1 Prelab Solutions

8/12/2019 1 Prelab Solutions

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Fluids Prelab

1 Gauge Pressure

Ignoring air pressure, show that the pressure caused by a column of water

1.00 inches tall is 248.9 Pa.

The difference in pressure between the top and the bottom of a column of fluid of height

h is

P= g h

The density of water is

water= 1000kg

m3

h= 1.00 inches = 2.54 cm = 2.54 102 m

P = g h=

103

kg

m3

9.8

m

s2

(2.54 102 m)

P= 248.9

kg m

s2

1

m2

= 248.9

N

m2= 248.9 Pa

The difference in pressure P is

P= 248.9 Pa

2 Atmospheric Pressure

A smooth, thin, square piece of rubber, eight inches by eight inches, is sitting

on a smooth table at sea level. What is the total force of the atmosphere on

the material?

Pressure is force per unit area,

P =F

A

1


2/4

so the force F is

F=P A

Atmospheric pressure is

P = 1 atm = 14.7 lbs

in2

The area A of the piece of rubber is

A= (8 in) (8 in) = 64in2

F =P A=

14.7

lbs

in2

(64in2) = 940 lbs

So the force on the piece of rubber is

F= 940 lbs

Do you think you could lift this weight?

3 The Weight of Air

What is the change in the weight of a hollow cylinder of height 10 cm and

radius 4 cm when the air is pumped out of it?

Density is mass per unit volume,

=m

V

so the mass of the air in a cylinder of volume V is

mair =airV

The density of air is

air = 1.20kgm3

The radius r of the cylinder is

r= 4 cm = 0.04 m

The area A of the base of the cylinder is

A= r2 = 0.00502 m2

The height h of the cylinder is

2


3/4

h= 10 cm = 0.1 m

Since you know the area of the base of the cylinder and the height of the cylinder,

you can calculate the volume in cubic meters,

V =A h= r2 h= 0.000502 m3

Now that you know the density of air and the volume of the air in the cylinder, youcan calculate the mass of the air in the cylinder.

The change in the mass of the cylinder is thus

mair =airV =

1.20

kg

m3

(0.000502 m3) = 0.0006 kg = 0.6 g

So the change in the mass of the cylinder is

mair = 0.0006 kg = 0.6 g

Since weightisw= mg, the change in the weight of the cylinder is

wair =mairg = (0.0006 kg)

9.8m

s2

= 0.006

kg m

s2 = 0.006 N

The change in the weightof the cylinder is thus

w= 0.006 N

4 Pressure versus Elevation

Assume there is a 30 foot difference in elevation between the first and the

third floors of Thimann laboratories. What is the difference in atmospheric

pressure between the two floors, expressed in inches of water?

The change in elevation between the first and the third floors is H = 30 ft. The

difference in atmospheric pressure is

P=airg H

The density of air is

air = 1.2kg

m3

1 foot = 0.3048 m

3


4/4

H= 30 feet = 30 (0.3048 m) = 9.1 m

The difference in atmospheric pressure is thus

P=airg H

P =airg H=

1.20

kg

m3

9.8

m

s2

(9.1 m)

P= 116

kg m

s2

1

m2

= 116

N

m2= 116 Pa

So the difference in pressure P is

P= 116 Pa

The problem asks you to express this difference in pressures in terms of the height y

of a column of water. You can apply the same equation,P =waterg y

You now know the difference P in the pressure between the first and third floors.Calculate the height y of the column of water.

P =waterg y

P =airg H=waterg y

airH=watery

y = airwater

H

You know the density water of water,

water= 1000kg

m3

y = airwater

H=

1.20kg / m3

1000kg / m3

(9.1 m) =

1.2

10009.1 m = 0.011 m = 1.1 cm

1 inch = 2.54 cm

1 cm = 12.54

in = 0.39 in

Then

y= airwater

H= 1.1 cm = 1.1 (0.39 in) = 0.47 in

So the change in elevation between the first and the third floors represents a differencein atmospheric pressure P= 116 Pa, which can be expressed as a risey in a column ofwater,

y= airwater

H= 0.47 in

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1 Prelab Solutions