Homework 6 Assignment Print View

1. Award: 33.33 out of 33.33 points Liquid ammonia at −20°C is flowing through a 24-m-long section of a 5-mm-diameter copper tube at a rate of 0.09 kg/s. Determine the pressure drop, the head loss, and the pumping power required to overcome the frictional losses in the tube. The density and dynamic viscosity of liquid ammonia at –20°C are ρ = 665.1 kg/m3 and μ = 2.361 × 10-4 kg/m·s. The roughness of copper tubing is 1.5 × 10-6 m. The pressure drop is 1520 kPa. The head loss is 233 m. The pumping power required to overcome the frictional losses is 0.206 kW. References Worksheet Difficulty: Medium Liquid ammonia at −20°C is flowing through a 24-m-long section of a 5-mm-diameter copper tube at a rate of 0.09 kg/s. Determine the pressure drop, the head loss, and the pumping power required to overcome the frictional losses in the tube. The density and dynamic viscosity of liquid ammonia at –20°C are ρ = 665.1 kg/m3 and μ = 2.361 × 10-4 kg/m·s. The roughness of copper tubing is 1.5 × 10-6 m. The pressure drop is 1488 ± 10% kPa. The head loss is 227 ± 10% m. The pumping power required to overcome the frictional losses is 0.201 ± 10% kW. Explanation: The following assumptions have been made here: 1. The flow is steady and incompressible. 2. The entrance effects are negligible, and thus the flow is fully developed. 3. The pipe involves no components such as bends, valves, and connectors. 4. The piping section involves no work devices such as pumps and turbines. First we calculate the average velocity and the Reynolds number to determine the flow regime: V= ṁ ρA c ṁ = ρ ( ) πD 2 4 = 0.09 kg/s (665.1 kg/m 3) ( π(0.005 m) 2 4 ) = 6.892 m/s Re = ρVD μ = (665.1 kg/m 3)(6.892 m/s)(0.005 m) 2.361 × 10 − 4 kg/m ⋅ s = 0.9707 × 10 5 which is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the pipe is ε D = 1.5 × 10 − 6 m = 3 × 10 − 4 0.005 m The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme): ( ) () ϵ 1 √f D = − 2.0log 3.7 + 2 . 51 Re√f → 1 √f = − 2.0log ( 3 × 10 − 4 3.7 + 2.51 0.9707 × 10 5√f ) It gives f = 0.01956. Then, the pressure drop, the head loss, and the useful pumping power required become L ρV 2 ΔP = ΔP L = f D hL = ΔP L ρg 2 24 m = 0.01956 0.005 m L V2 24 m = f D 2g = 0.01956 0.005 m Ẇ pump = V̇ΔP = ṁΔP ρ = (665.1 kg/m 3)(6.892 m/s) 2 (6.892 m/s) 2 2(9.81 m/s 2) (0.09 kg/s)(1488 kPa ) 665.1 2 kg/m 3 ( ( 1 kN 1000 kg ⋅ m/s = 227 m 1 kW 1 kPa ⋅ m 3/s ) = 0.201 kW )( 1 kPa 1 kN/m 2 ) = 1488 kPa 2. Award: 33.33 out of 33.33 points The water needs of a small farm are to be met by pumping water from a well that can supply water continuously at a rate of 5 L/s. The water level in the well is 20 m below the ground level, and water is to be pumped to a large tank on a hill, which is 58 m above the ground level of the well, using 6-cm internal diameter plastic pipes. The required length of piping is measured to be 510 m, and the total minor loss coefficient due to the use of elbows, vanes, etc. is estimated to be 12. Taking the efficiency of the pump to be 77 percent, determine the rated power of the pump that needs to be purchased, in kW. The density and viscosity of water at anticipated operation conditions are taken to be 1000 kg/m3 and 0.00131 kg/m⋅s, respectively. The plastic pipe is smooth, and thus ε = 0. The rated power of the pump that needs to be purchased is 6.72 kW. References Worksheet Difficulty: Medium The water needs of a small farm are to be met by pumping water from a well that can supply water continuously at a rate of 5 L/s. The water level in the well is 20 m below the ground level, and water is to be pumped to a large tank on a hill, which is 58 m above the ground level of the well, using 6-cm internal diameter plastic pipes. The required length of piping is measured to be 510 m, and the total minor loss coefficient due to the use of elbows, vanes, etc. is estimated to be 12. Taking the efficiency of the pump to be 77 percent, determine the rated power of the pump that needs to be purchased, in kW. The density and viscosity of water at anticipated operation conditions are taken to be 1000 kg/m3 and 0.00131 kg/m⋅s, respectively. The plastic pipe is smooth, and thus ε = 0. The rated power of the pump that needs to be purchased is 6.71 ± 10% kW. Explanation: The following assumptions have been made here: 1. The flow is steady and incompressible. 2. The entrance effects are negligible, and thus the flow is fully developed. 3. The total minor loss coefficient due to the use of elbows, vanes, etc is estimated to be 12. 4. Water level in the well remains constant. 5. Both the well and the reservoir are open to the atmosphere. 6. The kinetic energy correction factors are the same at both the inlet and exit. We choose point 1 at the free surface of well water and point 2 at the free surface of the reservoir in the farm. We note the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and the fluid velocities at both points are low (V1 ≅ V2 ≅ 0). We take the free surface of the well as the reference level (z1 = 0). Then, the energy equation for a control volume between these two points simplifies to P1 ρg + α1 V 21 2g + z 1 + h pump, u = P2 ρg + α2 V 22 2g + z 2 + h turbine, e + h L → h pump, u = z 2 + h L where z2 = 58+20 = 78 m and ( L h L = h L , total = h L , major + h L , minor = f D + ∑ K L ) 2 V2 2g The average velocity in the pipe and the Reynolds number are V= V̇ Ac Re = = ρVD μ V̇ 0.005 m 3/s = ( ) ( πD 2 π(0.06 m) 2 4 4 = ) = 1.768 m/s (1000 kg/m 3)(1.768 m/s)(0.06 m) 0.00131 kg/m ⋅ s = 80977 which is greater than 4000. Therefore, the flow is turbulent. The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the Colebrook equation: ( ) () ϵ 1 √f D = − 2.0log 3.7 + 2 . 51 Re√f → 1 √f ( = − 2.0log 0 + 2.51 80977√f ) It gives f = 0.01881. The sum of the loss coefficients is given to be 12. Then, the total head loss, the useful pump head, and the required pumping power become ( hL = f L D ) ( 2 + ∑ KL V2 2g = (0.01881) 510 m 0.06 m + 12 ) (1.768 m/s) 2 2(9.81 m/s 2) = 27.38 m h pump, u = z 2 + h L = 78 + 27.38 m = 105.38 m Ẇ pump = V̇ρgh pump, u η pump (0.005 m 3/s)(1000 kg/m 3)(9.81 m/s 2)(105.38 m = 0.77 ) ( 1 kN 1000 kg ⋅ m/s 2 )( 1 kW 1 kN ⋅ m/s ) = 6.71 kW 3. Award: 33.34 out of 33.34 points A certain part of cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a diameter of 30 cm, and the flow is fully turbulent. One of the branches (pipe A) is 1500 m long, while the other branch (pipe B) is 2300 m long. If the flow rate through pipe A is 0.4 m3/s, determine the flow rate through pipe B. Disregard minor losses and assume the water temperature to be 15°C. Is the flow turbulent or laminar? Is the friction factor dependent or independent of the Reynold's number? The density and dynamic viscosity of water at 15°C are ρ = 999.1 kg/m3 and μ = 1.138 × 10–3 kg/m·s. The roughness of the cast iron pipe is ε = 0.00026 m. The flow rate through pipe B is 0.323 m3/s. The flow is turbulent , and thus the friction factor is independent of the Reynolds number. References Worksheet Difficulty: Medium A certain part of cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a diameter of 30 cm, and the flow is fully turbulent. One of the branches (pipe A) is 1500 m long, while the other branch (pipe B) is 2300 m long. If the flow rate through pipe A is 0.4 m3/s, determine the flow rate through pipe B. Disregard minor losses and assume the water temperature to be 15°C. Is the flow turbulent or laminar? Is the friction factor dependent or independent of the Reynold's number? The density and dynamic viscosity of water at 15°C are ρ = 999.1 kg/m3 and μ = 1.138 × 10–3 kg/m·s. The roughness of the cast iron pipe is ε = 0.00026 m. The flow rate through pipe B is 0.323 ± 10% m3/s. The flow is turbulent , and thus the friction factor is independent of the Reynolds number. Explanation: The following assumptions have been made here: 1. The flow is steady and incompressible. 2. The entrance effects are negligible, and thus the flow is fully developed. 3. The minor losses are negligible. 4. The flow is fully turbulent and thus the friction factor is independent of the Reynolds number (to be verified). The average velocity in pipe A is VA = V̇ Ac = V̇ = ( ) ( 0.4 m 3/s πD 2 π(0.30 m) 2 4 4 ) = 5.659 m/s When two pipes are parallel in a piping system, the head loss for each pipe must be same. When the minor losses are disregarded, the head loss for fully developed flow in a pipe of length L and diameter D is ˙ ˙ V 1 = 0.91 m 3/s, V 2 = 2.09 m 3/s Writing this for both pipes and setting them equal to each other, and noting that DA = DB (given) and fA = fB (to be verified) gives 2 LA VA fA D A 2g 2 LB VB = fB D B 2g → VB = VA √ LA LB √ = (5.659 m/s) 1500 m 2300 m = 4.57 m/s Then, the flow rate in pipe B becomes V̇ B = A cV B = ( ) ( πD 2 4 VB = π(0.30 m) 2 4 ) (4.57 m/s) = 0.323 m 3/s Prove that flow is fully turbulent, and hence the friction factor is independent of Reynolds number, The velocity in pipe B is lower. Therefore, if the flow is fully turbulent in pipe B, then it is also fully turbulent in pipe A. The Reynolds number in pipe B is Re B = ρV BD μ = (999.1 kg/m 3)(4.57 m/s)(0.30 m) 1.138 × 10 − 3 kg/m ⋅ s = 1.2 × 10 6 which is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the pipe is ε D = 0.00026 m 0.30 m = 0.00087 From Moody’s chart, we observe that for a relative roughness of 0.00087, the flow is fully turbulent for the Reynolds number greater than about 106. Therefore, the flow in both pipes is fully turbulent, and thus the assumption that the friction factor is the same for both pipes is valid.