A liquid of density 1354 kg/m^3

Question

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)?

Answer 1

(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
h₂-h₁ =-7.82 m
Bernoulli Equation
p₁+ρ‧ν₁²/2 +ρ‧gvh₁ = p₂+ρ‧ν₂²/2 +ρ‧gvh₂
p₁ = p₂+ρ(ν₂²-ν₁²)/2 +ρ‧g(h₂-h₁)=
=141855 +1354‧(82.4² -2.45²)/2 -1354‧9.8‧7.82 =
= 4838223.9 Pa = 47.7 atm

Answer 2

p₁+ρ‧ν₁²/2 +ρ‧g‧h₁ = p₂+ρ‧ν₂²/2 +ρ‧g‧h₂
p₁ = p₂+ρ(ν₂²-ν₁²)/2 +ρ‧g(h₂-h₁)=
=141855 +1354‧(82.4² -2.45²)/2 -1354‧9.8‧7.82 =
= 4838223.9 Pa = 47.7 atm

Answer 3

it tells me that its incorrect