11:2 Pipe Flow - Type I, II and III Problems
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Overview
Derek Elsworth reviews Type I, II, and III pipe-flow problems, showing how geometry, flow rate, and head loss determine which quantities must be calculated iteratively. Examples include a 25 kW pump feeding a 0.04 m³/s flow through a pipe and nozzle, a 1 hp pump problem in English units, and sizing a smooth rectangular air duct from a specified pressure drop. An opening discussion connects pipe losses to Norwegian pumped-storage hydropower, wells, and hydraulic fracturing.
Key takeaways
- The three pipe-flow types are classified by the unknown: Type I finds head loss from known geometry and flow, Type II finds flow from known geometry and head loss, and Type III sizes geometry from a target flow and head loss.
- For Type II calculations, friction factor depends on Reynolds number, which depends on the unknown velocity; estimating, recalculating, and checking against the Moody chart resolves this dependency.
- For Type III sizing, an assumed diameter or duct dimension gives an initial Reynolds number and friction factor, but the resulting geometry must be used to update both quantities until they agree.
- In the metric pump example, a 0.04 m nozzle and 0.06 m pipe have different velocities even though they carry the same 0.04 m³/s flow, so continuity must connect their velocity terms.
- Pump power contributes head according to fluid unit weight and volumetric flow rate; in the English-unit example, raising the outlet by 2 ft adds a term that turns a direct solution into a cubic equation.
- Pipe diameter trades capital cost against operating efficiency: larger pipes reduce velocity and friction losses, a relevant design choice in pumped storage and other high-flow systems.
Chapters
- An opening segment describes the European Union’s plans to cut greenhouse-gas emissions and increase renewable generation, while noting wind output varies over time.
- Norwegian reservoirs could store surplus wind power by pumping water uphill, then generate electricity when wind is low and demand is high.
- The proposal depends on greater transmission capacity between Norway and European consumption centers, while minimizing environmental impacts.
- Derek Elsworth explains that pumped-storage systems lose energy in pipes during both pumping and generation, so the cycle cannot return all the input power.
- Since pipe losses rise strongly with velocity, larger-diameter pipes can reduce operating losses, but they raise construction costs.
- Submersible pumps push water up deep wells because suction is limited by cavitation; hydraulic fracturing similarly creates conductive pathways and props them open with sand.
- The problem categories vary according to which of three groups is unknown: geometry, flow rate, or head loss.
- Geometry includes pipe diameter, roughness, and length; flow rate can be expressed as velocity or volumetric flow rate.
- Head loss can be represented as HL or as a pressure difference, ΔP.
- Type I problems provide the pipe geometry and flow conditions, leaving head loss as the quantity to find.
- The Darcy-Weisbach relation uses friction factor, pipe length, hydraulic diameter, and velocity head: HL = f(L/DH)(V²/2g).
- Calculate Reynolds number from velocity, characteristic diameter, density, and viscosity, use it to obtain the friction factor, then compute head loss.
- Type II problems provide pipe geometry and head loss but leave velocity or flow rate unknown, so the friction factor cannot be known at the outset.
- Start with an estimated friction factor, rearrange Darcy-Weisbach to estimate velocity, and use that velocity to calculate Reynolds number.
- Check the resulting Reynolds number and relative roughness against the Moody chart, revise the friction factor, and repeat until the estimate is consistent.
- Type III problems specify a target flow condition and head loss but require the pipe dimensions, so an initial diameter must be assumed.
- For the illustrated case, the target velocity and head loss are each 1 m in their respective units, with an assumed 1 cm pipe used to estimate Reynolds number.
- Use the estimated friction factor to solve for diameter, then recalculate Reynolds number and friction factor with the revised diameter.
- A revised diameter can change Reynolds number enough to change the appropriate friction factor, requiring another pass through the calculation.
- The example’s estimates shift between nearby diameter values, illustrating that iteration may oscillate rather than converge immediately.
- When the flow is laminar, use the relation f = 64/Re directly and substitute it into the head-loss equation as an alternative to relying on a turbulent Moody-chart estimate.
- The system has a flow rate of 0.04 m³/s, a 0.06 m pipe, a 0.04 m nozzle, and a pump supplying 25 kW.
- Apply the energy equation between a large upstream tank and the downstream outlet, accounting for pressure, elevation, pump head, exit velocity, and pipe friction.
- Because the nozzle and pipe diameters differ, calculate their velocities separately using continuity; express pump head through power, fluid unit weight, and volumetric flow rate.
- A second system uses a 1 hp pump, friction factor 0.025, 25 ft of pipe, a 0.12 ft diameter, and an outlet initially 8 ft above the selected datum.
- The large pressurized tank has negligible velocity; the upstream pressure head is related to the 8 ft water elevation, while the outlet pressure is atmospheric.
- Converting pump power to head introduces a velocity dependence, which combines with the velocity-head and friction terms to determine the flow rate.
- Raising the outlet by 2 ft changes its elevation from 8 ft to 10 ft, so the tank pressure head and outlet elevation no longer cancel.
- The added 2 ft head term combines with the pump-power relation, producing an equation with both velocity and velocity cubed.
- Unlike the original case, which can be rearranged and solved directly with a cube root, the modified case requires solving a cubic and selecting the physically meaningful positive root.
- The duct example specifies 100 m³/s of air, a smooth 50 m duct, and a pressure drop given as roughly 40 to 44 mm of water.
- The rectangular duct has width-to-height ratio 3:1, giving area 3H², perimeter 8H, and hydraulic diameter DH = 1.5H.
- Convert the water-column pressure drop to pressure, then compare it with Darcy-Weisbach head loss using air density and the duct’s hydraulic diameter.
- For the duct, assume a friction factor to solve for height, use that height to calculate velocity and Reynolds number, and revise the factor for the smooth duct as needed.
- Once height is determined, the 3:1 aspect ratio fixes the width; repeated iterations narrow the geometry to a consistent solution.
- Elsworth closes by previewing parallel and branching pipe networks, plus flow-rate measurement, as the next topics.
Transcript excerpt
0:02 the European Union has agreed to undertake major Cuts in greenhouse gas emissions and to increase power generation from renewable energy sources good as anything wind power will make up much of Europe's future power generation several major projects are already being constructed and planned on the continent in the North Sea and along the UK Coastline wind power can only be and variations are considerable over time these variations do not always correspond with the consumption a large share of wind power would represent a major challenge to the stability of the transmission systems and the security of supply to the consumers so I guess this is a not a public message but a um a partisan message from a a company that owns dams
0:50 in uh Norway as you'll see there is a need for a backup capacity in order to generate power on days with no wind and high demand there is also a need for storage capacity for surface power on days with strong wind and low demand in Norway with its long history of Hydro power generation we find half of Europe's Reservoir capacity new pump storage power plants in connection with existing reservoirs could be part of the solution in securing a reliable Energy System on Windy days Surplus power could be used to pump water from low to high altitude reservoirs this water could in turn be released to generate power on days when demand is high and wind levels are low this cycle can be repeated over and over
1:41 [Music] [Music] again in order to use Norwegian hydr power reservoirs as rechargeable batteries for the European power supply more transmission capacity must be EST between Norway and the major European consumption areas pump storage Power Solutions must be developed in such a way that the
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Derek Elsworth.