Q₁ = 20cfsP₁ = 15 psiLFT36in435Pa•> 24 inFNAN45⁰°= 14.78 Psi Q=20 cfsP₁ = 15 psiF₁RX36.m-245Fixy+V₂F22/P₂=14.78 psiSolve for both FMM, FMyFNX, FNY.a = avro (Far)d arctan24 indλ:•FNxRNeglect the head loss in the bend,calculate the force acted on the bend.HW.Σ BmV₂X component ax companect of(Q (u - U.₁) - 2Fasx = F₁ - F₂5145² - Frecs45°ext,(Q(v₁ - v. ) - ΣFarty - Fay - Fis-4²=Sin 45°

Answered: Image /qna-images/answer/078c3d52-80e7-42bd-80a0-32f7747f2c9f.jpg

Answered: Q₁ = 20cfs P₁ = 15 psi LFE 36in k P₂²…

Engineering

Mechanical Engineering

Q₁ = 20cfs P₁ = 15 psi LFE 36in k P₂² FNA •> 24 in 45 0 = 14.78 Psi

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Generally, there are three types of vibration: longitudinal vibration, transverse vibration, and torsional vibration.

Question

Q₁ = 20cfs
P₁ = 15 psi
LFT
36in
435
Pa
•> 24 in
FNA
N
45⁰°
= 14.78 Psi

Q=20 cfs
P₁ = 15 psi
F₁
R
X
36.m
-245
Fixy
+
V₂
F2
2/P₂=14.78 psi
Solve for both FMM, FMy
FNX, FNY.
a = avro (Far)
d arctan
24 in
dλ:
•FNx
R
Neglect the head loss in the bend,
calculate the force acted on the bend.
HW.
Σ Bm
V₂
X component a
x companect of
(Q (u - U.₁) - 2Fasx = F₁ - F₂5145² - Fre
cs45°
ext,
(Q(v₁ - v. ) - ΣFarty - Fay - Fis-4²
=
Sin 45°

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