Answer:
P=11 kW
Explanation:
Given that
Number of poles= 8
I.E.C. 180L motor frame
From data book , for 8 poles motor at 50 Hz
Speed = 730 rpm
Power factor = 0.75
Efficiency at 100 % load= 89.3 %
Efficiency at 50 % load= 89.1 %
Output power = 11 kW
Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.
P=11 kW
Answer:
In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.
Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are the alloy element of iron ans carbon.Wrought iron is the purest from of iron.
Heat treatment process for ferrous materials :
1.Normalizing
2.Annealing
3.Quenching
4.Surface hardening
Heat treatment process for non ferrous materials :
Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.
When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.
The use of bimetallic structure -In clock ,thermometers ,engines.
Answer:
time = 5.22 hr
Explanation:
Given data:
Energy of battery = 9400 J
Power consumed by three led bulb is 0.5 watt
we know Power is give as
plugging all value and solve for time
time = 18,800 sec
in hour
1 hour = 3600 sec
therefore in 18,800 sec
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Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.
