Find: Q [L/s] L h1 h1 = 225 [m] h2 h2 = 175 [m] Q

Find: Q [L/s] L h1 h1 = 225 [m] h2 h2 = 175 [m] Q 2.3 5.7 9.2 13.8
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 2.3 5.7 9.2 13.8 Find the flowrate, Q, in liters per second. [pause] In this problem, 2, non-pressurized tanks, ---

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 2.3 5.7 9.2 13.8 of known total head values, are connected by a ---

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 2.3 5.7 9.2 13.8 smooth pipe, of known length, and diameter.

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 2.3 5.7 9.2 13.8 [pause] Since the flowrate, Q, equals ---

Find: Q [L/s] L Q = V * A h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 the average velocity of flow times the cross-sectional area, the flowrate can be determined by calculating --- Q = V * A

Find: Q [L/s] L Q = V * A ƒ(d) h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 the area, which is a function of the diameter, and a fairly straight forward calculation, and by determining the --- Q = V * A ƒ(d)

Find: Q [L/s] L Q = V * A ƒ(d) ε ƒ(d, h1, h2, L, h1 h1 = 225 [m] h2
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 =1.02*10-6 s the velocity, which is a function of all the variables, and a bit more involved. [pause] Q = V * A ƒ(d) ε D ƒ(d, h1, h2, L, , , …)

Find: Q [L/s] L Q = V * A ƒ(d) h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 Let’s begin with the area, which equals --- Q = V * A ƒ(d)

Find: Q [L/s] π L Q = V * A ƒ(d)= d2 h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 PI divided by 4, times the diameter of the pipe, squared. If the problem doesn’t specifically state --- π Q = V * A ƒ(d)= d2 * 4

Find: Q [L/s] π L Q = V * A ƒ(d)= d2 t do di h1 h1 = 225 [m] h2
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 an inside diameter, outside side diameter, or mention the wall thickness, --- π Q = V * A ƒ(d)= d2 * t do 4 di pipe

Find: Q [L/s] π L Q = V * A ƒ(d)= d2 t do d di h1 h1 = 225 [m] h2
d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 we’ll assume the given diameter, d, equals the inside diameter, --- π Q = V * A ƒ(d)= d2 * t do 4 d di pipe

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 and can be used to find the flow area of a pressurized pipe. π Q = V * A ƒ(d)= d2 * t do 4 d di pipe

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 The area equals times 10 to the –3, meters squared. [pause] π Q = V * A ƒ(d)= d2 * t do 4 d di A=7.854*10-3 [m2] pipe

d = 10 [cm] h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 L Q “smooth pipe” m2 s =1.02*10-6 Next we’ll determine the velocity, v. π Q = V * A ƒ(d)= d2 * t do 4 d di A=7.854*10-3 [m2] pipe

Find: Q [L/s] L Q = V * A h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 Q “smooth pipe” m2 s =1.02*10-6 Our strategy will be to utilize --- Q = V * A

Find: Q [L/s] ? L Q = V * A h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 2 equations, common to pressure flow problems.

Find: Q [L/s] ? L Q = V * A hL=f h1 h1 = 225 [m] h2 h2 = 175 [m] Q * *
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f The Darcy-Weisbach equation for headloss, --- * * d 2*g

Find: Q [L/s] ? L Q = V * A hL=f Re= h1 h1 = 225 [m] h2 h2 = 175 [m] Q
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f and the equation for Reynold’s number ---- * * d 2*g V*d Re=

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f used with the Moody Diagram. [pause] Let’s begin with the --- * * d 2*g f V*d Re= ε d Re

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f headloss equation, and determine our known variables. [pause] * * d 2*g f V*d Re= ε d Re

Find: Q [L/s] ? L Q = V * A hL=f h1 h1 = 225 [m] h2 h2 = 175 [m] Q * *
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f The total headloss through the pipe equals --- * * d 2*g

Find: Q [L/s] ? L Q = V * A hL=f h1-h2 h1 h1 = 225 [m] h2 h2 = 175 [m]
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f the total head at Tank 1, minus the total head at Tank 2. * * d 2*g h1-h2

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f 225 meters, minus 175 meters, equals --- * * d 2*g h1-h2 225 [m]-175 [m]

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f 50 meters. * * d 2*g h1-h2 225 [m]-175 [m] 50 [m]

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f hL=50 [m] The problem statement provides, the length, L, --- * * d 2*g h1-h2 225 [m]-175 [m] 50 [m]

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f hL=50 [m] the diameter, d, [pause] and we know the --- * * d 2*g h1-h2 225 [m]-175 [m] 50 [m]

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f hL=50 [m] m gravitational acceleration, g. The variables we don’t know are --- * * 9.81 d 2*g s2 h1-h2 225 [m]-175 [m] 50 [m]

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f hL=50 [m] m the friction factor, f, and the velocity, v. [pause] If these unknown variables are isolated, the equation becomes, --- * * 9.81 d 2*g s2 h1-h2 225 [m]-175 [m] 50 [m]

Find: Q [L/s] ? L Q = V * A hL=f f * V2 = h1 h1 = 225 [m] h2
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s =1.02*10-6 L V2 hL=f hL=50 [m] the friction factor time the velocity squared, equals, the headloss times the diameter, times 2g, all divided by the length. * * d 2*g hL*D*2*g f * V2 = L

Find: Q [L/s] ? L Q = V * A hL=f f * V2 = h1 h1 = 225 [m] h2
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] Plugging in the values we just solved for, the right hand side --- * * d 2*g hL*d*2*g f * V2 = L

Find: Q [L/s] ? L Q = V * A hL=f f * V2 = f * V2 =0.04905 h1
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] of the equation equals meters squared per second squared. [pause] The second equation is --- * * d 2*g hL*d*2*g f * V2 = L m2 f * V2 = s2

Find: Q [L/s] ? L Q = V * A hL=f Re= f * V2 = f * V2 =0.04905 h1
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] Reynolds's number equals the velocity times the diameter, divided by the kinematic viscosity. * * d 2*g hL*d*2*g V*d Re= f * V2 = L m2 f * V2 = s2

Find: Q [L/s] ? L Q = V * A hL=f Re= f * V2 = f * V2 =0.04905 h1
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] Since the problem provides the diameter, d, and kinematic viscosity, nu, we’ll again isolate the --- * * d 2*g hL*d*2*g V*d Re= f * V2 = L m2 f * V2 = s2

Find: Q [L/s] ? L Q = V * A hL=f Re= = f * V2 = f * V2 =0.04905 h1
d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] unknown variables on one side of the equation, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L m2 f * V2 = s2

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 s m =1.02*10-6 9.81 L V2 s2 hL=f hL=50 [m] plug in the known variables, and get the velocity --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L m2 f * V2 = s2

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 m =1.02*10-6 9.81 s L V2 s2 hL=f hL=50 [m] divided by the Reynolds’s number, equals 1.02 times 10 to the –5 meters per second. * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m s m2 f * V2 = =1.02*10-5 s2 Re

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 m =1.02*10-6 9.81 s L V2 s2 hL=f hL=50 [m] Right now, we effective have 2 equations, and 3 unknown variables. Those unknown variables are the --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m s m2 f * V2 = =1.02*10-5 s2 Re

d = 10 [cm] L h1 h1 = 225 [m] Tank d h2 h2 = 175 [m] 1 Tank L = 2,000 [m] 2 ? Q “smooth pipe” Q = V * A m2 m =1.02*10-6 9.81 s L V2 s2 hL=f hL=50 [m] friction factor, velocity, and Reynolds’s number. The last piece of information we’ll need is the --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

Find: Q [L/s] f Re hL=f Re= = f * V2 = f * V2 =0.04905 h1 = 225 [m]
d = 10 [cm] f Moody h1 = 225 [m] Diagram h2 = 175 [m] ε d L = 2,000 [m] Re “smooth pipe” m2 m =1.02*10-6 9.81 s L V2 s2 hL=f hL=50 [m] Moody Diagram, which relates the friction factor to the Reynolds’s number, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f Moody h1 = 225 [m] Diagram h2 = 175 [m] ε d L = 2,000 [m] Re “smooth pipe” m2 =1.02*10-6 s L V2 hL=f hL=50 [m] based on the relative roughness, epsilon over d. The problem states the pipe --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f Moody h1 = 225 [m] Diagram h2 = 175 [m] ε d L = 2,000 [m] Re “smooth pipe” m2 =1.02*10-6 s L V2 hL=f hL=50 [m] is a smooth pipe, which means the roughness can be thought of, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f Moody h1 = 225 [m] Diagram h2 = 175 [m] ε =0 d L = 2,000 [m] Re “smooth pipe” m2 =1.02*10-6 s L V2 hL=f hL=50 [m] as zero. Therefore, we’ll use the bottom-most relative roughness line, for turbulent flow, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f Moody h1 = 225 [m] Diagram h2 = 175 [m] ε =0 d L = 2,000 [m] smooth Re “smooth pipe” m2 =1.02*10-6 s L V2 hL=f hL=50 [m] on the Moody Diagram, as our third and final relationship needed to solve for the three unknown variables. [pause] The following method, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f h1 = 225 [m] h2 = 175 [m] ε L = 2,000 [m] smooth =0 d “smooth pipe” Re m2 =1.02*10-6 s L V2 hL=f hL=50 [m] is just one method, among many, to iteratively solve for the unknown variables. * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

d = 10 [cm] f h1 = 225 [m] h2 = 175 [m] ε L = 2,000 [m] smooth =0 d “smooth pipe” Re m2 =1.02*10-6 s L V2 hL=f hL=50 [m] Looking back at our first equation, we’ll solve for the velocity term, which equals, ---- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

Find: Q [L/s] f Re hL=f Re= = f * V2 = f * V2 =0.04905 ε =0 hL=50 [m]
smooth =0 d Re m2 =1.02*10-6 s L V2 hL=f hL=50 [m] The square root of the quantity meters squared per seconds squared, divided by the friction factor. [pause] Next, the second equation --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

smooth =0 d Re m2 =1.02*10-6 s L V2 hL=f hL=50 [m] is solved for the Reynolds’s number, which equals, --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

smooth =0 m d 1.02*10-5 s Re m2 =1.02*10-6 s L V2 hL=f hL=50 [m] the velocity divided by 1.02 times 10 to the –5, meters per second. [pause] Then, a table will be set up --- * * d 2*g V hL*d*2*g V*d Re= = f * V2 = Re d L V m2 m f * V2 = =1.02*10-5 s2 Re s

Find: Q [L/s] f Re f V(f) [m/s] Re(f) - Re(v) 0.04905 V= f V Re=
1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f to track the calculations as we search for the most accurate friction factor, to the nearest thousandth. The method we’ll use, is as follows. First, ---

Find: Q [L/s] f Re f f1 V(f) [m/s] Re(f) - Re(v) 0.04905 V= f V Re=
1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f f1 we’ll guess a reasonable value for the friction factor, f sub 1. [pause] Then, ---

Find: Q [L/s] f Re f f1 V(f) [m/s] Re(f) - Re(v) 0.04905 V= f V Re=
1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f f1 we’ll solve for the corresponding velocity, using the first equation, ---

Find: Q [L/s] f Re f f1 v1 V(f) [m/s] Re(f) - Re(v) 0.04905 V= f V Re=
1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f f1 add this velocity value to the table, and then use it to solve for the --- v1

Find: Q [L/s] f Re f f1 v1 V(f) [m/s] Re(f) - Re(v) 0.04905 V= f V Re=
1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f f1 Reynolds number, using the second equation, --- v1

Find: Q [L/s] f Re f f1 v1 V(f) [m/s] Re(f) - Re(v) Re(f)1 - Re(v)1
s2 f V= f V Re= m s 1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f f1 and add that value to the table. [pause] Knowing we’re dealing with a smooth pipe, --- v1 Re(f)1 - Re(v)1

Find: Q [L/s] f Re f f1 v1 f1 V(f) [m/s] Re(f) - Re(v) Re(f)1 - Re(v)1
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 a second Reynolds number will be determined using the friction factor and Moody diagram. v1 Re(f)1 - Re(v)1

s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 The difference between these two Reynolds numbers represents the --- v1 Re(f)1 - Re(v)1

s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 magnitude of error between the actual friction factor, f, and the evaluated friction factor, f sub 1. And the sign of this difference indicates --- v1 Re(f)1 - Re(v)1

s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 whether the next friction factor should be greater than or less than the friction factor just evaluated. v1 Re(f)1 - Re(v)1 or

Find: Q [L/s] f Re f f1 v1 f2 f1 V(f) [m/s] Re(f) - Re(v)
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 Then the process repeats, using a second friction factor, --- v1 Re(f)1 - Re(v)1 or f2

Find: Q [L/s] f Re f f1 v1 f2 v2 f1 V(f) [m/s] Re(f) - Re(v)
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 which generates a second velocity, v1 Re(f)1 - Re(v)1 or f2 v2

Find: Q [L/s] f Re f f1 v1 f2 v2 f1 V(f) [m/s] Re(f) - Re(v)
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 and a tighter bound on the most accurate value of f. v1 Re(f)1 - Re(v)1 or f2 v2 Re(f)2 - Re(v)2 or

Find: Q [L/s] f Re f f1 v1 f2 v2 f3 v3 f1 V(f) [m/s] Re(f) - Re(v)
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 The more iterations, the closer we get to the actual friction factor. v1 Re(f)1 - Re(v)1 or f2 v2 Re(f)2 - Re(v)2 or f3 v3 Re(f)3 - Re(v)3 or

Find: Q [L/s] f Re f f1 v1 f2 v2 f3 v3 f4 v4 f1 V(f) [m/s]
s2 f V= f V Re= f1 m s 1.02*10-5 Re Re(f)1 f V(f) [m/s] Re(f) - Re(v) next f f1 [pause] Now for some actual numbers, --- v1 Re(f)1 - Re(v)1 or f2 v2 Re(f)2 - Re(v)2 or f3 v3 Re(f)3 - Re(v)3 or f4 v4 Re(f)4 - Re(v)4

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 we’ll first try a friction factor of 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 Solving for velocity, we get, --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 1.401 meters per second. Next, Reynolds number is calculated --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 using this velocity [pause] And the second Reynolds number is determined, --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 from the friction factor and the diagram. This second Reynolds number is an approximation. 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 The difference between Reynolds numbers, is a large negative number, --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 -117,352 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 so the second friction factor will be much smaller than the first. Next we’ll try --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 -117,352 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 0.015, for f. The velocity and Reynolds numbers --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 are calculated. And the third value of f should be --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 slightly larger than the second. [pause] Splitting the difference, ---- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 22,715 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 0.020 is evaluated, --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 1.566 60,000 - 153,533 0.020 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 and is too large, --- 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 0.020 1.566 60,000 - 153,533 1.699 135,000 - 166,530 0.017 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 f equal to is also too large, but we’re getting closer. 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 0.020 1.566 60,000 - 153,533 0.017 1.699 135,000 - 166,530 0.016 1.751 183,000 - 171,656

1.02*10-5 Re f V(f) [m/s] Re(f) - Re(v) next f 0.025 0.016 is the most accurate value for f, rounded to the nearest thousandth. [pause] 1.401 20,000 - 137,352 1.808 200,000 - 177,285 0.015 0.020 1.566 60,000 - 153,533 0.017 1.699 135,000 - 166,530 0.016 1.751 183,000 - 171,656

Find: Q [L/s] f f V(f) [m/s] 0.04905 V= f V Re= 1.02*10-5 0.025 1.401
By knowing the velocity in the pipe, we can now solve --- 1.401 1.808 0.015 0.020 1.566 0.017 1.699 0.016 1.751

Find: Q [L/s] f π f Q = V * A A= d2 V(f) [m/s] * 4 7.854*10-3 [m2]
s2 f V= f V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 the original equation, for flowrate, by multiplying ---- 1.401 7.854*10-3 [m2] 1.808 0.015 0.020 1.566 0.017 1.699 0.016 1.751

s2 f V= f V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 1.751, meters per second, by times 10 to the –3, meters squared. The flowrate equals, --- 1.401 7.854*10-3 [m2] 1.808 0.015 0.020 1.566 0.017 1.699 0.016 1.751

s2 f V= f V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 meters cubed per second, or --- 1.401 7.854*10-3 [m2] 1.808 0.015 m3 0.020 1.566 Q = s 0.017 1.699 0.016 1.751

s2 f V= f V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 13.75 liters per second. [pause] 1.401 7.854*10-3 [m2] 1.808 0.015 m3 L 0.020 1.566 Q = * 1,000 s m3 0.017 1.699 Q = [L/s] 0.016 1.751

Find: Q [L/s] π f Q = V * A A= d2 2.3 5.7 9.2 13.8 V(f) [m/s] * 4
s2 2.3 5.7 9.2 13.8 V= f V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 When reviewing the possible solutions, --- 1.401 7.854*10-3 [m2] 1.808 0.015 m3 L 0.020 1.566 Q = * 1,000 s m3 0.017 1.699 Q = [L/s] 0.016 1.751

Find: Q [L/s] π AnswerD f Q = V * A A= d2 2.3 5.7 9.2 13.8 V(f) [m/s]
s2 2.3 5.7 9.2 13.8 V= f AnswerD V Re= m s 1.02*10-5 π f V(f) [m/s] Q = V * A A= d2 * 4 0.025 the answer is D. 1.401 7.854*10-3 [m2] 1.808 0.015 m3 L 0.020 1.566 Q = * 1,000 s m3 0.017 1.699 Q = [L/s] 0.016 1.751

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4

Find: Q [L/s] L h1 h1 = 225 [m] h2 h2 = 175 [m] Q

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Find: Q [L/s] L h1 h1 = 225 [m] h2 h2 = 175 [m] Q

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Presentation on theme: "Find: Q [L/s] L h1 h1 = 225 [m] h2 h2 = 175 [m] Q"— Presentation transcript:

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