ل Q = 20cfs P - 15 : F, 36in محياه E N { P₂² - 24 in 5° با = 14.7 8 Ps

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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Knowing that the pressure is gauge pressure . Also, I want you to solve it , by using the equation I provided .
Q=20cfs
P-15psi
LFE
36in
~270
ㅅ
P2=14.78 Psi
.. 24 in
FNA
450
Transcribed Image Text:Q=20cfs P-15psi LFE 36in ~270 ㅅ P2=14.78 Psi .. 24 in FNA 450
Q=20 cfs
P₁ = 15 psi
F₁
R
X
36.m
->45
Fivy
+
Solve for both FMX, FMy
FN.Y.
FNY
a= auch (Fax)
V₂
2₂=14.78 psi
F2
2₁/P₂=
24 in
•FNIx
R
Neglect the head loss in the bend,
calculate the force acted on the bend.
HW.
E Bm
V₂
X component a
x componect of
(Q(u - U₁) - 2For x = F₁-F₂ con48" - Fi
и,
cs45²
ext,
(Q(v₁ - v₁ ) = EFerty = Fay-F2544²-
Sin
Transcribed Image Text:Q=20 cfs P₁ = 15 psi F₁ R X 36.m ->45 Fivy + Solve for both FMX, FMy FN.Y. FNY a= auch (Fax) V₂ 2₂=14.78 psi F2 2₁/P₂= 24 in •FNIx R Neglect the head loss in the bend, calculate the force acted on the bend. HW. E Bm V₂ X component a x componect of (Q(u - U₁) - 2For x = F₁-F₂ con48" - Fi и, cs45² ext, (Q(v₁ - v₁ ) = EFerty = Fay-F2544²- Sin
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