Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheets

Tutor Sheets: WATS 7.

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 ThermodynamicsWeekly Assessed Tutorial Sheet 7 (WATS 7)

TUTOR SHEET – Data used in the Worked Solution

Q1. A quantity of a fluid, of mass 0.78 kg, is contained in a vertical heat insulated cylinder, sealed at the top by a frictionless piston. The piston has a diameter of 1.38 m and its weight produces a pressure of 1.25 MN/m2 in the fluid. The fluid is stirred by a paddle for 53 minutes and during this time the period the paddle absorbs 0.155 kW of power and the piston rises a distance of 0.40 m. Calculate

i) The change in internal energy of the fluid during the process (kJ). [1 dp](3 mark).ii) The change in the specific internal energy during the process (kJ/kg) [1 dp](1 mark).

Q2. A compression ignition engine has as single cylinder with a bore of 17.35cm and a stroke of 24.00cm. The compression ratio is 14 to 1. At the start of the expansion stroke the gases in the clearance volume have a pressure of 9.90 Bars and it is assumed that the expansion is a reversible polytropic process to the law PV1.29 = C. Calculate

i) The pressure at the end of the stroke (Bar). [1 dp] (2 mark).ii) The work done (kJ). [4 dp] (1 mark).

Q3. Provide a multiple-choice type question that could be used to help revision. The question can be from any of the material covered in this subject. Naturally you should present the question and the possible responses together with the correct answer. Whilst there are many possible question types I am only asking for a multiple-choice type question that has one correct answer. You should also write the question such that you seek to get the correct answer from 5 possible alternatives. i.e. Yes/No True false questions are not allowed. Note it is my intention to use these as part of the revision process. (3 marks)

_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

WATS 7 Worked solution

This sheet is solved using the TUTOR data set.

Q1 i) The change in internal energy of the fluid during the process.

The first thing to note is that this problem is a non-flow problem hence the first law for a closed system is relevant. i.e.

Since the question asks you to calculate the change in internal energy, ( ), you simply have to focus on finding the magnitudes and directions of any work and heat transfers.

Note that the question states the cylinder is ‘heat insulated’. Hence for this problem it

should be apparent that the heat transfer, ( ), to or from the cylinder is zero. Of course you are assuming that it is perfectly insulated and in the absence of any other data, i.e. properties and thicknesses’ of the insulating material, for now, that is a fair assumption. So for this case you are only solving

Again, a read of the question will highlight the fact that there are TWO work transfers. One of which comes from the paddle stirring the fluid whereas the other comes from the fact that the piston is moving – read below.

The question states the cylinder is sealed at the top by a frictionless piston. This means that the volume of the cylinder may change as the piston slides up and down but the contents of the cylinder will be held at a constant pressure. Think this through - assume the pressure inside the cylinder is somehow increased, this pressure increase would simply move the piston to ensure that its ‘self weight’ balances that force which is inside the cylinder. Since the piston has a fixed size, fixed mass and gravity doesn’t change the force imposed by the piston on the fluid won’t change. Hence when it settles it will always be in equilibrium with the pressure below it which must, by definition, match that arising from the piston itself.

This information, fixed pressure and changing volume, is needed for you to calculate the work done due to the moving piston.

_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Taking the two ‘work done’ terms individually

a) Work done due to moving piston

Since you now need to find the relationship between pressure and volume. In this case, however, there is no relationship because pressure, ( ), is constant, so the integral becomes

After integration this becomes

On application of the ‘limits of integration’ the result is simply

Unfortunately the initial volume, (V1), is not known. Nor is there enough data to be able to calculate it. The final volume, (V2), however, is clearly the initial volume plus the increase

in volume due to the moving piston. i.e.

Using this relationship we can write

which is the same as

i.e. the V1 terms cancel

The pressure, (P), is given and all that remains is to find the additional_volume.

Since the piston is cylindrical this is simply

Using the student specific data gives

= 0.598m3.

Hence = 747500 J = 748 kJ.

Note: Since the gas is expanding, i.e. the piston is pushed outwards, this is work done by the system. Recall also the sign convention. Work done by the system is positive.

_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

b) the work associated by the paddle stirring.

The second component of work associated with this system is that due to the fluid being stirred by the paddle.

The question states that the paddle draws 0.155 kW and that it does this for 53 minutes. Recall that a Watt is simply a Joule/second, (1W = 1J/s), hence

0.155 kW = 0.155 kJ/s

Therefore over the 53 minute duration the paddle draws 0.155 * (53*60) = 493 kJ

In this instance, the fluid is being stirred by a paddle. i.e. this is work done on the system and hence, applying the sign convention, this is negative. i.e. -493 kJ.

The total, net work, done by the system is, therefore 748 – 493 = 255 kJ

Remembering we are trying to find the changing internal energy i.e.

In this case this is

kJ.

Hoping you didn’t forget the minus sign. This means that internal energy in the system is decreasing.

ii) The change in the specific internal energy during the process.

A specific property is the magnitude of the property per unit mass. Therefore the second part of the question is asking for the change in internal energy per unit mass i.e. (kJ/kg).

in terms of the change in internal energy is so the change in specific energy

Using the calculated data from i) above and the student specific data gives

= -326 kJ/kg

_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Q2.

i) The pressure at the end of the stroke.

From the question, it is known that and also that Bar.

The compression ratio is 14:1 which is

For completeness this can also be written as therefore

and also it is obvious that

Using the data from the question allows Vswept to be calculated. i.e.

= 0.005675 m3.

Hence = 0.000437 m3. Note also that this is V1.

Remembering that gives = 0.006112 m3

Inserting all this student specific data into the relationship gives

from which P2 can be found i.e.

= 32936 Pa = 0.329 Bar

_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

ii) The work done.

Previous sheets have shown the working for this integral when and when . As such this working is not repeated here. The resulting expression is however,

.

Application of the student specific data gives

= 797J = 0.797kJ

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

University of Hertfordshire

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© University of Hertfordshire 2009

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_______________________________________________________________________________________________WATS 7. Mark Russell (2005)Student number 51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire