Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheets

Tutor Sheets: WATS 6.

The WATS form a collection of weekly homework type problems in the form of out-of-class tutorial sheets.

Each WATS typically comprises of a couple of main questions of which each has around four/five linked supplementary questions. They were developed as part of an LTSN Engineering Mini-Project, funded at the University of Hertfordshire which aimed to develop a set of 'student unique' tutorial sheets to actively encourage and improve student participation within a first year first ‘fluid mechanics and thermodynamics’ module. Please see the accompanying Mini-Project Report “Improving student success and retention through greater participation and tackling student-unique tutorial sheets” for more information.

The WATS cover core Fluid Mechanics and Thermodynamics topics at first year undergraduate level. 11 tutorial sheets and their worked solutions are provided here for you to utilise in your teaching. The variables within each question can be altered so that each student answers the same question but will need to produce a unique solution.

FURTHER INFORMATION

Please see http://tinyurl.com/2wf2lfh to access the WATS Random Factor Generating Wizard.

There are also explanatory videos on how to use the Wizard and how to implement WATS available at http://www.youtube.com/user/MBRBLU#p/u/7/0wgC4wy1cV0 and http://www.youtube.com/user/MBRBLU#p/u/6/MGpueiPHpqk.

For more information on WATS, its use and impact on students please contact Mark Russell, School of Aerospace, Automotive and Design Engineering at University of Hertfordshire.

© University of Hertfordshire 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

14.20 m

Pipe length 229 m

1.70 m

Valve.Pressure loss = 33 Pa

Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheet 6 (WATS 6)

TUTOR SHEET – Data used in the Worked Solution

Q1. Consider the pipe and tank layout shown in figure 1. Assuming a fluid with a relative density of 1.06 flows through a 68 mm diameter pipe from the large tank to the small tank - calculate -

i) the velocity of the fluid flowing through the pipe (m/s) (3 marks)ii) the Reynolds Number of the flow (1 mark).iii) the likely nature of the flow regime i.e. laminar, transitional or turbulent (1 mark).iv) the mass flow rate of fluid flowing through the pipe system (kg/s) (1 mark)v) the volume flow rate of fluid flowing through the pipe system (m3/s) (1 mark).

Assume now that the velocity for part i) has been calculated to be 2.960 m/s calculate

vi) the head loss associated with the pipe line only (m) (1 mark)vii) the pressure loss associated with the pipe line only (Pa) (1 mark)viii) the head loss due to all the minor losses (m) (2 mark)ix) the pressure loss due to all the minor losses (Pa) (1 mark)x) the loss coefficient of the valve and (2 mark)xi) the ratio, as a percentage, of the minor to the pipe losses.(%) (1 mark)

You may assume the following :The friction factor associated with the interaction of the fluid and the pipe surface is 0.00600.The fluids kinematic viscosity is 1.13 x 10-6 m2/sThe loss coefficients associated with the fluid as it leaves and enters the tanks are 0.78 and 1.02 respectively.

Figure 1. Drawing for Q1.

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

WATS 6 Worked solution

This sheet is solved using the TUTOR data set.

Q1 i) the velocity of the fluid flowing through the pipe

The actual total head loss must equal the available head. For this case the available head is 14.2 – 1.7m = 12.5m. It is assumed that the flow of water somehow does not drain the large tank nor does it change the height of the fluid of the low tank.

For the system shown the actual total head loss is the sum of the head losses due to –

a) The exit from the large tank in-to the pipe line.b) The pipe itself.c) The valve and d) The inlet from the pipeline into the small tank.

For fittings, the head loss is usually calculated via

(m)

Where-as for straight pipe the head loss is usually calculated via

(m)

Collecting all the terms together allows us to write

Since the pressure loss of the valve is given we need to re-write this as a head loss. i.e.

Writing this for the valve and collecting the student specific data gives.

which is

12 .5=0 .03976 Cm2 +4 .1195Cm

2 +0 .003177+0 .05199 Cm2

Hence

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

= 1.72 m/sii) the Reynolds Number of the flow.

Re=ρC m d

μ In this case you are given kinematic viscosity (υ ) and not dynamic viscosity (μ ) hence

Re=Cm d

υ because υ= μ

ρ

Application of student specific data gives.

Re=1 .72∗0 . 068

1 .13∗10−6 = 103504

iii) the likely nature of the flow regime i.e. laminar, transitional or turbulent.

For flow in pipes turbulence is likely when Re > 4000. In this case, therefore, the flow is likely to be turbulent. Note we do not say the fluid is turbulent but the flow is turbulent.

iv) the mass flow rate of fluid flowing through the pipe system

m=ρ V and V=ACm therefore m=ρ AC i.e. for this case

m=(1. 059∗1000)∗π∗0 .0682

4∗1 .72

= 6.62 kg/s

v) the volume flow rate of fluid flowing through the pipe system

Volume flow rate V=ACm therefore for this case

V= π∗0. 0682

4∗1 .72

= 0.00625 m3/s

vi) the head loss associated with the pipe line only

Straight pipe loss= 4∗0 .006∗229∗2 .962

2∗9. 81∗0 .068 = 36.09 m

vii) Pressure loss associated with the pipe only

Pressurelossstraightpipe

=head loss∗ρg

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Pressurelossstraight pipe

=36 . 09∗(1 .059∗1000 )∗9 . 81 = 374931 Pa

viii) Head losses due to all the minor losses.In this case the minor losses are derived from the exit from the tank into the pipe, the valve and the exit from the pipe into the small tank. Remembering

Minorhead loss=

K fitting∗Cm2

2 g

For tank to pipe and pipe to tank losses.

Minorheadloss=

K leavestank∗Cm

2

2g+

Kenterstank∗Cm

2

2 g

Minorhead loss=(0 .78+1.02) 2 .962

2∗9 .81 = 0.804m

For the valve

Valveheadloss=

Valvepressure loss

ρg

=

33(1 .059∗1000 )∗9 .81 0.0032m

Therefore total minor losses = 0.804 + 0.003 = 0.807 m

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

ix) Pressure losses due to all the minor losses.

Minor pressureloss=Minorheadloss

∗ρg therefore

Minor pressureloss=0 .807∗(1.059∗1000)∗9 .81 = 8344 Pa

x) the loss coefficient of the valve.

Recall the minor head losses, (i.e. fittings etc), are found via -

Minorheadloss=K Fitting∗

Cm2

2 g or in terms of pressure loss is

Minor pressureloss=K Fitting∗

ρC m2

2 writing for the loss coefficient (K valve) gives

K valve=Valvepressureloss∗ 2

ρCm2

Which, using the data, gives,

K valve=33∗ 2

(1 . 059∗1000 )∗2 . 962 = 0.0071

xi) The ratio, as a percentage, of the minor to the pipe losses.

The minor losses = 8344 PaThe pipe losses = 374931Pa

Hence the minor losses to the pipe losses =

8344374931 = 0.0225 which is 2.25 %

Hope this helps.

If you see any errors or can offer any suggestions for improvements then please e-mail me at m.b.russell@herts.ac.uk

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

CreditsThis resource was created by the University of Hertfordshire and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.

© University of Hertfordshire 2009

This work is licensed under a Creative Commons Attribution 2.0 License.

The name of the University of Hertfordshire, UH and the UH logo are the name and registered marks of the University of Hertfordshire. To the fullest extent permitted by law the University of Hertfordshire reserves all its rights in its name and marks which may not be used except with its written permission.

The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence.  All reproductions must comply with the terms of that licence.

The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher.

_______________________________________________________________________________________________WATS 6. Mark Russell (2005)Student number 51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire