Help calculate pipe outlet pressure

[BIGEYE]

Where am I going wrong? Appreciate if someone can take a look at this and show me where I am going wrong.
A conical pipe with inlet diameter 80mm and inlet pressure 0.5 bar, is inclined so that the outlet is 2m higher than the inlet. The outlet diameter is 140mm, the fluid density is 1000 kg/m3, and the mass flow rate is 40 kg/s.
Calculate:
a) The inlet and outlet velocities
b) The outlet pressure
My workings are giving the outlet pressure higher than the inlet pressure,which must be wrong due to the head and increased diameter. Here is how I arrived at my answers:

First calculate the inlet velocity V1
Calculate CSA of inlet (A1)
A1 = (PI × 0.08^2)/4
= 5.02 × 10^-3 m2

Calculate CSA of outlet (A2)
A2 = (PI × 0.15^2)/4
= 0.0154 m2

Inlet Velocity V1
V1 = Q/(A1 × ρ)
V1 = 40/(0.00502 × 1000)
= 7.968 m/s

Outlet Velocity V2
V2 = Q/(A2 × ρ)
V2 = 40/(0.015 × 1000)
= 2.67 m/s

Outlet Pressure Calc
Inlet Pressure P1 = 0.5 bar = 50000 pa
Inlet head h1 = 0 m
Outlet head h2 = 2m
Gravity = 9.81
gh1+ 1/2 V1^2+ (P1/ρ) = gh2+ 1/2 V2^2+ (P2/ρ)
= (9.81 × 0) + 1/2 x 7.968^2 + (50000/1000) = (9.81 × 2) + 1/2 x 2.67^2 + (P2/1000)
81.74 = 23.18 + (P2/1000)
P2 = 1000(81.74-23.18)
P2 = 58560 pa
P2 = 0.58 bar

Attached is a diagram of the pipe.

 

[patj]

Ah, I see.

You have the wrong sign on the outlet head level. It is higher and therefore has a lower head (with positive being in the downwards direction on your drawing).

What you always want to do is start from the outlet where you know the fluid's pressure (generally atmospheric) and set that as the datum level then the h part for the inlet will be a positive value if lower and negative if higher.

PS i use the terms head and pressure to mean the same thing with water flow. The other thing to do to make calculations easier if dealing with water is to ignore the density (and gravity in the static part of the bernoulli equation) unless you really need to and just work in meters of head where 10m of water is approximately equal to 1 bar

 

[nlightner]

Hello,

Thank you for sharing your calculations for the pipe outlet pressure. I have reviewed your work and it seems like you have made some small errors in your calculations.

For the inlet velocity, you have correctly calculated the cross-sectional area of the inlet and used the correct density. However, your calculation for the velocity should be V1 = 40/(0.00502 x 1000) = 7.968 m/s, which you have correctly used in your subsequent calculations.

For the outlet velocity, you have made a small error in your calculation for the cross-sectional area of the outlet. It should be A2 = (PI x 0.07^2)/4 = 0.00385 m2. Using this value, the correct outlet velocity should be V2 = 40/(0.00385 x 1000) = 10.39 m/s.

For the outlet pressure calculation, you have correctly used the equation for energy conservation, but you have made a small error in your units. The pressure should be in Pa, not bar. Therefore, the correct equation should be: gh1 + 1/2 x V1^2 + (P1/ρ) = gh2 + 1/2 x V2^2 + (P2/ρ) = (9.81 x 0) + 1/2 x 7.968^2 + (50000/1000) = (9.81 x 2) + 1/2 x 10.39^2 + (P2/1000) = 81.74 = 20.43 + (P2/1000). Solving for P2, we get P2 = 1000(81.74 - 20.43) = 61430 Pa = 0.61 bar.

I hope this clears up any confusion and helps you arrive at the correct outlet pressure. Keep up the good work!