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Fluid Mechanics and
Thermodynamics
Weekly Assessed Tutorial Sheets

Tutor Sheets: WATS 9.
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.
Fluid Mechanics and Thermodynamics
Weekly Assessed Tutorial Sheet 9 (WATS 9)

TUTOR SHEET – Data used in the Worked Solution


Q1). The torque required to rotate a 5.90 m diameter flat disc in a gas is to be found by
measuring the torque required to rotate a geometrically similar disc of 468 mm diameter
in a liquid at 567 rad/s.

i) Calculate the dynamic viscosity (N s /m2) of a suitable liquid. [9 dp](2 marks)
ii) If the torque required for the smaller disc is 215 Nm calculate the torque (Nm) required
    to rotate the full size disc at 5.70 rad/s. [2 dp]                 (2 marks)

For the gas you may assume that the density is 3.70 kg/m3 and its dynamic viscosity is
2.77 x 10-5 N s/m2. For the liquid you may assume that the density is 840.00 kg/m3.




Q2)    Alternating, oscillating vortices are usually shed from a cylinder when it is exposed
to flow conditions having a ratio of inertial forces to viscous forces in the region of 90 –
1000.

Assuming a 9.40 mm diameter cylinder is exposed to a fluid with a dynamic viscosity of
17.81 x 10-6 N s/m2 and a density of 1.29 kg/m3 calculate -

i) the lowest speed (m/s) likely to cause vortex shedding to occur and [4 dp](1 marks)
ii) the highest speed likely (m/s) to cause vortex shedding occur. [4 dp](1 marks)




_______________________________________________________________________________________________
WATS 9.
                                                                                  Mark Russell (2005)
Student number 51                              School of Aerospace, Automotive and Design Engineering

                                                                            University of Hertfordshire
WATS 9
Worked solution
This sheet is solved using the TUTOR data set.

It is assumed for this that you already have the correct dimensionless groups. In some
instances, however, these groups may be given, whereas in others you may have to
construct them yourself. This exercise tries to get you recall and use them.


Q1    i) This part of the question is aimed at getting you to use the dimensionless groups
and use this information to relate each case to each other. Using your notes the
                                       ρD 2 N
dimensionless group for this case is
                                         µ

              ρ g Dr2 N g
                        ρ l Dl2 N l
Therefore                   =       where the subscripts g and l relate to the gas and liquid
              µg            µl
configurations respectively.

Since all the data is know for the gas case

ρ g Dr2 N g     3.7 * 5.9 2 * 5.7
              =                   = 26503354
   µg             2.77 *10 − 5

Which, by definition, must also be the same value that arises from solving the
dimensionless group for the liquid case. i.e.

           ρ l Dl2 N l
26503354 =
               µl

                                                         840 * 0.468 2 * 567
Substituting the known values gives 26503354 =
                                                                 µl

                     840 * 0.468 2 * 567
Therefore, µ l =                           = 0.0039358 N s/m2 = 3.936 * 10-3 N s/m2
                            26503354




Q1 ii) The second part of the question uses the following relationship


_______________________________________________________________________________________________
WATS 9.
                                                                                  Mark Russell (2005)
Student number 51                              School of Aerospace, Automotive and Design Engineering

                                                                               University of Hertfordshire
τg                             τg                 τl
             and as before                   =
ρg D N
     5
     g
         2
         g                     ρg D N
                                    5
                                    g
                                        2
                                        g        ρ l Dl5 N l2

                                                 τg
For the gas group we can write
                                        3.7 * 5.9 5 * 5.7 2

                                                    215
For the liquid group we can write
                                            840 * 0.468 5 * 567 2

                                                                τg                        215
Relating these two groups together gives                                      =
                                                                 5
                                                        3.7 * 5.9 * 5.7   2
                                                                                  840 * 0.468 5 * 567 2

From which we can find the torque required to rotate the big disc in gas i.e. ( τ g )

                                     215
In this case this is τ g =                         * 3.7 * 5.9 5 * 5.7 2 = 30.477 Nm
                             840 * 0.468 5 * 567 2




_______________________________________________________________________________________________
WATS 9.
                                                                                  Mark Russell (2005)
Student number 51                              School of Aerospace, Automotive and Design Engineering

                                                                                          University of Hertfordshire
Q2. This question requires you to recall the fact the ratio of inertial forces to viscous
forces is the much used ‘dimensionless group’ knows as the Reynolds Number.

The question also reinforces the idea of dimensional analysis and shows how
dimensionless groups can be used in engineering studies. i.e. in this case it is not the
fluid velocity or the fluids viscosity that are singularly important. For this vortex shedding
pattern to occur then it is the combination of these variables, with others, that determine
the likely-hood of this phenomenon.


For the lower velocity RE = 90. Recalling the definition of the Reynolds Number i.e.

     ρ CL
RE =       and for this flow configuration the characteristic length, L, is the cylinder
       µ
diameter. Applying the student unique data into the above gives


     1.29* C *9.4*10−3
90 =                   Hence the lower velocity, Cmin, is 0.132 m/s
        17.81*10−6


For the higher velocity, i.e. RE = 1000 and application of the student unique data gives –

       1.29* C *9.4*10−3
1000 =                   Hence the higher velocity, Cmax, is 0.1.469 m/s
          17.81*10−6




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 9.
                                                                                  Mark Russell (2005)
Student number 51                              School of Aerospace, Automotive and Design Engineering

                                                                            University of Hertfordshire
Credits
This 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 9.
                                                                                  Mark Russell (2005)
Student number 51                              School of Aerospace, Automotive and Design Engineering

                                                                                    University of Hertfordshire

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WATS 9 Fluid Mechanics and Thermodynamics- Master And Solution

  • 1. Fluid Mechanics and Thermodynamics Weekly Assessed Tutorial Sheets Tutor Sheets: WATS 9. 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.
  • 2. Fluid Mechanics and Thermodynamics Weekly Assessed Tutorial Sheet 9 (WATS 9) TUTOR SHEET – Data used in the Worked Solution Q1). The torque required to rotate a 5.90 m diameter flat disc in a gas is to be found by measuring the torque required to rotate a geometrically similar disc of 468 mm diameter in a liquid at 567 rad/s. i) Calculate the dynamic viscosity (N s /m2) of a suitable liquid. [9 dp](2 marks) ii) If the torque required for the smaller disc is 215 Nm calculate the torque (Nm) required to rotate the full size disc at 5.70 rad/s. [2 dp] (2 marks) For the gas you may assume that the density is 3.70 kg/m3 and its dynamic viscosity is 2.77 x 10-5 N s/m2. For the liquid you may assume that the density is 840.00 kg/m3. Q2) Alternating, oscillating vortices are usually shed from a cylinder when it is exposed to flow conditions having a ratio of inertial forces to viscous forces in the region of 90 – 1000. Assuming a 9.40 mm diameter cylinder is exposed to a fluid with a dynamic viscosity of 17.81 x 10-6 N s/m2 and a density of 1.29 kg/m3 calculate - i) the lowest speed (m/s) likely to cause vortex shedding to occur and [4 dp](1 marks) ii) the highest speed likely (m/s) to cause vortex shedding occur. [4 dp](1 marks) _______________________________________________________________________________________________ WATS 9. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 3. WATS 9 Worked solution This sheet is solved using the TUTOR data set. It is assumed for this that you already have the correct dimensionless groups. In some instances, however, these groups may be given, whereas in others you may have to construct them yourself. This exercise tries to get you recall and use them. Q1 i) This part of the question is aimed at getting you to use the dimensionless groups and use this information to relate each case to each other. Using your notes the ρD 2 N dimensionless group for this case is µ ρ g Dr2 N g ρ l Dl2 N l Therefore = where the subscripts g and l relate to the gas and liquid µg µl configurations respectively. Since all the data is know for the gas case ρ g Dr2 N g 3.7 * 5.9 2 * 5.7 = = 26503354 µg 2.77 *10 − 5 Which, by definition, must also be the same value that arises from solving the dimensionless group for the liquid case. i.e. ρ l Dl2 N l 26503354 = µl 840 * 0.468 2 * 567 Substituting the known values gives 26503354 = µl 840 * 0.468 2 * 567 Therefore, µ l = = 0.0039358 N s/m2 = 3.936 * 10-3 N s/m2 26503354 Q1 ii) The second part of the question uses the following relationship _______________________________________________________________________________________________ WATS 9. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 4. τg τg τl and as before = ρg D N 5 g 2 g ρg D N 5 g 2 g ρ l Dl5 N l2 τg For the gas group we can write 3.7 * 5.9 5 * 5.7 2 215 For the liquid group we can write 840 * 0.468 5 * 567 2 τg 215 Relating these two groups together gives = 5 3.7 * 5.9 * 5.7 2 840 * 0.468 5 * 567 2 From which we can find the torque required to rotate the big disc in gas i.e. ( τ g ) 215 In this case this is τ g = * 3.7 * 5.9 5 * 5.7 2 = 30.477 Nm 840 * 0.468 5 * 567 2 _______________________________________________________________________________________________ WATS 9. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 5. Q2. This question requires you to recall the fact the ratio of inertial forces to viscous forces is the much used ‘dimensionless group’ knows as the Reynolds Number. The question also reinforces the idea of dimensional analysis and shows how dimensionless groups can be used in engineering studies. i.e. in this case it is not the fluid velocity or the fluids viscosity that are singularly important. For this vortex shedding pattern to occur then it is the combination of these variables, with others, that determine the likely-hood of this phenomenon. For the lower velocity RE = 90. Recalling the definition of the Reynolds Number i.e. ρ CL RE = and for this flow configuration the characteristic length, L, is the cylinder µ diameter. Applying the student unique data into the above gives 1.29* C *9.4*10−3 90 = Hence the lower velocity, Cmin, is 0.132 m/s 17.81*10−6 For the higher velocity, i.e. RE = 1000 and application of the student unique data gives – 1.29* C *9.4*10−3 1000 = Hence the higher velocity, Cmax, is 0.1.469 m/s 17.81*10−6 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 9. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 6. Credits This 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 9. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire