A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
1 answer:
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Answer:
BOD concentration at the outflow = 17.83 mg/L
Explanation:
given data
flow rate of Q = 4,000 m³/day
BOD1 concentration of Cin = 25 mg/L
volume of the pond = 20,000 m³
first-order rate constant equal = 0.25/day
to find out
What is the BOD concentration at the outflow of the pond
solution
first we find the detention time that is
detention time t =
detention time t =
detention time = 5 days
so
BOD concentration at the outflow of pond is express as
BOD concentration at the outflow =
here k is first-order rate constant and t is detention time and Cin is BOD1 concentration
so
BOD concentration at the outflow =
BOD concentration at the outflow = 17.83 mg/L
Answer:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Explanation:
Given:
- The diameter of the pipe d = 5-in
- The pipe is supported every L = 9 ft of pipe in length
- The weight if the pipe + contents W = 10 lb/ft
Find:
determine the magnitude and location of the maximum bending moment in members AC and BC.
Solution:
- The figure (missing) is given in the attachment.
- We will first determine the external forces acting on each member:
Section: 9-ft section of pipe.
Sum of forces perpendicular to member AC = 0
F_d - 0.8*W*L = 0
F_d = 0.8*10*9 = 72 lb
Sum of forces perpendicular to member BC = 0
F_e - 0.6*W*L = 0
F_e = 0.6*10*9 = 54 lb
F_d = 72 lb , F_e = 54 lb
- Then we will determine the support reactions for each member AC point A and BC point B.
Section: Entire Frame.
Sum of moments about point B = 0
-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0
-A_y*(1.5625) + 15 + 39.375 = 0
A_y = 34.8 lb
Sum of forces in vertical direction = 0
A_y + B_y - 0.8*F_d - 0.6*F_e = 0
B_y = 0.8*(72) + 0.6*(54) - 34.8
B_y = 55.2 lb
Sum of forces in horizontal direction = 0
A_x + B_x - 0.6*F_d + 0.8*F_e = 0
A_x + B_x = 0
Section: Member AC
Sum of moments about point C = 0
F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0
72*2.5 - 34.8*12 - 9*A_x = 0
A_x = -237.6 / 9 = - 26.4 lb
B_x = - A_x = 26.4 lb
A_x = -26.4 lb , B_x = 26.4 lb
- Now we will calculate bending moment for each member at different sections.
Member AC:
From point A till just before point D
-0.6*A_x*x - A_y*0.8*x + M = 0
15.84*x - 27.84*x + M = 0
M = 12*x ..... max value at D, x = 12.25 in
M_max = 12*12.25/12 = 12.25 lb-ft
Member BC:
From point B till just before point E
-0.8*B_x*x + B_y*0.6*x + M = 0
-21.12*x + 33.12*x + M = 0
M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in
M_max = -12*8.75/12 = -8.75 lb-ft
- The maximum bending moments and their locations are:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
