thermal project # 2

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
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  Abstract: The mainpurpose 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.)
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  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
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  Theory and Background: Newton’sSecond 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.
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  A steam turbine’soperation 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)
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  Calculations and Results: Step1 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 = ( *v)/(2*pi*W3*sin(β3)) * (1/n) = 0.087004 feetṁ 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
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  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 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 20 32.2 150000 0.86 1800 q (in) q (out) (lbm/hr)ṁ 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
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  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 blades) ψ 85.75628 1.64308 5.476932 247.7705542 2.271334
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  85.9398 1.644837 5.482789248.0355335 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:
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  Based on thedata 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: