Answer:
Considering the guidelines of this exercise.
The pieces produced per month are 504 000
The productivity ratio is 75%
Explanation:
To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.
Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc =
=
=
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = ![\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500](https://tex.z-dn.net/?f=%5Csqrt%7B0.03%20i%20%20%2B%20j%200.35%2A%20%20j4.4%2A10%5E-6%7D%20%20%2A%20500)
= ![\sqrt{1.5456*10^{-6} < 175.1^0} * 500](https://tex.z-dn.net/?f=%5Csqrt%7B1.5456%2A10%5E%7B-6%7D%20%3C%20175.1%5E0%7D%20%2A%20500)
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = ![\frac{e^{gl} + e^{-gl} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7Be%5E%7Bgl%7D%20%2B%20e%5E%7B-gl%7D%20%20%7D%7B2%7D)
sin h (gl) = ![\frac{e^{gl}-e^{-gl} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7Be%5E%7Bgl%7D-e%5E%7B-gl%7D%20%20%7D%7B2%7D)
Answer: 150m
Explanation:
The following can be depicted from the question:
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Therefore, the plinth area of the building will be:
= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
= 10 × 15
= 150m