PIPING SYSTEMS In this chapter we will review some of the basic concepts associated with piping systems. Topics that will be considered in this chapter are - Pipe and tubing standards - Effective and hydraulic diameters - Head loss (pressure drop) calculation - Calculation of flow rate - Determination of pipe diameter Pipe and Tubing Standards Sizes for pipes and tubes are standardized. Pipes are specified by a nominal diameter and a schedule number. Example: 2-nominal schedule 40 Nominal diameter specifies the diameter of the pipe. However, it does not necessarily equal to the inside or outside diameter of the pipe. The schedule of the pipe specifies the wall thickness of the pipe. Based on this standardization, pipe dimensions are given in Appendix D as below Nominal Diameter Outside Dia. (ft) (cm) Schedule Inside Dia. (ft) (cm) Flow Area (ft 2 ) (cm 2 ) 1/8 0.405 1.029 40 (std) 0.2242 0.683 0.0003947 0.3664 (0.03375) 80 (xs) 0.01792 0.547 0.0002522 0.2350 Each nominal diameter specify only one outside diameter. ME415 4 1
Equivalent Diameters for Noncircular Ducts Non circular ducts are characterized by equivalent diameters. Two equivalent diameters used commonly are effective diameter and hydraulic diameter. Effective Diameter The effective diameter D eff is the diameter of circular duct that has the same area as the noncircular duct of interest. Thus D 4 2 eff 4 A noncircduct D eff A For example, for a rectangular duct of dimensions h w, the effective diameter is found as D eff 4A 4h w 2 h w Hydraulic Diameter, D h 4 flow area 4A D h wetted perimeter P ME415 4 2
HEAD LOSS (PRESSURE DROP) CALCULATION Physically, the head loss (pressure drop) is the energy lost by fluid due to viscous effects (friction), and area change along the flow (pipe). Loss due to the viscous effects is called major (or frictional) loss; loss due to the change in the flow cross sectional area is called local (minor) loss. Total head loss in a pipe is sum of the major and minor losses, i.e. h f h f hf major min or Major Loss Major loss is due to friction and it is calculated from the following expression: Where h f major f L D h 2 V 2g or P L f D h 2 V 2g c f: friction factor D h : Hydraulic diameter L: Pipe length V: Average velocity For laminar flow, friction factor is dependent only on Reynolds number. f 64 Re This expression is derived using the analytically obtained velocity distribution of laminar flow in a duct. ME415 4 3
In turbulent flows, friction factor is dependent on Reynolds number and pipe surface relative roughness, i.e. f Re, / s D h s /D h is relative roughness of pipe surface. Since there is no analytical solution for turbulent flows, experimentally obtained data is used for determination of friction coefficient in turbulent flows. The functional relation between friction factor and Re and s /D h is very complicated in turbulent flows. Therefore, it is very difficult to represent this relation for all ranges of turbulent flows by a single expression. Because of this, friction factor for turbulent flows are given as correlations and/or diagrams. Most commonly used chart is Moody diagram. In the literature, there are many correlations for friction coefficient in turbulent flows. These correlations are especially useful in using computer programs for the analysis of pipe flows. One of these correlations (more often used one) is given by Chen as follows: ME415 4 4
f Chen equation 5.0452 1 2.0ln ln 3.7065D Re 2.8257 h D h 1.1098 5.8506 0.8981 Re 2 for Re 2100 There are many other correlations. See page 90 in the book by Janna Churchill equation f 8 8 Re 12 1 B C 1.5 1/12 for Re 2100 where 1 B 2.457 ln 0.9 7 / Re 0.27 / D 37530 C Re 16 16 Haaland equation 6.9 f 0.782 ln Re 3.7D 1.11 2 Swamee-Jain equation f 0.250 5.74 log 0.9 3.7D Re 2 ME415 4 5
NOTES 1) When the flow rate and pipe diameters are known, moody diagram or the above correlations can be used to obtain friction factor and calculate head loss (h f ) and pressure drop (P). 2) When the volume flow rate Q is unknown, Moody diagram and the correlations require an iterative procedure to obtain a solution. If a graph of f vs Re(f) 1/2 is available, then the unknown Q can be solved straight forward without any iteration. Insert f vs Re(f)1/2 graph ME415 4 6
3) When the diameter D is unknown, Moody diagram and the correlations require an iterative procedure to obtain a solution. This is due to the fact that the relative roughness term contains diameter D, i.e. s =/D. Introducing a new parameter, roughness number, this problem can be eliminated. / D / Dg c Roughness number Ro Re VD V Q A 4Q 2 D Hence, roughness number becomes, g c Ro 4Q For unknown diameters, a graph of f vs f 1/5 Re with an independent parameter Ro can be constructed. This simplifies the problems. ME415 4 7
MINOR (LOCAL) LOSSES Minor loss pressure (head) is a loss due to the pipe elements (fittings, valve, elbow, etc.) that causes area changes in the flow. The minor loss in such pipe elements are calculated by assigning a loss factor K for each fitting. Pressure loss P 1 P 2 V P K 2 g c 2 Head loss h f 2 V K 2g K is loss factor, which is generally determined experimentally. For some pipe elements loss factor K is given below. ME415 4 8
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Example: A 4-nominal schedule 40 pipe conveys castor oil at a flow rate of 0.012 m 3 /s. The pipe is made of asphalt coated cast iron and is 100 m long. Determine the pressure drop experienced by the fluid. Solution From various property tables, we get following data For castor oil =960 kg/m 3 =650x10-3 Ns/m 2 (App. Table B.1) For 4 nominal sch 40 pipe D=10.23 cm A=82.19 cm 2 (App. Table D.1) For asphalt coated iron =0.012 cm (Table 3.1) To be completd in class ME415 4 12
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Example: Chloroform flows at a rate of 0.01 m 3 / through a 4-nominal schedule 40 wrought iron pipe. The pipe is laid out horizontally and is 250 m long. Calculate the pressure drop of the chloroform. Solution: From tables; For chloroform =1470 kg/m 3 =0.53x10-3 Ns/m 2 (App. Table B.1) For 4 nominal sch 40 pipe ID=D=10.23 cm A=82.19 cm 2 (App. Table D.1) For wrought iron =0.0046 cm (Table 3.1) ME415 4 14
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Example: A 6-nominal schedule 80 cast iron pipe is 11,270 ft long. It is to convey octane. The available pump can provide a pressure drop of 25 psi. Determine the expected flow rate of octane in the pipe. Solution: From table; For octane =43.74 lbm/ft 3 =1.07x10-5 lbfs/ft 2 (App. Table B.1) For 6 nominal sch 80 pipe ID=D=0.4801 ft A=0.1810 ft 2 (App. Table D.1) For cast iron =0.00085 ft (Table 3.1) ME415 4 16
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Example: A PVC plastic pipeline is to convey 50 lt/s of ethylene glycol over a distance of 2000 m. The available pump can overcome a frictional loss of 200 kpa. Select a suitable schedule 40 size for the pipe. Solution: From tables; For ethylene glycol =1100 kg/m 3 =16.2x10-3 Ns/m 2 (App. Table B.1) For plastic pipe =0.0=smooth (Table 3.1) ME415 4 18
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Example: A huge water reservoir is drained with a 4-nominal schedule 80 copper pipe. The piping system shown in the figure. The fittings are regular and threaded. Determine volume flow rate through the system. 20 m water 60 m of 4-nom sch 80 copper pipe 4-90 o elbows 1 globe valve 1 basket strainer Q=? 2 m Solution: From tables; For water =1000 kg/m 3 =10.89x10-3 Ns/m 2 (App. Table B.1) For 4 nominal sch 80 pipe ID=D=9.718 cm A=74.14 cm 2 (App. Table D.1) For copper pipe =0.00015 cm (Table 3.1) ME415 4 20
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Example: Air flows through an horizontal duct that is 3 m long. The duct is rectangular (30 cm x 1 5 cm) and made of galvanized sheet metal. Air flows through the duct at a velocity of 6.1 m/s. Determine the pressure drop in the duct. Solution: From tables; For air =1.17 kg/m 3 =18x10-6 Ns/m 2 (App. Table B.1) For galvanized metal surface =0.00015 cm (Table 3.1) ME415 4 22
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