Problem solving in fluid mechanics

There is a simplest explaination of fluid mechanics

1. CVE 215 2016 - 2017

2. The water pump in the figure maintains a pressure of 6.5 psi (gage) at point 1. There is a filter, a half-open valve ( k = 2.8 ), and two elbows (k = 0.64) and there are 80 ft of 4-inch diameter commercial steel pipe (f = 0.019). If the flow rate is 0.4 ft 3 /s, what is the loss coefficient of the filter? If the valve is wide open (k = 0) and K filter = 7, what is the resulting flow rate ? What is the percentage of increase or decrease of the flow rate? Draw the TEL and HGL for the system.

3. Pump Valve Filter Elbow  9.0 ft (1)

4. Taking the energy equation is written- from point 1 to point 2 “the surface of the tank”- as     Enter the given values into the above equation gives,   Since the pipe has a constant diameter, the continuity equation     Solution ….. (1)

5.   Substituting into “1” gives,   If       Substituting into “1”, with the valve wide opening and assuming f = 0.019       ….. (2)

6.  = 5.48 ft/s and the flow rate Q =     The flow rate is increased by a percentage of =  

7. A sub-marine moves horizontally in sea and has its axis 15m below the surface of water. A Pitot-tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U- tube containing mercury. The difference of mercury level is found to be 170mm. Find the speed of the sub-marine in kilometers per hour, knowing that the sp. Gr. of mercury is 13.6 and that of sea – water is 1.026 with respect to fresh water.

8. Given :   Diff . of mercury level, x = 170 mm = 0.17m Sp. G. of mercury, S.G =13.6 Sp. Gr. of sea water , ` S w =1.026     Where C V is a coefficient which is approximately unity for large pipes and smooth Pitot but is appreciably less for low Reynolds number flow

9. 3m 12m Invert Air Pressure = 45 kpa Water O ≈ Hinge The figure below shows a vertical circular gate in a 3-m diameter tunnel with water on one side and air on the other side.  

10.   Find the horizontal reaction at the hinge? A .   412 kN B.   408 kN C .   410 kN D.   414 kN How far from the invert on the of the tunnel is the net hydrostatic force acting gate ? A .   1.45 m B.   1.43 m C .   1.47 m D.   1.41 m Where will the hinge support be located (measured from the invert) to hold the gate in position? A .   1.48 m B.   1.46 m C .   1.44 m D.   1.41 m

11. Force on the gate due to air pressure “the diameter of the cylindrical gate D = 3 m”:   Force on the gate due to water:   The net force on the gate:   The horizontal reaction at the hinge = net force and opposite to its direction: F horizontal = 410 kN  ©

12. The horizontal water force on the gate acts at: “ e” is a distance down the center line:   The net hydrostatic force acts on the gate at a distance above the invert of the tunnel (A)   To find the location of the hinge support  

13. 3m 12m Air Pressure = 45 kpa Water O ≈ Hinge F air 1.5m F water Z e invert   ≈ F  

14. To find the location of the hinge support   The hinge support will be located to hold the gate in position at “measured from the invert ” (D)  

15. Starting with the Bernoulli’s equation, show that the volumetric flow rate of an incompressible fluid through a horizontal Venturi-meter where the throat and the upstream positions are connected by a U-tube containing a mercury of density  m can be given by:   Where: ,  

16. An orifice – meter and a Venturi-meter are connected in parallel in a horizontal pipe of diameter 50mm. The orifice has a throat diameter of 25 mm and discharge coefficient of 0.65, while the Venturi has a throat diameter of 38 mm and discharge coefficient of 0.95. “The pressure drop across both instruments must be the same” Determine the proportion of flow through either meter. Q Venturi Q Orifice Q total Q total

17. It is assumed that: The liquid flows along a frictionless horizontal pipe and that losses due to fittings can be neglected, The drop across both instruments is the same. Thus for both meters, the flow is:     Where t he ratio of the pipe to the Venturi area in terms of diameter is 1 2

18.   and t he ratio of the pipe to the orifice area in terms of diameter   Since t he differential pressure across both meters is the same Eq. (1) and (2) give  

19. For the given data,     3 4   Eq. (4) reduces to   From continuity, the total flow Q T is the sum of the flow through the two meters. That is

20. Thus 20% of flow passes through the orifice-meter while the rest (80%) of the flow passes through the Venturi-meter.

21. A cylindrical tank of diameter 1m mounted on its axis contains a liquid which drains through a 2 cm diameter orifice in the base of the tank. If the tank originally contains 1000 liters, determine: The time taken for total drainage “Take the coefficient for the orifice as 0.60. A H H 1 H 2 a orifice dH Flow out

22. A circular water tank of 4 m diameter contains 5 m deep water. An orifice of 400 mm diameter is provided at its bottom. Find the time taken for water level fall from 5 m to 2 m. Take C d = 0.60 . Solution: The area of the tank and The circular orifice The time “T” taken to fall water level from H 1 = 5.0 m to depth H 2 = 2.0 m .   Given: A circular tank with diameter D = 4.0 m Orifice with diameter d orifice = 0.40 m Initial water depth in the tank H 1 = 5.0 m Final water depth in the tank H 2 = 2.0 m.

23. A jet of water of diameter d = 50 mm issues (تدفق) with a velocity of 4.9 m/s from a hole in the vertical side of an open tank which kept filled with water to a height of 1.5 m above the center of the hole. Calculate the reaction of the jet on the tank and its content when: It is stationary, It is moving with a velocity U = 12 m/s in the opposite direction to the jet while the velocity of the jet relative to the tank remains unchanged. Jet Dia. 50 mm & V = 4.9 m/s 1.5 m U

24. 1.0 m Scale Inflow with an area = 0.01 m 2 and V = 8 m/s Area = 0.1 m 2 A container weighting 100 N “empty” has cross– sectional area of 0.1 m 2 and two equal sized openings on opposite sides as shown in the figure. The container is kept on a weighting balance and water flows into it from the top through a jet of 0.01 m 2 area at 8.0 m/s velocity. If under steady flow condition the depth of water in the tank is 1.0 m, determine the scale reading. Ans. (1720N)

25. For the two orifices shown in the figure, determine y 2 such that:   Orifice 2 Orifice 1 X 2 X 1     10 m

26. Consider the jet from the firs orifice,   Since ,   =   =     Similarly, for the jet from the second orifice      

27. Rearranging gives,   Solving for h 2  

28. A weir 36 meters long is divided into 12 equal bays by a vertical posts, each 60 cm wide. Determine the discharge over he weir if the head over the crest is 1.2 m. Number of end contraction, n = 2 x12 = 24 “Each bay has two end contraction” Discharge by using Francis Formula with end contraction “neglecting velocity of approach”   Substituting the given values gives   Given Length of the weir ”L” = 36 m Number of bays “n” = 12 Number of 12 bays, number of vertical post = 11 Width of each post = 60 cm = 0.60 m Effective length = L – 11x 0.60 = 29.4 m, Head on weir = 1.2 m  

29. A B C R = 1 m Tank full of water 1.5 m 0.2 kg f / m 2

30. A B C R = 1 m Tank full of water 1.5 m 2.0 m G H O E F 

31. Extreme suction head (for centrifugal pump) Centrifugal pump suction occurs due to pressure differential between vessel from which pumped medium is taken, and impeller blades. Excessive increase of pressure differential may result in the occurrence of cavitation – process when pressure drops down to value at which fluid boiling temperature lowers below pumped medium temperature and it starts to evaporate in flow space forming multiple bubbles. Bubbles are carried away by stream further downstream where under action of building up pressure they are condensed and collapse, accompanied with multiple hydraulic shocks that negatively tell on pump service life. In order to avoid negative influence of cavitation the suction head of centrifugal pump has to be limited .

32. Pump Suction pipeline Pump impeller axis Pressure pipeline Pressure pipeline Receiving tank Suction head Total suction head Pressure head Intake tank

33. Geometric suction head can be determined by the formula: h г  = ( P -P 1 )/( ρ· g) - h св  - w²/(2·g) - σ· H h Г  – geometric suction head, m P  – intake tank pressure, Pa P 1  – pressure of the impeller blades, Pa ρ – pumped medium density, kg/m 3 g – gravity acceleration, m/s 2 h св  – losses due to overcoming suction pipeline hydraulic resistance, m w²/(2·g) – suction pipeline velocity head, m σ* H – added resistance losses proportional to head, m where σ – cavitation factor, H – head created by pump Cavitation factor can be calculated by the empirical formula: σ = [( n·√Q) / (126H 4/3 )] 4/3 σ – cavitation factor n – impeller rotation speed, s -1 Q – pump performance capacity, m 3 /s Н – created head, m There is also formula for centrifugal pumps for calculation of head margin providing absence of cavitation: H кв  = 0,3·( Q·n²) 2/3  H кв  – head margin, m Q – centrifugal pump performance capacity, m 3 /s n – impeller rotation speed, s  - 1

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35. The theory is to build a dam on a large river that has a large drop in. The dam stores lots of water behind it in the reservoir. Near the bottom of the dam wall there is the water intake. Gravity causes it to fall through the penstock inside the dam. At the end of the penstock there is a turbine, which is turned by the moving water . The shaft from the turbine goes up into the generator, which produces the power. Power lines are connected to the generator that carry electricity to your home and mine. The water continues past the propeller through the tailrace into the river past the dam. By the way, it is not a good idea to be playing in the water right below a dam when water is released!

36. Egypt forms part of the world's most water-scarce region; more than 95% of the country is desert. This satellite photograph shows Lake Nasser (centre) and the Toshka Lakes (centre left).

37. Toshka project

38. The shown figure shows a free overfall in a horizontal frictionless rectangular channel. Assuming the flow to be horizontal at section “1” and the pressure at the brink of section “2” to be atmospheric throughout the depth, prove that:   Where q = discharge per unit width, and   y 2 y 1 q Brink  

39. Open channel flow is covered in essentially all civil and environmental engineering programs, usually by final-year undergraduate or graduate students studying water resources. Developed by an author who has been teaching open channel flow to university students for the past fifteen years,  Fundamentals of Open Channel Flow  provides the students with a detailed explanation of the basics of open channel flow using examples and animation, and offers expert guidance on the practical application of graphical and computational tools. It highlights the practical computational tools students can use to solve problems, such as spreadsheet applications.   It assumes a foundation in fluid mechanics, then adopts a deliberately logical sequence through (تتبنى التسلسل المنطقي عمدا من خلال) energy , momentum, friction, gradually varied flow (first qualitative, then quantitative), and the basics of sediment transport.

40. The Hydraulic Institute defines NPSH as the total suction head in feet absolute, determined at the suction nozzle and corrected to datum, less the vapor pressure of the liquid in feet absolute. Simply stated, it is an analysis of energy conditions on the suction side of a pump to determine if the liquid will vaporize at the lowest pressure point in the pump.  The pressure which a liquid exerts on its surroundings is dependent upon its temperature. This pressure, called vapor pressure, is a unique characteristic of every fluid and increased with increasing temperature. When the vapor pressure within the fluid reaches the pressure of the surrounding medium, the fluid begins to vaporize or boil. The temperature at which this vaporization occurs will decrease as the pressure of the surrounding medium decreases . A liquid increases greatly in volume when it vaporizes. One cubic foot of water at room temperature becomes 1700 cu. ft. of vapor at the same temperature. 

41. It is obvious from the above that if we are to pump a fluid effectively, we must keep it in liquid form. NPSH is simply a measure of the amount of suction head present to prevent this vaporization at the lowest pressure point in the pump . NPSH Required is a function of the pump design. As the liquid passes from the pump suction to the eye of the impeller, the velocity increases and the pressure decreases. There are also pressure losses due to shock and turbulence as the liquid strikes the impeller. The centrifugal force of the impeller vanes further increases the velocity and decreases the pressure of the liquid. The NPSH Required is the positive head in feet absolute required at the pump suction to overcome these pressure drops in the pump and maintain the majority of the liquid above its vapor pressure. The NPSH Required varies with speed and capacity within any particular pump. Pump manufacturer's curves normally provide this information .   Net Positive Suction Head (NPSH) NPSH Available is a function of the system in which the pump operates. It is the excess pressure of the liquid in feet absolute over its vapor pressure as it arrives at the pump suction. Fig. 4 shows four typical suction systems with the NPSH Available formulas applicable to each. It is important to correct for the specific gravity of the liquid and to convert all terms to units of "feet absolute" in using the formulas .

42. Fig. 4 Calculation of system Net Positive Suction Head Available for typical suction conditions.

43. P B  = Barometric pressure in feet absolute.  V P  = Vapor pressure of the liquid at maximum pumping temperature , in feet absolute.  P = Pressure on surface of liquid in closed suction tank, in feet absolute .  L s  = Maximum static suction lift in feet.  L H  = Minimum static suction head in feet.  h f = Friction loss in feet in suction pipe at required capacity In an existing system, the NPSH Available can be determined by a gauge on the pump suction. The following formula applies: NPSH A =P B –V B  Gr + h V Where Gr = Gauge reading at the pump suction expressed in feet (plus if above atmospheric , minus if below atmospheric) corrected to the pump centerline . hv = Velocity head in the suction pipe at the gauge connection, expressed in feet.

44. Cavitation is a term used to describe the phenomenon, which occurs in a pump when there is insufficient NPSH Available. When the pressure of the liquid is reduced to a value equal to or below its vapor pressure the liquid begins to boil and small vapor bubbles or pockets begin to form. As these vapor bubbles move along the impeller vanes to a higher pressure area above the vapor pressure, they rapidly collapse. The collapse, or "implosion" is so rapid that it may be heard as a rumbling noise, as if you were pumping gravel. In high suction energy pumps, the collapses are generally high enough to cause minute pockets of fatigue failure on the impeller vane surfaces. This action may be progressive, and under severe (very high suction energy) conditions can cause serious pitting damage to the impeller.

45. Capacity Capacity (Q) is normally expressed in gallons per minute ( gpm ). Since liquids are essentially incompressible, there is a direct relationship between the capacity in a pipe and the velocity of flow. This relationship is as follows: Where A = area of pipe or conduit in square feet. V = velocity of flow in feet per second. Q = Capacity in gallons per minute NOTE : On vertical pumps the correction should be made to the eye of the suction or lowest impeller.

46. Centrifugal Pump Fundamentals - Head Section   The pressure at any point in a liquid can be thought of as being caused by a vertical column of the liquid which, due to its weight, exerts a pressure equal to the pressure at the point in question. The height of this column is called the static head and is expressed in terms of feet of liquid.  The static head corresponding to any specific pressure is dependent upon the weight of the liquid according to the following formula. A Centrifugal pump imparts velocity to a liquid. This velocity energy is then transformed largely into pressure energy as the liquid leaves the pump. Therefore, the head developed is approximately equal to the velocity energy at the periphery of the impeller This relationship is expressed by the following well-known formula: 

47. Where H  = Total head developed in feet.  v  = Velocity at periphery of impeller in feet per sec.  g  = 32.2 Feet/Sec 2 We can predict the approximate head of any centrifugal pump by calculating the peripheral velocity of the impeller and substituting into the above formula. A handy formula for peripheral velocity is:  D  = Impeller diameter in inches  V  = Velocity in ft./sec  The above demonstrates why we must always think in terms of feet of liquid rather than pressure when working with centrifugal pumps. A given pump with a given impeller diameter and speed will raise a liquid to a certain height regardless of the weight of the liquid, as shown in Fig. 1. 

48. Fig. 1 Identical Pumps Handling Liquids of Different Specific Gravities.

49. All of the forms of energy involved in a liquid flow system can be expressed in terms of feet of liquid. The total of these various heads determines the total system head or the work which a pump must perform in the system. The various forms of head are defined as follows.  SUCTION LIFT exists when the source of supply is below the center line of the pump. Thus the STATIC SUCTION LIFT is the vertical distance in feet from the centerline of the pump to the free level of the liquid to be pumped.   

50. Fig. 2-a Suction Lift ? Showing Static Heads in a Pumping System Where the Pump is Located Above the Suction Tank. (Static Suction Head) SUCTION HEAD exists when the source of supply is above the centerline of the pump. Thus the STATIC SUCTION HEAD is the vertical distance in feet from the centerline of the pump to the free level of the liquid to be pumped.

51. STATIC DISCHARGE HEAD is the vertical distance in feet between the pump centerline and the point of free discharge or the surface of the liquid in the discharge tank. TOTAL STATIC HEAD is the vertical distance in feet between the free level of the source of supply and the point of free discharge or the free surface of the discharge liquid. FRICTION HEAD ( hf ) is the head required to overcome the resistance to flow in the pipe and fittings. It is dependent upon the size, condition and type of pipe, number and type of pipe fittings, flow rate, and nature of the liquid. Frictional tables are included in Water Data. VELOCITY HEAD ( hv ) is the energy of a liquid as a result of its motion at some velocity V. It is the equivalent head in feet through which the water would have to fall to acquire the same velocity, or in other words, the head necessary to accelerate the water. Velocity head can be calculated from the following formula: The velocity head is usually insignificant and can be ignored in most high head systems. However, it can be a large factor and must be considered in low head systems.

52. PRESSURE HEAD must be considered when a pumping system either begins or terminates in a tank which is under some pressure other than atmospheric. The pressure in such a tank must first be converted to feet of liquid. A vacuum in the suction tank or a positive pressure in the discharge tank must be added to the system head, whereas a positive pressure in the suction tank or vacuum in the dis-charge tank would be subtracted. The following is a handy formula for converting inches of mercury vacuum into feet of liquid. Vacuum, feet of liquid = (Vacuum, in of Hg x 1.13)/ Sp. Gr. The above forms of head, namely static, friction, velocity, and pressure, are combined to make up the total system head at any particular flow rate. Following are definitions of these combined or "Dynamic" head terms as they apply to the pump. TOTAL DYNAMIC SUCTION LIFT ( hs ) is the static suction lift minus the velocity head at the pump suction flange plus the total friction head in the suction line. The total dynamic suction lift, as determined on pump test, is the reading of a gauge on the suction flange, converted to feet of liquid and corrected to the pump centerline*, minus the velocity head at the point of gauge attachment.

53. TOTAL DYNAMIC SUCTION HEAD  ( h s ) is the static suction head plus the velocity head at the pump suction flange minus the total friction head in the suction line. The total dynamic suction head, as determined on pump test, is the reading of the gauge on the suction flange, converted to feet of liquid and corrected to the pump centerline*, plus the velocity head at the point of gauge attachment.  TOTAL DYNAMIC DISCHARGE HEAD  ( h d ) is the static discharge head plus the velocity head at the pump discharge flange plus the total friction head in the discharge line. The total dynamic discharge head, as determined on pump test, is the reading of a gauge at the discharge flange, converted to feet of liquid and corrected to the pump centerline*, plus the velocity head at the point of gauge attachment.  TOTAL HEAD  ( H) or TOTAL Dynamic HEAD (TDH) is the total dynamic discharge head minus the total dynamic suction head or  TDH = h d + h s  (with a suction lift)  TDH = h d - h s  (with a suction head)

54. Representation of static discharge head, static suction lift and total static head

55. Cavitation and NPSH If the suction pressure is only slightly grater than the vapor pressure, some liquid may flash to vapor inside the pump. This process is called as “Cavitation” which greatly reduce the pump capacity and causes severe erosion. If the suction pressure is actually less than the vapor pressure there will be vaporization in the suction line and no liquid can be drawn into the pump. To avoid the cavitation the pressure at pump inlet must exceed the vapor pressure by a certain value called as the N et P ositive S uction H ead. The required recommended value of NPSH is about 2 to 3m (5 to 10ft) for small centrifugal pumps but it increases to 15m (50ft) for larger pumps.

56. The figure above shows a vertical circular gate in a 3-m diameter tunnel with water on one side and air on the other side.  

57. A container weighting 100 N “empty” has cross – sectional area of 0.1 m 2 and two equal sized openings on opposite sides as shown in the figure. The container is kept on a weighting balance and water flows into it from the top through a jet of 0.01 m 2 area at 8.0 m/s velocity. If under steady flow condition the depth of water in the tank is 1.0 m, determine the scale reading.

58. Discharge striking the vane = 0.0554 m 3 /s

59. A Pelton turbine develops 3000 kw under a head of 300m. The overall efficiency of the turbine is 83%. If the speed ratio = 0.46, C v = 0.98 and specific speed is 16.5, then find: Diameter of the turbine, and Diameter of the jet. Solution Given: Power, Net head, , Overall efficiency Speed ratio C v value, Specific speed  

60. Using equation, The velocity “ V n ” at the outlet of nozzle is given by: = Since the speed ratio The speed of the bucket ratio  Diameter of the wheel “D”  

61. To determine the nozzle diameter, let Q = Discharge through the turbine in m 3 /s, which can be calculated using, Substituting the given values To find the jet diameter, =  

62. In turbines, water enters from very high head i.e. very high pressure . As the water passes through turbine vanes, it gives its energy to vanes and pressure of water decreases to sub-atmospheric values so that we can have high energy output because of higher pressure difference across turbine. Now we just can't make turbine open to atm. because of low pressures otherwise air entrapment could cause problems like cavitation . Now in order to again have water exiting at atmospheric pressure, we need to raise water pressure. To achieve this, we use draft tube and because of its tapering outward shape, it acts as diffuser since area of flow increases. Thus we have water coming out from turbine at atm. pressure . Another function of draft tube is that it also acts as collector for water coming out of turbine and directs it to reservoir.

63. Surge Tank Closed aqueduct to absorb sudden pressure rise and provide during pressure drop It is mostly installed when distance between dam and power house is large Functions When load decreases water moves backward & get stored, Addition supply of water will be provided, when load increases, Avoid vacuum during water supply closed.