2. 8.2 OBJECTIVES  Describe the appearance of laminar flow and turbulent flow  State the relationship used to compute the Reynolds number  Identify the limiting values of the Reynolds number by which you can predict whether flow is laminar of turbulent.  Compute the Reynolds number for the flow of fluids in round pipes and tubes.  State Darcy`s equation for computing the energy loss due to friction for either laminar or turbulent flow.

3. The next objectives will be discuss by the next group..  State the Hagen-Poiseuille equation for computing the energy loss due to friction in laminar flow.  Define the friction factor as used in Darcy`s equation.  Determine the friction factor using Moody`s diagram for specific values of Reynolds number and the relative roughness of the pipe.  Compute the friction factor using equations developed by Swamee and Jain.  Compute the energy loss due to friction for fluid flow in circular pipes, hoses, and tubes and use the energy loss in the general energy equation.  Use the Hazen-Williams formula to compute energy loss due to friction for the special case of the flow of water in circular pipes.

4. 8.3 REYNOLDS NUMBER  Osborne Reynolds was the first to demonstrate that laminar or turbulent flow can be predicted if the magnitude of a dimensionless number, that called the Reynolds number (N R ), is known.  It can be shown experimentally and verified analytically that the character of flow in a round pipe depends on four variables which are fluid density, fluid viscosity, pipe diameter and average velocity of flow  The equation of Reynolds number N R = v D p = v D (8-1) N R = v D p = v D (8-1) μ v μ v These two forms of the equation are equivalent because These two forms of the equation are equivalent because v = μ/p v = μ/p

5. where ; where ; v = velocity v = velocity D = diameter of pipe D = diameter of pipe p = fluid density p = fluid density μ = fluid viscosity μ = fluid viscosity  We can demonstrate that the Reynolds number is dimensionless by substituting standard SI units into Eq. (8-1): N R = v D p = v x D x p x 1 N R = v D p = v x D x p x 1 μ μ μ μ N R = m x m x kg x m.s N R = m x m x kg x m.s s m 3 kg s m 3 kg  Because all units can be cancelled, N R is dimensionless.  The Reynolds number is one of the several dimensionless number useful in the study of fluid mechanics and heat transfer.  The process called dimensional analysis can be used to determine dimensionless numbers.

6. Table 8.1 lists the standard units for quantities used in calculation of Reynolds number to ensure that it is dimensionless. Table 8.1 lists the standard units for quantities used in calculation of Reynolds number to ensure that it is dimensionless. Flows having large Reynolds number, typically because of high velocity and/or low viscosity, tend to be turbulent. Flows having large Reynolds number, typically because of high velocity and/or low viscosity, tend to be turbulent. Those fluids having high viscosity and/or moving at low velocities will have low Reynolds numbers and will tent to be laminar. Those fluids having high viscosity and/or moving at low velocities will have low Reynolds numbers and will tent to be laminar. Quantity SI units U.S Customary Units Velocitym/sft/s Diametermft Density kg/m 3 or N.s 2 /m 4 slug/ft 3 or Ib.s 2 /ft 4 Dynamic viscosity N.s/m 2 or Pa.s or kg/m.s Ib.s/ft 2 or slug/ft.s Kinematic viscosity m 2 /s ft 2 /s

7. Example Problem 8.2 Example Problem 8.2 Determine whether the flow is laminar or turbulent if water at 70 0 C flows in a 1-in Type K cooper tube with a flow rate of 285 L/min. Determine whether the flow is laminar or turbulent if water at 70 0 C flows in a 1-in Type K cooper tube with a flow rate of 285 L/min.Solution Evaluate the Reynolds number, using Eq. (8-1): N R = v D p = v D N R = v D p = v D μ v μ v For a 1-in Type K cooper tube, D = 0.02527 m and A = 5.017x10 -4 m 2 (from Appendix H). Then we have v = Q = 285L/min x 1m 3 /s = 9.47 m/s v = Q = 285L/min x 1m 3 /s = 9.47 m/s A 5.017x10 -4 m 2 60 000L/min A 5.017x10 -4 m 2 60 000L/min v = 4.11x10 -7 m 2 /s (from Appendix A) v = 4.11x10 -7 m 2 /s (from Appendix A) N R = (9.47)(0.02527) = 5.82x10 5 N R = (9.47)(0.02527) = 5.82x10 5 4.11x10 -7 4.11x10 -7 Because the Reynolds number is greater than 4000, the flow is turbulent.

8. 8.4 Critical Reynolds Number For practical application in pipe flow, we find if the Reynolds number is a) Less than 2000 ( N r < 2000), the flow will be laminar b) Greater than 4000 (N r >4000), the flow will be assumed as turbulent But if falls in the range of Reynolds number between 2000 and 4000, therefore this region is called critical region.

9. This typical application is well within the laminar flow range and well within the turbulent flow range, so the existence of this region of uncertainty does not cause great difficulty. This typical application is well within the laminar flow range and well within the turbulent flow range, so the existence of this region of uncertainty does not cause great difficulty. If the system is found to be in critical region, usually practice is to change the flow rate or pipe diameter to cause the flow definitely laminar or turbulent. More precise analysis is then possible. If the system is found to be in critical region, usually practice is to change the flow rate or pipe diameter to cause the flow definitely laminar or turbulent. More precise analysis is then possible.

10. Example 8.3 Example 8.3 Determine the range of average velocity of flow for which the flow would be in the critical region if SAE 10 oil at 15°C is flowing in a 2-in Schedule 40 steel pipe. The oil has specific gravity of 0.89 Answer: The flow would be in critical region if 2000<N R <4000. The values for μ, D, and p: D = 52.5mm (Appendix F) D = 52.5mm (Appendix F) μ = 1 x 10 -1 N.s/m 2 ( From Appendix D) μ = 1 x 10 -1 N.s/m 2 ( From Appendix D) p = 0.89(1000kg/m 3 ) = 890kg/m 3 p = 0.89(1000kg/m 3 ) = 890kg/m 3

11. Substituting these values: For N R = 2000 v = (2.14 x 10 -3 )(2 x 10 3 ) = 4.3m/s v = (2.14 x 10 -3 )(2 x 10 3 ) = 4.3m/s For N R =, v = (2.14 x 10 -3 )(4 x 10 3 ) = 8.56m/s v = (2.14 x 10 -3 )(4 x 10 3 ) = 8.56m/s Therefore, if 4.3< v < 8.56 m/s, the flow will be in the critical region.

12. 8.5 DARCY EQUATION In general energy equation : In general energy equation : the term h L is defined as the energy loss from system the term h L is defined as the energy loss from system One component of the energy loss is due to friction in the flowing fluid. Friction is proportional to the velocity head of the flow and to the ratio of the length to the diameter of the flow stream, for the case in pipes and tubes. One component of the energy loss is due to friction in the flowing fluid. Friction is proportional to the velocity head of the flow and to the ratio of the length to the diameter of the flow stream, for the case in pipes and tubes.

13. This can expressed in Darcy equation: This can expressed in Darcy equation: where : h L = energy loss due to friction (N.m/N, m, Ib-ft/Ib, or ft) L = length of flow stream ( m or ft ) D = pipe diameter ( m or ft) v = average velocity of flow ( m or ft) F = friction factor ( dimensionless) It can be used to calculate the energy loss due to friction in long straight section of round pipe for both laminar and turbulent flow.