A 3-phase, 50 Hz, 110 KV overhead line has conductors placed in a horizontal plane 3 m apart. Conductor diameter is 2.5 cm. If t
1 answer:
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Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
The mass flow rate of the mixture in the manifold is 6.654 kg/min
Explanation;
In this question, we are asked to calculate mass flow rate of the mixture in the manifold
Please check attachment for complete solution and step by step explanation.
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m =
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn =
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs