Calculating water pressure from bernouliis equation

[victoriafello]

A power station is supplied with water from a reservoir. A pipeline connects the reservoir to the turbine hall
The flow of water through the pipeline is controlled by a valve which is located exactly 500 metres below the surface of the water in the reservoir. The pipeline is 0.30 m in diameter.

Calculate the static pressure at the lower end of the pipeline when the valve is in the closed position

relivant equation

bernouliis equation

P + 1/2 rho v^2 + pgh = constant

my attempt so far -

for the surface of the reservior
P1 - atmospheric pressure 1.00x10^5
rho1 = density of water 1.00x10^3
g = 9.81
h1 = 0
v1 = 0 (water is not moving)

for the bottom of the pipe
p2 - unknown
rho 2 = density of water 1.00x10^3
g = 9.81
h2 = 500
v2 - 0 (water not moving)

rearrange for p2

p2 = 1/2 Rho (V1-V2)g(h1-h2)-P1

this gives

1/2*1.00x10^3+1.00x10^3*9.81*(0-500)-1.00x10^5

so p2 = 4904500

or 4.9x10^6

can someone let me know if i went wrong anywhere as this doesn't look correct? i don't think the diamter of the pipe is important for this part of the equation,

 

[billmcalpine7]

I've got this question too victoriafello...am currently working on it and will have something, at any rate, posted soon this morning..

 

[billmcalpine7]

Someone I know worked out the answer to the first question as 4.90e6 N/m² too...so I reckon you may well be right on that one

I got this of the same person for the second answer:

2.
pipe area = πD²/4 = π(0.3)²/4 = 0.0707 m²
water flow = 5.00 m³/s
V = water velocity = 5.00/0.0707 = 70.7 m/s
V²/2g = (70.7)²/2g = 255 m of H2O
gauge pressure = 500 - 255 = 245 m of H2O => (4.90e6)(245/500) = 2.40e6 N/m² ANS-2