How Does Bernoulli's Equation Apply in a Bourbon Distillery's Plumbing System?

[Tycho]

In a bourbon distillery plant, the refined product, approximated as 40% by weight ethanol, 60% by weight water, is pumped through a plumbing system. At the OUTLET, where the inner diameter of the pipe is 5.08cm, the product fills an aluminum open cylinder (Diameter .442m, height 1.26m) in 98.9 seconds. At the inlet of this pipe the diameter is 7.62 cm and the pipe is 3.85m above the cylinder. the density of the product is 922kg/m^3

a) what is the outlet speed of the product?

b) what is the gauge pressure at the inlet of the pipe (diameter 7.62cm)

okay, i got the outlet speed. and i know that i need to use bernoulli's equation to solve for the gauge pressure. I'm not liking bernoulli or his equation right now. I'm having trouble figuring out all of the variables that go in it. a little help, pu-leeze? thanks, my friends!

 

[Nate810]

a) The outlet speed of the product can be calculated using the continuity equation:V = (A1 x T1)/(A2 x T2)Where V is the velocity, A1 and A2 are the cross-sectional areas of the pipe at the inlet and outlet, and T1 and T2 are the times taken for the product to travel from the inlet to the outlet.For this problem, A1 = (π × (7.62/2)^2), A2 = (π × (5.08/2)^2), T1 = 98.9 seconds and T2 = 0 (since the outlet is a container). Therefore, V = (π × (7.62/2)^2 x 98.9)/(π × (5.08/2)^2 x 0) = 5.9 m/s.b) To calculate the gauge pressure at the inlet of the pipe, we must use Bernoulli's equation: P + (ρ × g × h) + (1/2) × ρ × v^2 = constantWhere P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid above the inlet, and v is the velocity at the inlet.Therefore, P = constant - (ρ × g × h) - (1/2) × ρ × v^2For this problem, ρ = 922 kg/m^3, g = 9.81 m/s^2, h = 3.85 m, and v = 5.9 m/s.Therefore, P = constant - (922 kg/m^3 x 9.81 m/s^2 x 3.85 m) - (1/2) × 922 kg/m^3 × 5.9 m/s^2 = 7.4 x 10^4 Pa (or 74 kPa).

 

[wais]

Sure, I'd be happy to help you understand Bernoulli's equation and how to apply it in this scenario.

First, let's review Bernoulli's equation: P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Where:
P1 and P2 are the pressures at points 1 and 2
ρ is the density of the fluid
v1 and v2 are the velocities at points 1 and 2
g is the acceleration due to gravity
h1 and h2 are the heights at points 1 and 2

In this scenario, we are interested in finding the gauge pressure at the inlet of the pipe (point 1) and the outlet speed (v2) of the product.

a) To find the outlet speed, we can use the fact that the product fills the cylinder in 98.9 seconds, which means that the volume of the product entering the cylinder is equal to the volume of the cylinder. So, we can use the formula for the volume of a cylinder to find the outlet speed:

V = πr^2h = (π*0.442^2*1.26) = 0.739 m^3

Since this volume is filled in 98.9 seconds, the flow rate (Q) can be calculated as:

Q = V/t = 0.739/98.9 = 0.00747 m^3/s

Now, we can use the continuity equation, which states that the flow rate at any point in a system is constant, to find the outlet speed:

Q = A1v1 = A2v2

Where:
A1 and A2 are the cross-sectional areas at points 1 and 2
v1 and v2 are the velocities at points 1 and 2

We know the area at the outlet (A2) is equal to the area of the pipe (π*0.0508^2), so we can rearrange the equation to solve for v2:

v2 = (A1v1)/A2 = (π*0.0762^2*0.00747)/π*0.0508^2 = 0.0499 m/s

b) To find the gauge pressure at the inlet of the pipe, we can use Bernou