- #1
atlamillia
- 4
- 0
After typing all of that out, I think I got it, thanks!
I am working on a project to build a biodiesel reactor. The heating system of said reactor is provided by a loop of pipe containing a pump (1 HP, 36 litres per minute (9.5 gal), 50m max head (164 ft)), a hot water heater, and approximately 141 in of pipe including 5 90 degree bends.
The loop of pipe begins at the pump, which is at y = 0, and travels through the hot water heater (also y = 0), then goes up about 50in to the inside of the oil heating tank (y = 50), and down 50in to the pump again.
I am trying to solve for the best diameter for the pipe in this situation.
Form of Bernoulli equation:
(p1/rho + alpha1*v1^2/2 + gz1) - (p2/rho + alpha2*v2^2/2 + gz2) = head loss total
head loss is sum of major and minor losses
major = f*L/D*v^2/2
minor = f*L(equivalent)/D*v^2/2
Where
p1, p2 are pressures in psi
rho = density of water
v1, v2, v = velocity (found from V = Q/A, where Q is pump flow rate and A is area of pipe cross section) in in/s
g = 386 in/s^2
z1, z2 = change in height
alpha1, alpha2 = kinetic energy coefficient
D = diameter (in)
f = friction factor
The diameter must be solved for using iteration.
Choosing a diameter of 1 inch (which is what is expected to work based on other systems I've observed) I calculate the various information I will need to put in the Bernoulli and head loss equations. The objective is to compare pressure difference given by the Bernoulli equation to the maximum pressure the pump can provide (therefore the equation is being solved for the pressure difference across the pump, and the entire loop of pipe).
For the Bernoulli equation, the velocity terms cancel out because the velocity does not change throughout. The height terms also cancel out because the fluid goes up the same amount as it goes down during the course of the loop (this is the part I'm not sure about!). This leaves the pressure difference over density on the Bernoulli side.
For the head loss side, I need f, V, L, and L(equivalent).
I attempt to use a Moody chart to find f. The chart suggests I'd be ok using the laminar approximation, so my f is Re/64, or about 0.0022.
V is Q/A, so Q = 9.5gal/min*1min/60s*231in^3/gal and A = pi/4*(1)^2
V = 46.6in/s
L is given to be about 141 inches.
The L(equivalent) is the number of bends times the equivalent length of a 90 degree turn, or 5*30 = 150 inches.
Therefore
delta-p/rho = f/D * V^2/2(L + L(equivalent))
delta-p/.036 = 0.0022/1 * 46.6^2/2(291)
delta-p = 25 psi
The maximum pressure provided by the pump is given by the equation
delta-p = rho*delta-head = 1.8*164 = 5.9 psi
I had to increase the size of the pipe vastly through iteration in order to get the pressures to match. Where could I be going wrong? I know that a 1 in pipe is working in a similar device already.
Homework Statement
I am working on a project to build a biodiesel reactor. The heating system of said reactor is provided by a loop of pipe containing a pump (1 HP, 36 litres per minute (9.5 gal), 50m max head (164 ft)), a hot water heater, and approximately 141 in of pipe including 5 90 degree bends.
The loop of pipe begins at the pump, which is at y = 0, and travels through the hot water heater (also y = 0), then goes up about 50in to the inside of the oil heating tank (y = 50), and down 50in to the pump again.
I am trying to solve for the best diameter for the pipe in this situation.
Homework Equations
Form of Bernoulli equation:
(p1/rho + alpha1*v1^2/2 + gz1) - (p2/rho + alpha2*v2^2/2 + gz2) = head loss total
head loss is sum of major and minor losses
major = f*L/D*v^2/2
minor = f*L(equivalent)/D*v^2/2
Where
p1, p2 are pressures in psi
rho = density of water
v1, v2, v = velocity (found from V = Q/A, where Q is pump flow rate and A is area of pipe cross section) in in/s
g = 386 in/s^2
z1, z2 = change in height
alpha1, alpha2 = kinetic energy coefficient
D = diameter (in)
f = friction factor
The Attempt at a Solution
The diameter must be solved for using iteration.
Choosing a diameter of 1 inch (which is what is expected to work based on other systems I've observed) I calculate the various information I will need to put in the Bernoulli and head loss equations. The objective is to compare pressure difference given by the Bernoulli equation to the maximum pressure the pump can provide (therefore the equation is being solved for the pressure difference across the pump, and the entire loop of pipe).
For the Bernoulli equation, the velocity terms cancel out because the velocity does not change throughout. The height terms also cancel out because the fluid goes up the same amount as it goes down during the course of the loop (this is the part I'm not sure about!). This leaves the pressure difference over density on the Bernoulli side.
For the head loss side, I need f, V, L, and L(equivalent).
I attempt to use a Moody chart to find f. The chart suggests I'd be ok using the laminar approximation, so my f is Re/64, or about 0.0022.
V is Q/A, so Q = 9.5gal/min*1min/60s*231in^3/gal and A = pi/4*(1)^2
V = 46.6in/s
L is given to be about 141 inches.
The L(equivalent) is the number of bends times the equivalent length of a 90 degree turn, or 5*30 = 150 inches.
Therefore
delta-p/rho = f/D * V^2/2(L + L(equivalent))
delta-p/.036 = 0.0022/1 * 46.6^2/2(291)
delta-p = 25 psi
The maximum pressure provided by the pump is given by the equation
delta-p = rho*delta-head = 1.8*164 = 5.9 psi
I had to increase the size of the pipe vastly through iteration in order to get the pressures to match. Where could I be going wrong? I know that a 1 in pipe is working in a similar device already.
Last edited: