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.
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