2. (5) Why is the critical Reynolds number for the transition to turbulent flow only 500 for open channels, but 2000 for pipes flowing full?

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1 CEE 4 Fnal Exam, Aut pts. possble. Some nformaton that you mght fnd useful s provded below, and a Moody dagram s attached at the end of the exam. Propertes of Water ρ = 1000 kg/m = 1.94 slugs/ft γ = 9800 N/m = 6.4 lb/ft µ 70F = 1.00 x 10 N-s/m =.05 x 10 5 lb-s/ft ν 70F = 1.00 x 10 6 m /s = 1.06 x 10 5 ft /s p vap =.4 kpa = 0.06 ps Constants, Conversons, etc. g = 9.81 m/s =. ft/s 1 kw = 1000 N-m/s p atm = kpa = 14.7 ps 1 hp = 550 ft-lb/s Area Volume Centrod I c Rectangle bh h/ bh /1 Trangle bh/ h/ bh /6 Crcle π d /4 d/ π d 4 /64 Sphere π d π d / 6 d/ Cone π d h/ 1 h/4 (from flat surface of cone) 1. (10) Dscuss brefly the smlartes and dfferences between the Darcy- Wesbach and Hazen-Wllams equatons. Focus on when and how the equatons are used, not on the mathematcal dfferences.. (5) Why s the crtcal Reynolds number for the transton to turbulent flow only 500 for open channels, but 000 for ppes flowng full?. (15) Calculate the horzontal component of the force on the sold, concal plug shown below. 1

2 4. (15) Calculate the horzontal force of the free jet on the crcular plate shown n the followng schematc. The flow rate s 4.5 L/s, the jet s strkng the fxed, crcular plate, and the pressure gage s measurng the pressure at the centerlne of where the flow strkes the plate. 5. (0) Calculate the horsepower that the pump must supply to pump the water at a flow rate of.5 ft /s n the system shown below, gnorng headlosses other than those through the ppes. The ppes have an absolute roughness of 6.67x10 4 ft. Draw the HGL and EL on the dagram, from plane A to plane B. In drawng ths sketch, you should consder mnor headlosses (.e., headlosses caused by processes other than frcton wth the ppe walls). Draw and label the lnes clearly, and dentfy all locatons where both lnes are at the same elevaton. Do you expect the pressure to be negatve anywhere n the system?

3 B El. 150 ft A El. 15 ft 000 ft; 8 n. El. 50 ft El. 40 ft 1000 ft; 8 n. P 6. (0) Calculate the flow rate from ths water tank f the 6-n. ppelne has a frcton factor of 0.00 and s 50 ft long, and the water s at 70 o F. The water n the tank s 5 ft deep. The mnor loss coeffcents (k) for headlosses at the sharpedged entrance to the ppe and at the submerged dscharge are 0.5 and 1.0, respectvely. Where s the locaton of lowest pressure n the system? Mght cavtaton be a problem at ths locaton?

4 7. (15) In the flow network shown below, Loop I ncludes ppes 1,, and ; Loop II ncludes ppes, 4, and 5; and Loop III ncludes ppes, 6, and 7 (ppe numbers are ndcated by # n the dagram). A frst guess has been made for the flow rate n each ppe (n cfs) and s shown on the dagram; the correspondng frctonal headlosses have been calculated usng the Darcy-Wesbach equaton and are summarzed below. What should the next guesses be for the drecton and magntude of the flows n ppes and? Ppe Q h f

5 CIVE 4 Aut 004 Fnal Exam Solutons 1. The DW and HW equatons both relate headloss n full-flowng conduts to geometrc factors, flud propertes, and operatng parameters. The DW equaton apples to any flud flowng wth any velocty (.e., any Reynolds number). On the other hand, the HW equaton s specfcally for water, n turbulent flow under for a lmted range of Reynolds numbers that correspond to condtons typcal for water supply ppes.. In open channel flow, the Reynolds number s typcally defned as R h Vρ/µ, whereas n crcular ppes flowng full, t s defned as DVρ/µ. Snce the hydraulc radus s one-fourth the dameter of a crcular ppe, a Reynolds number of 500 n an open channel corresponds to a Reynolds number of 000 n a full ppe. Thus, the reason for the dfferent Reynolds numbers at the transton from lamnar to turbulent flow s the dfferent ways that the Reynolds number s defned; the actual propertes of the flow at the transton are (vrtually) dentcal n the two types of systems.. The net horzontal force on the cone equals the horzontal force to the rght on the flat, vertcal (crcular) surface mnus the horzontal force to the left on the concal sdes. The horzontal force on any dfferental area of the concal sdes equals the force on a projecton of that area on a vertcal plane. Therefore, that force exactly balances the force on the correspondng vertcal area on the flat surface (.e., at the same depth). As a result, the net horzontal force on the cone s the force on the porton of the flat sde of the cone that has only ar, and not water, to the rght. Ths area s a crcle wth a dameter of ft. The average pressure on the crcle s γ h c, so: ( ft) lb F = pca= γhca= 6.4 ( 9 ft) π =,970 lb ft 4 4. At the plate, the velocty head s converted entrely to pressure head. The pressure before the water hts the plate s zero, and s the velocty at the pont where the pressure s beng measured. In addton, the elevaton s dentcal at those two locatons. Therefore, usng Bernoull s equaton, we can fnd the upstream velocty, and then use ths value n conjuncton wth the momentum equaton to fnd the force on the plate: p 1 γ + z 1 + v1 p g = γ + z + v g gp 9.81 m/s 690,000 Pa v1 = = = 7. m/s γ 9810 kg/m 5

6 ( 4.5 L/s)( 1000 kg/m )( 7. m/s) F = Qρv= = 1579 N to the rght 1000 L/m 5. To fnd the requred horsepower, we need to know the headloss n the ppe. We can fnd the headloss wth the DW equaton, f we frst estmate the frcton factor from ether the Moody dagram or the Haaland equaton (assumng the flow s turbulent). In ether case, to estmate f, we need to know the Reynolds number, whch we can fnd as follows: Q.5 ft /s 1 V = 7.16 ft/s π d /4 = π 8/1 ft /4 = ( 7.16 ft/s)( 8/1 ft) 5 Vd Re = = = 4.50x10 ν x10 ft /s The value of e/d can be computed drectly from the gven nformaton: e d x10 ft = = /1 ft For these values of Re and e/d, the frcton factor can be found from the Moody dagram as approxmately The two ppes have dentcal dameter, roughness, and velocty, so the headloss per unt length s the same n them, and the total headloss can be computed based on ther combned length of 000 ft. The total headloss through the two ppe sectons s therefore: ( 7.16 ft/s) LV 000 ft hl = f = 0.01 = 75. ft d g 8/1 ft. ft/s The total head that the pump must supply equals the amount of head lost due to frcton and the elevaton gan of 100 ft, for a total of 175. ft. The power requrement s therefore: lb ft 1 hp P= γ Q h tot = ( 175. ft) = 49.7 hp ft s 550 ft-lb/s The HGL begns at the water surface and drops below the surface as soon as the water enters the ppe (due to the energy loss at the entrance and the ncreased velocty). The drop n the HGL due to the velocty ncrease s V /g, or: V 7.16 ft/s 0.80 ft g =. ft/s = 6

7 If the entrance has a sharp edge, an addtonal one-half velocty head (~0.4 ft) s lost. From there, the HGL drops farther as the water flows toward the pump, due to frcton. The frctonal loss n ths porton of the ppe s one-thrd of the total h L (1000 ft/000 ft), or 5.1 ft, so the HGL s 6. ft (computed as ) below the reservor surface when the water reaches the pump. At the pump, 175. ft of head s added to the water, so the HGL rses to approxmately 50.1 ft above the second (upper) reservor. From there, t drops steadly to the surface of the second reservor. The EL s at the same level as the HGL n the frst reservor and drops when the water frst enters the ppe, but not as much as the HGL (snce the only drop n the EL s due to the entrance effect)., so t s 0.80 ft above the HGL at that pont. From there on, t remans slghtly 0.8 ft above the HGL untl the water dscharges nto the upper reservor. At that pont, energy s dsspated and the EL drops to the level of the reservor, whch s also the level of the HGL. The HGL and EL concde at both reservor surfaces, slghtly away from the regon where the water enters/leaves. Snce the ppe s just 10 ft below the surface of the lower reservor, the HGL drops substantally below the ppe approachng the pump, so there s negatve pressure n approxmately the latter two-thrds of ths ppe. HGL EL P 7

8 6. Applyng the energy equaton between the surfaces of the two pools of water, we fnd: p 1 γ + V1 z + p 1 = g γ V + z + + hl g z1 z = hl The headloss s: so: V LV V hl = hl, entrance + hl, ppe + hl, ext = f g d g g 50 ft V 50 ft = /1 ft. ft/s. ft/s 50 ft V = = 0. ft/s 50 ft /1 ft The pressure s zero at the top of the upper pool of water and then ncreases as the elevaton decreases. At the entrance to the ppe, some pressure head s lost to frcton, and some s converted to velocty head. The amounts of pressure head converted to velocty head and lost due to frcton at that pont are: ( 0. ft/s) Velocty head = = 14.6 ft. ft/s V hl, entrance = 0.5 = 0.5( 14.6 ft) = 7.1 ft g Total loss of pressure head = ( ) ft = 1.9 ft Because the pressure head s 5 ft before the water enters the ppe, and t drops by 1.9 ft mmedately thereafter, the net pressure head s 16.9 ft just nsde the ppe, at the top. The headloss due to frcton n the ppe s: LV 50 ft = = 0.0 ( 14.6 ft) = 8.5 ft 0.5 ft hlppe, f d g 8

9 Ths frctonal headloss corresponds to a loss of a lttle more than 0.5 ft/ft. Snce the water s smultaneously ganng 1 ft of pressure head per ft traveled due to the elevaton loss, the net pressure ncreases as elevaton decreases. Therefore, the pont of mnmum energy s just after the water has entered the ppe. The pressure head at that pont s 16.9 ft, as noted above. Convertng ths value to an absolute pressure, we fnd: p p = p + p = γ + p γ mn mn, absolute mn, gage atm atm lb 1 ft = 6.4 ( 16.9 ft) 14.7 ps 7.60 ps + =+ ft 144 n Snce the vapor pressure of water at 70 o F s 0.06 ps, cavtaton wll not occur. 7. Ppe # s a part of all three loops, and ppe # s a part of loop I only. The correctons to the flow rates n the three loops can therefore be computed as follows: hl, QI = = = hl, Q 1 hl, QII = = = 1.50 hl, Q hl, QIII = = = hl, Q Defnng postve flow n ppe # to be from rght to left (clockwse for loop #1), the correcton to the flow n that ppe equals: Q =+ QI QII QIII = = Defnng postve flow n ppe # to be from lower left to upper rght (also clockwse for loop #1), the correcton to the flow n that ppe equals: Q =+ Q I =

10 The new guesses for the flow rates are therefore: In ppe : In ppe : = cfs (flow from rght to left) =.9 cfs (flow from upper rght to lower left) 10