Homework 4 Assignment Print View

1. Award: 33.33 out of 33.33 points While traveling on a dirt road, the bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank. If the height of the gasoline in the tank is 26 cm, determine the initial velocity of the gasoline at the hole. Determine the velocity change with time and how the flow will be affected if the lid of the tank is closed tightly. The initial velocity is 2.26 m/s. References Worksheet Difficulty: Medium While traveling on a dirt road, the bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank. If the height of the gasoline in the tank is 26 cm, determine the initial velocity of the gasoline at the hole. Determine the velocity change with time and how the flow will be affected if the lid of the tank is closed tightly. The initial velocity is 2.26 ± 10% m/s. Explanation: The following assumptions have been made here: 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The air space in the tank is at atmospheric pressure, and the splashing of the gasoline in the tank during travel is not considered. This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. We take point 1 to be at the free surface of gasoline in the tank so that P1 = Patm (open to the atmosphere), V1 ≅ 0 (the tank is large relative to the outlet), and z1 = 0.26 m and z2 = 0 (we take the reference level at the hole). Also, P2 = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to P1 ρg 2 + V1 2g + z1 = P2 ρg 2 + V2 2g 2 + z2 → z1 = V2 2g → V2 = √2gz1 = √2 (9. 81 m/s 2 )(0.26) = 2.26 m/s 2. Award: 33.33 out of 33.33 points Air at 105 kPa and 310 K flows upward through a 6-cm-diameter inclined duct at a rate of 65 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure change across the reducer is measured by a water manometer.The elevation difference between the two points on the pipe where the two arms of the manometer are attached is 0.20 m. Take the density of water to be ρ = 1000 kg/m3. The gas constant of air is R = 0.287 kPa·m3/kg·K. Determine the differential height between the fluid levels of the two arms of the manometer in centimeters. The differential height between the fluid levels of the two arms of the manometer is 12.9 cm. References Worksheet Difficulty: Medium Air at 105 kPa and 310 K flows upward through a 6-cm-diameter inclined duct at a rate of 65 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure change across the reducer is measured by a water manometer.The elevation difference between the two points on the pipe where the two arms of the manometer are attached is 0.20 m. Take the density of water to be ρ = 1000 kg/m3. The gas constant of air is R = 0.287 kPa·m3/kg·K. Determine the differential height between the fluid levels of the two arms of the manometer in centimeters. The differential height between the fluid levels of the two arms of the manometer is 12.9 ± 10% cm. Explanation: The following assumptions have been made here: 1. The flow through the duct is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2. Air is an ideal gas. 3. The effect of the air column on the pressure change is negligible because of its low density. 4. The air flow is parallel to the entrance of each arm of the manometer, and thus no dynamic effects are involved. We take points 1 and 2 at the lower and upper connection points, respectively, of the two arms of the manometer and take the lower connection point as the reference level. Noting that the effect of elevation on the pressure change of a gas is negligible, the application of the Bernoulli equation between points 1 and 2 gives P1 ρg + V12 2g + z1 = P2 ρg + V22 2g + z 2 → P 1 - P 2 = ρ air V22 - V12 2 where ρ air = V1 = P RT V̇ A1 = = 105 kPa ( 0.287 kPa·m /kg·K ) ( 310 K ) 3 V̇ πD 1 2 = V̇ A2 = V̇ πD 2 2 4 π ( 0.06 m ) 2 =22.989 m/s 4 4 V2 = 0.065 m 3/s = 1.180 kg/m 3 = 0.065 m 3/s π ( 0.04 m ) 2 =51.73 m/s 4 Substituting, ( P 1 - P 2 = 1.180 kg/m 3 ) ( 51.73 m/s ) 2 - ( 22.989 m/s ) 2 2 ( 1N 1 kg·m/s 2 ) = 1267.03 N/m 2 = 1267.03 Pa The differential height of water in the manometer corresponding to this pressure change is determined from ΔP = ρwatergh to be h= P1 - P2 ρ waterg = 1267.03 N/m 2 ( 1000 kg/m ) ( 9.81 m/s ) 3 2 ( 1 kg·m/s 2 1N ) = 0.1292 m = 12.9 cm 3. Award: 33.34 out of 33.34 points Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.4 m3/s and exits through a 25-cmdiameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 m. For a combined turbine–generator efficiency of 83 percent, determine the net electric power output. Disregard the effect of the kinetic energy correction factors. Take the density of water to be 1000 kg/m3 and the density of mercury to be 13,560 kg/m3. The net electric power output is 44 kW. References Worksheet Difficulty: Medium Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.4 m3/s and exits through a 25-cmdiameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 m. For a combined turbine–generator efficiency of 83 percent, determine the net electric power output. Disregard the effect of the kinetic energy correction factors. Take the density of water to be 1000 kg/m3 and the density of mercury to be 13,560 kg/m3. The net electric power output is 43 ± 10% kW. Explanation: The following assumptions have been made here: 1. The flow is steady and incompressible. 2. All losses in the turbine are accounted for by turbine efficiency, and thus hL = 0. 3. The elevation difference across the turbine is negligible. 4. The effect of the kinetic energy correction factors is negligible, that is, α1 = α2 = α = 1. The density of water is given as 1000 kg/m3 and the density of mercury as 13,560 kg/m3. Points 1 and 2 are taken at the inlet and the exit of the turbine, respectively. Noting that the elevation effects are negligible, the energy equation in terms of heads for the turbine reduces to P1 ρg + α1 V12 + z 1 + h pump, u = 2g P2 ρg + α2 V22 2g + z 2 + h turbine, e + h L → h turbine, e = P1 − P2 ρ water g ( α V12 − V22 + 2g ) () → 1 where V1 = V̇ A1 = V̇ = ( ) ( πD 1 2 4 V2 = V̇ A2 = V̇ = ( ) ( 0.4 m 3/s π ( 0 . 25 m) 2 4 ) 0.4 m 3/s πD 2 2 π ( 0 . 25 m) 2 4 4 ) = 5.66 m/s = 8.15 m/s The pressure drop corresponding to a differential height of 1.2 m in the mercury manometer is determined as follows: P1 − P2 = ( (ρHg ) − ρ water gh ) ( P 1 - P 2 = [ 13, 560 − 1000 kg/m 3] 9. 81 m/s 2 )( ( 1. 2 m) 1 kN 1000 kg ⋅ m / s 2 P 1 - P 2 = = 148 kN/m 2 = 148 kPa Substituting in Eq. (1), the turbine head is determined as follows: ) h turbine, e = 148 kN/m 2 ( 1000 kg/m 3 )( 9 . 81 m/s 2 1000 kg ⋅ m/s 2 ) 1 kN + (1. 0) (5.66m/s) 2 − (8.15 m/s) 2 ( 2 9.81 m/s 2 ) = 13.3 m Then, the net electric power output of this hydroelectric turbine is determined as follows: Ẇ turbine = η turbine-gen ṁgh turbine, e = η turbine-gen ρV̇gh turbine, e ( Ẇ turbine = 0.83 1000kg/m 3 )(0.4 m /s )(9.81m/s )(13.3 m) 3 2 ( 1N 1 kg ⋅ m/s 2 )( 1 kW 1000 N ⋅ m/s ) = 43 kW