Bundle: Understanding Motor Controls, 4th + Mindtap Electrical, 2 Terms (12 Months) Printed Access Card
4th Edition
ISBN: 9780357097755
Author: Stephen L. Herman
Publisher: Cengage Learning
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Chapter 51, Problem 2RQ
Draw the two schematic symbols used to represent the diac.
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- Section View - practice Homework 0.5000 3.0000 2,0000 1.0000arrow_forwardDrawing the section view for the following multiview drawing AutoCAD you see the section pratice I need to show how to autocadarrow_forwardA boiler with 80% efficiency produces steam at 40bar and 500 C at a rate of 1.128kg/s. The temperature of the feed water is raised from 25 C to 125 C in the economizer and the ambient air is drawn to the boiler at a rate of 2.70 kg/s at 16 C. The flue gases leave the chimney at rate of 3 kg/s at 150 C with specific heat of 1.01 kJ/kg.K. The dryness fraction of steam collected in the steam drum is 0.95. 1- Determine the heat value of the fuel. 2- The equivalence evaporation. 3- Draw the heat balance sheet.arrow_forward
- A rotating shaft is made of 42 mm by 4 mm thick cold-drawn round steel tubing and has a 6 mm diameter hole drilled transversely through it. The shaft is subjected to a pulsating torque fluctuating from 20 to 160 Nm and a completely reversed bending moment of 200 Nm. The steel tubing has a minimum strength of Sut = 410 MPa (60 ksi). The static stress-concentration factor for the hole is 2.4 for bending and 1.9 for torsion. The maximum operating temperature is 400˚C and a reliability of 99.9% is to be assumed. Find the factor of safety for infinite life using the modified Goodman failure criterion.arrow_forwardI need help with a MATLAB code. This code just keeps running and does not give me any plots. I even reduced the tolerance from 1e-9 to 1e-6. Can you help me fix this? Please make sure your solution runs. % Initial Conditions rev = 0:0.001:2; g1 = deg2rad(1); g2 = deg2rad(3); g3 = deg2rad(6); g4 = deg2rad(30); g0 = deg2rad(0); Z0 = 0; w0 = [0; Z0*cos(g0); -Z0*sin(g0)]; Z1 = 5; w1 = [0; Z1*cos(g1); -Z1*sin(g1)]; Z2 = 11; w2 = [0; Z2*cos(g2); -Z2*sin(g2)]; [v3, psi3, eta3] = Nut_angle(Z2, g2, w2); plot(v3, psi3) function dwedt = K_DDE(~, w_en) % Extracting the initial condtions to a variable % Extracting the initial condtions to a variable w = w_en(1:3); e = w_en(4:7); Z = w_en(8); I = 0.060214; J = 0.015707; x = (J/I) - 1; y = Z - 1; s = Z; % Kinematic Differential Equations dedt = zeros(4,1); dedt(1) = pi*(e(3)*(s-w(2)-1) + e(2)*w(3) + e(4)*w(1)); dedt(2) = pi*(e(4)*(w(2)-1-s) + e(3)*w(1) - e(1)*w(3)); dedt(3) = pi*(-e(1)*(s-w(2)-1) - e(2)*w(1) + e(4)*w(3));…arrow_forwardalpha 1 is not zero alpha 1 can equal alpha 2 use velocity triangle to solve for alpha 1 USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',...…arrow_forward
- 3. Find a basis of eigenvectors and diagonalize. 4 0 -19 7 a. b. 1-42 16 12-20 [21-61arrow_forward2. Find the eigenvalues. Find the corresponding eigenvectors. 6 2 -21 [0 -3 1 3 31 a. 2 5 0 b. 3 0 -6 C. 1 1 0 -2 0 7 L6 6 0 1 1 2. (Hint: λ = = 3)arrow_forwardUSE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),...…arrow_forward
- USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ...…arrow_forwardThe answer should equal to 1157. Please sent me the solution. Thank you!arrow_forwardBONUS: If the volume of the 8cm x 6.5cm x 6cm Block of Aluminum was 312cm3 before machining, find how much material was removed when the fixture below was machined. +2 2.00 cm 6.00 cm 2.50 cm 6.50 cm 1.00 cm 2.50 cm 11.00 cm 8.00 cm 30 CP 9411 FL.4) (m² 1157 Area of triangle = 1/2*B*H Area of circle = лR² Circumference of a circle = 2πR 6.00 cm 6.50 cm 1.50 cm Radius 1.50 cm 1.00 cmarrow_forward
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