Answer:
the nation will suffer terrible consequences
Explanation:
I did that and got it right
Answer:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Explanation:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Answer:
(a) Effectiveness of the regenerator= 0.433
(b) The rate of heat removal=21.38 kW
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
Catalog A can carry the larger load
Explanation:
Refer to the attachment for details to the answer.
Answer:
Following are the solution to the given question:
Explanation:
Line voltage:
![V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v](https://tex.z-dn.net/?f=V_L%3D%5Csqrt%7B3%7DV_%7Bph%7D%3D%5Csqrt%7B3%7D%28120%29%20%5C%20v)
Power supplied to the load:
![P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi](https://tex.z-dn.net/?f=P_%7BL%7D%3D%5Csqrt%7B3%7DV_%7BL%7DI_%7BL%7D%20%5Ccos%20%5Cphi)
![10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A](https://tex.z-dn.net/?f=10%5Ctimes%2010%5E3%3D%5Csqrt%7B3%7D%28120%20%5Csqrt%7B3%7D%29%20I_%7BL%7D%5C%20%280.85%29%5C%5C%5C%5CI_%7BL%7D%3D%2032.68%5C%20A)
Check wye-connection, for the phase current:
![I_{ph}=I_L= 32.68\ A](https://tex.z-dn.net/?f=I_%7Bph%7D%3DI_L%3D%2032.68%5C%20A)
Therefore,
Phasor currents:
Magnitude of the per-phase load impedance:
Phase angle:
![\phi = \cos^{-1} \ (0.85) =31.79^{\circ}](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Ccos%5E%7B-1%7D%20%5C%20%280.85%29%20%3D31.79%5E%7B%5Ccirc%7D)
Please find the phasor diagram in the attached file.