Problem 10.3 P=2 psig P=0 psig R₂ = 21 in weld D=13.25 in (inner diameter) A carbon steel, 14-in O.D (13.25-in I.D), schedule 30, 90° elbow is to be welded to a pipe carrying water at = 4000 gal/min. Find the force required in the weld to support the elbow. The elbow weighs 150 lbf. The density of water is 1.94 slug/ft³. (Hint: The mass of water is approximately equal to mw p(D²/4)L where D is the inside diameter of the pipe and L is the arc length associated with Re.) Ans: 150 lbf ≤ |Rx ≤ 165 lbf 120 lbf ≤ |R₂| ≤ 125 lbf

Elements Of Electromagnetics
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Problem 10.3
P=2 psig
P=0 psig
R₂ = 21 in
weld
D=13.25 in (inner diameter)
A carbon steel, 14-in O.D (13.25-in I.D),
schedule 30, 90° elbow is to be welded
to a pipe carrying water at = 4000
gal/min.
Find the force required in the weld to
support the elbow. The elbow weighs
150 lbf. The density of water is 1.94
slug/ft³.
(Hint: The mass of water is approximately
equal to mw p(D²/4)L where D is the
inside diameter of the pipe and L is the arc
length associated with Re.)
Ans: 150 lbf ≤ |Rx ≤ 165 lbf
120 lbf ≤ |R₂| ≤ 125 lbf
Transcribed Image Text:Problem 10.3 P=2 psig P=0 psig R₂ = 21 in weld D=13.25 in (inner diameter) A carbon steel, 14-in O.D (13.25-in I.D), schedule 30, 90° elbow is to be welded to a pipe carrying water at = 4000 gal/min. Find the force required in the weld to support the elbow. The elbow weighs 150 lbf. The density of water is 1.94 slug/ft³. (Hint: The mass of water is approximately equal to mw p(D²/4)L where D is the inside diameter of the pipe and L is the arc length associated with Re.) Ans: 150 lbf ≤ |Rx ≤ 165 lbf 120 lbf ≤ |R₂| ≤ 125 lbf
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