LECTURE 9. Hydraulic machines III and EM machines 2002 MIT PSDAM LAB

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1 LECTURE 9 Hydraulic machines III and EM machines

2 .000 DC Permanent magnet electric motors Topics of today s lecture: Project I schedule revisions Test Bernoulli s equation Electric motors Review I x B Electric motor contest rules (optional contest) Class evaluations

3 Project schedule updates Approx START WHAT DUE PTS 07 March Project mgmt spread sheet 14 March [ 0 ] 1 March HMK 6: 1 page concept & equations + SIMPLE 1 page explanation 0 April [ 80 ] 19 March Gear characteristics 1 page explanation 0 April [ 10 ] 19 March CAD files & DXF files 09 April (via zip disk) [ 90 ] Σ: 00

4 BERNOULLI S EQUATION

5 Streamlines Streamline: Line which is everywhere tangent to a fluid particle s velocity. z g V A Stream Line V B For a steady flow, stream lines do not move/change A stream line is the path along which a fluid particle travels during steady flow. For one dimensional flow, we can assume that pressure (p) and velocity (v) have the same value for all stream lines passing through a given cross section v 1 Bernoulli s equation for steady flow, constant density: + ρ p + g z = Constant

6 Bernoulli derivation For two cross section (ends of control volume) located dx apart: Something in dx Something out = Something in + d(something) dx dx dv dx vout= v in + dx A out= A in + da dx dx Differntial changes with dx p out= p in + dp dx dx dz dx zout= z in + dx

7 Bernoulli derivation Stead flow momentum equation for Control Volume From F = m a following a fluid mass mc in v in + ΣF on CV = mc out v out For a stead, there is no stored mass dv v v + dx dx dx mc in = mc out = mc mc v in + ΣF on CV = mc v out ΣF on CV = mc (v out - v in ) dv v out = v in + dx dx ΣF on CV = ρ A v dv dx dx

8 Bernoulli derivation Pressure force: z p A p( da/ dx )dx [p + ( dp/ dx )dx ] [A + ( da / dx )dx ]

9 Bernoulli derivation Gravity: z m g dz m g dx = F x

10 Bernoulli derivation Summation of pressure and gravity forces: ΣF on CV = p A + p da dx dx - p A - p da dx ~ 0 dx A dp dx (terms) dx m g dz dx dx ΣF on CV = A dp dx m g dz dx dx m = ρ A dx ΣF on CV = A dp dx ρ A g dz dx dx dx

11 Bernoulli derivation Equating momentum flow and applied forces: ΣF on CV = A dp dx ρ A g dz dx = ΣF on CV = ρ A v dv dx dx dx dx A dp dx ρ A g dz dx = ρ A v dv dx dx dx dx A ( dp) ρ A g ( dz) = ρ A v ( dv ) Exact differentials for constant ρ dp g ( ρ dz ) = v (dv) v + 1 p + g z = Constant ρ

12 Bernoulli s Equation: Assumptions Flow along streamline B.E. can only be used between points on the SAME streamline. Inviscid flow: Loss due to viscous effects is negligible compared to the magnitudes of the other terms in Bernoulli s equation. Bernoulli s equation can t be used through regions where fluids mix: Mixed jets & wakes (flow want to break up, swirl resulting shear dissipates energy) Pumps & motors Other areas where the fluid is turbulent or mixing. Stream lines Mixed Jet Mixing = Can t Use Bernoulli s Equation You can not use Bernoulli s Equation through jets or turbulent areas Stead state Velocity of the flow is not a function of time, BUT!!! it can be a function of position. Incompressible

13 Bernoulli s Example Pipe of variable diameter Given v A, find the pressure difference between A & B as a function of A A, A B, and v A. Is this a rise or drop in pressure? z g A B C Assumptions (along streamline, inviscid, stead state, incompressible) Bernoulli s equation between points A and B: v A p A + g z A = v B + p B + g z B ρ ρ Note : z B = z A v A ρ p A = v B + ρ p B (p B p A ) = ρ (v A v B )

14 Bernoulli s Example Pipe of variable diameter z g A B C (p B p A ) = ρ (v A v ) B Volume flow rate equality: Q A = Q B = Q C so A A v A = A B v B = A C v C v B = v A A A A B (p B p A ) = ρ v A v A A A = A B ρ v A A 1 A A B A A > 1 so 1 A A < 1 so the pressure drops! A B A B

15 DC PERMANENT MAGNET MOTOR

16 Vector cross product review Vector cross products: A B = C Mutual perpendicularity: C is mutually to A and B B z y x Magnitude: C = A B sin(θ A-B ) θ A-B A C When: A & B are PARALLEL, the magnitude of (A X B) is 0 θ A-B = 0 o C = A B sin(0 o )= 0 A & B are PERPENDICULAR, the magnitude (A X B) is maximized θ A-B = 90 o C = A B sin(90 o )= A B

17 Magnetic force on a conductor (wire) Force on a conductor carrying a current through a magnetic field: F m = I B Where: F m = Magnetic Force [N or lbf ] I wire = B = Current Magnetic flux density m Amps or C s N s Weber C m or m MAGNETIC FLUX DENSITY, B y CONDUCTOR / WIRE CURRENT, I wire z x MAGNETIC FORCE, F m

18 Magnetic torque on a simple electric machine Force on a conductor carrying a current through a magnetic field: For wire 3, θ I-B always = 90 o so sin(θθ I-B ) always = 1 Force on wires 1 (to left) and (to right) do not act to make wire rotate F M = I B F M = I B sin(θ I B ) = I B T = r F = r (I B) T = r I B sin(θ R F ) B y y 3 θ I-B I wire z x F m θ R-F z F m 1 T m SIDE VIEW OF MACHINE

19 Magnetic torque as a function of position B F m y θ R-F I wire F m z x z R y Battery T m SIDE VIEW Maximum Torque θ R-F = 90 o F m θ R-F θ R-F Torque Decreases as sin(θ R-F ) Decreases R & Fm Parallel so Torque = 0 Torque Reverses Direction as sin(θ R-F ) is now negative. Note θ R-F is in opposite direction. θ T m Time 0 R F m R R F m R T m T m F m T m = 0 θ R-F Time 1 Time Time 3

20 What does the torque vs θ curve look like? [ mins] θ R-F 0 o y F m z R T m 0 SIDE VIEW Seconds

21 Torque on simple wire loop carrying current D +T MAX T = 0 D A B C D A C -T MAX Torque curve of simple loop B Side view of simple loop

22 Keeping the machine in motion How to keep the machine moving Once the wire passes horizontal, T m tries to stop the wire from rotating. To keep the wire rotating, we must either shut off the current or reverse the current. If we turn off the current, the wire will continue to rotate due to its inertia. If we reverse the current direction when the wire reaches horizontal, T m will act to keep the wire spinning If current continues in the same direction, Tm tries to stop wire from spinning. Changing current direction will change the direction of Fm. This in turn switches the direction of Tm. Tm will now act to keep the wire spinning. T m F m R F m R θ R-F θ R-F T m At time 3 Without Current Switch At time 3 With Current Switch

23 Torque on switched wire loop carrying current D +T MAX T = 0 D A B C D A C -T MAX Torque curve of switched loop B Side view of switched loop

24 DC Permanent magnet electric motor build & contest In your kit you will find materials to build a simple electric motor How it works: Motor current shuts off when torque becomes negative Rotor inertia carries rotor until current turn on Repeated cycle keeps the motor spinning. We will have a contest (in your lab sessions) to determine winner Prizes How do you maximize energy input? How do you minimize losses? Friction You may need to try various things Class record = 1800 RPM! Fastest motor = $0 Cheesecake Factory gift certificate Record breaker = Keep Lego kit at end of class

25 Torque on switched wire loop carrying current D +T MAX T = 0 D A B C D A C -T MAX Torque curve of switched loop B Side view of switched loop

26 DC Permanent magnet electric motor build & contest Contest rules: The contest is OPTIONAL Motor may only contain the materials in your kit and a roll of life savers You may use the contents of your tool kits to help shape/make the motor You may not use any other tools/machines to make the motor Any wire coil must be wound around the battery or the roll of life savers You may obtain up to 3 ft of additional wire from a TA if you need it Your motor may only be powered by our power source (fresh D battery) We will test them in lab next week

Where does Bernoulli's Equation come from?

Where does Bernoulli's Equation come from? Introduction By now, you have seen the following equation many times, using it to solve simple fluid problems. P ρ + v + gz = constant (along a streamline) This