R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
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Explanation:
Answer:
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Explanation:
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density

Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Answer:
W = 2695.14 lb
Explanation:
given,
side of cube = 3 ft
density of the cube = 3.1 slugs/ft³
we know,

mass = density x volume
volume = 3³ = 27 ft³
mass = 3.1 x 27
m = 83.7 slugs.
weight calculation
converting mass from slug to pound
weight of 1 slug is equal to 32.2 lb
now,
weight of the cube is equal to
W = 83.7 slugs x 32.2 lb/slug
W = 2695.14 lb
hence, weight is equal to W = 2695.14 lb