thermal project # 2
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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:
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