# 01 The Nature of Fluids

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### 01 The Nature of Fluids

(Water Resources I)

Dave Morgan

Prepared using Lyx, and the Beamer class in LATEX2ε,

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### Recommended Text

A recommended text to accompany these notes is Applied Fluid Mechanics by Mott:

1.3, 1.4, 1.6 (omit US units), 1.7, 1.8, 1.9, 1.11 Study Example Problems: 1.5 - 1.9

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### Elementary Properties of Fluids

Fluids can be either liquid or gas

A liquid tends to ow and conform to the shape of its container

Liquids are not readily compressible (for the purpose of this course, we consider them to be imcompressible)

A gas tends to expand to ll the closed container it is in (or to disperse if not contained).

Gases are readily compressible

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### Primary Units

SI units are used. Four primary units will be used extensively in this course:

Quantity SI unit Dimension Length metre, m L

Mass kilogram, kg M Time second, s T Temperature Kelvin, K θ

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### Derived Units

Quantity SI Unit Dimensions velocity m/s LT−1 acceleration m/s2 LT2 force N, newton MLT−2 kg·m/s2 energy J, joule (work) N · m ML2T−2 kg m2/s2 power N · m/s ML2T3 J/s

pressure Pa, pascal ML−1T2

(stress) N/m2

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### Derived Units:

Quantity SI Unit Dimensions Volume ow rate, Q m3/s L3T−1

L/s

Weight ow rate, W N/s ML−1S−2

kg/m/s2

Mass ow rate, M kg/s MT−1

Specic weight, γ N/m3 ML2T2

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### Pressure

Pressure is given by:

p = F A

It is the force per unit area on a surface, where 1 N/m2=1 Pa (pascal)

Blaise Pascal (1623 - 1662), after whom the Pascal programming language was named, determined the following principles:

1 Pressure acts uniformly in all

directions on a small volume of a uid at rest

2 In a uid conned by solid

boundaries, pressure acts

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### Pressure

Pressure is given by:

p = F A

It is the force per unit area on a surface, where 1 N/m2=1 Pa (pascal)

Blaise Pascal (1623 - 1662), after whom the Pascal programming language was named, determined the following principles:

1 Pressure acts uniformly in all

directions on a small volume of a uid at rest

2 In a uid conned by solid

boundaries, pressure acts

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### Pascal's Laws

Pressure acts uniformly in all directions on a small volume of a uid at rest. The forces must balance out

(i.e. Σ Fx= ΣFy=0); otherwise the volume of uid will not be in equilibrium and cannot remain at rest.

Also, the volume must be suciently small that we do not have to consider the mass, and therefore the weight, of the volume of uid. If the weight is not negligible, the upward pressure on the bottom of the volume will have to be greater than the downward pressure on the top of the volume so that

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### Pascal's Laws

In a uid conned by solid boundaries, pressure acts

perpendicularly to the boundaries. Why is this true?

(Consider a small volume of uid at rest against one of the boundaries? If this volume remains at rest, what are the forces that act upon it?)

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### Density

Density is mass per unit volume: ρ = m

V

The density of water between 0◦C and 15C is close to

1000 kg/m3.

It has a maximum density at 4◦C.

Above 15◦C, the density drops steadily to a density of 958 kg/m3

at 100◦C.

(There is a table of values for the properties of water at the back of Applied Fluid Mechanics by Mott, or from numerous other sources)

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### Density

Density is mass per unit volume: ρ = m

V

The density of water between 0◦C and 15C is close to

1000 kg/m3.

It has a maximum density at 4◦C.

Above 15◦C, the density drops steadily to a density of 958 kg/m3

at 100◦C.

(There is a table of values for the properties of water at the back of Applied Fluid Mechanics by Mott, or from numerous other sources)

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### Specic Weight

Specic weight is weight per unit volume: γ = w

V

Water has a specic weight of 9.81 kN/m3 between 0C and 15C.

Since w = mg, it follows that: γ = w

V = mg

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### Specic Weight

Specic weight is weight per unit volume: γ = w

V

Water has a specic weight of 9.81 kN/m3 between 0C and 15C.

Since w = mg, it follows that: γ = w

V = mg

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### Specic Weight

Specic weight is weight per unit volume: γ = w

V

Water has a specic weight of 9.81 kN/m3 between 0C and 15C.

Since w = mg, it follows that: γ = w

V = mg

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### Specic Gravity

Specic gravity is the ratio of the density (or specic weight) of a substance to the density (or specic weight) of water at 4◦C.

Then, the specic gravity of a substance s is given by sg = γs

γw@4C =

ρs ρw@4C

The density of gasoline at 25◦C is 680 kg/m3 and the density of

water at 4◦C is 1000 kg/m3 . Therefore, the specic gravity of

gasoline at 25◦C is sg= 680/1000 = 0.68.

The specic weight of mercury at 25◦C is 132.8 kN/m3 and the

specic weight of water at 4◦C is 9.81 kN/m3 so the specic

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### Specic Gravity

Specic gravity is the ratio of the density (or specic weight) of a substance to the density (or specic weight) of water at 4◦C.

Then, the specic gravity of a substance s is given by sg = γs

γw@4C =

ρs ρw@4C

The density of gasoline at 25◦C is 680 kg/m3 and the density of

water at 4◦C is 1000 kg/m3 . Therefore, the specic gravity of

gasoline at 25◦C is sg= 680/1000 = 0.68.

The specic weight of mercury at 25◦C is 132.8 kN/m3 and the

specic weight of water at 4◦C is 9.81 kN/m3 so the specic

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### Specic Gravity

Specic gravity is the ratio of the density (or specic weight) of a substance to the density (or specic weight) of water at 4◦C.

Then, the specic gravity of a substance s is given by sg = γs

γw@4C =

ρs ρw@4C

The density of gasoline at 25◦C is 680 kg/m3 and the density of

water at 4◦C is 1000 kg/m3 . Therefore, the specic gravity of

gasoline at 25◦C is sg= 680/1000 = 0.68.

The specic weight of mercury at 25◦C is 132.8 kN/m3 and the

specic weight of water at 4◦C is 9.81 kN/m3 so the specic

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### Nature of Fluids

Example

Calculate the pressure produced in the oil in a closed cylinder by a piston with diameter 7.5 cm exerting a force of 11175 N

7.5 cm 11175 N Solution p = FA = 11175 N π (0.075)2/4 m2 = 2529500 Pa = 2.53 MPa

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### Nature of Fluids

Example

Calculate the pressure produced in the oil in a closed cylinder by a piston with diameter 7.5 cm exerting a force of 11175 N

7.5 cm 11175 N Solution p = FA = 11175 N π (0.075)2/4 m2 = 2529500 Pa = 2.53 MPa

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### Nature of Fluids

Example

Calculate the pressure produced in the oil in a closed cylinder by a piston with diameter 7.5 cm exerting a force of 11175 N

7.5 cm 11175 N Solution p = FA = 11175 N π (0.075)2/4 m2 = 2529500 Pa = 2.53 MPa

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### Nature of Fluids

Example

Calculate the weight of 1 m3 of kerosene if it has a mass of 823 kg

Solution

W = mg

= 823 kg × 9.81 m/s2 = 8070 N

Note: In general, use 5 signicant gures for interim calculations and 3 signicant gures for displayed solutions.

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### Nature of Fluids

Example

Calculate the weight of 1 m3 of kerosene if it has a mass of 823 kg

Solution

W = mg

= 823 kg × 9.81 m/s2 = 8070 N

Note: In general, use 5 signicant gures for interim calculations and 3 signicant gures for displayed solutions.

(24)

### Nature of Fluids

Example

Calculate the density and the specic weight of benzene if its specic gravity is 0.876. Solution 0.876 = ρb ρwater@4C ρb = 0.876 × 1000 kg/m3 = 876 kg/m3 0.876 = γb γwater@4C γb = 0.876 × 9.81 kN/m3 = 8.59 kN/m3

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### Nature of Fluids

Example

Calculate the density and the specic weight of benzene if its specic gravity is 0.876. Solution 0.876 = ρb ρwater@4C ρb = 0.876 × 1000 kg/m3 = 876 kg/m3 0.876 = γb γwater@4C γb = 0.876 × 9.81 kN/m3 = 8.59 kN/m3

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### Nature of Fluids

Example

A cylindrical tank with diameter 12.0 m contains water at 20◦C to

a depth of 4.0 m. If the water is heated to 65◦C, what is the depth

of the water? (Assume that the tank dimensions remain constant and that there are no losses due to evaporation.)

Solution Volume at 20◦C: V20=πd 2h20 4 = π (12.0 m)2(4.0 m) 4 =452.39 m3 Mass of water in the tank:

m = ρV20=998 kg/m3×452.39 m3=451490 kg

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### Nature of Fluids

Example

A cylindrical tank with diameter 12.0 m contains water at 20◦C to

a depth of 4.0 m. If the water is heated to 65◦C, what is the depth

of the water? (Assume that the tank dimensions remain constant and that there are no losses due to evaporation.)

Solution Volume at 20◦C: V20= πd 2h20 4 = π (12.0 m)2(4.0 m) 4 =452.39 m3 Mass of water in the tank:

m = ρV20=998 kg/m3×452.39 m3=451490 kg

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### Nature of Fluids

Solution (continued) V20= πd 2h20 4 = π (12.0 m)2(4.0 m) 4 =452.39 m3 m = ρ20V20=998 kg/m3×452.39 m3=451490 kg Volume at 65◦C: V65= m ρ65= 451490 kg 981 kg/m3 =460.23 m3 Depth at 65◦C: h65=4V65 πd2 = 4 × 460.23 m3 π (12.0)2 =4.0639 m The depth at 65◦C is 4.06 m

References

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