Operações

Problem 8.1

Problem 8.2

Given: Data on air flow in duct
Find: Volume flow rate for turbulence; entrance length
Solution
The given data is

D = 0.25⋅ m

From Fig. A.3

ν = 1.46⋅ 10

2
−5 m

⋅

s

The governing equations are
Re =

V⋅ D ν Llaminar = 0.06⋅ Recrit⋅ D

Q π Hence

Recrit =

4

π

2

Recrit = 2300

⋅D

Q=

or, for turbulent,

Lturb = 25⋅ D − 40⋅ D

4

⋅D ⋅V

⋅D
2

ν

or

Q=

Recrit⋅ π ⋅ ν ⋅ D
4

3

Q = 0.396

m min Llaminar = 0.06⋅ Recrit⋅ D

or, for turbulent,

Llaminar = 34.5 m

Lmin = 25⋅ D

Lmin = 6.25 m

Lmax = 40⋅ D

Lmax = 10 m

Problem 8.3

Given: That transition to turbulence occurs at about Re = 2300
Find: Plots of average velocity and volume and mass flow rates for turbulence for air and water
Solution

From Tables A.8 and A.10

ρ air = 1.23⋅

ρ w = 999⋅

The governing equations are

Re =

For the average velocity

V=

2
−5 m

kg
3

m

ν air = 1.45 × 10

ν w = 1.14 × 10

3

m

Recrit⋅ ν
D

2
−5 m

Vair =

⋅

Recrit = 2300

ν

D

⋅

s

2
−6 m

kg

V⋅ D

2300 × 1.45 × 10
Hence for air

⋅

s

2

0.0334⋅
Vair =

D

m s s

2
−6 m

2300 × 1.14 × 10
For water

Vw =

⋅

2

0.00262⋅

s

Vw =

D

m s D

For the volume flow rates

Q = A⋅ V =

Hence for air

For water

Qair =

Qw =

π
4

π
4

π
4

2

⋅D ⋅V =

π

2

⋅D ⋅

4

Recrit⋅ ν
D

2
−5 m

× 2300 × 1.45⋅ 10

⋅

s

2
−6 m

× 2300 × 1.14⋅ 10

⋅

s

=

π ⋅ Recrit⋅ ν
4

⋅D

2

⋅D

Qair = 0.0262⋅

m
×D
s

2

⋅D

m
×D
Qw = 0.00206⋅ s Finally, the mass flow rates are obtained from volume flow rates mair = ρ air⋅ Qair

kg
×D
mair = 0.0322⋅ m⋅ s

mw = ρ w⋅ Qw

kg
×D
mw = 2.06⋅ m⋅ s

These results are plotted in the associated Excel workbook

Problem 8.3 (In Excel)

Given: That transition to turbulence occurs