James Li # 26

5/8/2015

Raj, Rishi

ME 43000

Spring 2015

Thermal Project # 2: Impulse stage design for turbine blade

Abstract: The main purpose of this project is to be able to understand turbines by designing a

midsection for the impulse stage of the steam turbine. The impulse stage is the steam turbine’s

first stage with no degree of reaction. The midsection’s angle is calculated using the formula for

the absolute and relative velocity and the blade height is calculated using the velocity ratio.

Design constraints for this design project were in place as listed below in order to make the

calculations. Concepts such as energy conservation and continuity are vital for this design project

because they are needed to maintain the turbine’s thermodynamic balance in order for it to work

properly. The objective is to figure out how many blades and their lengths are needed in order to

operate the impulse stage.

Table of Contents

1. Nomenclature

2. Theory and background

3. Introduction

4. Calculations and Results

5. Discussion and Conclusion

6. References

7. Appendix

Nomenclature

ΔH = Change in enthalpy (BTU/lbm)

P = Pressure (Psi)

T = Temperature (degrees Fahrenheit)

Vo = Adiabatic velocity (ft/s.)

U = Blade velocity (ft/s.)

gc = Gravitational constant (ft/s^2)

v = specific volume (ft^3/lbm)

= Mass flow rate (lbm/hr)

W = Real fluid flow velocity (ft/s.)

ρ = density (lb/ft^3)

Vx = Axial Velocity (ft/s)

= Thermal efficiency

rt = radius of tip (ft)

rh = radius of hub (ft)

rm = Mid-Radius (ft)

H = Power (Kwt)

c = chord length (in.)

s = spacing between blades (ft)

A = Blade Area (ft^2)

w = specific work (btu/lbm)

dm = mean diameter of the rotor (ft)

n = rotational speed of the shaft (Rpm)

m = main flow rate 3* 10^5 (lbm/hr)

α = angle at which adiabatic velocity hits blade

β = angle at which Real fluid velocity hits blade

Z = number of blades

Ψ = loading factor

b = axial cord length (in.)

Ω = angular velocity of the rotor

Theory and Background:

Newton’s Second law plays an important role for the design of the impulse stage of the

turbine. The definition of Newton’s second law states that the sum of all the forces acting on a

body in motion is equal to the body’s mass times its acceleration. In the impulse stage, the steam

gets blasted through a nozzle jet at a high velocity, impacting the turbine’s blade with a force that

is proportional to its mass multiplied by its acceleration.

a) Σ F = m*a: The Newton’s second law of motion Equation

The conservation of momentum is also vital to the design of the impulse stage because

the law of momentum conversation states that a system can not have an overall net change of

momentum. This means that a system’s momentum at the beginning of its process is equal to its

momentum at the end of its process. Since the overall change in momentum of the impulse

stage’s system is conserved, the forces acting on the blade will encounter an equal and opposite

reaction. This causes the blades to move.

b) m1*v1 = m2*v2 : Equation for the conservation of momentum

c) F = m * (dv/dt) : Equation for the force as a function of momentum

The conservation of mass is vital in understanding the impulse stage of a turbine as well.

A turbine is a closed system therefore; it has a controlled volume and surface. The application of

mass conservation here shows that the mass that enters the system is equal to the mass that exits

the system. This concept can be used to calculate the area of the turbine using the mass flow rate

as shown in following formula.

d) : Equation for mass conservation

Introduction:

A turbine is mechanical rotary device that extracts heat energy from fluid flow and

converts it into mechanical work. In this project, steam turbines will be the main focus with the

steam being the fluid, in this case, to be converted into work. Steam turbines can be used as

power plant generators because they are capable of producing enough energy to power an entire

city.

A steam turbine’s operation consists of two stages: the impulse stage and the reaction

stage. In the impulse stage, the turbine has a fixed nozzle that accelerates the fluid, steam in this

case, in order to increase the kinetic energy that moves the blades and rotates the shaft in order to

create useful mechanical work. The reaction stage is when the reaction force generated by the

nozzle expelled steam’s acceleration passes into the rotor. As the steam passes through the rotor

and into the stator, it changes direction and generates a change in pressure, temperature, and

velocity to create work. The focus will primarily be on the impulse stage for this particular

design.

The impulse stage on the report will have the following parameter constraints listed

below in order to determine the turbine blade dimensions, space between each blade, and the

number of blades needed to successfully operate the turbine.

Parameters

Power: 150,000 KW

Turbine inlet pressure: 2000 Psi

Inlet temperature: 1000 degrees Fahrenheit

Shaft rotational speed: N = 1800 Rpm

Heat to be converted into work: ΔH = 20 Btu/lbm

U/Vo = 0.38 – 0.5

α2 = 8 -12 degrees

Main flow rate: 3 * 10^5 lbm/hr

Thermal efficiency: ηt = 86

Entry velocity: V = 200 ft/sec.

Specific volume: v = 0.39479ft^3/lbm

Mass flow rate = 855060.4047

q(in) = 1223.23348 btu/lbm (First Project Cycle 3 Step 5)

q(out) = 624.6792759 btu/lbm (First Project Cycle 3 Step 5)

Mass flow rate: ṁ = 855060.4047 lbm/hr (First Project Cycle 3 Step 5)

Calculations and Results:

Step 1 Enthalpy change: ΔH (BTU/lbm) = 20

Step 2 Velocity converted from converting Enthalpy change:

V2 = Vo = Sqrt(2*gc*ΔH* (778 Btu/lbm)) = 1001.031 ft/s

Step 3 U/Vo = sqrt(ηt)/2 = 0.46368

Step 4 Blade velocity: U = (U/Vo)*(Vo) = 464.1592 (ft/s)

Step 5 n = 1800 rpm

U = (2*pi*n*rm/60)

rm = (60*U)/(2*pi*n) = 2.462 ft

Step 6 Set : α2 = 8 degrees

β2 = arctan[(V2*sin(α2))/(V2*cos(α2)-U)] = 14.8043 degrees or 0.2584 radians

W2 = V2*sin(α2)/sin(β2) = 545.2298 ft/s

Step 7 β3 = β2 = 14.8043 degrees or 0.2584 radians

W3 = W2 = 545.2298 ft/s

Step 8 α3 = arctan[(W3*sin(β3))/ (W3*cos(β3)-U)] = 65.6770 degrees or 1.1462 radians

V3 = W3*sin(β3)/sin(α3) = 152.8872 ft/s

Step 9 Specific work: w = (W2*cos(β2) + W3*cos(β3))*(U/gc) = 15197.04 btu/lbm

Step 10 efficiency η = (w)/[(Vo^2)/(2*gc)] = 0.9767

Step 11 For the Impulse design: we do not need the stator height because the stator is a nozzle.

L2 = (ṁ*v)/(2*pi*W3*sin(β3)) * (1/n) = 0.087004 feet

Step 12 c/S = 1

c = 0.5 inches

Step 13 0.85 = 2*(S/b)*[(tan(α2) + tan(α3)] * cos^2(β3)

b = 2*(S/0.85)*[(tan(α2) + tan(α3)] * cos^2(β3) = 0.4696 inches

In order for the design to work b < S, in this case 0.4696 < 0.5 so it checks out

N*S = 2*pi*rm

N = (2*pi*rm)/S = 247.7706 248 blades

rh/rt = 0.3

A = pi*(rt^2-rh^2) = pi * (2*rm) * (rm – L2) *(1/(rh/rt)-1) = 85.7563 ft^2

Step 14 ψ = w/(U^2/gc) = 2.271334 < 3 Loading factor is acceptable

Step 15 The design is acceptable because it meets the design criteria.

Excel spread sheet

Δh (btu/lbm) gc (ft/s^2) Power (Kw) ηt n

20 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 180020 32.2 150000 0.86 1800

q (in) q (out) ṁ (lbm/hr)v (ft^3/lbm) @

2000 psi Vo = V2 (ft/s)

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

1223.23348 624.6792759 855060.4047 0.39479 1001.031468

U/Vo U (ft/s) α2 (degrees) α2 (radians) β2 = β3 (degrees)

0.463680925 464.159197 8 0.13962634 14.8043505

0.463680925 464.159197 8.5 0.148352986 15.71459673

0.463680925 464.159197 9 0.157079633 16.62219667

0.463680925 464.159197 9.5 0.165806279 17.5270189

0.463680925 464.159197 10 0.174532925 18.42893583

0.463680925 464.159197 10.5 0.183259571 19.32782374

0.463680925 464.159197 11 0.191986218 20.22356298

0.463680925 464.159197 11.5 0.200712864 21.11603798

0.463680925 464.159197 12 0.20943951 22.0051374

β2 = β3 (radians) W2 = W3 (ft/s) α3 (radians) α3 (degrees) V3 (ft/s)

0.25838466 545.2297543 1.146280578 65.67703927 152.8872

0.274271454 546.2958258 1.175620425 67.35808865 160.3177

0.290112061 547.423752 1.202735209 68.91165133 167.8364

0.305904188 548.6130657 1.227889649 70.35289457 175.4312

0.321645608 549.8632784 1.251311482 71.6948668 183.0922

0.337334162 551.1738808 1.273197 72.94881461 190.8109

0.35296776 552.5443436 1.293715793 74.12445482 198.58

0.368544388 553.9741188 1.313014783 75.23020552 206.3932

0.3840621 555.4626397 1.331221635 76.27338126 214.2452

w η rm (ft) dm (ft) L2 (ft)

15197.04 0.976674 2.462441 4.924882 0.087004

15160.9 0.974351 2.462441 4.924882 0.081921

15122.59 0.971889 2.462441 4.924882 0.077404

15082.11 0.969287 2.462441 4.924882 0.073365

15039.46 0.966546 2.462441 4.924882 0.069731

14994.65 0.963666 2.462441 4.924882 0.066445

14947.67 0.960647 2.462441 4.924882 0.06346

14898.54 0.95749 2.462441 4.924882 0.060735

14847.25 0.954194 2.462441 4.924882 0.05824

c (in.) c/S S (in.) b (in) rh/rt

0.5 1 0.5 0.469603 0.3

0.5 1 0.5 0.444058 0.3

0.5 1 0.5 0.419076 0.3

0.5 1 0.5 0.394787 0.3

0.5 1 0.5 0.371268 0.3

0.5 1 0.5 0.348556 0.3

0.5 1 0.5 0.326659 0.3

0.5 1 0.5 0.305571 0.3

0.5 1 0.5 0.285271 0.3

A (ft^2) rh (ft) rt (ft) N (number of ψ

blades)

85.75628 1.64308 5.476932 247.7705542 2.271334

85.9398 1.644837 5.482789 248.0355335 2.265933

86.10286 1.646396 5.487988 248.2707298 2.260207

86.24869 1.64779 5.492633 248.4808818 2.254156

86.37987 1.649043 5.496809 248.6697735 2.247782

86.49849 1.650175 5.500582 248.840463 2.241084

86.60627 1.651202 5.504008 248.9954486 2.234063

86.70463 1.65214 5.507132 249.1367916 2.22672

86.79473 1.652998 5.509993 249.2662072 2.219055

Result ( α2= 8 degrees) The axial cord b = 0.4696 inches which is less than S = 0.5 inches so it works out.

The number of blades (N) for this design is 248 blades.

The loading Factor (ψ) = 2.271334 which is less than 3 so it is acceptable

The efficiency of the impulse stage is 0.976674 or around 97.67 percent

The Blade length/height is around 0.087004 feet or about 1.044 inches.

Discussion and Conclusion:

Based on the data and calculation, the design for the impulse turbine at α2 = 8 degrees is

acceptable because it meets all the design requirements. The previous sections showed us how to

design an impulse stage of the turbine. The equations used were based on the law of energy

conservation which is needed in order to analyze how the turbine transforms heat into

mechanical work via rotational motion.

The relationship between mass flow rate and velocity is vital for the design of the

impulse stage of the turbine because they are directly proportional. Since the mass flow rate and

velocity is proportional, changes in either area will impact the final outcome of the work. It is

should be noted that the axial cord length and spacing in between the blades are inversely

proportional to the number of blades in the turbines.

The laws of Physics and Thermodynamics are crucial in understanding how to design a

turbine because they provide us with the equations to analyze how the turbine works and how to

make the turbine safely and efficiently in order to power the cities we live in and make our lives

easier.

References

[1]: Thermodynamics, an Engineering Approach. Yunus A. Cengel & Michael A. Boles. Seventh

edition. McGraw Hill Publications

[2]: Lecture notes: ME: 43000 Thermo-Fluid Systems Analysis and Design, Professor Rishi Raj.

[3]: Thermo-Fluid Systems Analysis and Design. Rishi S. Raj. Sixth edition. Publication of G.I.

Corporation.

[4]: Me 43000 Project 1: Optimal Thermal Cycle for steam turbine Power Plant

Appendix: