Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheets

Tutor Sheets: WATS 5.

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.

WATS 5 Worked solution

This sheet is solved using the TUTOR data set.

Q1. Given the table below calculate –

i) the density (kg/m3) of air at 20.10°C (3dp) (1 mark)ii) the dynamic viscosity (N s /m2) of air at 6.10°C (9dp) (1 mark)iii) the kinematic viscosity (m2/s) of air at 11.10°C (9dp) (1 mark)iv) the density (kg/m3) of air at 22.80°C (3dp) (1 mark)v) the dynamic viscosity (N s /m2)of air at 10.00°C (9dp) (1 mark)vi) the kinematic viscosity (m2/s) of air at 7.80°C (9dp) (1 mark)vii) the air temperature (ºC) when its density is 1.270kg/m3 (2dp) (1 mark)viii) the air temperature (ºC) when its dynamic viscosity is 17.70 x 10-6 N s/m2 (2dp)(1 mark)

Table 1. Properties of air at standard sea level atmospheric conditions.Temperature (°C) Density (kg/m3) Dynamic Viscosity (10-6 N s/m2)

0 1.293 17.110 1.248 17.620 1.205 18.130 1.165 18.6

Q2. If Calculate C when - i) AB = constant, A1 = 208947, B1= 2.30 and A2 = 99586 (1dp) (1 mark)ii) AB1.20 = constant, A1 = 226573.00, B1= 3.30 and A2 = 419610 (1dp) (1 mark)iii) A = constant = 382057.00, B1= 2.50 and B2 = 2.70 (1dp) (1 mark)iv) B = constant = 7.30, A1= 298599 and A2 = 381272 (1dp) (1 mark)

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheet 5 (WATS 5)

TUTOR SHEET – Data used in the Worked Solution

Q1 The first part of the question is oriented around developing your skills and confidence in interpolation. There will be many instances in both this module and other modules when such a skill is required. Using table 1 sheet 160 asks to find.

i) What is the density of air at 20.1°C ?ii) What is the dynamic viscosity of air at 6.1°C ?iii) What is the kinematic viscosity of air at 11.1°C ?iv) What is the density of air at 22.8°C ?v) What is the dynamic viscosity of air at 10.0°C ?vi) What is the kinematic viscosity of air at 7.8°C ?vii) At what temperature is the density of air 1.27kg/m3 ?viii) At what temperature is the dynamic viscosity of air 17.7 x 10-6 N s/m2 ?

Table 1. Properties of air at standard sea level atmospheric conditions.

Temperature (°C) Density (kg/m3) Dynamic Viscosity (10-6 N s/m2)0 1.293 17.1

10 1.248 17.620 1.205 18.130 1.165 18.6

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

i) Table 1 suggests that the constraints for this interpolation are 0°C and 10°C. A chart of the data between these constraints is shown below.

The graph shows -i) the lower value is 20°C and the corresponding density is 1.205 kg/m3.ii) the higher value is 30°C and the corresponding density is 1.165 kg/m3.Hence the Δt = (t2-t1) = 30 – 20 = 10°C & Δρ = (ρ2-ρ1) = 1.165 – 1.205 = -0.04 kg/m3.

Since the question asks for the density at 20.1°C we have to find how far along the relationship our data point is. i.e. -

(temp’ of req’d density– temp’ of lower density) / (temp’ of high – temp’ of low).

Which in this case is

(20.1 – 20) / (30 – 20) = 0.1°C /10°C = 0.01.

So the actual density is found via 0.01 * (-0.04) + 1.205. Where the first two parts, (the product), is the proportion of the density difference needed to get to the required value and the final part, the term added, is the value of the lower value.

Hence the density at 20.1 °C is 1.2046 kg/m3.

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

1.16

1.165

1.17

1.175

1.18

1.185

1.19

1.195

1.2

1.205

1.21

15 20 25 30 35Temperature

Den

sity

ii) Table 1 suggests that the constraints for this interpolation are 0°C and 10°C. A chart of the data between these constraints is shown below.

The graph shows -i) the lower value is 0°C and the corresponding dynamic viscosity is 17.1 *10-6 N s /m2 ii) the higher value is 10°C and the corresponding dynamic viscosity is 17.6 *10-6 N s /m2 Hence the Δt = (t2-t1) = 10 – 0 = 10°C & Δμ = (μ2-μ1) = 17.6 – 17.1 = 0.5 *10-6

N s /m2

Since the question asks for the dynamic viscosity 6.1°C we have to find how far along the relationship our data point is. i.e. -

(temp’ of req’d viscosity– temp’ of lower viscosity) / (temp’ of high – temp’ of low).

Which in this case is

(6.1 – 0) / (10 – 0) = 6.1°C /10°C = 0.61

So the actual dynamic viscosity is found via 0.61 * 0.5 *10-6 + 1.71 10-6 . Where the first two parts, (the product), is the proportion of the viscosity difference needed to get to the required value and the final part, the term added, is the value of the lower value.

Hence the dynamic viscosity at 6.1 °C is 17.4 * 10-6 N s /m2

The same procedure can be used to find the other data.

A point to note, however, is the fact that kinematic viscosity = dynamic viscosity / density.

As a check the remaining answers are -

iii) 1.42 * 10-5 m2/siv) 1.194 kg/m3

v) 17.6 *10-6 N s /m2 vi) 1.39 * 10-5 m2/svii) 5.11°Cviii) 12.0 °C

Q2. This question first requires you to undertake the integration and then substitute known or calculated values into the resulting expression.

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

17

17.1

17.2

17.3

17.4

17.5

17.6

17.7

-5 0 5 10 15Temperature

Dyn

amic

vV

isco

sity

You are doing this because it will be shown later in your studies that

.

i.e. to calculate work done you will have to undertake some integration hence this work is good practice. In some instances it may help to draw the relationship. That way you may get a better understanding of what you are doing. It is worth noting also that you are resolving definite integrals. Hence you are undertaking the integration between fixed limits. In these cases whilst you may not have all the limit data you do have enough information to allow you to find any unknown limit data. This will be clearer as the worked solution notes proceeds.

Recall for all cases we are resolving C=∫

1

2

Adb

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

i) AB = constant, A1 = 208947, B1= 2.3 and A2 = 99586 ?

Since AB = constant and you have data for A1, B1 and A2 we should first progress to get the data for B2.

A1 = 208947, B1= 2.3 therefore AB = 480578 therefore for B2 = 480578/A2 = 4.83

This now gives enough data for the upper and lower limits of integration.

To undertake the integration C=∫

1

2

Adb.

A cannot be integrated with respect to b but the relation AB = constant is helpful.

Since we can write A = constant/B and substitute this into the integral. i.e.

which can be written . This can now be integrated to give.

applying the limits to expression gives

which is the same as

.This expression can now be evaluated to find the value of C. Remembering also that the

‘const’ term is AB gives i .e. In this case C = 356137

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

ii) AB1.20 = constant, A1 = 226573, B1= 3.3 and A2 = 419610 ?

B2 = (constant/A2)1/1.2 = 1.975

The integration is C=∫

1

2

Adb but again A cannot be integrated wrt b. In this case

AB1.2 = Constant therefore which is the same as

substitute this into the integral. i.e. which can be written

. This can now be integrated to give. Application of

the limits gives which is the same as

Recall also that allows us to write

Which can be tidied to give

This can also be written

Solving for our values gives

= - 404388

_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

iii) A = constant = 382057, B1= 2.5 and B2 = 2.7 ?

In this case A = constant therefore, by definition, A1 = A2 = 325270.

C=∫1

2

Adb which for the case of constant A can be written as

Which when inetrgated is simply Using the limits gives .

And substituting the values gives

i.e. C = 764111.

iv) B = constant = 7.3, A1= 298599 and A2 = 381272

For the final case B = constant. Hence looking at term to be integrated it can be seen that db = 0 , by definition. Hence here C must also equal 0

.

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 5. 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.

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_______________________________________________________________________________________________WATS 5. Mark Russell (2005)Student number 51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire