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程序代写案例-CIVE97034

时间：2021-01-03

Page 1 of 5

CIVE97034 - EE & HWRM

IMPERIAL COLLEGE LONDON

MSc EXAMINATION 20XX

ENVIRONMENTAL ENGINEERING &

HYDROLOGY AND WATER RESOURCE MANAGEMENT

This paper also forms part of the relevant examination for the

Answer TWO questions

All questions carry equal marks

Please answer each question in a SEPARATE answer book

© 20XX Imperial College London

Page 2 of 5

CIVE97034 - EE & HWRM

Note that a formulae sheet is provided at the end of the paper

1. Answer all parts of this question

A closed structure with a sloped side wall shown below in Figure 1.1 is under pressure from

the fluid with density of 1=800 kg m

-3 and the air. The width of the structure is L=3 m. If the

pressure in the air, pair = -10.2 kPa, calculate the following:

(a) Hydraulic head for the fluid.

(1 mark)

(b) The magnitude, direction and point of application of horizontal components of the

hydrostatic force acting on the sloped side wall.

(10 marks)

(c) The magnitude, direction and point of application of the resulting horizontal force.

(2 marks)

(d) The magnitude and direction of vertical components of the hydrostatic force acting on

the sloped side wall.

(6 marks)

(e) The magnitude and direction of the resulting vertical force.

Figure 1.1

Equations needed:

Moment of inertia for rectangle: =

ℎ3

12

Volume of triangular prism: =

1

2

ℎ, where b and h are the width and height of the

prism respectively and L is the length of the prism

(1 mark)

Page 3 of 5

CIVE97034 - EE & HWRM

2. Answer all parts of this question

A channel with a rectangular cross-section has a flume, which restricts the channel width from

1 m upstream of the flume to 0.75 m within the throat of the flume (see Figure 2.1 below).

Downstream of the flume the width expands again to 1 m. The flow in the throat of the flume

is critical. Downstream of the flume a hydraulic jump forms, with subcritical depth y4=0.9 m.

The flow rate in the channel is Q=1 m3 s-1.

(a) Calculate the critical depth (y2) and the specific energy in the throat.

(4 marks)

(b) Calculate the flow depth upstream of the flume (y1).

(5 marks)

(c) Calculate the supercritical depth prior to the hydraulic jump (y3).

(2 marks)

(d) Sketch specific energy-depth curves for both the 0.75 m wide and 1 m wide sections

and mark the four depths (y1 to y4) on the corresponding curves.

(6 marks)

(e) List three assumptions made in the above calculations.

(3 marks)

Figure 2.1

Page 4 of 5

CIVE97034 - EE & HWRM

3. Answer all parts of this question

A large diameter circular pipe (D1=400mm) is extended with a smaller diameter pipe

(D2=200mm) and fixed to the bottom of the reservoir (shown in Figure 3.1 below). Both pipes

are 5m long, and the water flows out into the atmosphere from the pipe 2 in the cross-section

2. Local energy loss coefficients for the entry cross-section and pipe contraction are 2=0.5

and 2=0.3, respectively. The Darcy-Weisbach friction factors for pipes 1 and 2 are 1=0.03

and 2=0.02, respectively.

Note that datum level corresponds to the centre of the pipes.

(a) If the water level in the reservoir is HR=1.25 m, calculate the flow rate Q through the

system.

(6 marks)

(b) Calculate the pressure in the cross-section 1.

(3 marks)

(c) Calculate and sketch the total force acting on both pipes, taking account of the fact that

the relevant volume of the fluid is between cross-sections 1 and 2.

(9 marks)

(d) Sketch the energy and hydraulic head lines along the pipes from the reservoir to the

outlet from the pipe 2.

Figure 3.1

Equation needed:

Volume of the cylinder: = (2/4) ∙

(2 marks)

Page 5 of 5

CIVE97034 - EE & HWRM

Environmental fluid mechanics: formulae sheet

Eccentricity:

= −

Euler’s equation

( + ) +

+

= 0

Manning’s equation:

=

2 2

4/3

Chezy’s equation:

=

2

2

Darcy-Weisbach

equation:

=

4

2

2

Conjugate depths:

2

1

=

1

2

(√1 + 81

2 − 1)

Gradually varied

flow:

=

0 −

1 − 2

Bed shear stress:

0 =

St Venant equations:

+

=

1

+

+

+ =

Muskingum:

+1 =

∆

−2

∆

+2−2

−1

+1 +

∆

+2

∆

+2−2

−1

+

−

∆

+2=2

∆

+2−2

CIVE97034 - EE & HWRM

IMPERIAL COLLEGE LONDON

MSc EXAMINATION 20XX

ENVIRONMENTAL ENGINEERING &

HYDROLOGY AND WATER RESOURCE MANAGEMENT

This paper also forms part of the relevant examination for the

Diploma of Imperial College

CIVE97034 – Environmental Fluid Mechanics

Monday: January 20XX Duration: 1h Answer TWO questions

All questions carry equal marks

Please answer each question in a SEPARATE answer book

© 20XX Imperial College London

Page 2 of 5

CIVE97034 - EE & HWRM

Note that a formulae sheet is provided at the end of the paper

1. Answer all parts of this question

A closed structure with a sloped side wall shown below in Figure 1.1 is under pressure from

the fluid with density of 1=800 kg m

-3 and the air. The width of the structure is L=3 m. If the

pressure in the air, pair = -10.2 kPa, calculate the following:

(a) Hydraulic head for the fluid.

(1 mark)

(b) The magnitude, direction and point of application of horizontal components of the

hydrostatic force acting on the sloped side wall.

(10 marks)

(c) The magnitude, direction and point of application of the resulting horizontal force.

(2 marks)

(d) The magnitude and direction of vertical components of the hydrostatic force acting on

the sloped side wall.

(6 marks)

(e) The magnitude and direction of the resulting vertical force.

Figure 1.1

Equations needed:

Moment of inertia for rectangle: =

ℎ3

12

Volume of triangular prism: =

1

2

ℎ, where b and h are the width and height of the

prism respectively and L is the length of the prism

(1 mark)

Page 3 of 5

CIVE97034 - EE & HWRM

2. Answer all parts of this question

A channel with a rectangular cross-section has a flume, which restricts the channel width from

1 m upstream of the flume to 0.75 m within the throat of the flume (see Figure 2.1 below).

Downstream of the flume the width expands again to 1 m. The flow in the throat of the flume

is critical. Downstream of the flume a hydraulic jump forms, with subcritical depth y4=0.9 m.

The flow rate in the channel is Q=1 m3 s-1.

(a) Calculate the critical depth (y2) and the specific energy in the throat.

(4 marks)

(b) Calculate the flow depth upstream of the flume (y1).

(5 marks)

(c) Calculate the supercritical depth prior to the hydraulic jump (y3).

(2 marks)

(d) Sketch specific energy-depth curves for both the 0.75 m wide and 1 m wide sections

and mark the four depths (y1 to y4) on the corresponding curves.

(6 marks)

(e) List three assumptions made in the above calculations.

(3 marks)

Figure 2.1

Page 4 of 5

CIVE97034 - EE & HWRM

3. Answer all parts of this question

A large diameter circular pipe (D1=400mm) is extended with a smaller diameter pipe

(D2=200mm) and fixed to the bottom of the reservoir (shown in Figure 3.1 below). Both pipes

are 5m long, and the water flows out into the atmosphere from the pipe 2 in the cross-section

2. Local energy loss coefficients for the entry cross-section and pipe contraction are 2=0.5

and 2=0.3, respectively. The Darcy-Weisbach friction factors for pipes 1 and 2 are 1=0.03

and 2=0.02, respectively.

Note that datum level corresponds to the centre of the pipes.

(a) If the water level in the reservoir is HR=1.25 m, calculate the flow rate Q through the

system.

(6 marks)

(b) Calculate the pressure in the cross-section 1.

(3 marks)

(c) Calculate and sketch the total force acting on both pipes, taking account of the fact that

the relevant volume of the fluid is between cross-sections 1 and 2.

(9 marks)

(d) Sketch the energy and hydraulic head lines along the pipes from the reservoir to the

outlet from the pipe 2.

Figure 3.1

Equation needed:

Volume of the cylinder: = (2/4) ∙

(2 marks)

Page 5 of 5

CIVE97034 - EE & HWRM

Environmental fluid mechanics: formulae sheet

Eccentricity:

= −

Euler’s equation

( + ) +

+

= 0

Manning’s equation:

=

2 2

4/3

Chezy’s equation:

=

2

2

Darcy-Weisbach

equation:

=

4

2

2

Conjugate depths:

2

1

=

1

2

(√1 + 81

2 − 1)

Gradually varied

flow:

=

0 −

1 − 2

Bed shear stress:

0 =

St Venant equations:

+

=

1

+

+

+ =

Muskingum:

+1 =

∆

−2

∆

+2−2

−1

+1 +

∆

+2

∆

+2−2

−1

+

−

∆

+2=2

∆

+2−2