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FLUID FLOW FOR CHEMICAL ENGINEERING Dr Mohd Azmier Ahmad Tel: +60 (4) 5996459 EKC 212 CHAPTER 8 (Part 5) TRANSPORTATION SYSTEM.

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Presentation on theme: "FLUID FLOW FOR CHEMICAL ENGINEERING Dr Mohd Azmier Ahmad Tel: +60 (4) 5996459 EKC 212 CHAPTER 8 (Part 5) TRANSPORTATION SYSTEM."— Presentation transcript:

1 FLUID FLOW FOR CHEMICAL ENGINEERING Dr Mohd Azmier Ahmad Tel: +60 (4) 5996459 Email: chazmier@eng.usm.my EKC 212 CHAPTER 8 (Part 5) TRANSPORTATION SYSTEM & FLUID METERING Design of pumping system

2  Fluid flows from areas of high pressure to areas of low pressure.  Pump operates by creating low pressure at the inlet which allows the liquid to be pushed into the pump by atmospheric or “head” pressure (pressure due to the liquid’s surface being above the centerline of the pump).  Head refers to gains or losses in pressure caused by gravity and friction as fluid moves through the system. It can be measured in unit of length (not N/m 2, lbs/in 2 or PSI).  Significance of using the “head” term instead of the “pressure” term : pressure from a pump will change if the specific gravity (weight) of the liquid changes, but the head will not change. Terminology

3   Static Suction Head, h ss or Z S : Vertical distance from the water line to the centerline of the impeller. If the liquid level is above pump centerline, Z s is positive. If the liquid level is below pump centerline, Z s is negative (suction lift condition).   Static Discharge Head, h sd or Z d : Vertical distance from the discharge outlet to the point of discharge or liquid level when discharging into the bottom of a water tank.   Vapor Pressure Head, h vapor : Vapor pressure at which a liquid exist in equilibrium at a given temperature.   Surface Pressure Head, h p = P/ρ : P absolute on the surface   Velocity head, h v = V 2 /2g : Liquid energy due to motion Terminology

4 Friction Head, h f : Head required to overcome the resistance to flow in the pipe/fittings. Dependent on size/type of pipe and fittings. Total or Dynamic Suction Head, H s = Z s + h ps + h vs – h fs Total or Dynamic Discharge Head, H d = Z d + h pd + h vd + h fd Total Head, H T = H d – H s = ∆Z + ∆h p + ∆h v + h f Terminology

5 Converting pressure to head for SI unit Converting pressure to head for British unit Pressure in N/m 2 and head in m Pressure in lb/in 2 (psi) and head in ft

6 Suction Head

7 Suction Lift

8

9  For cavitation, first understand vapor pressure, P vapor.  P vapor is the pressure required to boil a liquid at a given temperature.  For example, water will not boil at T room since its P vapor is lower than the surrounding pressure (P vap < P fluid ).  But, raise the water’s temperature 100 o C, the vapors are released because at 100 o C the P vapor is greater than the atmospheric pressure (P fluid < P vapor ).  In the same manner, pump cavitation occurs as the pressure in the pump inlet drops below the P vapor of the liquid (P fluid < P vapor ).  Vapor bubbles form at the pump inlet & moved to the discharge of the pump where they collapse, often taking small pieces of the pump with them. Vapor pressure and cavitation

10  Pump can pump only liquids, not vapors.  However, centrifugal force of the impeller vanes further rise in the temperature & fall in pressure at the pump suction, thus induces vaporization (P fluid < P vap ).  Pump always needs to have a sufficient amount of suction head present to prevent this vaporization.  NPSH as a measure to prevent liquid vaporization, overcome the pressure drop in the pump and maintain the majority of the liquid above its P vapor.  NPSH is expressed in terms of absolute fluid column height. Note:  Any discussion of NPSH or cavitation is only concerned about the suction side of the pump.  For total head, both the suction & discharge sides are involved.

11 Cavitation is often characterized by:  Loud noise often described as a grinding or “marbles” in the pump  Loss of capacity (bubbles are now taking up space)  Pitting damage to parts as material is removed by the collapsing bubbles  Unstable and the power consumption may be erratic.  Vibration and mechanical damage such as bearing failure can also occur.

12  Understanding the significance of NPSH is very much essential during installation as well as operation of the pumps.  NPSH can be defined as two parts:  NPSH Available (NPSH A ): It is the excess pressure of the liquid over its P vapor at the pump suction, to avoid cavitate & maintain a liquid state. NPSH A must be calculated.  NPSH Required (NPSH R ): Minimum pressure required at the suction port of the pump to keep the pump from cavitating. NPSH R curves are provided by the pump manufacturer. NPSH R varies with speed and capacity within any particular pump. NPSH R increases as the pump capacity & liquid velocity increase. Insufficient NPSH R will cause a low pressure to exist at the pump intake. This will cause the liquid to boil and cause cavitation.  NPSH A >NPSH R for the pump system to operate without cavitating. Net Positive Suction Head (NPSH)

13 How to improve NPSH available   Consult with the pump manufacturer about reducing the NPSH R.   Use larger pump model and slowing it down. A gear pump generating 100 gpm at 780 rpm has an NPSH R of 9.1 ft of water. By switching to a larger pump running at 350 rpm for same 100 gpm, the NPSH R drops to 3.8 ft of water   Increase static head (height of liquid in feed tank)   Use cool liquid at pump   Increase feed diameter   Reduces frictional losses

14  Small centrifugal pumps, NPSH = 2-3 m.  Very large pumps: it can increase up to 15 m.  NPSH increases with increase in (i) pump capacity, (ii) impeller speed & (iii) discharge pressure  NPSH A can be calculated using: where, p ps = absolute pressure at surface of reservoir p vapor = vapour pressure of the liquid h fs = friction in suction line Z s = height from the water line to pump inlet ------- (8.7)

15 Equation Power supplied, P B to any particular pump from an external source can be determined using; where is the mass flow rate. The power delivered, P f to the fluid can be calculated using; ------------ (8.4) ------------ (8.5)

16 Example 8.1: Given NPSH R of 2.5 m, calculate the pump NPSH A for a tank with a liquid level 2 m above the pump intake Take the atmospheric pressure = 10m, vapor pressure head = 3m, friction loss = 2m Solution : NPSR A =(h ps –h vapor ) + Z s – h fs = (10m–3m) + 2m – 2m = 7m NPSH A > NPSH R for the pump system to operate well

17 Benzene at 37.8 o C is pumped through the system of Fig. 8.1 at the rate of 0.1515 m 3 /min. The reservoir is at atmospheric pressure. The gauge pressure at the end of the discharge line is 345 kN/m 2. The discharge is 3.048m & the pump suction 1.22m above the level in the reservoir. The discharge line is 1.5-in. Schedule 40 pipe. The friction in the suction line is known to be 3.45 kN/m 2 & that in the discharge line is 37.9 kN/m 2. The mechanical efficiency of the pump is 0.6 (60%). The density of benzene is 865 kg/m 3, and its vapour pressure at 37.8 o C is 26.2 kN/m 2. Example 8.2: Fig. 8.1

18 Solution a)The upstream station a is at the datum height. is found from Appendix 5. For a 1.5 in Sch. 40 pipe, a velocity of 1 ft/s (0.3048 m/s) corresponds to a flowrate of 6.34 gal/min (0.024m 3 /min), Calculate (a) Developed pump head; (b) Pump work; (c) Total power input and (d) If the pump manufacturer specifies a required NPSH of 3.05 m, will the pump be suitable for this service? The velocity at point a is negligible because of the larger diameter of the tank in comparison with that of the pipe. Also = 1.0 (turbulent)

19 W p = 478.4 / 0.6 = 797.34 J/kg Developed or Total Head, H T = H d – H s = ∆Z + ∆h p + ∆h v + h f Work pump :

20 b) Total power input The mass flow rate is = (865 kg/m 3 ) (0.1515 m 3 /min) (1min/60s) = 2.184 kg/s The power input is :

21 c) Use Eq. 8.7 for NPSH The available NPSH A > NPSH R, so the pump should be suitable for the proposed service.

22 WORK IN PAIR 10 Benzene at 37.8 o C is pumped through the system of Fig. 8.1 at the rate of 0.012 m 3 /min. The reservoir is at atmospheric pressure. The gauge pressure at the end of the discharge line is 345 kN/m 2. The discharge is 4.48m & the pump suction 2.12m above the level in the reservoir. The discharge line is 0.5-in. Schedule 40 pipe. The friction in the suction line is known to be 5.45 kN/m 2 & that in the discharge line is 27.9 kN/m 2. The mechanical efficiency of the pump is 0.75 (75%). The density of benzene is 865 kg/m 3, and its vapour pressure at 37.8 o C is 26.2 kN/m 2. Calculate : (a) Developed pump head; (b) Pump work; (c) Total power input and (d) If the pump manufacturer specifies a required NPSH of 3.05 m, will the pump be suitable for this service?

23 Example 8.2 Pumping fresh water of 100 gpm at 68°F. Atmospheric pressure = 14.7 psi Liquid level above pump centerline = 5 feet Friction head, h fs = 2.34 feet Vapor pressure of water at 68°F = 0.27 psi. Specific gravity = 1 NPSH R = 29 feet Solution: Z s = 5 feet h p = pressure x 2.31/sg. = 14.7 psi x 2.31/1 = 34 feet h vapor = pressure x 2.31/sg = 0.27 x 2.31/1 = 0.62 feet h f =2.34 feet. NPSH A = (h p – h vapor ) – h f + Z s = (34 - 0.62) - 2.34 + 5 = 36.04 feet NPSH A > NPSH R, so the pump should be suitable for the proposed service.

24 WORK IN PAIR 11 1.Calculate NPSH A. Given: Pumping fresh fluid of 100 gpm at 85°F. Atmospheric pressure = 14.7 psi Liquid level below pump centerline = 5.2 feet Friction head, h fs = 3.54 feet Fluid vapor pressure 85°F = 1.44 psi. Specific gravity = 1.22 NPSH R = 24 feet Guideline: h p = pressure x 2.31/sg. h vapor = pressure x 2.31/sg NPSH A = (h p – h vapor ) – h f + Z s

25 References Appendix III.3 (pg 642-46) in Seider et al., Process Design Principals (our text for this class). Chapter 12 in Turton et al., Analysis, Synthesis, and Design of Chemical Processes. Chapter 13 in Peters and Timmerhaus, Plant Design and Economics for Chemical Engineers. Chapter 8 in McCabe, Smith and Harriott, Unit Operations of Chemical Engineering.


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