Answer / Explanation:
To proper understand the answers that is given to the question, we need to understand some basic terms that has been used in the question.
Energy: This can be refereed to as the quantitative property that is transferred to an object for the purpose of the object working or to heat up the object. It can also be referred to as conserved quantity that is energy can be converted from one form or state to another but cannot destroyed.
Power: This can be defined as the rate of doing work or transferring heat per unit time from one state to another. The SI Units of power is watt which is equal to one joule per second.
Hence, the formula that links energy and power is:
Energy = Power x Time
Now. referring back to the question (a) asking how much energy do we save if we execute at the current speed and turn off the system when the computation is complete: The answer is = 50%. That is 50% of the energy is saved.
(b) If we recall the formula for calculating energy,
we have:
Energy = 1 /2 Load x V²
Changing the frequency does not affect the energy.However, it affects the power.
So therefore, the new energy is 1 / 2 Load x ( 1/2 V)² ,
reducing it to about 1 /4 of the old energy.
28384 *x soít cos estematema
The specific heats of gases are given as Cp and Cv at constant pressure and constant volume respectively while solids and liquids are having only single value for specific heat.
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
Answer:
Angle of discharge make at the edge of tube=64.9 degrees.
