A liquid of density 1354 kg/m^3
ﬂows with speed 2.45 m/s into a pipe of diameter 0.29 m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 7.82 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.4 atm.
A) What is the velocity v2 of the liquid ﬂowing out of the exit end of the pipe? Assume the viscosity of the ﬂuid is negligible and the ﬂuid is incompressible. The acceleration of gravity is 9.8 m/s^2
and Patm = 1.013 × 10^5 Pa. Answer in units of m/s.
B) Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe? Answer in units of Pa
I found A) to be 82.41800 m/s and this is correct. Can someone help me with B)?
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3 answers

(a) continuity equation:
A₁v₁=A₂v₂
π‧d₁²/4‧v₁=π‧d₂²/4‧v₂
v₂=v₁(d₁/d₂)² =82.4 m/s.
(b)
p₁=1.4 atm = 141855 Pa
h₂h₁ =7.82 m
Bernoulli Equation
p₁+ρ‧ν₁²/2 +ρ‧gvh₁ = p₂+ρ‧ν₂²/2 +ρ‧gvh₂
p₁ = p₂+ρ(ν₂²ν₁²)/2 +ρ‧g(h₂h₁)=
=141855 +1354‧(82.4² 2.45²)/2 1354‧9.8‧7.82 =
= 4838223.9 Pa = 47.7 atm 👍
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Elena 
p₁+ρ‧ν₁²/2 +ρ‧g‧h₁ = p₂+ρ‧ν₂²/2 +ρ‧g‧h₂
p₁ = p₂+ρ(ν₂²ν₁²)/2 +ρ‧g(h₂h₁)=
=141855 +1354‧(82.4² 2.45²)/2 1354‧9.8‧7.82 =
= 4838223.9 Pa = 47.7 atm 👍
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Elena 
it tells me that its incorrect
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maura
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