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
A compass is a analog sensor
Answer:
2074.2 KW
Explanation:
<u>Determine power developed at steady state </u>
First step : Determine mass flow rate ( m )
m / Mmax = ( AV )₁ P₁ / RT₁ -------------------- ( 1 )
<em> where : ( AV )₁ = 8.2 kg/s, P₁ = 0.35 * 10^6 N/m^2, R = 8.314 N.M / kmol , </em>
<em> T₁ = 720 K . </em>
insert values into equation 1
m = 0.1871 kmol/s ( mix )
Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )
W( power developed at steady state )
W = m [ Yco2 ( h1 - h2 )co2
Attached below is the remaining part of the detailed solution
Answer:
Explanation:
find attached the solution to the question
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.