(a)Complete the table below for the three slope and flow classifications Type Re
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Question
(a)Complete the table below for the three slope and flow classifications Type Relationship between Normal depth Type of Type of flow d,) and critical dc slo dn> de Critical Critical (1 mark) (b)Calculate the discharge for a flow depth of 1.2 m in a trapezoidal channel with bottom width 1.5 m and side slope (batters) 1V:3H. The channel bed slope is 0.8 m/100 m and Manning's coefficient (n) is 0.03 The flow from the above part enters a 2.0 m diameter storm water sewer laid at a gradient of 0.007 m/m for which Appendix 2 applies. Use Appendix 3 to obtain the depth of the uniform flow in the sewer and also the velocity of the flow (10 marks) (c)The head difference across a 65 mm circular orifice, operating in a submerged condition, is 56 mm of water. The co-efficient of discharge (Ca) for flow condition is 0.60. Compute the discharge through the orifice (2 marks) (d) Uniform flow in a rectangular channel is 0.65 m deep and has a velocity of 3.8 m/s. Classify the flow and bed slope (2 marks)Explanation / Answer
a) For dn>dc,Slope will be Mild and flow will be Subcritical
For dn<dc, Slope will be Steep and flow will be Supercritical
b) Discharge by Manning's equation is given by
Q=1/n A R2/3S1/2
Where n is Mannings coefficient =0.03
R is Hydraulic Radius =Area/Perimeter
Area=y(B+my), Here B=1.5m,y=1.2m,m=3
Area=1.2(1.5+3x1.2)=6.12m2
Perimeter=B+2y(m2+1) =1.5+2x1.2(32+1)=9.09m
S is slope=0.8/100
Q=1/0.03x6.12x(6.12/9.09)2/3(0.8/100)1/2=14.016m3/s
You will have to use Appendix 3 which is not given for the next part
Velocity is given by the relation v=1/n R2/3S1/2
c) All valus are given only formula is to be applied
Q=Cd2gh=0.6(2x9.81x56x10-3)=0.629m3/s
d)First we shall find the FroudeNumber Fr
Fr=v/(gy)=3.8/(9.81x0.65)= 1.5
For Fr>1, flow is supercritical
yn=0.65m
For finding critical depth we shall use the relation
Q2/g=A3/T
This is same equation for maximum discharge condition. Hence for rectangular channel width equals twice of delth of flow
Here T is top width equal to 2xcrticial depth of flow
Area= 1/2 x 2yc x yc=yc2
(0.65x3.8)2/9.81=yc6/2yc
yc=1.04m
yc>yn
Hence slope is Steep
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