Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.21P
To determine

The stream velocity.

Expert Solution
Check Mark

Answer to Problem 7.21P

Stream velocity U=20ms

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that:

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation:

We get U=2(8251.2)9.81×0.02971.2U=20ms.

To determine

The boundary layer thickness.

Expert Solution
Check Mark

Answer to Problem 7.21P

The boundary layer thickness δ=4.5mm

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that,

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation,

We get U=2(8251.2)9.81×0.02971.2U=20ms ANSWER (a)

As in above discussion boundary layer ends at y=4.5mm

Then this value is equal to boundary layer thickness.

y=δ=4.5mm.

To determine

The wall shear stress.

The wall shear stress τw=0.14pa

Given information:

The readings of y in mm and h in mm as mention in question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that,

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation,

We get U=2(8251.2)9.81×0.02971.2U=20ms ANSWER (a)

As in above discussion boundary layer ends at y=4.5mm

Then this value is equal to boundary layer thickness.

y=δ=4.5mm

We know that:

The wall shear stress τw=μuy|y=0

We can write τwμΔuΔyμuy10y10 ---------equation 1

where y1=0.5×103m is the first value of y from table.

uy1=2Δpρair=2(ρoilρair)ghmano(y1)ρair

=4.02ms

Put required values in equation 1

τw1.8E5(4.0200.00050)τw0.14pa.

The total friction drag.

The total friction drag FD=0.25N

Given information:

The readings of y in mm and h in mm as mentioned in the question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that,

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation:

We get U=2(8251.2)9.81×0.02971.2U=20ms ANSWER (a)

As in above discussion boundary layer ends at y=4.5mm

Then this value is equal to boundary layer thickness.

y=δ=4.5mm

We know that:

The wall shear stress τw=μuy|y=0

We can write τwμΔuΔyμuy10y10 ---------equation 1

where y1=0.5×103m is the first value of y from table.

uy1=2Δpρair=2(ρoilρair)ghmano(y1)ρair

=4.02ms

Put required values in equation 1:

τw1.8E5(4.0200.00050)τw0.14pa

We know that the friction drag at wall is twice of shear stress (τw) times area (xb).

We will use the value of x estimated in question 7.20 (x=0.908m) .and take width b=1.0m

Now FD=2τw(xb)

Put all required values:

We get FD=2×0.14×0.908×1.0FD=0.25N..

Expert Solution
Check Mark

Answer to Problem 7.21P

The wall shear stress τw=0.14pa

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that,

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation,

We get U=2(8251.2)9.81×0.02971.2U=20ms ANSWER (a)

As in above discussion boundary layer ends at y=4.5mm

Then this value is equal to boundary layer thickness.

y=δ=4.5mm

We know that:

The wall shear stress τw=μuy|y=0

We can write τwμΔuΔyμuy10y10 ---------equation 1

where y1=0.5×103m is the first value of y from table.

uy1=2Δpρair=2(ρoilρair)ghmano(y1)ρair

=4.02ms

Put required values in equation 1

τw1.8E5(4.0200.00050)τw0.14pa.

To determine

The total friction drag.

Expert Solution
Check Mark

Answer to Problem 7.21P

The total friction drag FD=0.25N

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mentioned in the question.

For air at 20°C,1atm take

Density of air ρair=1.2kgm3

Density of oil ρoil=S.G×ρwater=0.827×998=825kgm3

Dynamic viscosity μ=1.8E5kgms

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after y=4.5mm, this shows the boundary layer ends here and free stream velocity starts from this location y=4.5mm.

So, we will consider manometer head hmano=0.0297m from table.

We know that,

The stream velocity U=2Δpρair=2(ρoilρair)ghmanoρair

Put the required value in this equation:

We get U=2(8251.2)9.81×0.02971.2U=20ms ANSWER (a)

As in above discussion boundary layer ends at y=4.5mm

Then this value is equal to boundary layer thickness.

y=δ=4.5mm

We know that:

The wall shear stress τw=μuy|y=0

We can write τwμΔuΔyμuy10y10 ---------equation 1

where y1=0.5×103m is the first value of y from table.

uy1=2Δpρair=2(ρoilρair)ghmano(y1)ρair

=4.02ms

Put required values in equation 1:

τw1.8E5(4.0200.00050)τw0.14pa

We know that the friction drag at wall is twice of shear stress (τw) times area (xb).

We will use the value of x estimated in question 7.20 (x=0.908m) .and take width b=1.0m

Now FD=2τw(xb)

Put all required values:

We get FD=2×0.14×0.908×1.0FD=0.25N..

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Chapter 7 Solutions

Fluid Mechanics

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