Question

A rectangular channel which is laid on a bottom slope of 0.0064 is to carry 20 m3/sec of water . Determine the width of channel when the flow is in critical condition. Take n =0.015

Answer 1

Solution:- the values given in the question are as follows:

bottom slope(S)=0.0064

discharge(Q)=20 m^3/s

n=0.015

width of channel(B)=?

According to manning's equation-

v=(1/n)*R^(2/3)*S^0.5 , [Eq-1]

where, R=hydraulic radius=R=A/P

A= area of channel and P=perimeter

flow is in critical condition

A=B*yc

yc=critical depth

P=perimeter=B+2yc , (B >>> yc)

P=B

R=(B*yc)/B=yc

R=yc

values put in above equation-(1)

v=(1/0.015)*yc^(2/3)*(0.0064)^0.5

v=5.333*yc^(2/3) , [Eq-2]

we know that-

Q=A*v , [Eq-3]

values put in above equation equation-(3)

20=(B*yc)*5.33yc^(2/3)

20=5.333*B*yc^(5/3) , [Eq-4]

critical depth(yc)-

yc=[q^2/g]^(1/3)

where, q=Q/B=20/B

yc=[{(20/B)^2}/g]^(1/3)

value of yc put in above equation-(4)

20=5.333*B*[{{(20/B)^2}/g}^(1/3)]^(5/3)

20=5.333*B*[{(20/B)^2}/g]^(6/3)

20=5.333*B*[{(20/B)^2}/g]^2

20=5.333*B*(20^4)/(B^4*g^2)

20*g^2*B^3=5.333*20^4

B^3=5.333*20^4/(9.81^2*20)   

where, g=9.81 m^2/s

B^3=443.326

B=7.625 m

width of channel(B)=7.625 m

[Ans]