Draw a radial power circuit arrangement containing 3 single fused sockets and 2 double unfused sockets showing the live, earth a
1 answer:
Answer:
Attached below is the Radial power circuit arrangement
Explanation:
Radial power circuit arrangement is done in a way that a single cable starts from the fuse box and connects to all the outlet socket contained in the circuit also the cable contains wires ( live , neutral and earth )
The advantage of a radial power circuit arrangement is that it enables easy identification of electrical faults on the circuit.
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Answer:
472,826 lb/hr
Explanation:
As per the given data:
Solids in feed= 2%
Solids in concentrate= 25%
HTA1 = HTA2 = 1000 ft^2
U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F
Overall material balance: Feed= Distillate + concentrate ----> Eq-1
Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25
Feed = 12.5 * concentrate ---> Eq-2
Boiling point rise = negligible, so solution & solvent vapor temperature will be same.
Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).
As per the data:
Latent heat 212F = 300 Btu/lb
Latent heat 100F = 320 Btu/lb
As per material balance:
Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT
1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr
2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr
Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr
Substituting the above in Eq-1
Feed = 435,000 + concentrate
Substitute Eq-2 in the above
12.5 * concentrate = 435,000 + concentrate
Concentrate, L1 = 435,000/11.5 = 37826 lb /hr
Feed, F = 435,000 + 37826 = 472,826 lb/hr
1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.
[ Find the figure in the attachment ]
Answer:
Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib
Answer : moment of inertia = 186.7 Ib - in
Explanation:
Given data
weight of the mailbox = 3.2 Ib
weight of the uniform cross member = 10.3 Ib
The origin is of mailbox and cross member is 0
The perpendicular distance from Y axis of centroid of the mailbox
= 4 + (25/2) = 16.5"
The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"
therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)
= 52.8 + 133.9 = 186.7 Ib-in
