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