Eric Stone Fluid Mechanics Homework 3.1 Problem 15-5 Purpose: To determine the deflection of manometer fluid for orifice sizes of 1 in. and 7 in. Pipe Reynolds number, NR Diagrams: Sources: Mott, R. Untener, J.A., "Applied Fluid Mechanics", 7 th ed. Pearson Education, Inc, (2015). Design Considerations: • Incompressible fluids • Constant properties • A sharp-edged orifice is used, thus C-values will come from figure 15.7 Data/variables: Q = 25 gal/min(l ft A 3/7.4805 gal)(l min/60 s) = 0.0557 ft A 3/s Aioin = [(pi)(10 in./12 in.) A 2]/4 = 0.5454 ft A 2 Vi = Q / A 10ln = (0.0557) / (0.5454) = 0.102 ft/s [S.G]Ammonia = [0.83] [Rho] water(S40F - 1.98slugs/ft A3 Dynamic viscosity = 2.5 x 10 A (-6) lb.s/ft A2 NRe = [(Vi)(D)[S.G] AmmOnia[Rho]water@40F] / (Dynamic Vis.) = [(0.102)(10/12)(0.83)(1.98)] / (0.0000025) = 54,747 (Betadm = (lin/lOin) = 0.1 Ci, n = 0.595 (Beta) 7i n .= (7in/10in) = 0.7 C7,n = 0.618 Procedure: This problem will utilize equation 15-6, since I have direct information about the flowrate in the pipe. I have already calculated my pertinent information and will re-arrange equation 15-6 in order to solve for the pressure difference. Then, using the manometry relation I will compute the deflection, h, of the manometer fluid, water @ 40F. Calculations: (To follow) Summary: The pressure differential for case a, the 1 in. orifice, will cause a 22.3 ft deflection of the water in manometer. The pressure differential for case b, the 7 in. orifice, will only cause a 0.00655 ft deflection of the manometer water. Materials: Sharp-edged orifice within a 10 in. inner diameter pipe. Analysis: The hardest part of this problem was the rearranging of equation 15-6. The 1 in. orifice will increase fluid velocity greatly, exhibiting a much larger pressure differential across the orifice than that of the 7 in. orifice. This example helped to visualize the impact of geometry.
Transcript
Eric Stone
Fluid Mechanics
Homework 3.1
Problem 15-5
Purpose: To determine the deflection of manometer fluid for orifice sizes of 1 in. and 7 in.
Pipe Reynolds number, NR
Diagrams:
Sources: Mott, R. Untener, J.A., "Applied Fluid Mechanics", 7 t h ed. Pearson Education, Inc, (2015).
Design Considerations:
• Incompressible fluids • Constant properties
• A sharp-edged orifice is used, thus C-values will come from figure 15.7
Data/variables:
Q = 25 gal/min(l ft A3/7.4805 gal)(l min/60 s) = 0.0557 ft A 3/s Aioin = [(pi)(10 in./12 in.) A2]/4 = 0.5454 f t A 2 Vi = Q / A 1 0 l n = (0.0557) / (0.5454) = 0.102 ft/s [S.G]Ammonia = [0.83] [Rho] water(S40F - 1.98slugs/ftA3 Dynamic viscosity = 2.5 x 10 A(-6) lb.s/ftA2 NRe = [(Vi)(D)[S.G]AmmOnia[Rho]water@40F] / (Dynamic Vis.) = [(0.102)(10/12)(0.83)(1.98)] / (0.0000025) = 54,747 (Betadm = (lin/lOin) = 0.1 Ci,n = 0.595 (Beta)7in .= (7in/10in) = 0.7 C7,n = 0.618
Procedure: This problem will utilize equation 15-6, since I have direct information about the flowrate in the pipe. I have already calculated my pertinent information and will re-arrange equation 15-6 in order to solve for the pressure difference. Then, using the manometry relation I will compute the deflection, h, of the manometer fluid, water @ 40F.
Calculations: (To follow)
Summary: The pressure differential for case a, the 1 in. orifice, will cause a 22.3 ft deflection of the water in manometer.
The pressure differential for case b, the 7 in. orifice, will only cause a 0.00655 ft deflection of the manometer water.
Materials: Sharp-edged orifice within a 10 in. inner diameter pipe.
Analysis: The hardest part of this problem was the rearranging of equation 15-6. The 1 in. orifice will increase fluid velocity greatly, exhibiting a much larger pressure differential across the orifice than that of the 7 in. orifice. This example helped to visualize the impact of geometry.
Calculations [15-11, continued
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Problem 15-9
Purpose: To determine an appropriate diameter, d, of a flow nozzle in a copper pipe with an expected flow-rate range.
Sources: Mott, R. Untener, J.A., "Applied Fluid Mechanics", 7 t h ed. Pearson Education, Inc, (2015).
Design Considerations:
• Constant properties • Incompressible fluids • Flow nozzle is to be installed in a 5 in. Type K copper tube • Linseed oil @ 11 F • A mercury manometer is used for pressure differential across the nozzle
o Scale ranges from 0 to 8 in. mercury • Expected range of flow rate from 700 gal/min to 1000 gal/min
Data/variables:
• Qi = 1000 gal/min = 2.23 ft A 3/s Vi = Qi / A5i„ = 17.7 ft/s • Dsm = 0.4004 ft (Gamma)mercury = 844.9 #/ft A3 • A5,n = 0.1259 f t A 2 (Gamma)Linseedoil = 58 #/ft A3 • Kinematic viscosity of linseed oil = 3.84 x 10 A(-4) ft A 2/s • Reynolds = (vD) / kinematic = 17.1(0.4004)/[0.000384] = 18,456
Procedure: Rearranging equation 15-6,1 will solve for the flow area at the nozzle, A2. I will use figure 15.5 in the book in order to approximate a C value using the Beta-line shown (Beta = d/D = 0.5).
Calculations: To follow
Summary: The appropriate flow nozzle diameter would be about 0.326 ft, or 3.91 in, to show mercury level of 6 inches atQmax = 100 gal/min.
Materials: An installed flow nozzle, moving linseed oil @ 77 F, through a 5 in nominal size Type K copper tubing.
Analysis: This calculation is of course an approximation, since I pulled the C-value from figure 15.5. Using the equation for C in flow nozzles would allow me to iterate from this point in order to converge on the exact value of nozzle diameter.
Calculatjgns [15-91. continued
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Eric Stone
Chapter 14 problems
Problem 14-6
Purpose: To compute the Hydraulic Radius for the the figure of a rain gutter shown if water depth is 2 inches.
Diagrams:
Sources: Mott, R. Untener, J.A., "Applied Fluid Mechanics", 7 t h ed. Pearson Education, Inc, (2015).
Procedure: I will compute R directly using the simple formula above by finding first, the area of the cross-section of the flow and second, the wetted perimeter.
Procedure: I will utilize the Manning equation, for U.S. customary units, to solve for the discharge rates. I will do this by calculating the appropriate area and wetter perimeters for the two depths of interest.
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Calculations:
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Summary: For the flow depth of 3 ft, Cbn = [32.2 ft A 3/s] , and for the flow depth of 6 ft, Q 6 f t = [141.1 ft A 3/s]
Materials: Water flowing in a trapezoidal open channel with levees on either side.
Analysis: The difficulty of this problem comes from the 'complex' geometry of the cross-sections. The key to this problem was correctly representing the sums of parts for area, as well as the correct addition of the levees to the base trapezoid shape for calculation of the wetted perimeter.
