Question

A rectangular suppressed weir, 1 m in length is used in a reservoir with an area of 50 square yards. During dry season, it was requested for 40 cubic meters per second of water to be discharged using the weir. If the discharge coefficient is 0.72 and the initial head of the weir was 1.20 m, answer the following:

a, What will be the final head acting on the weir after the said volume was discharged?

b. How long will it take for the weir to discharge the demand as stated in the problem?

Answer #1

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

length of weir(L)=1 m

dischrge(Q)=40 m^3/s

dischrage coefficient(Cd)=0.72

initial head(Hi)=1.2 m

area of reservoir(Ar)=50 yard^2 , OR 41.8064 m^2

**(a)**

dischrge formula of rectangular weir

**Q=(2/3)*Cd*L*(2g)^0.5*H^(2/3)**

where, L= length of weir , Cd= coefficient of dischrge

H= head of weir , g=acceleration due to gravity

40=(2/3)*0.72*1*(2*9.81)^0.5*(H)^(2/3)

18.8135=H^(2/3)

H=18.8135^(3/2)

H=7.0736 m

**final head of weir during dischrge 40
m^3/s (Hf)=7.0736 m**

**(b)**

**Q=volume/time**

**Q={area(final head-initial
head)}/time**

40=(41.8064*7.0736-1.2)/t

t=41.8064*5.8636/40=6.1389=6.139 seconds

**time take for weir to dischrge the demand
flow(t)=6.139 s ,** **[Ans]**

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 49 minutes ago

asked 49 minutes ago

asked 54 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago