Problem set #6 1) Water flows by gravity from one lake to another as sketched in Figure 1 at the steady rate of 82 gpm. (a) What is the loss in available energy associated with this flow? (b) If this same amount of loss is associated with pumping the fluid from the lower lake to the higher one at the same flowrate, estimate the amount of pumping power required. Figure 1 energy loss = gΔz = 32.17 ft ft lb 47 ft = 1511 2 s slug (b) Energy required to pump at the same rate counter to this flow. Ws = 1.94 slug ft 3 ft lb ft lb 0.182 32.17 47+1511 =1067 3 ft sec slug sec 2) Water is to be pumped from the large tank shown in Figure with an exit velocity of 5.5 m/s. It was determined that the original pump (pump 1) that supplies 1 kW of power to the water did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How much power must pump 2 add to the water? The head loss for this flow is hL = 250Q2, where hL is in m when Q is in m3/s. Wp =Wp1 +Wp2 Wp2 =Wp Wp1 = 2.258 kW 1 kW = 1.258 kW 3) Oil (SAE 30) at 15.6oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the channel is 35 kPa/m, and the distance between the plates is 4 mm. The flow is laminar. Determine: (a) the volume rate of flow (per meter of width), (b) the magnitude of the shearing stress acting on the bottom plate, and (c) the velocity along the centerline of the channel. 2 .002 m N m2 q= 35000 3 = 4.912 104 N s m s 3 0.38 2 m 3 (b) Tyx = p Δp N N y= h = 35000 3 .002 m = 70 2 x l m m (c) 3 q 3 q 3 vmax = V,V = vmax = = 2 2h 2 2h 2 m2 s = 0.372 m 2 .002 m s 4.912 104 4) Two fixed, horizontal, parallel plates are spaced 0.6 in. apart. A viscous liquid (μ = 8 × 10-3 lb•s/ft2, SG = 0.9) flows between the plates with a mean velocity of 0.6 ft/s. The flow is laminar. (a) Determine the pressure drop per foot in the direction of flow. (b) What is the maximum velocity in the channel? (a) Δp 3μv = 2 = l h lb s ft .6 2 ft s = 23.04 lb 2 ft 3 .3 f t 12 3 8 103 (b) 3 3 ft ft vmax = V = .6 = .9 2 2 s s 5) A viscous fluid (specific weight = 80 lb/ft3; viscosity = 0.03 lb•s/ft2) is contained between two infinite, horizontal parallel plates as shown in the Figure. The fluid moves between the plates under the action of a pressure gradient, and the upper plate moves with a velocity U while the bottom plate is fixed. A U-tube manometer connected between two points along the bottom indicates a differential reading of 0.1 in. If the upper plate moves with a velocity of 0.03 ft/s, at what distance from the bottom plate does the maximum velocity in the gap between the two plates occur? Assume laminar flow. lb s ft 1 .03 ft 2 ft s + 12 = .0741 ft .0741 12 ft = .8887 in yumax = 1 lb 2 1 in ft .3334 3 12 ft 6) A vertical shaft passes through a bearing and is lubricated with an oil having a viscosity of 0.2 N•s/m2 as shown in the Figure. Assume that the flow characteristics in the gap between the shaft and bearing are the same as those for laminar flow between infinite parallel plates with zero pressure gradient in the direction of flow. Estimate the torque required to overcome viscous resistance when the shaft is turning at 90 rev/min. .03 l .075 T = 2πri3 μω = 2π m b 2 3 .160 m N s = 0.399 Nm .2 2 3π -3 m .25×10 m 7) A viscous liquid (μ = 0.012 lb•s/ft2, ρ = 1.79 slugs/ft3) flows through the annular space between two horizontal, fixed, concentric cylinders. If the radius of the inner cylinder is 3.5 in. and the radius of the outer cylinder is 4.5 in., what is the pressure drop along the axis of the annulus per foot when the volume flowrate is 0.15 ft3/s? 8μQ Δp π = 2 l ro2 ri 2 4 4 ro ri r ln o ri lb s ft 3 .15 ft 2 sec lb π = = 17.80 2 2 2 2 ft 4.5 3.5 ft - ft 4 4 12 12 4.5 3.5 ft- ft4.5 in 12 12 ln 3.5 in 8 .012 8) A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pipe, by how much does the presence of the wire reduce the flowrate if (a) d/D = 0.4; (b) d/D = 0.05? (a) 2 1 .4 2 4 = 0.7957 = 79.57% flow reduction = 1 1 .4 + ln .4 (b) 2 1 .052 4 = 0.3321 = 33.21% flow reduction = 1 1 .05 + ln .05