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:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 - 
............1
efficiency ( reversible ) η1 = 1 - 
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency = 
second-law efficiency = 
second-law efficiency = 62.42 %
Answer:
the third statement is true
Explanation:
given data
Lenovos cost more than Dells
Lenovos cost less than Apples
solution
we have given 1st statement that is express as
cost (Lenovo) > cost (Dell) ..................1
and
2nd statement that is express as
cost (Lenovo) < cost (Apple)
so we can say it as
cost (Apple) > cost (Lenovo) ......................2
and
now above Both equation 1 and 2 can be written as
cost (Apple) > cost (Lenovo) > cost (Dell) .........................3
so we can say cost of Apples is more than the cost of Lenovos and the cost of Dells
so as that given 3rd statement is true
