This **preview** has intentionally **blurred** parts. Sign up to view the full document

**Unformatted Document Excerpt**

1. A tank is filled with seawater to a depth of 12 ft. If the specific gravity of seawater is 1.03 and the atmospheric pressure at this location is 14.8 psi, the absolute pressure (psi) at the bottom of the tank is most nearly A. 5.4 B. 20.2 C. 26.8 D. 27.2 Hint: Absolute pressure = gage pressure + atmospheric pressure where, gage pressure = zh and z = specific weight of seawater = SG x w Solution: z = 1.03 (62.4 lb/cu ft) = 64.27 lb/cu ft Gage pressure at the bottom of the tank = (12 ft) x (64.37 lb/cu ft) = 772.44 lb/sq ft = (772.44 lb/sq ft)/(144 sq in/sq ft) = 5.36 psi Absolute pressure = 14.8 psi + 5.36 psi= 20.16 psi Therefore, the key is (B). 2. An open tank contains brine to a depth of 2 m and a 3-m layer of oil on top of the brine. Density of brine is 1,030 kg/m3 and the density of oil is 880 kg/m3. The gage pressure (kPa) at the bottom of the tank is most nearly A. 4.7 B. 20.2 C. 25.6 D. 46.1 Hint: Pressure due to each layer of liquid = h where h is the height of the liquid layer and is the specific weight of the liquid. Solution: Since it is required to determine the gage pressure, pressure at the free surface = 0 Pressure due to the oil layer = (3 m) {(880 kg/m3)(9.81 m/s2)}/1,000= 25.9 kPa Pressure due to the brine layer= (2 m) {(1,030 kg/m3) (9.81 m/s2)}/1,000= 20.2 kPa Hence pressure at the bottom= 25.9 + 20.2 = 46.1 kP Note: 1 kg-ms2 = 1 N; 1 N/m2 = 1 Pa Therefore, the key is (D). 3 The figure shows two cylinders of diameter D and 2D, connected to each other and . containing an incompressible fluid. The two cylinders are fitted with leak-proof pistons of weight W1 and W2 as shown. Which of the following is a correct expression? A. W2 = W1/2 B. W2 = W1 C. W2 = 2 W1 D. W2 = 4 W1 Hint: The pressure is the same in both cylinders. Pressure = Weight/area Solution: Considering the weight W1, pressure Considering the weight W2, pressure Equating the two, Hence, W2 = 4 W1 Therefore, the key is (D). 4. The figure shows the relationship between shear stress and velocity gradient for two fluids, A and B. Which of the following is a true statement? A. Absolute viscosity of A is greater than that of B B. Absolute viscosity of A is less than that of B C. Kinematic viscosity of A is greater than that of B D. Kinematic viscosity of A is less than that of B Hint: By definition, absolute viscosity Thus, slope of the lines in the plot is absolute viscosity. Kinematic viscosity = absolute viscosity/density. Solution: Since the slope of the line for A is greater than that for B, viscosity of A is greater than that of B. Therefore, the key is (A). 5. A flat plate is sliding at a constant velocity of 5 m/s on a large horizontal table. A thin layer of oil (of absolute viscosity = 0.40 N-s/m2) separates the plat from the table. To limit the shear stress in the oil layer to 1 kPa, the thickness of the oil film (mm) should be most nearly A. 0.2 B. 1.6 C. 2.0 D. 3.5 Hint: By definition, absolute viscosity, where, velocity gradient = DU/d, DU = difference in velocity across the oil film; and d = the thickness of the oil film. Solution: DU = (5 0) m/s = 5 m/s Hence, Therefore, the key is (C). 6. A 2-in. diameter shaft is supported by two sleeves, each of length = 2 in. as shown. The internal diameter of the sleeves is 2.1 in. The radial space between the shaft and the sleeves is filled with an oil of viscosity = 8x10-3 lb-s/ft2. If the shaft is rotated at a speed of 600 rpm, the viscous torque (ftlb) on the shaft is most nearly A. 0.15 B. 0.64 C. 3.20 D. 6.40 Hint: Torque, T = (Shear force, F) x (Radius, R) Shear force = (Shear stress, t) x (Area, A) Shear stress, t = dU/dy Solution: DU/Dy = R1w/(R2 R1) = {(1/12 ft) (600 x 2p/60)}/(1.05/12 1.0/12)= 1256.64 s-1 Shear stress, t = (8x10-3 lb-s/ ft2)(1256.64 s-1)= 10.05 lb/sq ft Shear force per sleeve = t x A = t x [(2pR1)(L)] = (10.05 lb/sq ft) [(2p(1/12)(2/12)]= 0.88 lb Torque per sleeve= (0.88 lb) (1/12 ft)= 0.07 ft-lb Torque for 2 sleeves= 0.15 ft-lb Therefore, the key is (A). 7. A 2-in. diameter cylinder is floating vertically in seawater with 75% of its volume submerged. If the specific gravity of seawater is 1.03, the specific weight (lb/cu ft) of the cylinder is most nearly A. 48.2 B. 64.2 C. 83.2 D. 85.7 Hint: When a body is floating in a fluid, weight of the body, W (= V b ) is balanced by the buoyancy force, Fb (= Vd f) where, V = volume of the body b = specific weight of the body Vd= volume displaced f = specific weight of the fluid Solution: Equating W to Fb, V b = V d f b = (Vd) f where, f = 1.03 gw = 1.03 (62.4 lb/cu ft) = 64.27 lb/cu ft Hence, b = 0.75 (64.27 lb/cu ft) = 48.20 lb/cu ft Therefore, the key is (A). 8. A clean glass tube is to be selected in the design of a manometer to measure the pressure of kerosene. Specific gravity of kerosene = 0.82 and surface tension of kerosene = 0.025 N/m. If the capillary rise is to be limited to 1 mm, the smallest diameter (cm) of the glass tube should be most nearly A. 1.25 B. 1.50 C. 1.75 D. 2.00 Hint: The capillary rise, where, s = surface tension of the fluid; b = angle of contact; g = specific weight of the fluid; d = diameter of tube. Solution: Rearranging the equation for the capillary rise and substituting the given data, Therefore, the key is (A). 9. An object weighs 275 N when fully immersed in water and 325 N when fully immersed in oil of specific gravity 0.9. The volume of the object (m3) is most nearly A. 0.02 B. 0.05 C. 0.20 D. 0.50 Hint: When an object is fully immersed in a fluid, the apparent weight = W Fb, where, W = true weight of the object Fb = buoyancy force = Vf V= volume of object f = specific weight of the fluid Solution: When the object is fully immersed in water, 275 N = W Vgw or, W = 275 N + Vgw(1) When the object is fully immersed in oil, 325 N = W V or, W = 325 N + Vgoil (2) Subtracting (1) from (2), 0 = 50 N V(gw goil ) or, Therefore, the key is (B). 10. A block of volume V and specific gravity, SG, is anchored by a light cable to the bottom of a lake as shown. If the specific weight of the water in the lake is gw, the tension, T, in the cable is given by A. B. C. D. Hint: The force balance on the block is: Fb = T + W Fb = Vw W = weight of the block= Vb= V(SG w) Solution: From the force balance, T = Fb W= Vw V(SG w) Hence, T = Vw (1 SG) Therefore, the key is (C). 1 When a uniform flat plate is placed horizontally at a depth of h as shown in Figure 1 1, the magnitude of the force exerted by the fluid on the plate is 20 kN. When this . plate is tilted about its center of gravity through 30 as shown in Figure 2, the magnitude of the force (kN) exerted by the fluid on the plate is most nearly A. 20 sin 30 B. 20 cos 30 C. 20 tan 30 D. 20 Hint: The force balance on a flat plate = f A dc Where, f is the specific weight of the fluid; A is the area of the plate; and dc is the depth of center of g Solution: Since the depth of center of gravity is the same in both cases, and the area is the same, the m Therefore, the key is (D). 1 The figure shows an L-shaped gate ABC hinged at B. Ignoring the weight of the 2 gate, the value of h (m) when the gate is about to open is most nearly . A. 2.0 B. 2.5 C. 3.0 D. 3.5 Hint: Let the hydraulic force on side AB = F1 and the hydraulic force on side BC = F the moment of F1 about B should equal the moment of F2 about B. Hydraulic force on any surface = A = area of the surface; dc = depth of center of gravity of the surface. Solution: Considering unit width of the gate, F1 = (9.81 kN/m3)(2 m x 1 m) (h m) = 19.62h kN and, F2 = (9.81 kN/m3)(h m x 1 m) (h/2 m) = 4.9h2 kN Moment of F1 about B = F1 x 1 m = (19.62h kN) (1 m) = 19.62h kN-m Moment of F2 about B = F2 x h/3 = (4.9h2 kN) (h/3 m) = 1.63h3 kN-m Equating the above two, 19.62h kN-m = 1.63h3 kN-m Hence, h = (19.62/1.63) Therefore, the key is (D). = 3.47 m 1 The figure shows a gate made of a uniform plate of height 4 m and width 1 m, 3 hinged about it a horizontal axis through its center of gravity, G. Depth of water on . the upstream side of the plate, h is 3 m. The magnitude of the minimum force (kN) that must be applied at A to keep the gate in the vertical position is most nearly A. 22 B. 26 C. 41 D. 66 Hint: The force at A will be minimum when it is applied horizontally. The moment of this force about to the hydraulic force on the plate. Hydraulic force on the plate = Adc where, A = area of the submer is the depth of center of gravity of the submerged portion. Solution: Hydraulic force on gate, F = (3 m x 1 m)(9.8 kN/m3)(3 m/2) = 44.1 kN Moment of the hydraulic force about the hinge, F x (2 m h/3) = 44.1 kN-m Hence the minimum force at A to hold the plate in the vertical position = 44.1 kN-m/2 m = 22.05 kN. Therefore, the key is (A). 14. The figure shows the cross section of a dam. Assume the specific weight of water as 62.4 lb/cu ft. Considering unit width of the dam, the maximum moment of the hydraulic force on the dam about point P (ft-lb) is most nearly A. 0.8 x 106 B. 1.3 x 106 C. 2.6 x 106 D. 3.9 x 106 Hint: The moment will be maximum when the water depth rises to the top of the dam. Moment about P = Force, F x distance of point of action from P, y Force, F = f A dc where, f = specific weight of fluid = 62.4 lb/cu ft A = submerged area dc = depth of center of gravity of submerged area Distance from P, y = 1/3 of submerged depth Solution: Considering unit width of the dam, A = 50 ft x 1 ft = 50 ft2 dc of submerged depth = 1/2 (50 ft)= 25 ft Maximum hydraulic force on the dam, F= (62.4 lb/cu ft)(50 ft2) (25 ft)= 78,000 lb Distance of point of action from P= 1/3 (50 ft)= 16.67 ft Maximum moment= (78,000 lb) (16.67 ft)= 1.3 x 106 ft-lb Therefore, the key is (B). 1 The figure shows two chambers A and B, where two manometers are being used to 5 measure the pressure. The manometeric fluid is the same in both manometers, with . a specific gravity of 1.4. When ha = 6 cm and hb = 4 cm, the gage pressure in chamber A (kPa) is most nearly A. 0.98 B. 1.37 C. 98 D. 137 Hint: Let pA = gage pressure in chamber A and pB = gage pressure in chamber B. Relate p deflection, ha, and relate pB to patm using manometric deflection, hb. Solution: pB = pA ha m = pA ha (SG w) patm = pB hb m = pB hb (SG w) Combing the above two equations, pA - patm = pA,gage = (ha + hb)m = (ha + hb)(SG w Therefore, the key is (B). 16. A manometer is being used as shown to measure the pressure in a pressurized tank. The tank is partially filled to a depth of 25 cm with a fluid of specific gravity (SG) = 0.78. The specific gravity (SG) of the manometric gage fluid is 3.5. The gage pressure in the headspace (kPa) when h = 8 cm is most nearly A. 0.02 B. 0.22 C. 2.21 D. 22.15 Hint: Let the gage pressure in headspace to be p. The pressure at depth h = p at the surface + h where is the specific weight of the fluid. Find a horizontal plane and equate the pressure exerted by the two fluids at that plane. Solution: With the interface of the two fluids as a reference plane, considering the fluid in tank, Pressure at the interface = p + (0.78 x 9.8 kN/m3) (25 cm + h cm)/100 Considering the gage fluid, Pressure at the interface = patm. + (3.5 x 9.8 kN/m3) (h cm)/100 Equating the two, with patm. = 0 p = (3.5 x 9.8 kN/m3) (h cm)/100 - (0.78 x 9.8 kN/m3) (25 cm + h cm)/100 substituting h = 8/100 m, p = 0.22 kPa. Therefore, the key is (B). 1 An inclined-tube manometer is being used as shown to measure the pressure in a 7 pressurized tank. The tank is partially filled to a depth of 20 cm with a fluid of . specific gravity = 0.78. The specific gravity of the manometric gage fluid is 3.5. The gage pressure in the headspace (kPa) when h = 8 cm is most nearly A. 0.3 B. 0.9 C. 1.2 D. 1.8 Hint: Let the gage pressure in headspace to be p. Find the pressure at the interface in terms of the heig of the gage fluid and equate that to the pressure in the tank at that horizontal level to find p. Solution: Gage pressure at the interface due to gage fluid = ggage (h sin 30 ) = (3.5 x 9.8 kN/m3)(0.08 p + [(20 6)/100 m](0.78 x 9.8 kN/m3) = 1.372 kPa Therefore, p = 1.372 kPa [(20 6 Therefore, the key is (A). 18. For a body partially submerged in a fluid and at equilibrium, which of the following is a true statement? A. The weight of the body is equal to the weight of the volume of fluid displaced B. The weight of the body is less than the weight of the volume of fluid displaced C. The weight of the body is greater than the weight of the volume of fluid displaced D. The specific gravity of the body is greater than the specific gravity of the fluid Hint: When a body is partially or fully submerged in a fluid, the fluid exerts a vertical force on the body known as the buoyancy force. This buoyancy force is equal to the weight of the fluid volume displaced by the body. Solution: Since the body is at equilibrium, the weight of the body should equal the buoyancy force, which in turn is equal to the weight of the volume of fluid displaced. Thus, the weight of the body is equal to the weight of the volume of fluid displaced. Therefore, the key is (A). 19. An open separation tank contains brine to a depth of 2 m and a 3-m layer of oil on top of the brine. A uniform sphere is floating with at the brine-oil interface with 80% of its volume submerged in brine. Density of brine is 1,030 kg/m3 and the density of oil is 880 kg/m3. The density of the sphere (kg/ m3) is most nearly A. 825 B. 910 C. 955 D. 1,000 Hint: When a body is at the interface of two fluids, the buoyancy force equals the sum of the weights of the volumes of the fluids displaced by the body. At equilibrium, the weight of the body equals the total buoyancy force. Solution: Let V be the volume of the sphere, Vd be the displaced volume, and the specific weight of the fluid = g. Buoyancy force due to brine, Fb= Vd = (80% of V) ( 1,030 kg/m3 x g) Buoyancy force due to oil, Fo= Vd = (20% of V) ( 880 kg/m3 x g) Weight of sphere, W= V = V Equating W to (Fb + Fo), V = (80% of V) ( 1,030 kg/m3 x g) + (20% of V) ( 880 kg/m3 x g) = 0.8 ( 1,030 kg/m3 ) + 0.2 ( 880 kg/m3) = 1,000 kg/m3 Therefore, the key is (D). 20. At a certain section in a pipeline, a reducer is used to reduce the diameter from 2D gradually to diameter D. When an incompressible fluid flows through this pipeline, the velocity is U1 in the first section and U2 in the second section. Which of the following is a true conclusion? A. U2 = 4U1 B. U2 = 2U1 C. U2 = U1/2 D. U2 = U1/4 Hint: From continuity equation, flow rate before reduction = flow after reduction Flow rate = Area of flow x Velocity of flow Solution: Let Q be the flow rate. From continuity equation, Q = A1 U1 = A2 U2 where, A1 = area before reduction= /4(2D)2= D2 and, A2 = area after reduction= /4(D)2= D2/4 Thus, U2 = (A1/ A2) U1 = ( D2/ D2/4) U1 = 4 U1 Therefore, the key is (A). 21. When a Newtonian fluid flows under steady, laminar condition through a circular pipe of constant diameter, which of the following is NOT a correct conclusion? A. The shear stress at the centerline of the pipe is zero B. The maximum velocity at a section is twice the average velocity at that section C. The velocity will decrease along the length of the pipe D. The velocity gradient at the centerline of the pipe is zero Hint: Under laminar flow in circular pipes, the velocity distribution is parabolic and symmetrical about the centerline. In addition, the continuity equation also applies. Solution: Due to the symmetrical velocity distribution, velocity gradient at the centerline is zero. Hence, the shear stress at the centerline is also zero. From the parabolic velocity distribution, Vmax = 2 Vave. Since the pipe diameter is constant, by continuity equation- Q = AV, velocity should remain constant along the length of the pipe. Therefore, the key is (C). 22. A Newtonian fluid flows under steady, laminar conditions through a circular pipe of diameter 0.16 m at a volumetric rate of 0.05 m3/s. Under these conditions, the maximum local velocity (m/s) at a section is most nearly A. 2.0 B. 2.5 C. 3.0 D. 5.0 Hint: Under laminar flow in circular pipes, the velocity distribution is parabolic and symmetrical about the centerline. In such cases, the maximum velocity at a section is double the average velocity at that section. Solution: Average velocity at any section= Q/A= Q/[pD2/4] In this case, average velocity= (0.05 m3/s)/[p (0.16 m)2/4]= 2.5 m/s Hence the maximum velocity= 2 x 2.5 m/s= 5.0 m/s Therefore, the key is (D). 23. A 5-cm diameter pipeline is delivering water from a storage tank to an open canal. The water level in the storage tank can be assumed to be at a constant height of 12 m above the discharge point. Ignoring all losses, the discharge (m3/min) under these conditions is most nearly A. 0.03 B. 1.80 C. 7.35 D. 15.34 Hint: Discharge, Q = Area of pipe, A x Velocity of flow, U Determine U by applying Bernoulli equation. Solution: From Bernoulli equation between free surface (point 1) in tank and discharge point (point 2): where, p1 = 0; U1 = 0; z1 = 12 m; p2 = 0; and z2 = 0 if the discharge point is taken as the datum for elevation head. Substituting these values, or, Hence, Therefore, the key is (B). 24. A 5-cm diameter pipeline is delivering water from a storage tank to an open canal. The water level in the storage tank can be assumed to be at a constant height of 12 m above the discharge point. Ignoring all losses, the Reynolds Number in the pipeline under these conditions is most nearly A. 8.6 x 104 B. 8.6 x 105 C. 8.6 x 106 D. 8.6 x 107 Hint: Reynolds number = where, U is the velocity in the pipe. Determine U by applying Bernoulli equation between free surface in tank and the discharge point. Solution: From Bernoulli equation between free surface (point 1) in tank and discharge point (point 2): where, p1 = 0; U1 = 0; z1 = 12 m; p2 = 0; and z2 = 0 if the discharge point is taken as the datum for elevation head. Substituting these values, or, Hence, Reynolds number = Therefore, the key is (B). 2 The figure shows a horizontal pipeline with a sudden enlargement. The energy 5 grade line and the hydraulic grade line under a certain flow of an incompressible . fluid are also shown. The ratio of the diameter downstream to the diameter upstream of the enlargement is most nearly A. 1.26 B. 1.50 C. 1.68 D. 2.50 Hint: The vertical distance between the energy grade line and the hydraulic grade line is the velocity h From continuity equation, Q = A1 U1 = A2 U2 where, A1 and U1 are the area and velocity upstream of th A2 and U2 are the area and velocity downstream of the enlargement. Solution: From the energy grade line and the hydraulic grade line, giving giving From continuity equation, Therefore, the key is (A). giving 2 The figure shows a horizontal pipeline with a sudden enlargement. The energy 6 grade line and the hydraulic grade line under a certain flow of an incompressible . fluid of specific weight 10 kN/m3 are also shown. The pressure change due to the enlargement is most nearly A. an increase of 3 kPa B. a decrease of 3 kPa C. an increase of 30 kPa D. a decrease of 30 kPa Hint: The energy grade line indicates no energy loss. The decrease in velocity head (from 5 m to 2 m) is converted to an increase of pressure head. Solution: Velocity head upstream of enlargement = 5 m Velocity head downstream of enlargement = 2 m Decrease in velocity head= 5 m 2 m = 3 m Hence increase in pressure head= 3 m Or, increase in pressure= h= (10 kN/m3) (3 m) = 30 kPa Therefore, the key is (C). 27. The figure shows a horizontal pipeline with a sudden enlargement. The energy grade line and the hydraulic grade line under a certain flow of an incompressible fluid of specific weight 10 kN/m3 are also shown. The pressure change due to the enlargement is most nearly A. a decrease of 27 kPa B. an increase of 27 kPa C. a decrease of 30 kPa D. an increase of 30 kPa Hint: The energy grade line indicates a head loss of 0.3 m due to the enlargement. The decrease in velocity head (from 5 m to 2 m) is converted to an increase of pressure head. Solution: Velocity head upstream of enlargement = 5 m Velocity head downstream of enlargement = 2 m Decrease in velocity head= 5 m 2 m = 3 m Head loss due to enlargement= 0.3 m Hence, net increase in pressure head= 3 m 0.3 m= 2.7 m Or, increase in pressure = h= (10 kN/m3) (2.7 m) = 27 kPa Therefore, the key is (B). 2 The figure shows a 10-cm diameter, horizontal pipeline with two piezometers 8 installed 3 m apart. Under laminar flow of lubricating oil (specific gravity = 0.92 . and viscosity = 3.8x10-1 Pa-s), the difference in piezometer readings is 12 cm. The flow rate (m3/min) under the above conditions is most nearly A. 0.02 B. 0.09 C. 0.14 D. 0.21 Hint: The flow rate, Q, can be found from where, p is the pressure drop in a pipe of dia p = h, where h is the difference in the piezometer readings. Solution: p = 0.92 (9.8 kN/m3 x 103 N/kN) (12/100 m)= 1,082 Pa Hence, flow rate, Therefore, the key is (C). 2 When fluid flow is characterized as fully turbulent, which of the following is a true 9 statement? . A. Friction factor will increase with increase of Reynolds Number B. Friction factor will decrease with increase of Reynolds Number C. Friction factor is independent of Reynolds Number D. Friction factor is independent of relative roughness Hint: Refer to the Moody's Diagram, where the X-axis represents the Reynolds Number, the Y-axis re Solution: As can be observed in the Moody's Diagram shown above, in the fully turbulent region, the that as Reynolds Number increases, the friction factor remains constant. Therefore, the key is (C). 3 The figure shows the energy grade line when lubricating oil (specific gravity = 0 0.92 and viscosity = 3.8x10-1 Pa-s) is flowing under laminar conditions through a . horizontal pipe. If the flow rate is 0.15 m3/min, the diameter of the pipe (cm) is most nearly A. 8 B. 10 C. 12 D. 15 Hint: The diameter, D, can be found from the Hagen-Poiseuille equation: p = h, where h is the drop in the energy grade line. Solution: p = 0.92 (9.8 kN/m3 x 103 N/kN) (12/100 m)= 1,082 Pa Q = 0.15 m3/min = 0.0025 m3/s Rearranging the Hagen-Poiseuille equation, Substituting the known values, Diameter, Therefore, the key is (B). 31. Ethyl alcohol (specific gravity = 0.79 and viscosity = 1.19x10-3 Pa-s) is flowing through a 25-cm diameter, horizontal pipeline. When the flow rate is 0.5 m3/min, the Reynolds Number is most nearly A. 28,158 B. 31,424 C. 35,597 D. 42,632 Hint: Reynolds Number, Re, can be found from: where, U can be found from the continuity equation Q = UA. Solution: From continuity equation, Hence, Reynolds Number Therefore, the key is (A). 32. Considering the flow of an incompressible fluid through a horizontal pipe, which of the following is a correct statement? A. The energy grade line is always parallel to the centerline of the pipeline B. The energy grade line is always above the hydraulic grade line C. The energy grade line is always horizontal D. The energy grade line is always parallel to the hydraulic grade line Hint: Energy grade line is a plot of the total energy head of the fluid as a function of pipe length. Hydraulic grade line is a plot of the piezometric head as a function of pipe length. Piezometric head = elevation head + pressure head Solution: The energy grade line (EGL) is a plot of total head, h = The hydraulic grade line (HGL) is a plot of The total energy head of the fluid can decrease due to head loss due to friction or head removed by turbines; total energy head of the fluid can increase due to head addition by a pump. Thus, it has no relation to the centerline of the pipe, and is not always horizontal. The EGL is always at a distance of above the HGL. However, U can change due to changes in pipe diameter. Therefore, the key is (B). 3 The schematic of a pumping system to pump water from a canal to an overhead 3 storage tank is shown. At the design pumping rate of 0.5 m3/min, the total head . loss of the system is 10% of the total static head. The power added by the pump (kW) is most nearly A. 1.0 B. 1.5 C. 2.0 D. 3.0 Hint: Power, P, added to the fluid can be found from where, h is the total head added by pum Solution: Total static head = suction lift + delivery head Suction lift = 3 m Delivery head = 15 m + 4 m= 19 m Total static head = 3 m + 19 m = 22 m Total head loss = 10% of 22 m = 2.2 m Total head added = 22 m + 2.2 m = 24.2 m Hence, power = (0.5/60 m3/s) (9.8 x 1,000 N/m3) (24.2 m) = 1,976 W = 1.98 kW Therefore, the key is (C). 3 The schematic of a pumping system to pump water from a canal to an overhead 4 storage tank is shown. The total head loss of the system is to be 10% of the total . static head. If the pump is powered by a 5 kW motor at an efficiency of 85%, the pumping rate (m3/min) is most nearly A. 0.2 B. 0.6 C. 1.0 D. 2.0 Hint: Pump power, P, is related to the pumping rate by Solution: Total static head = suction lift + delivery head Suction lift = 3 m where, h is the total head added by p Delivery head = 15 m + 4 m= 19 m Total static head = 3 m + 19 m = 22 m Total head loss = 10% of 22 m = 2.2 m Total head added = 22 m + 2.2 m = 24.2 m Pumping rate, Therefore, the key is (C). 35. A 20-cm in diameter pipeline with a relative roughness of 0.01 has a total length 45 of m. water When is pumped through it at a rate of 5 m3/min, the major head loss (m) is most nearly A. 3 B. 10 C. 15 D. 20 Hint: Major head loss = where, U = Q/A. To find f, first determine whether flow is laminar or turbulent. Solution: From continuity equation, Reynolds Number = Hence, flow is turbulent. From Moody's diagram, at a Reynolds N of 5.94 x 105 and relative roughness of 0.01, f = 0.038 Therefore, major head loss = Therefore, the key is (A). 3 A hydropower turbine fed with water from a reservoir is shown. The total length 6 of the pipe line is 100 m, with a diameter of 0.5 m. Assuming the friction factor to . be 0.03 at a flow rate of 0.5 m3/s, and ignoring minor losses, the power generation potential of this system (kW) is most nearly A. 80 B. 100 C. 140 D. 160 Hint: Power generation can be found from can be found from Energy equation. Head loss where, Q is the flow rate and h is the net head ava Solution: From continuity equation, Head loss, Applying Bernoulli equation between free surface (point 1) in the reservoir and the discharge point (p where, p1 = 0; U1 = 0; z1 = 35 m; p2 = 0; U2 = 2.55 m/s; and hL = 1.98 m Therefore, hr = 32.7 m Hence, power generation potential = (0.5 m3/s) (9.8 kN/m3) (32.7 m) = 160.2 kW. Therefore, the key is (D). 37. The stagnation pressure at a point in a pipeline of diameter D carrying an incompressible fluid is ps while the static pressure is p. The density of the fluid is and its specific weight, , Which of the following is a correct expression for the flow rate at this point? A. B. C. D. Hint: From continuity equation, flow rate, Q = Velocity, U x Area, A From definition of stagnation pressure, ps and static pressure, p, Solution: Combining the continuity equation and the definitions of stagnation pressure, ps and static pressure, p, Therefore, the key is (B). 3 Figure shows a partition maintaining constant water levels on the two sides. The 8 discharge (m3/day) through a 2-cm diameter orifice (coefficient of contraction . 0.83) located as shown is most nearly A. 70 B. 100 C. 150 D. 180 Hint: Flow rate, where, h1 is the depth upstream of the orifice and h Solution: Using the orifice equation, with the given data, Therefore, the key is (A). 3 Consider the free jet of an incompressible fluid flowing through an orifice fitted to 9 a constant level tank as shown. Ignoring all losses, which of the following is a . correct statement about the magnitude of the initial velocity U of the jet? A. U is directly proportional to the orifice diameter B. U is inversely proportional to the fluid's density C. U is proportional to the square root of the depth h D. U is proportional to the square of the depth h Hint: Apply Bernoulli equation between the free surface of the tank and a point in the jet just downstr Solution: Applying Bernoulli equation between the free surface of the tank (point 1) and a point in the where, p1 = 0; U1 = 0; z1 = h; p2 = 0; and z2 = 0. Substituting the above values, ; thus, U is proportional to the square root of the depth h. Therefore, the key is (C). 40. A 1:12 scale model is to be built to study the flow over a spillway. The flow rate in the prototype 180 m3/s. The flow rate in the model (m3/s) should be most nearly A. 0.36 B. 1.25 C. 8.25 D. 15.0 Hint: Equate Froudes Number for the model and the prototype: Use continuity equation to find ratio of flow rates Solution: From continuity equation, From geometric scale factor, , From Froudes Number ratio, Hence, giving Qm = Qp x 0.002 = 180 m3/s x 0.002 = 0.36 m3/s. Therefore, the key is (A). 41. In the study of aerodynamic drag on a stationary body, an appropriate nondimensional grouping has been found to be where, P is the power lost, is the density of the fluid, A is a typical area, and U is the velocity of the fluid. In laboratory tests with a 1:10 scale model at 25 C, the power lost was measured as 5 W when the air velocity was 0.5 m/s. The power lost in the prototype (kW) at 25 C when the air velocity is 2 m/s will be most nearly A. 2.0 B. 3.2 C. 12.0 D. 32.0 Hint: Equate the nondimensional grouping for the model and the prototype: From the geometric scale factor, where, is the scale factor. Solution: By equating the non-dimensional grouping, . Therefore, the key is (D). 42. Which of the following is a non-dimensional grouping where, F is a force; is the density; A is the area; and U is a velocity? A. B. C. D. Hint: Since none of them is a standard non-dimensional number, check if any of them can be reduced to a familiar expression. Solution: Recalling the result for drag or lift: F = CD(AU2)/2, we can deduce that must be non-dimensional. Therefore, the key is (B). 43. A 0.3-m diameter pipeline terminates in a nozzle of outlet diameter = 0.15 m. When water flows through this pipe at a rate of 0.25 m3/s, the force required to hold the pipe (N) is most nearly A. 660 B. 1,320 C. 2,650 D. 5,300 Hint: Force, F = rate of change of momentum Rate of change of momentum = (mass flow rate, Q) x (change in velocity, U) Solution: Mass flow rate, Inlet velocity, U1 Outlet velocity, U2 Hence, force, F Therefore, the key is (C). 44. A 0.3-m diameter pipeline terminates in a nozzle of outlet diameter = 0.15 m. The free jet from the nozzle is deflected through 90 by a flat plat as shown. When water flows through this pipe at a rate of 0.25 m3/s, the force required to hold the plat (N) is most nearly A. 880 B. 1,760 C. 2,640 D. 3,530 Hint: Force, F on fluid = rate of change of momentum Rate of change of momentum = (mass flow rate, Q) x (change in velocity, U) Solution: Mass flow rate, Velocity before impact, U1 Velocity after impact, U2= 0 Force, F on fluid Hence, force on plate = 3,530 N. Therefore, the key is (D). 4 A pipeline terminates in a nozzle of outlet diameter = 0.15 m. The free jet from the 5 nozzle is deflected through 120 by a vane as shown. Water flow rate through this . pipe is 0.25 m3/s. Ignoring friction, the force exerted by the jet on the vane (N) is most nearly A. 5,295 B. 3,530 C. 1,775 D. 875 Hint: Force, F on vane = force on fluid = rate of change of momentum Rate of change of momentum = (mass flow rate, Q) x (change in velocity, ?U) Solution: Mass flow rate, Velocity before impact, U1,x Velocity after impact, U2,x Force, F on fluid Hence, force on vane = 5,295 N. Therefore, the key is (A). 4 An open water tank has an orifice of diameter 4 cm fitted on its side, 3 m below 6 the free surface. The tank is mounted on frictionless wheels as shown. Ignoring all . friction effects, the force (N) necessary to keep the tank at rest is most nearly A. 42 B. 740 C. 420 D. 740 Hint: Apply momentum equation: Force due to jet, F= Q(V2 0) where, V2 = ?(2gh); and Q = AV Solution: V2 = [2(9.81 m/s2) (3 m)] = 7.67 m/s A= (4/100 m)2/4= 0.00125 m2 Q= (7.67 m/s)(0.00125 m2)= 0.01 m3/s Hence, force = (0.01 m3/s)(998 kg/m3)(7.67 m/s 0) = 73.8 N Therefore, the key is (B). 47. The following data are available for a pumping system: Fluid water Static suction lift 3m Static delivery head 32 m Total dynamic head 5 m Pumping rate 0.5 m3/s The power added by the pump (kW) is most nearly A. 25 B. 185 C. 200 D. 300 Hint: Power added, P= (Specific weight) (Flow rate, Q) (Total head added) Total head added = Total static head + total dynamic head Solution: Total static head= suction lift + delivery head= 3 m + 32 m = 35 m Total head= 35 m + 5 m= 40 m Hence, power added = (9.8 kN/m3) (0.5 m3/s) (40 m) = 196 kW Therefore, the key is (C). 48. The following data are available for a pumping system: Fluid water Static suction lift 3m Static delivery head 35 m Suction side dynamic head 1 m Delivery side dynamic head 6 m Motor rating 30 kW Efficiency 80% The pumping rate (m3/min) under the above conditions is most nearly A. 2.6 B. 3.3 C. 4.0 D. 5.6 Hint: Power added, P= Eff. x Motor rating = (Specific weight) (Flow rate, Q) (Total head added) Total head added = Total static head + total dynamic head Total static head= Suction lift + delivery head Total dynamic head= Dynamic head on (suction side + delivery side) Solution: Total static head= 3 m + 35 m = 38 m Total dynamic head= 1 m + 6 m= 7 m Total head added= 38 m + 7 m= 45 m Hence, flow rate = Therefore, the key is (B). 4 The head-vs-capacity curves for two centrifugal pumps A and B are shown below: 9 . Which of the following is a correct statement at a flow rate of 600 ft3/min? A. Water horse-power of Pump A is 34 HP B. Water horse-power of Pump B is 34 HP C. Water horse-power of Pump A is 40 HP D. Water horse-power of Pump B is 60 HP Hint: Water horse-power, WHP= (Specific weight) (Flow, Q) (Head, h)/( 550 ft-lb/s-HP) where, Q is Solution: At 600 ft3/min, pump A delivers a head of 30 ft. So, WHP = (62.4 lb/cu ft)(10 ft3/s)(30 ft)/(550 ft-lb/s-HP)= 34.0 HP At 600 ft3/min, pump B delivers a head of 35 ft. So, WHP = (62.4 lb/cu ft)(10 ft3/s)(35 ft)/(550 ft-lb/s-HP)= 39.7 HP Therefore, the key is (A). 5 The head-vs-flow rate and efficiency-vs.-flow rate curves for a centrifugal pump 0 pumping water are shown below: . The horse-power (HP) required by this pump at a flow rate of 700 ft3/min is most nearly A. 22 B. 26 C. 29 D. 31 Hint: Water horse-power, WHP= (Specific weight, g) (Flow, Q) (Head, h)/ (550 ft-lb/s-HP) where, g i The brake horse power, BHP = WHP/Efficiency Solution: At 700 ft3/min, pump delivers a head of 20 ft. So, WHP = (62.4 lb/cu ft)((700/60) ft3/s)(20 ft)/ (550 ftlb/s-HP)= 26.4 HP At 700 ft3/min, efficiency of the pump is 85%. So, BHP = (26.4 HP)/0.85 = 31.1 HP Therefore, the key is (D). 5 The figure below shows the pump curve, the system curve, and the efficiency 1 curve for a system pumping water. The brake horse-power (HP) required to run . this pump under the condition shown is most nearly A. 35 B. 40 C. 55 D. 65 Hint: The brake horse power at the operating point= [Qgh/ h]? (550 ft-lb/s-HP) where, Q is the flow r the efficiency at the operating point. The operating point is where the pump curve intersects the system Solution: At the operating point, Q = 600 ft3/min= 10 ft3/ s h = 30 ft h = 85% Hence, brake horse power = (10 ft3/s)(62.4 lb/ft3)(30 ft)/[(0.85)(550 ft-lb/s-HP) = 40.0 HP Therefore, the key is (B). 5 The figure below shows a branched pipe network. A pressure gage just upstream 2 of A reads 60 psi and a pressure gage just downstream of D reads 54 psi. The flow . rates, diameters, the friction factors, and the lengths of the two branches are as follows: Branch ABD Branch ACD Flow rate Q 2Q Diameter D D Length L L Which of the following is a true conclusion? A. Pressure drop in branch ACD = 4 psi B. Pressure drop in branch ABD = 2 psi C. Pressure drop in branch ACD = Pressure drop in branch ABD = 6 psi D. Pressure drop in branch ACD = Pressure drop in branch ABD = 3 psi Hint: In branched pipe network such as the one shown, the head loss is the same in each branch. Solution: Pressure drop in branch ABD = 60 psi 54 psi = 6 psi Pressure drop in branch ACD = 60 psi 54 psi = 6 psi Hence pressure drop in ABD = pressure drop in ACD = 6 psi. Therefore, the key is (C). 5 The figure below shows a branched pipe network. The flow rates, diameters, the 3 friction factors, and the lengths of the two branches are as follows: . Branch ABD Branch ACD Flow rate Q 2Q Diameter D D Length L L Friction factor f1 f2 Which of the following is a true conclusion? A. f1 = 2f2 B. f1 = 4f2 C. f1 = f2 D. f1 = f2/2 Hint: In branched pipe network such as the one shown, the head loss is the same in each branch. Head Solution: Equating the head loss in branch ACD to the head loss in branch ABD, Since the pipes are of the same diameter, D, V1 = Q/A and V2 = 2Q/A Combining the above two, Hence, f1 = 4f2 Therefore, the key is (B). 54. A channel of rectangular cross section and width 4 m is carrying a discharge of 2 m3/s at a uniform depth of 1.5 m. If the Mannings n for the channel is 0.012, the slope (%) of the channel bed is most nearly A. 0.001 B. 0.002 C. 0.003 D. 0.004 Hint: According to Mannings Equation, the velocity, flow rate = VA, where Rh = (Area, A)/(Wetted perimeter, WP) Solution: Area of flow, A= width x depth= (4 m)(1.5 m)= 6.0 m Wetted perimeter, WP= width + 2 x depth= 4 m + 2(1.5 m)= 7.0 m Hydraulic radius, Rh= A/WP= (6.0 m)/(7.0 m)= 0.857 m Velocity, V= Q/A= (2 m3/s)/(6.0 m2)= 0.33 m/s Substituting the above into Mannings Equation, , and the = 0.0019% Therefore, the key is (B). 5 A circular sewer is of diameter 2 m. Under full flow without surcharge, the 5 velocity of flow is 0.8 m/s. Assuming the Mannings n to be a constant, the velocity . of flow (m/s) when the sewer is flowing half full will be most nearly A. 0.2 B. 0.4 C. 0.8 D. 1.6 Hint: According to Mannings Equation, the velocity, and the flow rate = VA, where R Solution: Hydraulic radius, Rh under full flow= [ D2/4]/ [ D]= D/4 Hydraulic radius, Rh under half full flow= {(1/2)[ D2/4]}/{(1/2)[ D]}= D/4 Since Rh, S0, and n are all the same under full flow and half full flow, the velocity will also remain the Therefore, the key is (C). 5 A circular sewer is of diameter 2 m. Under full flow without surcharge, the 6 velocity of flow is 0.8 m/s. Assuming the Mannings n to be a constant, the flow . rate (m3/s) when the sewer is flowing half full will be most nearly A. 0.25 B. 0.63 C. 1.26 D. 2.52 Hint: According to Mannings Equation, the velocity, Flow rate = VA where Rh = (Area, A)/(Wetted peri Solution: Hydraulic radius, Rh under full flow= [ D2/4]/[ D]= D/4 Hydraulic radius, Rh under half full flow= {(1/2)[ D2/4]}/{(1/2)[ D]}= D/4 Since Rh, S0, and n are all the same under full flow and half full flow, the velocity will also remain the Area of flow under half full flow= {(1/2) ( (2 m)2/4} = 1.57 m2 Hence, flow rate under half full flow= (1.57 m2)(0.8 m/s)= 1.26 m3/s Therefore, the key is (C). 5 When a pitot tube is placed in an open channel as shown. When the pitometer 7 reads 11.5 cm, the velocity (m/s) at the point P is most nearly . A. 1.5 B. 3.3 C. 4.8 D. 5.5 Hint: Let the depth of point P be h. Applying Bernoulli equation between an upstream point, 1, at a de where, p1 = h. and pP = (h + 11.5)? Solution: Combining the above equations, Hence, Therefore, the key is (A). 5 Under laminar flow conditions, the pressure drop per unit length at a given flow 8 rate, Q, is . A. proportional to D4 B. proportional to D2 C. proportional to D1/2 D. proportional to D1/4 Hint: Pressure drop = where, U = Q/A and, under laminar flow conditions, f = 64/R Solution: From continuity equation, U = Q/A = Q/[ D2/4] Hence, and, pressure drop = Hence, pressure drop per unit length Thus, p/L is proportional to D1/4 Therefore, the key is (D). 5 Crude oil is flowing through a pipe of diameter 1.25 m diameter and relative 9 roughness 0.002 at a Reynolds Number of 2x104. Specific gravity of the oil is 0.84 . viscosity of the oil is 0.4 N-s/m2. The head loss per 100 m (m) of this pipe is most nearly A. 5.3 B. 7.1 C. 9.2 D. 11.6 Hint: Head loss per 100 m= where U is the average velocity and f is the friction facto the flow is turbulent, and hence, f has to be read from the Moody's diagram. U can be found from the Solution: From Moodys diagram, at Reynolds Number of of 2x104 and relative roughness of 0.002, th Average velocity, U = Re / D= (2x104?(0.4 N-s/m2)/{(0.84 (998 kg/m3)(1.25 m)} = 7.63 m/s Hence, head loss per 100 m= (0.03) (100 m/1.25 m) {(7.63 m/s) 2/(2 x 9.81 m/s2)} = 7.13 m Therefore, the key is (B). 6 Crude oil of specific gravity 0.82 is flowing at a rate of 10 m3/s through a pipe of 0 diameter 1.25 m at a friction factor of 0.02. The head loss (m) due to viscous . effects over 100 m of the pipe is most nearly A. 0.55 B. 2.15 C. 4.35 D. 5.40 Hint: Head loss in 100 m of the pipe= Use continuity equation to find U: U = Q/A. where f is the friction factor and U is the avera Solution: From continuity equation, U = (Q)/(?D2/4) = (10 m3/s)/{ (1.25 m) 2/4)}= 8.15 m/s Hence, head loss in 100 m of the pipe= Therefore, the key is (D). 6 Crude oil of specific gravity 0.82 and viscosity of 0.4 N-s/m2 is flowing at a rate of 1 10 m3/s through a pipe of diameter 1.25 m and relative roughness = 0.002. The . head loss (m) due to viscous effect over 50 m of the pipe is most nearly A. 4.0 B. 7.0 C. 9.0 D. 12.0 Hint: Head loss in 50 m of the pipe= where f is the friction factor and U is the average To find f, need to determine Reynolds Number and check if flow is laminar or turbulent. Use continui Solution: From continuity equation, U = (Q)/(?D2/4) = (10 m3/s)/{? (1.25 m) 2/4)}= 8.15 m/s Reynolds Number = UD/ = (0.82x 998 kg/m3) (8.15 m/s) (1.25 m)/ (0.4 N-s/m2) = 2.08x10 Since Reynolds Number is greater than 4,000, flow is turbulent. Hence, from Moody's diagram, for Re = 2.08x104 and relative roughness = 0.002, friction factor, f = 0 Hence, head loss in 50 m of the pipe= Therefore, the key is (A). 6 The figure shows a system for pumping water from a canal to an overhead tank. 2 The pipe line is 0.1 m in diameter and has a total length of the pipeline is 60 m. . When the flow velocity is 3 m/s, the friction factor is 0.02. Ignoring minor losses, the head added by the pump (m) under these confitions is most nearly A. 40.0 B. 45.5 C. 65.5 D. 105.5 Hint: Apply Energy equation between the two free surfaces: where, ha is the head added by the pump and hL is the head loss in the pipeline = Solution: Head loss = From the Energy equation, ha = z2 + hL z1 = 140 m + 5.5 m 100 m= 45.5 m. Therefore, the key is (B). 6 The coefficient of drag on an automobile of frontal area A is CD. The power 3 required to overcome drag force when the automobile is traveling at a speed of U . in still air of density ? is given by A. B. C. D. Hint: Drag force, F, on a body is given by and the power required to overcome that for Solution: The power required to overcome drag force = F U= Therefore, the key is (D). 6 The figure shows a system for pumping water from a canal to an overhead tank. At 4 a flow rate of 10 m3/s, the major head loss in the system is 15 m and the minor . head loss is 4 m. Pump curves for four models of pumps are also shown. A. Model A B. Model B C. Model C D. Model D Hint: At the operating point of 10 m3/s, Total head= elevation gain + major loss + minor loss Solution: Substituting the given data, total head = (140 100) m + 15 m + 4 m = 59 m The model that can deliver 10 m3/s at 59 m is Model B. Therefore, the key is (B). 6 An impulse turbine is to be used at a hydroelectric power plant. The nozzle 5 diameter is 25 cm. If the water flow rate is 5 m3/s, the maximum power (MW) that . can be produced by this turbine is most nearly A. 2 B. 6 C. 12 D. 26 Hint: Maximum power that an impulse turbine can produce= Use continuity equation to find U: U = Q/A. where Q is the flow rate; is the Solution: From continuity equation, U = (Q)/( D2/4) = (5 m3/s)/{ (0.25 m) 2/4)}= 101.9 m/s Hence, maximum power that can be produced= Therefore, the key is (D). 6 The hydraulic diameter of a circualr sewer flowing half-full is equal to 6 A. half its diameter . B. its diameter C. double its diameter D. times its diameter Hint: Hydraulic diameter, Dh is defined as Solution: From the definition of Dh, for a sewer flowing half full, Therefore, the key is (B). 6 The drag coefficient for a car with a frontal area of 28 ft2 is 0.32. Assuming the 7 density of air to be 2.4x10-3 slugs/ft3, the drag force (lb) on this car when driven at . 60 mph against a head wind of 20 mph is most nearly A. 37 B. 83 C. 148 D. 185 Hint: Drag force = where, CD is the coefficient of drag; is the density of air; A is the front Solution: Relative velocity = 60 mph + 20 mph = 80 mph = 117.3 ft/s Hence drag force= Therefore, the key is (C). = 148 lb. 6 The drag coefficient for a car with a frontal area of 26 ft2 is being measured in a 8 8 ft x 8ft wind tunnel. The density of air under the test conditions is 2.4x10-3 . slugs/ft3, When the air flow rate is, 500,000 ft3/min, the drag force on the car was measured to be 170 lb. The drag coefficient under the test conditions is most nearly A. 0.28 B. 0.30 C. 0.32 D. 0.34 Hint: Drag force = where, CD is the coefficient of drag; is the density of air; A is the front the relative velocity. Find U from continuity equation: U = Q/Atunnel Solution: From continuity equation, U = (500,000/60 ft3/s)/(8 ft x 8 ft)= 130.2 ft/s Hence, drag coefficient = Therefore, the key is (C). = 6 A manometer is connected across the tapering section of a pipeline as shown. The 9 specific gravity of the manometric fluid is 1.8 and the specific gravity of the fluid . flowing through the pipe is 0.72. When the velocity at section 1-1 is 3 m/s, the manometric deflection, h = 6 cm. Ignoring all losses, the velocity (m/s) at section 2-2 under the above conditions is most nearly A. 2.6 B. 3.5 C. 5.2 D. 8.6 Hint: Ignoring all losses, and applying Bernoulli equation between 1-1 and 2-2: From the manometer, where, m is the specific weight of the manometric fluid. Solution: From Bernoulli equation: From manometer: Combining the above, Therefore, the key is (B). giving, U 7 The relative roughness of a new pipeline is 0.002. The flow is such that the 0 Reynolds Number is 20,000. If the relative roughness increases to 0.006, and the . Reynolds Number remains the same, which of the following is most likely? A. Friction factor will decrease B. Friction factor will increase C. Friction factor will remain the same D. Head loss will remain the same Hint: Use Moody diagram to determine the change in friction factor when relative roughness changes Solution: From Moody diagram, At Reynolds N = 20,000 and relative roughness = 0.002, friction factor = 0.030 At Reynolds N = 20,000 and relative roughness = 0.006, friction factor = 0.036 Hence, friction factor increases, and therefore head loss will increase. Therefore, the key is (B). 7 The figure shows a reservior A connected to another reservior B by a pipeline. The 1 elevations in A and B can be assumed to remain constant at the levels shown. . The total head loss of the system A. is equal to 60 m B. is equal to 65 m C. is equal to 80 m D. can not be determined with the available data Hint: Use Bernoulli equation between the two free surfaces: Solution: In Bernoulli equation, UA = 0; pA = 0; UB = 0; pB = 0 Hence, Therefore, the key is (C). 7 The figure shows a reservior A connected to another reservior B by a pipeline. The elevations 2 in A and B can be assumed to remain constant at the levels shown. A pump/turbine is installed . in the pipeline that can function either as a turbine producing power when water flows from A to B; or as a pump to pump water from B to A. When functioning as a turbine at a certain flow rate, the head removed by the turbine is 70 m. When functioning as a pump, the head (m) that should be added to pump water from B to A is most nearly A. 60 B. 70 C. 80 D. 90 Hint: Use energy equation between the two free surfaces: When functioning as a turbine: where, hr is the head removed by the turbine. When functioning as a pump: where, ha is the head added to the pump. Solution: In energy equation, UA = 0; pA = 0; UB = 0; pB = 0 Hence, when functioning as a turbine, When functioning as a pump, or, or, Therefore, the key 7 The velocity of sound, c, in an ideal gas can be found from: where, k = Cp 3 . and T is the absolute temperature. Which of the following will be an alternate form of the above expre where p is the pressure, v is the specific volume; and is the density? A. B. C. D. Hint: For Ideal Gases, pv = p/ = RT In the given equation for c, substitute for T in terms of p, v, or Solution: Substituting for RT from Ideal Gas Law, Therefore, the key is (C). 7 Air is flowing through a duct at a velocity of 500 ft/s at 70 F. Use the following data: 4 Specific heat of air at constant pressure= 6,000 ft-lb/slug- R . Specific heat of air at constant volume= 4,285 ftlb/slug- R Gas Constant= 1,715 ft2/s2- R Assuming air to be an ideal gas, the Mach Number under the above conditions is most nearly A. 0.37 B. 0.44 C. 1.32 D. 2.25 Hint: Mach Number = V/c where, V is the velocity of the fluid and c is the velocity of sound in the flu Velocity of sound, c = (kRT) where k = Cp/Cv Solution: Substituting the given data, k = 6,000/4,285 = 1.40 Velocity of sound, c = [(1.40) (1,715 ft2/s2- R)(460 + 70)] = 1,128.16 ft/s Hence, Mach Number = (500 ft/s)/(1,128 ft/s) = 0.44 Therefore, the key is (B). 7 Air is flowing through a 2 ft x 2 ft duct at a rate of 2,000 ft3/s at 70 F. The following data are available 5 Specific heat of air at constant pressure= 6,000 ft-lb/slug- R . Specific heat of air at constant volume= 4,285 ft-lb/slug- R Gas Constant= 1,715 ft2/s2- R Assuming air to be an ideal gas, the Mach Number under the above conditions is most nearly A. 0.37 B. 0.44 C. 1.32 D. 2.25 Hint: Mach Number = V/c where, V is the velocity of the fluid and c is the velocity of sound in the flu Velocity of sound, c = (kRT) where k = Cp/Cv Use continuity equation to find fluid velocity, U = Q/ A Solution: Substituting the given data, k = 6,000/4,285 = 1.40 Velocity of sound, c = [(1.40)(1,715 ft2/s2- R)(460 + 70)] = 1,128.16 ft/s Velocity of fluid= (2,000 ft3/s)/(2 ft x 2 ft) = 500 ft/s Hence, Mach Number = (500 ft/s)/(1,128.16 ft/s) = 0.44 Therefore, the key is (B). ... View Full Document