dareimert

Atomic Variables ...

Asked: dareimert 35 Product: Maple

May 21 2020

0 1

I'm trying to document some generator calculations that use terms line Xd'. In Maple that turns into a derivative. I converted several 'ed variables into atomic variables. They have made my work sheet unusable. They appear to randomly loose connecin to their assigned values and prevent numeric solutions by showing up in symbolic form. Once this happens, I have to check each instance where the variable isuse to find where the loss/lossses occured. (copy and past each instance of the variable and check its value).

Once found I can't find a consistant way to re-associate them with the assigned value.

Any suggestions. See botton of attaced work sheet.

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/WS_Gen_I_and_E_for_fauls_on_HS_of_GSU_-_Time_Response_&_51V_Trip_Times_-R5a.mw .

Download WS_Gen_I_and_E_for_fauls_on_HS_of_GSU_-_Time_Response_&_51V_Trip_Times_-R5a.mw

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/WS_Gen_I_and_E_for_fauls_on_HS_of_GSU_-_Time_Response_&_51V_Trip_Times_-R5a.mw .

Download WS_Gen_I_and_E_for_fauls_on_HS_of_GSU_-_Time_Response_&_51V_Trip_Times_-R5a.mw

Combining log10...

Asked: dareimert 35 Product: Maple

November 06 2018

1 1

How do I combine log10 terms. I can do it with natural logs but I have not been able do do it with log10.

Fourier Analysis of Piecewise Function ...

Asked: dareimert 35 Product: Maple

February 17 2018

2 1

How do I extract harmonics and plot a spectrum from a piecewise function?

The function is "Ia" shown below near the bottom in the "Plot Line Current" section.

(1)

Vmax = RmsMS phase to neutral voltage

=

The #1 diode is forward biased and switches on when Va>Vc (Va=Vc). The voltage VD will be equal to Va-Vb until diode #2 turns on when Vc< Vb. Then VD will equal Vc-Vc.

The switching angle for the diodes is given above. If phase control is used all commutation is delayed by the angle α.

Without phase delay the average (DC) voltage applied to the laod is the average value of the 6 commutation periods over one cycle:

Emax = RmsMS phase to neutral voltage

Eq1 := Vdo = 6*(int(Van-Vbn, theta = com1 .. com2))/(2*Pi)

Vdo = 3*3^(1/2)*2^(1/2)*Erms/Pi

(2)

Eq2 := Vdc = 6*(int(Van-Vbn, theta = com1+alpha .. com2+alpha))/(2*Pi)

Vdc = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi

(3)

Commutation

The switching between diodes can not occur instantaneously because the supply circuit is inductive. Current can not change instantaneously in and inductive circuit.
The transfer of load current from the out-going phase to the incomming phase is accomplshed by a period of symultaneous conduction of the two adjacent diodes the creation of a circulating current between these two phases that resembles short circuit between the two phases.

The circulating current produces equal voltage drop with opposid polarities in each phase. In the out-going phase the drop adds to the soruce voltage and in the incomming phase the drop subtracts from the source voltage. Only the drop in the incomming phase effects the DC output voltage.

During commutation intervil com1 to com2 phases a and c are shorted while Id is transfered from phase c to phase a. During this time voltage Va-Vb minus the voltage drop in a phase caused by the commutation. This voltage drop is 1/2 the Vac voltage.

(1/2)*2^(1/2)*Erms*sin(theta)-(1/2)*2^(1/2)*Erms*cos(theta+(1/6)*Pi)

(4)

and Va=Vc at π/6

`ΔE__Avg` = 6*(int(`Δe`, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu))/(2*Pi)

`ΔE__Avg` = 3*((1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha)+(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha)-(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu)-(1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu))/Pi

(5)

`ΔE__Avg` = (3/2)*2^(1/2)*Erms*(cos((1/6)*Pi+alpha)+sin((1/3)*Pi+alpha)-sin((1/3)*Pi+alpha+mu)-cos((1/6)*Pi+alpha+mu))/Pi

(6)

`ΔE__Avg` = (3/2)*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi-(3/2)*Erms*2^(1/2)*3^(1/2)*cos(alpha)*cos(mu)/Pi+(3/2)*Erms*2^(1/2)*3^(1/2)*sin(alpha)*sin(mu)/Pi

(7)

`ΔE__Avg` = (3/2)*Erms*2^(1/2)*((-cos(mu)+1)*cos(alpha)+sin(alpha)*sin(mu))*3^(1/2)/Pi

(8)

`ΔE__Avg` = (1/2)*(-3*Erms*2^(1/2)*3^(1/2)*cos(alpha+mu)+3*2^(1/2)*Erms*3^(1/2)*cos(alpha))/Pi

(9)

`ΔE__Avg` = -(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(10)

DC output voltage

(11)

V__D = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi+(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(12)

V__D = (3/2)*2^(1/2)*Erms*3^(1/2)*(cos(alpha)+cos(alpha+mu))/Pi

(13)

Equating the intigral of the voltage drop between the phase A and C to the intigral of the voltage rise across the two phase inductances (thes intigrals represent eqivalent flux).

int(Van-Vcn, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu) = int(2*L*omega, theta = 0 .. Id)

2^(1/2)*Erms*cos((1/6)*Pi+alpha)+2^(1/2)*Erms*sin((1/3)*Pi+alpha)-2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu)-2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu) = 2*L*omega*Id

(14)

(15)

(16)

(17)

(18)

(19)

V__D = (3/2)*2^(1/2)*Erms*3^(1/2)*(cos(alpha)-(1/6)*(2*L*omega*Id-2^(1/2)*Erms*3^(1/2)*cos(alpha))*3^(1/2)*2^(1/2)/Erms)/Pi

(20)

V__D = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi-3*L*omega*Id/Pi

(21)

Line Currents

	Commutation Intervals

Plot Line Currents

	Line Rms Current

Download Graetz_Bridg7b.mw

subs fails to for one variable...

Asked: dareimert 35 Product: Maple

February 12 2018

1 0

Can anyone explain why 'subs' will sbstitute 'a' and 'c' but not 'b' (equations 2.3 and 2.4) in the attached document.

I have tried everything I can think of but nothing works.

If I cut and paste 2.3 and 2.4 to a clean document it works. I tried coying the whole document t a clean sheet but that does not fix the problem.

Vmax = RmsMS phase to neutral voltage

=

The #1 diode is forward biased and switches on when Va>Vc (Va=Vc). The voltage VD will be equal to Va-Vb until diode #2 turns on when Vc< Vb. Then VD will equal Vc-Vc.

The switching angle for the diodes is given above. If phase control is used all commutation is delayed by the angle α.

Without phase delay the average (DC) voltage applied to the laod is the average value of the 6 commutation periods over one cycle:

Emax = RmsMS phase to neutral voltage

Eq1 := Vdo = 6*(int(Van-Vbn, theta = com1 .. com2))/(2*Pi)

Vdo = 3*3^(1/2)*2^(1/2)*Erms/Pi

(1)

Eq2 := Vdc = 6*(int(Van-Vbn, theta = com1+alpha .. com2+alpha))/(2*Pi)

Vdc = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi

(2)

Commutation

The switching between diodes can not occur instantaneously because the supply circuit is inductive. Current can not change instantaneously in and inductive circuit.
The transfer of load current from the out-going phase to the incomming phase is accomplshed by a period of symultaneous conduction of the two adjacent diodes the creation of a circulating current between these two phases that resembles short circuit between the two phases.

The circulating current produces equal voltage drop with opposid polarities in each phase. In the out-going phase the drop adds to the soruce voltage and in the incomming phase the drop subtracts from the source voltage. Only the drop in the incomming phase effects the DC output voltage.

During commutation intervil com1 to com2 phases a and c are shorted while Id is transfered from phase c to phase a. During this time voltage Va-Vb minus the voltage drop in a phase caused by the commutation. This voltage drop is 1/2 the Vac voltage.

(1/2)*2^(1/2)*Erms*sin(theta)-(1/2)*2^(1/2)*Erms*cos(theta+(1/6)*Pi)

(3)

and Va=Vc at π/6

`ΔE__Avg` = 6*(int(`Δe`, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu))/(2*Pi)

`ΔE__Avg` = 3*((1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha)+(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha)-(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu)-(1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu))/Pi

(4)

`ΔE__Avg` = (3/2)*Erms*2^(1/2)*(cos((1/6)*Pi+alpha)+sin((1/3)*Pi+alpha)-sin((1/3)*Pi+alpha+mu)-cos((1/6)*Pi+alpha+mu))/Pi

(5)

`ΔE__Avg` = (3/2)*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi-(3/2)*Erms*2^(1/2)*3^(1/2)*cos(alpha)*cos(mu)/Pi+(3/2)*Erms*2^(1/2)*3^(1/2)*sin(alpha)*sin(mu)/Pi

(6)

`ΔE__Avg` = (3/2)*Erms*2^(1/2)*3^(1/2)*((-cos(mu)+1)*cos(alpha)+sin(alpha)*sin(mu))/Pi

(7)

`ΔE__Avg` = (1/2)*(-3*3^(1/2)*2^(1/2)*Erms*cos(alpha+mu)+3*2^(1/2)*Erms*3^(1/2)*cos(alpha))/Pi

(8)

`ΔE__Avg` = -(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(9)

DC output voltage

(10)

V__D = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi+(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(11)

V__D = (3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha)+cos(alpha+mu))/Pi

(12)

Equating the intigral of the voltage drop between the phase A and C to the intigral of the voltage rise across the two phase inductances (thes intigrals represent eqivalent flux).

int(Van-Vcn, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu) = int(2*L*omega, theta = 0 .. Id)

2^(1/2)*Erms*cos((1/6)*Pi+alpha)+2^(1/2)*Erms*sin((1/3)*Pi+alpha)-2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu)-2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu) = 2*L*omega*Id

(13)

(14)

(15)

(16)

(17)

(18)

V__D = (3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha)-(1/6)*(2*L*omega*Id-2^(1/2)*Erms*3^(1/2)*cos(alpha))*3^(1/2)*2^(1/2)/Erms)/Pi

(19)

V__D = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi-3*L*omega*Id/Pi

(20)

Line Currents

	A phase current intervals

RMS Current Line

(2.1)

I__RMS = sqrt(A/Pi)

I__RMS = (A/Pi)^(1/2)

(2.2)

A = -(3/16)*Erms^2*((-8*cos(mu)*sin(mu)-8*mu+16*sin(mu))*cos(alpha)^2-8*sin(alpha)*(cos(mu)-1)^2*cos(alpha)+4*cos(mu)*sin(mu)-4*mu)/(L^2*omega^2)+Idc^2*((2/3)*Pi-mu)+(1/4)*(6*(cos(mu)*sin(mu)-2*sin(mu)+mu)*cos(alpha)^2+6*sin(alpha)*(cos(mu)-1)^2*cos(alpha)-3*cos(mu)*sin(mu)+3*mu)*Erms^2/(L^2*omega^2)+Idc*((sin(mu)-mu)*cos(alpha)+sin(alpha)*(cos(mu)-1))*3^(1/2)*2^(1/2)*Erms/(L*omega)+Idc^2*mu

(2.3)

A = -(3/16)*Erms^2*a/(L^2*omega^2)+Idc^2*((2/3)*Pi-mu)+(1/4)*(6*(cos(mu)*sin(mu)-2*sin(mu)+mu)*cos(alpha)^2+6*sin(alpha)*(cos(mu)-1)^2*cos(alpha)-3*cos(mu)*sin(mu)+3*mu)*Erms^2/(L^2*omega^2)+Idc*c*3^(1/2)*2^(1/2)*Erms/(L*omega)+Idc^2*mu

(2.4)

Download subs_failed.mw

How do I extracting values from RootOf ...

Asked: dareimert 35 Product: Maple 2017

January 01 2018

1 1

I am trying to assess stability from real roots of Eigenvalues that are returned in the RootOf from. I tried allvalues(), evalf() both return the same RootOf list. If memory serves me correctly any square matrix should have and Eigenvalue solution.

Any suggestions?

Eignevalues.mw