Water resource - Open channel flow

[cantaloup]

Critical depth, critical velocity, critical slope…are critical! In design hydraulic conveying system, the engineer needs to know the critical slope to calculate the limit the water can flow just like a transportation engineer designing a section of hilly road, it can’t be too steep nor flat (steep road is unconvenient for traffic and flat road is costly to excavate). Here is a simple problem to solve (this question is from Gupta Hydrology & Hydraulic System textbook):

A trapezoidal channel is planned to be built with a bottom width of 13.1 ft and side slope of 4 to 1 (horizontal to vertical) carries a discharge of 1060 cfs. The engineer wants to know the critical depth so that he can calculate the critical slope for designing a proper flow. What is the nearest value of critical depth in feet?

A/ 3.5

B/ 4.0

C/ 4.5

D/ 5.0

 

[Suns Den]

Critical depth, critical velocity, critical slope…are critical! In design hydraulic conveying system, the engineer needs to know the critical slope to calculate the limit the water can flow just like a transportation engineer designing a section of hilly road, it can’t be too steep nor flat (steep road is unconvenient for traffic and flat road is costly to excavate). Here is a simple problem to solve (this question is from Gupta Hydrology & Hydraulic System textbook):
A trapezoidal channel is planned to be built with a bottom width of 13.1 ft and side slope of 4 to 1 (horizontal to vertical) carries a discharge of 1060 cfs. The engineer wants to know the critical depth so that he can calculate the critical slope for designing a proper flow. What is the nearest value of critical depth in feet?

A/ 3.5

B/ 4.0

C/ 4.5

D/ 5.0

Ac = 13.1 + 4 Yc2

Tc = 13.1 + 8 Yc2

subing in critical depth Q2/g = Ac3/Tc

n solving iteratively

Yc = 7.87 ft

 

[cantaloup]

Ac = 13.1 + 4 Yc2Tc = 13.1 + 8 Yc2

subing in critical depth Q2/g = Ac3/Tc

n solving iteratively

Yc = 7.87 ft

B is the correct answer. (this problem is taken from Gupta Hydraulics book)

The approach is easy but the derivation of the base formula (Q^2*T/G*A^3=1) and the method of solving by iteration would take time, in this case after derive the formula to solve for Y, just plug in the 4 answers to see which one is correct.

I put this formula in the HP33s; to get the answer quick, solve for Y.

G=Q*Q*(B+2*Y*Z)/((B*Y+Z*Y*Y)^3)

In case rectangular channel, put Z=0 or Z=0.0001, G=32.2 for US unit.

 

[Suns Den]

B is the correct answer. (this problem is taken from Gupta Hydraulics book)The approach is easy but the derivation of the base formula (Q^2*T/G*A^3=1) and the method of solving by iteration would take time, in this case after derive the formula to solve for Y, just plug in the 4 answers to see which one is correct.

I put this formula in the HP33s; to get the answer quick, solve for Y.

G=Q*Q*(B+2*Y*Z)/((B*Y+Z*Y*Y)^3)

In case rectangular channel, put Z=0 or Z=0.0001, G=32.2 for US unit.

my bad

as you can see i misplaced my Yc

Ac = 13.1 + 4 Yc2 -> should be Ac = 13.1 Yc+ 4 Yc2

Tc = 13.1 + 8 Yc2 -> should be Tc = 13.1 + 8 Yc

new depth is 3.99785 :deadhorse: for my mistake I got to beat this to death n hoping my mason will keep it close to the fifth

 

[Suns Den]

B is the correct answer. (this problem is taken from Gupta Hydraulics book)The approach is easy but the derivation of the base formula (Q^2*T/G*A^3=1) and the method of solving by iteration would take time, in this case after derive the formula to solve for Y, just plug in the 4 answers to see which one is correct.

I put this formula in the HP33s; to get the answer quick, solve for Y.

G=Q*Q*(B+2*Y*Z)/((B*Y+Z*Y*Y)^3)

In case rectangular channel, put Z=0 or Z=0.0001, G=32.2 for US unit.

I was cheating a bit by using excel, and also messed up my equation in that process

Thanks for the Eqn, its in my Hp now

ck 944A/31