Q₁ = 20cfs P₁ = 15 psi LFE 36in k P₂² FNA •> 24 in 45 0 = 14.78 Psi
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
Related questions
Question
![Q₁ = 20cfs
P₁ = 15 psi
LFT
36in
435
Pa
•> 24 in
FNA
N
45⁰°
= 14.78 Psi](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ce8b870-93ef-497a-8c2a-bbbd1041dd5b%2F078c3d52-80e7-42bd-80a0-32f7747f2c9f%2F5kmfd7j_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ce8b870-93ef-497a-8c2a-bbbd1041dd5b%2F078c3d52-80e7-42bd-80a0-32f7747f2c9f%2Fe5gmdrm_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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