P2=13.78psiHow to get this answer by using the equation I provide, which is Bernoulli equation SXQ. -20cfsP-15ps:F, V.=0.86222P?236inFind용용 · ·2V+29+Z2same ebevation+ Ziㅅ김V2=1.329 msQuin동명대IN45˚Psi

The value are The diameter at inlet is The diameter at outlet is The elevation is same

Answered: Find Ра ४ X + $2 Q₁ = 20cfs P₁ = 15 psi…

Find Ра ४ X + $2 Q₁ = 20cfs P₁ = 15 psi FV-0.862m 29 P₂ ? 2 4.4.4.. P₁ + = ४ 36in Same 29 (ND. elevation + Zi โม 24 in FHA FN

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.

Question

P2=13.78psi How to get this answer by using the equation I provide, which is Bernoulli equation

S
X
Q. -20cfs
P-15ps:
F, V.=0.8622
2
P?
2
36in
Find
용용 · ·
2
V
+
29
+Z2
same ebevation
+ Zi
ㅅ
김
V2=1.329 ms
Quin
동명대
IN
45˚
Psi

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