P15 056

background image

56. We use the result of part (a) in the previous problem.

(a) In this case, we have ∆p = p

1

= 2.0 atm. Consequently,

v =



2∆p

ρ ((A/a)

2

1)

=



4(1.01

× 10

5

Pa)

(1000 kg/m

3

) ((5a/a)

2

1)

= 4.1 m/s .

(b) And the equation of continuity yields V = (A/a)v = (5a/a)v = 5v = 21 m/s.

(c) The flow rate is given by

Av =

π

4



5.0

× 10

4

m

2



(4.1 m/s) = 8.0

× 10

3

m

3

/s .


Document Outline


Wyszukiwarka

Podobne podstrony:
P15 033
P15 027
P15 034
p12 056
12 2005 048 056
P15 074
04 2005 056 057
P15 042
P15 045
P15 035
pc 02s054 056
P15 083
p03 056
02 2005 054 056
P15 PRIMOTECQ
p36 056
P15 018

więcej podobnych podstron