Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000 Find the minimum pipe diameter, capital D, in inches. [pause] In this problem, --- D d S=0.005 max = 0.70 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000 a concrete sewer pipe, positioned at a half percent slope, conveys the sewage generated by a population of 30,000 people. D d max = 0.70 S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 12 16 18 24 P=30,000 people 18+P/1000 PF= 4+P/1000 The maximum allowable d over D is 0.70, and the peaking factor and generation rate are provided. [pause] Since the depth of water in the pipe --- D d max = 0.70 S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 PF= is at most 70% of the pipe diameter, --- D d S=0.005 max = 0.70 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 PF= d<0.7*D 4+P/1000 D d then the diameter of the pipe is at least, ---- D d = 0.70 S=0.005 max concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = d P=30,000 people < 0.70 D 18+P/1000 d PF= d<0.7*D D> 4+P/1000 0.70 D d the depth of water in the pipe, divided by 0.70. [pause] Our strategy will be to utilize --- D d S=0.005 max = 0.70 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = Q= A * R2/3 * S 1/2 d P=30,000 people < 0.70 D PF= d<0.7*D D> 4+P/1000 0.70 k Q= A * R2/3 * S 1/2 * n D d Manning’s equation, solve for the area and --- concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = Q= A * R2/3 * S 1/2 d P=30,000 people < 0.70 D PF= d<0.7*D D> 4+P/1000 0.70 k Q= A * R2/3 * S 1/2 * n D d hydraulic radius, and then, --- concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = Q= A * R2/3 * S 1/2 A * R2/3 = d P=30,000 people < 0.70 D 18+P/1000 d PF= d<0.7*D D> 4+P/1000 0.70 k Q= A * R2/3 * S 1/2 * n D d develop an expression for the left hand side of this equation --- Q * n A * R2/3 = k * S 1/2 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = Q= A * R2/3 * S 1/2 A * R2/3 = A*R2/3=ƒ(D) d P=30,000 people < 0.70 D 18+P/1000 d PF= d<0.7*D D> 4+P/1000 0.70 k Q= A * R2/3 * S 1/2 * n D d in terms of the the diameter, D, and then solve for D directly. [pause] So let’s begin by solving --- Q * n A * R2/3 = k * S 1/2 A*R2/3=ƒ(D) concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 4+P/1000 D d for the right hand side of this equation, which equals, the flowrate times the roughness coefficient, divided by k times the square root of the slope. S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 4+P/1000 D d We’ve been given the slope, which is 0.005, or 1/2%. [pause] The problem states the --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 4+P/1000 D d pipe is concrete, so we can look up it’s roughness coefficient, n, --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 0.010 plastic brass concrete brick CMP 0.011 0.013 0.015 0.022 material n 4+P/1000 D d in a standard table of values. The roughness for concrete is --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] ? = A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 0.010 plastic brass concrete brick CMP 0.011 0.013 0.015 0.022 material n 4+P/1000 D d most always 0.013. S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= 0.010 plastic brass concrete brick CMP 0.011 0.013 0.015 0.022 material n 4+P/1000 D d [pause] When Manning’s equation is used for English units, --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= k=1.49 4+P/1000 D d the value for k equals 1.486, or 1.49. [pause] Lastly, we turn to our flowrate, Q. S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 0.013 A * R2/3 = Q * n P=30,000 people k * S 1/2 PF= k=1.49 4+P/1000 D d The average flowrate of sewage through the pipe equals, --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Q * n P=30,000 people k * S 1/2 18+P/1000 PF= k=1.49 4+P/1000 average sewage Qave = P * generation rate D d the total population, P, times the average sewage generation rate. For sake of space, I’ll abbreviate --- S=0.005 concrete gal average sewage sewer = 90 generation rate person*day pipe

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D d this generation rate with, R sub a-v-e. [pause] However, in sewer design, we’ll want to account for the peak flowrate through the pipe, --- gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF d which equals the average flowrate, times a peaking factor, PF. gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d This peaking factor is a function of the population, P, and assumed to account for both seasonal and daily variation in flowrates. gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d Plugging in our known values for the --- Qpeak= P * Rave* PF(P) gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d population, --- Qpeak= P * Rave* PF(P) gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d the generation rate, --- Qpeak= P * Rave* PF(P) gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d and the peaking factor, after it’s value --- Qpeak= P * Rave* PF(P) gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d is calculated using the population, P, the peak flowrate computes to, --- Qpeak= P * Rave* PF(P) gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d 4.96 cubic feet per second. [pause] 4.96 is then plugged into --- Qpeak= P * Rave* PF(P) Qpeak= 4.96 [cfs] gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Qave = P * Qave = P * Rave Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 average sewage Qave = P * generation rate Qave = P * Rave D Qpeak= Qave* PF(P) d the variable Q, in addition to the other three values, --- Qpeak= P * Rave* PF(P) Qpeak= 4.96 [cfs] gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = 0.013 A * R2/3 = Q= 4.96 [cfs] Q * n P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 Q= 4.96 [cfs] D d and the right hand side of the equation becomes --- gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs] Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 Q= 4.96 [cfs] A * R2/3 = 0.6137 D d 0.6137 [pause] The units have to been dropped from the equation at this step, so we’ll have to keep in mind that the --- gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs] Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 Q= 4.96 [cfs] A * R2/3 = 0.6137 hydraulic D d area is in units of feet squared, and the hydraulic radius is in units of feet. radius [ft] area [ft2] gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = A * R2/3 = 0.6137 0.013 A * R2/3 = Q= 4.96 [cfs] Q * n 0.013 P=30,000 people A * R2/3 = k * S 1/2 18+P/1000 PF= S=0.005 k=1.49 4+P/1000 Q= 4.96 [cfs] A * R2/3 = 0.6137 hydraulic D d Now it’s time to develop an expression for the left hand side of the equation, in terms of the pipe diameter, capitol D. radius [ft] area [ft2] gal average sewage = Rave=90 generation rate person*day

Find: min D [in] = A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D) P=30,000 people A * R2/3 = 0.6137 18+P/1000 PF= R = ƒ (D) 4+P/1000 A = ƒ (D) D d We can divide and conquer this problem by solving for the area and hydraulic radius separately. gal average sewage = Rave=90 generation rate person*day

Find: min D [in] A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D) A = ƒ (D) D d If we increase the pipe figure, ---

Find: min D [in] θ A * R2/3 = 0.6137 R = ƒ (D) A = ƒ (D) D d and define the angle, theta, from geometry, we know the cross-sectional area of the water equals, ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = 1 8 D d * 1/8 times the quantity, theta minus sin theta, times the diameter squared. [pause] And the hydraulic radius equals, ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ d θ one quarter times the quantity, 1 minus sin theta divided by theta, time the diameter. [pause] In both of these equations, ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ d θ radians theta is measured in radians. [pause] To find theta, ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ d θ radians we’ll take a closer look at this triangle, or more specifically, ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ d θ radians half the triangle. Since we’re solving for d over D ---

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ d θ radians equal to 0.70, we’ll substitute in --- d = 0.70 D

Find: min D [in] θ (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * D 4 θ θ 0.70 * D radians 0.70 times the diameter, for the depth. [pause] Since the radius is --- d = 0.70 D d= 0.70 * D

Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ R= 1- * D * 4 θ 0.7 *D radians 0.5 * D half the diameter, we know the small leg of the triangle is --- d = 0.70 D d= 0.70 * D

Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ 0.2 * D R= 1- * D * 4 θ 0.7 *D radians 0.5 * D 0.2, times the diameter, and the hypotenuse of the triangle is --- d = 0.70 D d= 0.70 * D

Find: min D [in] (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 1 sin θ 0.2 * D R= 1- * D * 4 θ 0.5 * D 0.7 *D radians 0.5 * D 0.5, times the diameter. [pause] Knowing this is a right triangle, --- d = 0.70 D d= 0.70 * D

Find: min D [in] α (θ-sin θ) * D2 A * R2/3 = 0.6137 A = sin θ R= 1- θ 8 α 1 sin θ 0.2 * D R= 1- * D * 4 θ 0.5 * D 0.7 *D radians 0.5 * D the angle defined as alpha, equals, --- d = 0.70 D d= 0.70 * D

Find: min D [in] α α = cos-1 (θ-sin θ) * D2 A * R2/3 = 0.6137 A = 8 α 1 sin θ 0.2 * D R= 1- * D * 4 θ 0.5 * D 0.7 *D radians 0.5 * D the arc cosine of 0.2 time D divided by 0.5 times D, which equals, --- d = 0.70 D 0.2 * D α = cos-1 d= 0.70 * D 0.5 * D

Find: min D [in] α α = cos-1 (θ-sin θ) * D2 A * R2/3 = 0.6137 A = 8 α 1 sin θ 0.2 * D R= 1- * D * 4 θ 0.5 * D 0.7 *D radians 0.5 * D 1.159 radians. [pause] From the diagram, we observe theta equals --- d = 0.70 D 0.2 * D α = cos-1 d= 0.70 * D = 1.159 rad 0.5 * D

Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 * 8 α α 1 sin θ R= 1- * D * 4 θ θ radians 2 PI radians minus 2 times alpha, or, θ = 2 * π - 2 * α d = 0.70 D 0.2 * D α = cos-1 d= 0.70 * D = 1.159 rad 0.5 * D

Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 * 8 α α 1 sin θ R= 1- * D * 4 θ θ radians theta equals, 3.964, radians. [pause] This value of theta is next plugged into the equations --- θ = 2 * π - 2 * α d θ = 3.964 rad = 0.70 D 0.2 * D α = cos-1 d= 0.70 * D = 1.159 rad 0.5 * D

Find: min D [in] θ α α α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 * 8 α α 1 sin θ R= 1- * D * 4 θ θ for area and hydraulic radius, and the area equals, --- θ = 2 * π - 2 * α θ = 3.964 0.2 * D α = cos-1 = 1.159 rad 0.5 * D

Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964 A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 = 0.5873 * D2 * 8 1 sin θ R= 1- * D * 4 θ 0.5873 times the diameter squared, and the hydraulic radius equals, --- θ = 2 * π - 2 * α θ = 3.964 0.2 * D α = cos-1 = 1.159 rad 0.5 * D

Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964 A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 = 0.5873 * D2 * 8 1 sin θ R= 1- * D = 0.2962 * D * 4 θ 0.2962, times the diameter. [pause] θ = 2 * π - 2 * α θ = 3.964 0.2 * D α = cos-1 = 1.159 rad 0.5 * D

Find: min D [in] α = cos-1 (θ-sin θ) * D2 θ = 2 * π - 2 * α θ = 3.964 A * R2/3 = 0.6137 1 A = (θ-sin θ) * D2 = 0.5873 * D2 * 8 1 sin θ R= 1- * D = 0.2962 * D * 4 θ Returning to the previous equation, these value for area and hydraulic radius are --- θ = 2 * π - 2 * α θ = 3.964 0.2 * D α = cos-1 = 1.159 rad 0.5 * D

Find: min D [in] A * R2/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D D d substituted in for A, and R, ---

Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D 0.2610 * D8/3 = 0.6137 D d and after a few lines of algebra, the find that the minimum allowable diameter, D, is, ---

Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D 0.2610 * D8/3 = 0.6137 D = 1.378 [ft] D d 1.378 feet, which equals, ---

Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D 0.2610 * D8/3 = 0.6137 D = 1.378 [ft] = 16.54 [in] D d 16.54 inches [pause] This diameters must be rounded up, ---

Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D 0.2610 * D8/3 = 0.6137 D = 1.378 [ft] 12 16 18 24 = 16.54 [in] D d to the nearest possible solution, ---

Find: min D [in] A * R2/3 = 0.6137 0.2610 * D8/3 = 0.6137 A=0.5873 * D2 R=0.2962 * D 0.2610 * D8/3 = 0.6137 D = 1.378 [ft] 12 16 18 24 = 16.54 [in] D d and the answer is C. answerC

? 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