Find the flow rate of water at 60 F in each pipe. The valve shown in completely closed, neglect minor losses.

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1 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 1 1. Purpose #1 Find the flow rate of water at 60 F in each pipe. The valve shown in completely closed, neglect minor losses. 2. Drawing & Diagrams 3. Sources Applied Fluid Mechanics 7 th Edition BlackBoard 4. Design considerations Valve is closed so now water can flow through it, neglect the minor losses. 5. Data and Variables L= 50 Le=0 E= v=1.21*10^-5 D= ft g=32.2 ft/s

2 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 2 6. Procedure First assume the Q in what direction it is going in what pipe. Then Calculate V=Q/A and Nr=V*D/v. Calculate F by taking (0.25)/(LOG(1/(3.7*D/E)+5.74/Nr^0.9)^2). We find k=8*l*f / (pi^2*g*(d)^5). Then h is just Q*k^2 and 2kQ for literally 2kQ. We add the sum up from h and 2kQ from the 5 pipes and divided h / 2kQ to find the Delta Q. Once we have Delta Q we subtract that from the Q assumed and that will give us the Q(n). To find the percentage change we do Q / Delta Q. 7. Calculations & Summary g ft/s 32.2 D ft A ft^ E v Pipe # L Le Q V Nr F k h 2kQ Delta Q Q(n) % Change Materials -Water -Schedule 40, 2 1/2 inch pipe 9. Analysis Use the fluid mechanics software that makes this process a whole lot easier in the future.

Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 3 1. Purpose #2 Determine the depth when the flow rate of the channel show is 34.7 ft^3/s and has a manning value of 0.

3 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 3 1. Purpose #2 Determine the depth when the flow rate of the channel show is 34.7 ft^3/s and has a manning value of 0.04, if the average slope is Drawing & Diagrams 3. Sources Applied Fluid Mechanics 7 th Edition BlackBoard 4. Design considerations The design seems to be safe with the average slope the length of the channel. If outside and available for public to access, would suggest adding a plate or a grid line pattern to protect. 5. Data and Variables n= 0.04 s= Q=34.7 ft^3/s

4 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 4 6. Procedure Calculate the depth by using equation AR^2/3=nQ/1.49S^1/2 then find height (y) in the Area, WP, and R and calculate by iterating the LHS=RHS in excel. 7. Calculations

5 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 5 8. Summary n Q S Constant RHS Y A WP R R^2/3 LHS Materials -Water -Channel 10. Analysis The calculations show that the height should be 3 feet. The design overall as a channel looks right but if you wanted to increase the slope this would make the height less and take up less space.

Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 6 1. Purpose #4 At the end of a steel pipe there is a valve. When the valve closes all of the sudden there is a pressure increment of 2.

6 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 6 1. Purpose #4 At the end of a steel pipe there is a valve. When the valve closes all of the sudden there is a pressure increment of 2.81 MPa in the pipe. Find the water velocity for such a pressure increment. 2. Drawing & Diagrams 3. Sources Applied Fluid Mechanics 7 th Edition BlackBoard 4. Design considerations Due to closing the pipe suddenly, this will cause water hammer. Instead slowly close the pipe over time so no water hammering occurs and prevents damage to the pipe or injury from explosion.

7 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 7 5. Data and Variables E=2*10^7 N/cm^2 = 2*10^11 N/m^2 E(o)=2.03*10^5 N/cm^2 = 2.03*10^9 N/m^2 Internal Diameter of pipe= 600 mm = 0.6m Thickness of Pipe= 100 mm = 0.01m Density of water = 1000 kg/m^3 Pressure=2.81 MPa = 2,810,000 Pa 6. Procedure Apply the calculation of P= ρ*v*c. In this case we can solve for C by calculate the square root of E/ ρ divided by the square root of (1+ E(o)*Diameter / E * thickness). After obtaining C we plug it in to the equation and solve for the velocity. Pressure in Pascal's is divided by from C and 1000 from ρ. This gives us meters per second for the velocity.

8 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 8 7. Calculations

9 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 9 8. Summary P/c/ρ= v = 2.5 m/s 2,810,000/ /1000= Materials -Water -Steel Pipe -Valve 10. Analysis The design could either be changed to have a robot or computer controlled valve closing the valve automatically and slow enough to prevent damage. So no accidents can occur due to humans not knowing about water hammer can cause an explosion if closed fast.

Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: 10 1. Purpose #5 A free steam of water at 180 F is being deflected by a stationary vane (pictured below) through a 130 angel.

10 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Purpose #5 A free steam of water at 180 F is being deflected by a stationary vane (pictured below) through a 130 angel. The cross-sectional area of the stream is constant at 2.95 in^2 throughout the system. When the entering stream has a velocity of 22 ft/sec the horizontal force exerted on the water by the vane is 30.6 lb. Find the stream velocity if the horizontal force is reduced by half? and what is the vertical force? 2. Drawing & Diagrams 3. Sources Applied Fluid Mechanics 7 th Edition BlackBoard

11 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Design considerations Design looks to be safe and accurate to what the diagram provides. 5. Data and Variables A= 2.95 in^2/144 = ft^2 F(Xo)= 30.6 F(Xnew)= 30.6 / 2 = 15.3 lb ρ = slugs/ft^3 6. Procedure First draw diagram of how the vane and forces are working, then set the Y axis going up and the X axis going to the left. I set the equation so Q is split between V and A. F(x) / ρ * A is the LHS and the RHS is V*(V(x(2))-V(x(1))). Convert Area into ft^2. Find LHS by plugging in. Set V(x(2))= V * cos (50) and V(x(1))= -V * cos(0). Assume a V value in excel and iterate until LHS=RHS. Calculate Q with new V= Set V(y(2))= 15.55*sin(130) and V(y(1)=15.55*sin(0). Then calculate F(y)=1.883*0.3185*11.91= 7.15 lb

12 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Calculations

13 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Summary F(Xo) F(Xnew) Q V A p V(x1) V(x2) V(y1) V(y2) LHS RHS Fy (lb) Materials 180 F -Vane 10. Analysis The design could be changed in making the angle more or less, but would consider not making it turn into a full U-turn shape. I would recommend putting a 90 degree vane and then another 90 degree vane inverted, so it can completely make a full turn. Depending on the change in angle you will have to change the velocity in order to keep it flowing correctly.

14 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Purpose #6 The airfoil with the performance characteristics (shown in the figure below) has a chord length of 1.4 m and a span of 6.8 m. The airfoil is part of an airplane moving at a speed of 200 km/hr in the standard atmosphere at 200 m. It was determined that the lift and drag forces are 15.9 kn and 883 N when the angle of attack is 10 degrees. What should the angle of attack if the airplane weighted 25% more? 2. Drawing & Diagrams 3. Sources Applied Fluid Mechanics 7 th Edition BlackBoard

15 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Design considerations The area is the Span of the blade times the Chord depth, as long as this is kept to a certain minimum on the plane it should not affect the cost. 5. Data and Variables b= 6.8 m c= 1.4 m A= 9.52 m^2 V= 200km/hr = 55.5 m/s F(l)= 15.9 kn Appendix E; Table E.3: 200 m ρ=1.202 kg/m^3 6. Procedure Calculate the area first and convert the velocity of km/hr to m/s and plug into equation solving for C(l) using F(l)=C(l)*( ρ*v^2 / 2) *A. Then convert the F(l) kn to N and multiply by 1.25% that will give you the new weight added. Divided that new total by the LHS without C(l) and that will give you the answer. Compare that C(l) calculated into the graph given and intersect it where the angle meets.

16 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Calculations

17 Kevin Lis Fluid Mechanics MET 330 Test #3 Page Number: Summary C(l)= and estimating that located on the Y axis for the graph given, I located the angle of attack to be 12.5 degrees. 9. Materials 200 meters -Plane Material & airfoil 10. Analysis Seemed like a NASA problem, if the area of the Airfoil is increased then the C(l) is dropped lower which makes the angle drop as well, so it is recommended that you either keep the Area constant and add weight or change the area of the airfoil and keep the weight constant in area to have a change in the angle of attack.

AEROSPACE ENGINEERING DEPARTMENT. Second Year - Second Term ( ) Fluid Mechanics & Gas Dynamics

AEROSPACE ENGINEERING DEPARTMENT Second Year - Second Term (2008-2009) Fluid Mechanics & Gas Dynamics Similitude,Dimensional Analysis &Modeling (1) [7.2R*] Some common variables in fluid mechanics include: