66. A circular sewer 2m diameter has to carry a discharge of 2m³ /s when flowing nearly full. What is the minimum required slope to initiate the flow? Assume Manning’s N = 0.015.

1. 1. 0.00023

   
   
   

   2. 0.000036

   
   
   

   3. 0.000091

   
   
   

   4. 0.000014

   
   
   

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   Manning’s equation
   

   V =	1	m2/3C.1/2
   	n

   
   

   ⇒	Q	=	1	m2/3C.1/2
   	A		n

   
   

   m = R =	A	=	(π/4)D²	m2/3C.1/2
   	P		πD

   
   

   =	(1/4) × 2²
   	2

   
   

   =	1
   	2

   
   N = 0.015
   G = 2 m³ /s
   

   A =	π	× 2² = π
   	4

   
   

   ⇒	2	=	1	−		1		2/3	(i)1/2
   	π		0.015			2

   
   ∴ i = 0.00023

   Correct Option: A

   Manning’s equation
   

   V =	1	m2/3C.1/2
   	n

   
   

   ⇒	Q	=	1	m2/3C.1/2
   	A		n

   
   

   m = R =	A	=	(π/4)D²	m2/3C.1/2
   	P		πD

   
   

   =	(1/4) × 2²
   	2

   
   

   =	1
   	2

   
   N = 0.015
   G = 2 m³ /s
   

   A =	π	× 2² = π
   	4

   
   

   ⇒	2	=	1	−		1		2/3	(i)1/2
   	π		0.015			2

   
   ∴ i = 0.00023

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67. A trapezoidal channel with bottom width of 3m and side slope of 1V I.5H carries a discharge of 8.0m³ /sec with the flow depth of 1.5m. The Froude number of the flow is

1. 1. 0.066

   
   
   

   2. 0.132

   
   
   

   3. 0.265

   
   
   

   4. 0.528

   
   
   

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   Fr =	V
   	√gD

   
   
   A = [B + My]y
   = (3 + 1.5 × 1.5) 1.5
   = 7.87 m²
   T = B + 2 my
   = 3 + (2 × 1.5 × 1.5)
   = 7.5 m
   

   D =	A	=	7.87	= 1.05 m
   	T		7.5

   
   

   V =	Q	=	8	= 1.016 m/s
   	A		7.87

   
   

   Fr =	V
   	√gD

   
   

   =	1.016	= 0.316 (none is correct)
   	√9.8 × 1.05

   
   If we take D ≈ y ,
   

   Fr =	1.016	= 0.0265
   	√9.81 × 1.5

   Correct Option: C

   Fr =	V
   	√gD

   
   
   A = [B + My]y
   = (3 + 1.5 × 1.5) 1.5
   = 7.87 m²
   T = B + 2 my
   = 3 + (2 × 1.5 × 1.5)
   = 7.5 m
   

   D =	A	=	7.87	= 1.05 m
   	T		7.5

   
   

   V =	Q	=	8	= 1.016 m/s
   	A		7.87

   
   

   Fr =	V
   	√gD

   
   

   =	1.016	= 0.316 (none is correct)
   	√9.8 × 1.05

   
   If we take D ≈ y ,
   

   Fr =	1.016	= 0.0265
   	√9.81 × 1.5

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68. The force ‘F’ required at equilibrium on the semi-cylindrical gate shown below is
    

1. 1. 9.81kN

   
   
   

   2. 0.0kN

   
   
   

   3. 19.62kN

   
   
   

   4. none of the above

   
   
   

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   Taking horizontal & vertical components of hydrostatic forces
   F_H = ρ.g.h A_projected
   

   = 1000 × 9.81 ×		H		× (H × 1)
   		2

   
   

   = 1000 × 9.81 ×		2		× (2 × 1)
   		2

   
   = 19.62 kN/m width.
   F_v = weight of water displaced
   = Mg
   = ρvg = ρg. v
   

   = ρ.g		π	R² × 1
   		2a

   
   

   = 1000 × 9.81 ×	π(1)²	× 1
   	2

   
   = 15.41 kN/m width
   

   Location of centre of pressure, h' =	2	H
   	3

   
   

   =	2	× 2 = 1.33 m
   	3

   
   = 0.424 m
   Moment about hinge,
   ⇒ –F_H × (h' - h) + F_v × x + F × 2/2 = 0
   = – 19.62 × (1.33 – 1) + 15.41 × 0.424 + F × 1 = 0
   ∴ F = 0 kN

   Correct Option: B

   Taking horizontal & vertical components of hydrostatic forces
   F_H = ρ.g.h A_projected
   

   = 1000 × 9.81 ×		H		× (H × 1)
   		2

   
   

   = 1000 × 9.81 ×		2		× (2 × 1)
   		2

   
   = 19.62 kN/m width.
   F_v = weight of water displaced
   = Mg
   = ρvg = ρg. v
   

   = ρ.g		π	R² × 1
   		2a

   
   

   = 1000 × 9.81 ×	π(1)²	× 1
   	2

   
   = 15.41 kN/m width
   

   Location of centre of pressure, h' =	2	H
   	3

   
   

   =	2	× 2 = 1.33 m
   	3

   
   = 0.424 m
   Moment about hinge,
   ⇒ –F_H × (h' - h) + F_v × x + F × 2/2 = 0
   = – 19.62 × (1.33 – 1) + 15.41 × 0.424 + F × 1 = 0
   ∴ F = 0 kN

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69. The circular water pipes shown in the sketch are flowing full. The velocity of flow (in m/s) in the branch pipe ‘R’ is
    

1. 1. 3

   
   
   

   2. 4

   
   
   

   3. 5

   
   
   

   4. 6

   
   
   

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   Q_p = Q_q + Q_r
   A_pV_p= A_qV_q + A_rV_r
   

   π	× 4² × 6 =	π	× 4² × 5 +	π	× 2² × V
   4		4		4

   
   ∴ r = 4 m/s.

   Correct Option: B

   Q_p = Q_q + Q_r
   A_pV_p= A_qV_q + A_rV_r
   

   π	× 4² × 6 =	π	× 4² × 5 +	π	× 2² × V
   4		4		4

   
   ∴ r = 4 m/s.

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70. The top width and the depth of flow in a triangular channel were measured as 4 m and 1 m, respectively. The measured velocities on the centre line at the water surface, 0.2 m and 0.8 m below the surface are 0.7 m/s, 0.6 m/s and 0.4 m/s, respectively. Using two-point method of velocity measurement, the discharge (in m³ /s) in the channel is

1. 1. 1.4

   
   
   

   2. 1.2

   
   
   

   3. 1.0

   
   
   

   4. 0.8

   
   
   

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   Vavg =	V.2y + V.8y
   	2

   
   

   =	0.6 + 0.4	= 0.5 m/s
   	2

   
   

   A =	1	× T × y
   	2

   
   

   A =	1	× 4 × 1 = 2 m²
   	2

   
   Q = Av = 2 × 0.5 = 1 m³ /s.

   Correct Option: C

   Vavg =	V.2y + V.8y
   	2

   
   

   =	0.6 + 0.4	= 0.5 m/s
   	2

   
   

   A =	1	× T × y
   	2

   
   

   A =	1	× 4 × 1 = 2 m²
   	2

   
   Q = Av = 2 × 0.5 = 1 m³ /s.

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