Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf
Nec Article 430 covers selection of time-delay fuses for motor- overload protection.
<h3>What article in the NEC covers motor overloads?</h3>
Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .
Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.
Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.
Learn more about motor- overload from
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Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress 
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress 
Making thickness t the subject; we have
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
