/cosβ1
∴W = U(V1Cosα1-U) [1+KC]
If we express in terms of speed ratio, �=U/V1
W = V12 � (Cosα1- �) [1+KC]
Blade efficiency = ηb =W / (V12/2 gc)
Fifth Semester B.E. III IA Test, 2014
PES INSTITUTE OF TECHNOLOGY, BANGALORE
(Bangalore South Campus)
Hosur Road, 1KM before Electronic City, Bangalore-560 100
Department of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
ηb =2 V12 � (Cosα1- �) [1+KC] / V1
2
For maximum Blade efficiency,dηb/d� =0
we get speed ratio �= Cos α1 /2
∴ we get,
6. In a Parson’s turbine running at 1500rpm, the available enthalpy drop for an expansion is 63KJ/kg. if
the mean diameter of the rotor is 100cm, find the number of moving rows required. Assume stage
efficiency=0.8, blade outlet angle =200 and speed ratio= 0.7
solution: - numbers of rows = 6 ------------------------10m
7. In a single stage Impulse turbine, the nozzle angle is 250.
absolute velocity of the steam at the exit is 300m/s at 1200
in a direction of motion of the blades. Assuming
equiangular blades and no axial thrust determine (i) Blade
angles (ii) Power developed per Kg and (iii) Diagram
efficiency.
Solution: -
(i) Blade angles =36
(ii) Power developed per Kg = 136.5kW
(iii) Diagram efficiency. = 77.7%-----------------10m
8. The velocity of the steam at the exit of the nozzle is
440m/s, which is compounded in an impulse turbine by passing successively through the moving
blade and fixed blade and finally through the second ring of moving blades. the tip angle of the
moving blade through the turbine are 300. Assuming the loss of 10% in velocity due to friction
throughout the blades as the steam passes over the blade ring Find the velocity of moving blades in
order to have the final discharge of steam as axial. Also determine the diagram efficiency.
Fifth Semester B.E. III IA Test, 2014
PES INSTITUTE OF TECHNOLOGY, BANGALORE
(Bangalore South Campus)
Hosur Road, 1KM before Electronic City, Bangalore-560 100
Department of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
Rotor efficiency = {U(Vu1+Vu2)+ U(Vu3-Vu4)}/V12 = 86.6% -------------------10m