Lec 26 : Frictionless flow with work, pipe flow

1. Lec 26: Frictionless flow with work, pipe flow

2. 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

3. 1 2 Fluid Mechanics • A venturi tube or meter--converging-diverging 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.

4. TEAMPLAY • For a venturi such as that shown before, the following data apply: dia1 = 6.0 in, dia2 = 4.0 in. The pressure difference P1 – P2 = 3 psi. Water with a density of 62.4 lbm/ft3 is flowing. Find the rate of flow in ft3/min. • Hint: use flowrate Qv = A1V1 = A2V2

5. 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.

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

7. Pipe flow Cross section 2 Cross section 1

8. 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.

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

10. 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 turbine, for example.

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

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

13. 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.

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

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

16. 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

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

18. 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 kW of power from its driver. The delta P (pressure change) across the pump is 250 kPa. • Determine the mechanical efficiency of the pump and the temperature rise of the fluid.

19. 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).

20. Pipe and duct flow • It can be shown that the 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.

21. 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.

22. 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 non-dimensional variables.

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

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

25. 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.

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

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

29. Pipe and duct flow • So, • And

31. 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.

32. 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. • Write and simplify the basic equation. • Find the density, velocity of flow, Reynold’s number and friction factor. • Solve the problem.