Week 10 Flashcards
If you have a discharging storage tank with a nozzle at the bottom and a nozzle above it, why is the nozzle above have a lower velocity?
- The pressure at the low nozzle is the same and it’s based off the height h1.
- The pressure at the high nozzle is different as it’s based off the height h3 (smaller than h1)
- The pressure outside the nozzles are both the same
The velocity out of each nozzle can be expressed in terms of the height or the pressure at the nozzle (also dependent on height).
If h3 is less than h1, p3 is less than p1 and v3 is also less than v1.
If we divert the flow up, how will the flow velocity be affected?
Exact same scenario as having a nozzle higher up on the tank:
The pressure at 1 (where the flow leaves the tank) is unchanged from last case. The pressure at 2 (outside nozzle) is still atmospheric (and equal 0).
The pressure at 3 tho (inside nozzle right before exit) is based off the height h(3) (which is the height from the top of the water column to the top of the pipe).
Remember, what determines the flow velocity out is the pressure across the exit (p3-p2). Put another way, we had pressure p1 at the bottom of the tank, but some is lost to increase the water elevation from h1 to h3.
Also, if we raise h3 to =0, the flow V will be 0!
How about losses through Restriction? What happens to the flow out if the nozzle is skinnier?
If we force the flow through a small nozzle, we have to work harder to move that fluid at the same exit speed.
The pressure at 1 and 2 (inside and outside the tank) are the same, but p3 (inside nozzle before exit) is lower than pressure 1 due to losses.
The equation for flow out is p3-p2, also expressed as:
(p1-p_loss)-p2
Because p3 is the pressure at 1 minus the loss of pressure!
Regarding the losses through restrictions (making the nozzle smaller), what is a common way of representing these losses?
Because these losses are proportional to the velocity squared, you could represent these losses as:
p_loss = K*(ρ(v_3)^2)/2
These types of losses are proportional to - - velocity squared,
water density,
and some loss coefficient (K)
Recap: If pressure 1 is inside the tank, pressure 2 is atmospheric pressure outside, and pressure 3 is the pressure inside the nozzle, what will the flow velocity depend on?
THe flow velocity depends on the pressure difference between p3 and p2, which is the pressure difference between 1 and 2 minus the pressure lost due to friction from 1 to 3
Flow moving along a pipe will experience losses due to friction inside the walls of the pipe. How can we reduce this?
We can reduce this by shortening the pipe (exit velocity will then become bigger)
For pipe flow, what are the friction losses proportional to? inversely proportional to? What is the formula for friction losses?
For pipe flow, the losses are proportional to:
- the length of the pipe (L)
- a coefficeint known as a friction factor (f)
inversely proportional to:
- The inner diameter of the pipe (D)
The formula is:
p_loss = f * (L/D) * (ρv^2)/2
Our formula for p_loss from friction is:
p_loss = f * (L/D) * (ρv^2)/2
What does this formula represent?
What happens when we increase our losses?
This expression means that:
- The longer the pipe we use, the greater the losses. If we double the length of pipe, we double the losses from friction.
Since the pressure difference from p1 to p2 is usually fixed, by increasing our losses, we reduce the outflow velocity!
How will the v_out be affected when the pipe has the same diameter along its entire length?
How about a smaller diameter nozzle?
If the entire inside has the same diameter, then v_out will equal v in the equation
But if we have a small nozzle at the end while maintaining a big diameter everywhere else, v_out will be greater than the flow velocity (v) inside the pipe!
For our module, what is considered the storage system and what is considered the supply system?
What does each system govern?
Storage system:
- rain
- Catchment
- collection tank
- pump (energy)
- chemical treatment (energy)
- filters
storage
Supply system:
- storage
- filters (Here is not in the storage system)
- UV treatment (power)
- Use of water (tap, toilet, etc)
The supply storage system governs the consumption (AVERAGE sustained flow) while the supply system governs the PEAK (on-demand) flow
Which part of the rainwater harvester system stores energy that drives the fluid to the house for consumption?
The kinetic energy stored in the elevated storage tank!
Where should the filtration system be placed?
The filtration system must be placed BEFORE the UV treatment, as UV needs clear water to work!
To move the water from the ground to the elevated storage tank, the pump needs to supply a pressure. What would p_pump look like?
P_pump(Q) =ρgh_storage + (p_friction + p_restrictions + p_filter)
ρgh_storage represents the pressure to overcome the elevation difference, but the other pressures are due to other losses!
Remember:
- p_friction = f * (L/D) * (rho*v^2)/2
- p_restrictions = K * (rho*v^2)/2
- P_filter = C_filter * v
The pressure lost across the filter depends linearly on flow velocity through the filter, and on some constant C_f, based on the filter design!
p_pump(Q) =ρgh_storage + (p_friction + p_restrictions + p_filter)
What does the (Q) represent?
The pressure the pump can supply depends on the flowrate (Q) through the pump.
Remember, we can relate the volumetric flowrate (Q) with the flow velocity (v) as long as we known the cross-sectional area the water is flowing through!
We know that the pressure the pump supplied is a function of flowrate. What can we relate this to?
We can relate p(Q) to p(v)!
We can relate the flowrate to flow velocity in the pipe!
