1

ENSC3003 – Fluid Mechanics

Semester 1, 2015

Week 1 Lecture Notes

Conservation of Mass Units and Dimensions

© Dr Jeremy W. Leggoe February 2015

2

1.1 Conservation of Mass In order to understand the behaviour of fluids, whether they be stationary or in motion, it is necessary to understand the way in which mass, momentum and energy are transported within a fluid. In this course, we will accordingly focus on understanding and applying three fundamental principles:

• The principle of conservation of mass • The equation of motion (or “momentum balance”) • The mechanical energy equation (or “energy balance”)

Each of these principles will be explored in its continuum form, where it may be used to determine the fluid state at points in a continuum, and in its system ( “macroscale”) form, where it may be used to determine the overall behaviour of a fluid flow system. Of these, conservation of mass is the most fundamental principle. Almost all fluid mechanics analyses include a step (or steps) based on applying the principle of conservation of mass. The principle may be stated simply as follows; In a fluid flow, mass can be moved about, but it can be neither created nor destroyed. Consider the arbitrary open system sketched in Figure 1, which has a boundary comprised of closed impermeable boundary sections (ie walls), and clearly defined open boundary sections (the inlet and outlet sections).

Figure 1: Fluid flow system with combination of closed boundary sections (walls) and clearly defined open (inlet and outlet) sections. A is cross-sectional area, ρ is fluid density, and 𝑈 is the average velocity perpendicular to the cross-section

3

For an open section, the volume flow rate of fluid across the section is given by: 𝑄 = 𝑈!𝐴 (1.1) where;

• Q is the volume flow rate (in m3/s, for SI Units) • A is the cross-sectional area (m2) • 𝑈! is the average flow velocity in the direction normal to the section (m/s)

The mass flow rate across a section is then given by 𝑚 = 𝜌𝑄 = 𝜌𝑈!𝐴 (1.2) where;

• 𝑚 is the mass flow rate (kg/s) • 𝜌 is the density of the flowing fluid at the section (kg/m3)

The principle of conservation of mass may be stated mathematically as !"

!"= 𝑚!" −𝑚!"# (1.3)

where; • M is the total mass within the system boundaries (kg) • 𝑚!"  ,𝑚!"# are the rate of mass flow into and out of the open sections of the

system boundary respectively Note that we only need to consider the open sections – no mass can flow across the closed sections of the boundary. So for a system whose boundary has open sections, the rate of change of the mass of fluid contained within the system is equal to the difference between the mass flow rates into and out of the system. Note the implicit sign convention – an increase in the mass of the system is regarded as positive, so a mass flow into the system is positive and a mass flow out of the system is negative. Also note that since we are dealing with the total mass of fluid in the system, and the total flow across each of the inlet and outlet sections, this is the system (or “macroscale”) form of the principle. For the system illustrated in Figure 1, with inlet section 1 and outlet section 2, equation 1.3 becomes !"

!"= 𝑚! −𝑚! = 𝜌!𝑄! − 𝜌!𝑄! = 𝜌!𝑈!𝐴! − 𝜌!𝑈!𝐴! (1.4)

Of course, a general system may have multiple inlet and outlet sections. For a system with Nin inlets and Nout outlets; !"

!"= 𝑚!

!"#!!! − 𝑚!

!"#$!!! = 𝜌!𝑄!!"#

!!! − 𝜌!𝑄!!"#$!!! (1.5)

4

1.1.1 Special Cases Steady State At steady state, a system does not change with respect to time – in other words, the values of all system parameters at all points within the system can be regarded as constant with respect to time. For a system, this means that the geometry (and thus volume) must be fixed, and the values of all properties of the fluid at each point in the system are constant. Naturally, this means that the mass of fluid within the system must also be constant with respect to time, and therefore: !"

!"= 0 (1.6)

For a general system, then, at steady state equation 1.5 becomes: 0 = 𝑚!

!"#!!! − 𝑚!

!"#$!!! (1.7)

and thus 𝑚!

!"#!!! = 𝑚!

!"#$!!! (1.8)

In other words, the total rate of mass flow into the system must be equal to the total rate of mass flow out of the system. Incompressible fluids In fluid mechanics, liquids may generally be regarded as effectively incompressible under isothermal conditions. This means that at constant temperature, the density ρ of a liquid may be regarded as a constant, independent of the pressure. This means that in a system with fixed boundaries (and therefore constant volume), the mass of liquid within the system must be a constant at both steady state and under transient conditions. This implies that for an incompressible liquid system with fixed boundaries. !"

!"= 0 always (1.9)

and thus 𝑚!

!"#!!! = 𝑚!

!"#$!!! always (1.10)

Furthermore, since the density is a constant (and thus uniform throughout the system) the equation 𝜌!𝑄!!"#

!!! = 𝜌!𝑄!!"#$!!! (1.11)

5

can be simplified to become 𝑄!!"#

!!! = 𝑄!!"#$!!! (1.12)

Equation 1.12 has several important implications. Let us consider a simple system with a single inlet at section 1, and a single outlet at section 2. A pipe would be a practical example of this type of system, though more complex systems may also fit this description. If an incompressible liquid is flowing in the system, then from equation 1.12; 𝑄! = 𝑄! (1.13) and thus; 𝑈!𝐴! = 𝑈!𝐴! (1.14) This means that; !!

!!= !!

!! (1.15)

This relationship is of great practical importance in the design of pipework for liquid transport. For a pipe of constant diameter transporting an incompressible fluid, it means that the average velocities at the pipe inlet and outlet are always the same (and, by extension, the average velocity at any section in a pipe of constant diameter must also be a constant). For circular pipes, equation 1.15 implies that

!!!!= !!

!!=

!!!!!

!!!!!= !!!

!!! (1.16)

This is another useful relationship in pipe system design. For a given flow rate, doubling the diameter of the outlet compared to the inlet will give an outlet velocity that is one quarter of the inlet velocity; conversely, halving the outlet diameter will increase the outlet velocity by a factor of 4.

Thought Experiment

If you place a pump in the system between the inlet and outlet of a liquid system, what effect does that have on the relationship between 𝑈!! and 𝑈!!?

The answer is that for an incompressible fluid, there is no change! Equations 1.9 to 1.12 demonstrate that for a system of fixed volume transporting an incompressible liquid, the mass of fluid within the system must remain fixed. The interior configuration of the system becomes irrelevant to the calculation, since conservation of mass must be satisfied at the inlet and outlet.

6

Compressible Fluid Systems at Steady State Gases are compressible fluids, and we often need to know the inlet and/or outlet conditions for systems transporting compressible fluids. For compressible fluids, equation 1.6 still holds at steady state, but now the density is not constant throughout the system. As before, for a system at steady state we have !"

!"= 0 (1.17)

and for a system with a single inlet at section 1, and single outlet at section 2, we have 𝑚! = 𝑚! (1.18) For a compressible fluid the density is not a constant, so now we need to use the equation in the forms 𝜌!𝑄! = 𝜌!𝑄! (1.19) 𝜌!𝑈!𝐴! = 𝜌!𝑈!𝐴! (1.20) In the analysis of compressible fluid systems, the ideal gas law is often a useful tool. In its complete form, the equation is given by 𝑃𝑉 = 𝑚𝑅′𝑇 (1.21) where

• P is the absolute pressure of the gas (Pa) • V is the volume of gas (m3) • m is the mass of gas (kg) • R’ is the specific gas constant (the universal gas constant R divided by the

molecular weight of the gas) • T is the absolute temperature of the gas (in K)

In many problems, the modified form of the ideal gas equation is most helpful: !

!"= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (1.22)

where • 𝜌 is the density of the flowing fluid at the section (kg/m3)

7

1.2 Units and Dimensions To characterize fluid behaviour (or indeed any physical phenomenon), it is usually necessary to quantify the key characteristics of the fluid state and motion. This requires us to define certain numerical measures of these characteristics. All physical measurements are ultimately based on a set of four primary quantities, known as DIMENSIONS. They are (accompanied by the symbol used to represent each in Dimensional Analysis); Length L Mass M Time T Temperature θ All other quantities are defined as combinations of these four fundamental dimensions; Velocity 𝑈 = !

!

Pressure 𝑃 = !

!!!

Density 𝜌 = !

!!

Absolute Viscosity 𝜇 = !

!"

Momentum 𝑝 = !"

!

Note that the symbol = means “has dimensions of”. In contrast, UNITS are measures based on arbitrary standards for the fundamental dimensions. For each of these quantities, arbitrary standards are kept to provide the basis for systems of units. In the case of the SI units for length and mass, physical samples representing the metre or the kilogram were once kept; now the standards are based on more universally consistent measures;

• The original metre was one ten-millionth of the distance from the equator to the north pole along a meridian passing through Paris – the standard metre was once actually marked on metal bars. The metre is now defined as the length of the path travelled by light in vacuum during a time interval of 1  ⁄  299,792,458 of a second [from The NIST reference on constants, units and uncertainty; NIST is the US National Institute of Standards and Technology]

8

• The kilogram was defined at the end of the 18th century as the mass of a cubic decimeter of water. In 1889, the international prototype of the kilogram was a block made of platinum-iridium [The NIST reference on constants, units and uncertainty]

A variety of systems of units are available, each being based on a particular set of standards for the standard dimensions. The SI unit system, the most commonly used system in Australia, and other common systems of units are listed in Table 1.1.

System SI CGS Imperial/English Length Metre m Centimetre cm Foot ft Mass Kilogram kg Gram gm Pound mass lbm Time Second s Second s Second s

Temperature Kelvin K Kelvin K Rankine R Force Newton N =

kg.m/s2 Dynes

dyn=gm.cm/s2 Pound Force lbf =

32.2 lbm.ft/s2

Table 1.1 Common systems of units. Each of these systems of units will be encountered in engineering practice, not to mention several more exotic industry specific variations. The Oil & Gas industry in particular works in unique and variable unit systems. Performing conversions will be an important part of the assignments, laboratories and exams in this course, just as it will be in professional practice. To emphasise the importance of accurately and consistently handling units and conversions, in this course there will be marks available simply for setting out data and correctly completing conversions. 1.2.1 Handling Conversions in Calculations It is good practice to complete any necessary conversions at the start of any calculation, to ensure that the calculations are performed in a consistent set of units throughout. Conversions will be easy if you follow a logical, consistent approach. Example: Convert a velocity of 160 miles per hour to SI units (meters per second)

Approach 1: U = 160 mph = 0.04444 miles/sec = 71.51 m/s (note: performing the calculation in two stages provides an extra opportunity to check the calculation later)

Approach 2: U = 160 mph = !"#  !"#$%

!  !!"#× !  !!"#

!"##  !× !"#$  !"#$"%

!  !"#$% = 71.51 m/s

Note: I prefer the 1st approach – but either approach is acceptable. The important thing is to be logical and consistent.

9

You will need to get used to using conversion tables in texts and handbooks (a variety of online sources are also available). After a while, you will find that certain conversions become second nature due to the frequency of use. Common conversions include:

1 foot (ft) = 0.3048 metres (m) 1 kilogram (kg) = 2.205 Pounds (lb)

1 Kelvin (K) = 1.8 Rankine (R) 1 mile (mi) = 1609 metres (m)

1 inch (in) = 0.0254 metres (m) 1 atmosphere (atm) = 101.5 x 103 Pascals (Pa)

1 pound per square inch (psi) = 6.8947 kilopascals (kPa)

Example 1.2.1.1 – Conservation of Mass in a compressible fluid system at steady state, with unit conversions

Consider the system illustrated below. Compressed air enters a receiver vessel at an absolute pressure of 106 Pa, and a temperature of 200˚C. The volume flow rate of air leaving the vessel is 35.31 ft3/s, and the air leaves the vessel at an absolute pressure of 105 Pa and a temperature of 77˚F. Determine the volume flow rate at which air enters the vessel.

Data Note: All unit conversions should be completed in the initial listing of the problem data. In this case, we have a mix of unit systems, but SI units are the most common (and often the easiest to work with) so we will convert all quantities to SI units.

10

At inlet

• Pressure P1 = 106 Pa • Temperature T1 = 200˚C;

Converting to absolute temperature 200+273 = 473 K At Outlet

• Volume Flow Rate Q2 = 35.31 ft3/s = 1.0 m3/s • Pressure P2 = 105 Pa • Temperature T2 = 77˚F = 25˚C

Then, converting to absolute temperature 25˚C + 273 K = 298 K Note: in compressible fluid problems you must always work in absolute pressure and temperature. Any necessary conversions need to be completed at the start of the calculation. In this case, pressures were given as absolute pressure, so no conversion is necessary

Calculation

Now, at steady state, the inlet and outlet mass flow rates must be equal 𝑚! = 𝑚! and thus 𝜌!𝑄! = 𝜌!𝑄! This may be rearranged to form 𝑄! =

!!!!!!

(1) Now, from the ideal gas law !

!"= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Therefore !!

!!!!= !!

!!!!

And !!

!!= !!!!

!!!! (2)

11

Substituting equation (2) into Equation (1) 𝑄! =

!!!!!!!!

𝑄! 𝑄! =

!"!×!"#!"!×!"#

×1.0 So that 𝑄! = 0.159  𝑚!/𝑠 It is relatively straightforward to continue the calculation to determine the mass flow rates for the sections – this is left as an exercise for the reader. The average velocities could also be calculated if the diameters of the inlet and outlet section were known.

1.2.2 Dimensional Analysis

Dimensions play a critical role in developing engineering analyses. As an example, in any equation the dimensions of both sides of the equation must match – this can be useful tool for checking whether either an equation is correct (if it is not dimensionally correct, it cannot be correct), or whether your understanding of the equation and its variables is accurate. As we shall explore further later in the course, dimensionless groups are important tools in the development of empirical models for important phenomena. They are also an essential tool in the design of scaled experiments in fluid flow. Two of the key dimensionless groups in fluid mechanics are:

Reynolds Number 𝑅𝑒 =   !"#!

(1.23) Froude Number 𝐹𝑟 =  !

!

!" (1.24)

Where

• ρ is density • U is the characteristic velocity of the flow • L is the characteristic length of the flow • µ is the absolute viscosity • g is the acceleration due to gravity

12

Example 1.2.2.1 – Demonstrate that a group of parameters is dimensionless

For this example, we will demonstrate that the Froude number is dimensionless. The Froude number is given by

𝐹𝑟 =𝑈!

𝑔𝐿

Now: the dimensions of each of the parameters are as follows;

Velocity 𝑈 = !!

Gravity 𝑔 = !

!!

Length 𝐿 = 𝐿

Therefore

𝐹𝑟 =𝑈!

𝑔𝐿   =  𝐿𝑇

! 𝑇!

𝐿1𝐿 = 1  

The Froude number is therefore dimensionless. Example 1.2.2.2 – Calculating Dimensionless Numbers For this example, we will calculate the Froude number for a flow in which the velocity is 10 mph, the characteristic length is 100 mm, and gravity is 9.81 m/s2. As in all problems, the first step is to set out the data, and convert all quantities to a consistent set of units – The SI unit system is often the easiest choice. Data Velocity U = 10 mph = 0.0027778 miles/s = 4.469 m/s Gravity g = 9.81 m/s2 (no conversion necessary) Length L = 100 mm = 0.1 m Therefore

𝐹𝑟 =𝑈!

𝑔𝐿 =  4.469!

9.81×0.1 = 20.4

Note that since the Froude number is dimensionless there are no units in the number.

Home exercise – perform the calculation above in Imperial and CGS units, and confirm that the same result is achieved. Under a given set of conditions, a dimensionless group will give the same number regardless of the system of units used, as long as a consistent set of units is used for all quantities in the calculation

13

References The NIST reference on constants, units and uncertainty : Unit of length (meter) , Available at http://physics.nist.gov/cuu/Units/meter.html , [24 February, 2012] The NIST reference on constants, units and uncertainty : Unit of mass (kilogram) , Available at http://physics.nist.gov/cuu/Units/kilogram.html , [24 February, 2012]