Lec 26 Frictionless flow with work pipe flow

Lec 26: Frictionless flow with work, pipe flow 1

• For next time: – Read: § 8 -14 to 8 -18 – HW 13 due Monday, December 1, 2003 • Outline: – Bernoulli equation in pipe flow – Accounting for pumps and turbines – Accounting for friction in pipes • Important points: – Don’t forget the conservation of mass equation – Be careful how you account for pump and turbine losses – Understand the Moody Diagram 2

Fluid Mechanics • A venturi tube or meter--convergingdiverging nozzle frequently used to measure the volumetric flowrate of a fluid. It must be inserted into a pipe or duct as a part of the pipe or duct. 1 2 3

TEAMPLAY • For a venturi such as that shown before, the following data apply: dia 1 = 6. 0 in, dia 2 = 4. 0 in. The pressure difference P 1 – P 2 = 3 psi. Water with a density of 62. 4 lbm/ft 3 is flowing. Find the rate of flow in ft 3/min. • Hint: use flowrate Qv = A 1 V 1 = A 2 V 2 4

Pipe flow • We previously had the Bernoulli equation in this form and we observed that the terms are energy per unit mass. • In a pipe (or duct), this equation represents the mechanical energy per unit mass at a flow cross section. 5

Pipe flow • For frictionless flow, the mechanical energy will be the same at every cross section of the pipe and 6

Pipe flow Cross section 2 Cross section 1 7

Pipe flow • In real pipe or duct flows, energy must be used to overcome friction, and so at subsequent cross sections the energy is less. • Pipe flow--examples are water systems and petroleum pipeline systems. • Duct flow--examples are air conditioning ducts in buildings. 8

Pipe and duct flow • We had the following equation for incompressible steady flow: • or, replacing v by and rearranging 9

Pipe and duct flow • The term represents the increase in internal energy that occurs due to friction as the fluid flows in a pipe or duct. The textbook calls this a mechanical loss term, emech, loss. • The text also splits the work term into two: • wpump represents pumping power input or the power required by a pump or fan to move the flow. • Wturbine represents the work done by a water 10 turbine, for example.

Pipe and duct flow • Thus the book arrives at the following equation for adiabatic flow: • or 11

Pipe and duct flow • The previous equation • is equation 11 -31 in the text book, where the equation (incorrectly) appears thusly 12

Pipe and duct flow • The shaft work includes the work to increase the mechanical energy of the fluid and to overcome losses. Eq 11 -31 and 11 -32 should read: • where if we forget the wturbine for the moment, wpump, u raises the mechanical energy of the flow. 13

Pipe and duct flow • Divide the previous equation by g to get an equation in terms of head: 14

Pipe and duct flow • The term (e/g)mech, loss is identified as the head loss term, and we calculate it later. 15

Pipe and duct flow • Not all of the shaft power going into the pump is converted into useful mechanical energy supplied to the fluid, such as raising water up into a water tower. • Some is lost in frictional heating that manifests itself as a slight temperature rise of the fluid. • Thus wpump, shaft= wpump, u+emech, losses 16

Pipe and duct flow • This loss gives rise to the following pump efficiency 17

TEAMPLAY • Consider a water pump in a horizontal, constant area pipe. The mass flow rate is 50 kg/s, and the pump receives 17. 0 k. W of power from its driver. The delta P (pressure change) across the pump is 250 k. Pa. • Determine the mechanical efficiency of the pump and the temperature rise of the fluid. 18

Pipe and duct flow • Using the equation in terms of head loss and omitting the work terms (meaning we do not have a pump, turbine or compressor in the system), • Where the previous equation in terms of flow energy per unit mass has been divided by g (not gc) to yield units of head (length). 19

Pipe and duct flow • It can be shown that the head loss term can be determined in terms of the shear stress at the wall, o. • Where is the length of the pipe and D is the diameter of the pipe. 20

Pipe and duct flow • The shear stress at the wall is a function of five independent variables. • o = o( , , V, D, e) • is the dynamic or absolute viscosity • V is the mean velocity of the flow • e (or ) is the surface roughness of the pipe, and can be described as a length. 21

Pipe and duct flow • With five independent variables, the solution is very complex. • However, the complexity is reduced to two variables by use of only two nondimensional variables. 22

Pipe and duct flow • The non-dimensional shear stress can be expressed as 23

Pipe and duct flow = Re, the Reynolds number (nondimensional) = ratio of inertia forces/viscous forces e/D = relative roughness 24

Pipe and duct flow • The non-dimensional shear stress is tabulated as a function of the Reynolds number and the surface roughness. • The non-dimensional shear stress is called a friction factor. 25

Pipe and duct flow • Fanning friction factor, , used in heat transfer: • Moody friction factor, f, used in fluid mechanics: 26

Pipe and duct flow • Moody friction factor, f, used in fluid mechanics is typically given as • and appears on page 974 of the textbook. 27

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Pipe and duct flow • So, • And 29

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TEAMPLAY • Determine the pressure drop over 1000 ft of pipe carrying water if the flowrate is 70 cfs and the pipe diameter is 30 inches. Assume that the pipe has a roughness, e, of 0. 1 inches and the temperature is 25 C. Water has a density of 62. 4 lbm/cubic foot. 31

TEAMPLAY A 2 ft diameter cylindrical aire conditioning duct hands horizontally from a ceiling and carries 10, 000 CFM of 55 F air. Relative roughness factor e/D = 0. 0002. Find the pressure drop der 100 ft of duct length in the units of inches of water if 27. 7 in of water is 1 psia. 1. Write and simplify the basic equation. 2. Find the density, velocity of flow, Reynold’s number and friction factor. 3. Solve the problem. 32