BUDA UNIVERSITY Dr Ferenc Szlivka Professor Dr Szlivka

ÓBUDA UNIVERSITY Dr. Ferenc Szlivka Professor Dr. Szlivka: Heat and Flow Technology I_9 1

Friction flow in tube Chapter 9. 2

Laminar and turbulent flow Laminar flow Turbulent flow 3

Laminar and turbulent flow 4

Laminar and turbulent flow a. / Laminar flow b. / Turbulent flow 5

Turbulent flow Smog flow Flow around a cylinder (Kármán vortex street) 6

Pressure loss in a straight pipe There is a friction loss Darcy-formula 7

Solution of Navier-Stokes equation in tube „l" is the length , with Dp’the pressure loss 8

Solution of Navier-Stokes equation in tube for laminar flow Darcy-formula where Re is the Reynolds’ number 9

Straight pipe pressure loss in turbulent flow Reynolds’ number "v" is the average velocity, "d" is the pipe diameter "r" density of fluid "m" dynamic viscosity. The pipe wall is not smooth because of the producing process or the corrosion. The average roughness is k and the relative roughness k/d, or the reciprocal of it is d/k. 10

Nikuradse-diagram 11

Moody-diagram 12

Roughness of different materials Material k [mm] Riveted iron pipe 0. 9 -9. 0 Concrete 0. 3 - 3. 0 wood channel 0. 18 - 0. 9 Cast-iron 0. 26 -0. 6 Zincked iron 0. 1 -0. 15 Asphalted cast-iron 0. 1 -0. 15 Iron, (little rusted) 0. 1 -0. 3 Iron 0. 02 -0. 046 Drawn iron 0. 0015 -0. 03 Glass smooth 13

Moody-diagram Haaland-formula 14

Three different pipe problems I. Given the diameter of the pipe "d", the length of pipe „l", the average diameter "v" , or the volume flow rate „ qv ", and the data of fluid: density "r" and viscosity "m", The question is the pressure loss "Dp". II. Given the diameter of the pipe "d", the length of pipe „ l ", and the pressure loss Dp', and the data of fluid: density "r" and viscosity "m", The question is the volume flow rate " qv ". III. Given the length of pipe „ l ", the pressure loss Dp', the volume flow rate „qv", and the data of fluid: density "r" and viscosity "m„. The question is the diameter of the pipe "d". 15

I. pipe problem Calculate the pressure loss in an asphalted cast-iron pipe. Water is flowing in it data: Solution: From the 8. 1 table look out the dynamic viscosity. 16

10 1 10 -2 1, 3*10 -3 10 -4 10 -5 17

Material k [mm] Riveted iron pipe 0. 9 -9. 0 Concrete 0. 3 - 3. 0 wood channel 0. 18 - 0. 9 Cast-iron 0. 26 -0. 6 Zincked iron 0. 1 -0. 15 Asphalted cast-iron 0. 1 -0. 15 Iron, (little rusted) 0. 1 -0. 3 Iron 0. 02 -0. 046 Drawn iron 0. 0015 -0. 03 Glass smooth 18

0, 028 1250 19

II. pipe problem Calculate the average velocity in an asphalted cast-iron pipe! data: Solution: We don’t know the average velocity (or the volume flow rate) so we can’t calculate the "l"and the Re number. So we should make an iteration process. Fortunately the process is fast. Look out the roughness and the. 20

We don’t know the Re number, the velocity is unknown. So we assume beginning l 0=0, 02 - 0. 03. In this case l 0=0, 02. 1250 Dr. Szlivka: Fluid Mechanics 9. 21

Make a formula from Darcy’s formula. 22

0, 026 4, 073*104 23

Check the relative error between two steps If the difference is biger than 10% , we calculate once more! 24

A new l from the Moody diagram. Check the relative error. The difference is smaller than 10%, so it is the final result. The average velocity is: 25

III. pipe problem (design problem) data: Solution: The question is the diameter „d”. "l", and the Re-number, are depending on the velocity (which is unknown) we should make an iteration process. To solve the problem we choose a standard pipe diameter, which can be bought. The usual average velocity is 1 -2 m/s. 26

Nominal diameter [in] The diameter was choosen 5 in d=128, 2 mm. The solution after is the I. problem. In these case: Effective diameter [mm] 2 52. 5 2 1/2 62. 7 3 77. 9 3 1/2 90. 1 4 102. 3 5 128. 2 6 154. 1 8 202. 7 10 254. 5 27

0, 026 1068 28

The calculated pressure loss is bigger than the given pressure loss. So we should choose a bigger diameter but only with one step bigger diameter ! (Unless the pipe is to expensive. ) The calculated pressure loss is smaller than the given one. With a trothling we it can be made the difference. Comment: Put the volume flow rate into the Darcy’s formula: Nominal diameter [in] Effecti ve diamet er [mm]] 2 52. 5 2 1/2 62. 7 3 77. 9 3 1/2 90. 1 4 102. 3 5 128. 2 6 154. 1 8 202. 7 10 254. 5 29

Comment: Puting the volume flow rate into the Darcy’s formula: The formula shows: If the diameter is 10% smaller the pressure loss rises approximately 50%! 30

Noncircular duct loss coefficience where "K" is the circumference connected with the fluid , (example : Open channel ), "A" is the cross section area filled with fluid. 31

Noncircular duct loss coefficience Let us see a circle pipe, which has the same pressure loss than in a noncircular duct on the same length and the same shear stress on the wall. Find the diameter of this circular pipe, This diameter is called equivalent diameter. 32

Fitting pressure losses 33

Valves Straight valve latch Butterfly valve One way valve 34

Valves, taps, etc. pressure loss coefficience, and equivalent pipe length Type Loss coefficience z Streight valve fully open 6. 8 340 Corner valve fully open 2. 9 145 Tolózár teljesen nyitva 0. 26 13 1/4 open 18 900 1/2 open 3. 2 160 3/4 open 0. 7 35 One way valve 2. 7 135 Ball way valve 3 150 Butterfly valve fully open 0. 3 15 Valves, taps, pressure loss coefficience, and equivalent pipe length 35

Elbows loss coefficience, and equivalent pipe length Elbows 36

Air filter pressure loss 37

Air heater pressure loss 38