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Expected Value - Advanced Hydrology - Problems, Exercises of Engineering Dynamics

These are the Problems of Advanced Hydrology which includes Excess Ordinate, Data, Area Method, Histogram, Calculation, Contribution etc.Key important points are: Expected Value, Pressure, Point, Monotonically, Increasing Function, Wind Velocity, Problem

Typology: Exercises

2012/2013

Uploaded on 03/28/2013

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ekanath 🇮🇳

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V is wind velocity with a pdf f(v) = 1/10; 0 v 10. The pressure at point is
given by = 0.003 v . Find the expected value of pressure.
Sol : 1
f(v) = ; 0 v 10; = 0.003 v
10
E( ) =
≤≤ ω
ω
≤≤ ω
ω
2
2
?
E( ) = g( )d , given = 0.003 v or d = 0.006 vdv
1
given, f(v) = & = 0.003 v g( )will be monotonically increasing function.
10
dv 1 0.0274
Therefore, g( ) = f(v) d 10
=
−α
ω ω ωω ω ω
ω ∴ω
ω=×
ωω
12
2
0.00274( ) as = 0.003 v
ωω
Module 7
Example Problem
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2

V is wind velocity with a pdf f(v) = 1/10; 0 v 10. The pressure at point is

given by = 0.003 v. Find the expected value of pressure.

Sol : 1 f(v) = ; 0 v 10; = 0.003 v 10

E( ) =

≤ ≤ ω

ω

≤ ≤ ω

ω

2

2

?

E( ) = g( )d , given = 0.003 v or d = 0.006 vdv

given, f(v) = 1 & = 0.003 v g( )will be monotonically increa sing function. 10

dv 1 0. Therefore, g( ) = f(v) d 10

=

  • α

− α

ω ω ω ω ω ω

ω ∴ ω

ω = × ω (^) ω

1 2 2 0.00274( ) as = 0.003 v

− ω ω

Module 7

Example Problem

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1 2

1 1 2 2

0.3 3 1 2 2

0 0

or v = 18.257 or 18.257 ( ) 0.003 (^2)

or g(w) = 18.257 ( ) = 0.9129 (w) 10 2

Now, E(w) = 0.9129 (w) = 0. 3 2

= 0.9129 [0.3]

− (^) −

= = × ×

× × ×

× ×

× ×

dv w (^) w w

dw

w

w w dw

3 2

0 w 0.

= 0.1 (Ans.)

 ≤^ ≤ 

v

or w

or

Module 7

Example Problem Contd…

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