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)?

(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

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

it tells me that its incorrect

To find the pressure at the entrance end of the pipe, we can use Bernoulli's principle. Bernoulli's principle states that the total mechanical energy of a fluid at any point in a pipe is constant.

Let's use the equation form of Bernoulli's principle:

P1 + 0.5 * ρ * v1^2 + ρ * g * h1 = P2 + 0.5 * ρ * v2^2 + ρ * g * h2

Where:
P1 and P2 are the pressures at the entrance and exit of the pipe, respectively.
ρ is the density of the liquid.
v1 and v2 are the velocities at the entrance and exit of the pipe, respectively.
g is the acceleration due to gravity.
h1 and h2 are the heights of the pipe at the entrance and exit, respectively.

We know the following values:
ρ = 1354 kg/m^3 (density)
v1 = 2.45 m/s (entrance velocity)
v2 = 82.418 m/s (exit velocity, from part A)
g = 9.8 m/s^2 (acceleration due to gravity)
h1 - h2 = 7.82 m (height difference between entrance and exit)
P2 = 1.4 atm = 1.4 * 1.013 × 10^5 Pa = 1.4194 × 10^5 Pa (exit pressure)

Now, we need to rearrange the equation to solve for P1:

P1 = P2 + 0.5 * ρ * v2^2 + ρ * g * (h2 - h1) - 0.5 * ρ * v1^2

Substituting the known values:

P1 = 1.4194 × 10^5 Pa + 0.5 * 1354 kg/m^3 * (82.418 m/s)^2 + 1354 kg/m^3 * 9.8 m/s^2 * 7.82 m - 0.5 * 1354 kg/m^3 * (2.45 m/s)^2

Calculating this expression will give you the pressure at the entrance end of the pipe (P1) in units of Pascal (Pa).