Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheets

Tutor Sheets: WATS 2.

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 2 (WATS 2)

TUTOR SHEET – Data used in the Worked Solution

Q1. i) A rectangular door in the side of a fluid filled tank is 0.50 m wide and 4.80 mhigh. If the top edge of the door is 3.10 m below the fluids free surface and the fluidhas a density of 850 kg/m3 calculate

i) the total force on the door (kN) (1 mark)ii) the position of the line of action (below the free surface) (m). (1 mark)

Q2. Assuming now that the tank has been drained of the original fluid and thatanother fluid of relative density 0.89 has partially filled the tank so that its free surfaceis 0.52 m below the top of the door. Calculate

i) the total force now acting on the door (kN) (1 mark)ii) the position of the line of action (below the free surface) (m). (1 mark)iii) If the door is now turned through 90° (i.e. it is still in the side of the tank but itswidth and height have now been transposed) at what depth would the top edge of thedoor have to be for the resulting force to be the same as that calculated for Q2 i)? (m)

(1 mark)

iv) Assuming now that the door has been moved so that it is located in the horizontalfloor of the tank. Calculate the required depth of fluid in the tank if the door is only tobe exposed to the same force as calculated in Q2 i) (m).

(2 marks)

Q3. Please provide one hint or tip that would be useful to help future students tacklesuccessfully the problems on this sheet. You will be awarded 3 bonus marks forproviding a sensible hint or tip.

_______________________________________________________________________________________________WATS 2. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

WATS 2 Worked solution

This sheet is solved using the data set for student 51.

Q1 ForceOnDoor

= ∫Topof door

Bottomof door

ρ gwydy

Assuming density (ρ), acceleration due to gravity (g), and door width (w), do not change with respect to fluid depth (y) then the above becomes –

ForceOnDoor=ρ gw ∫

Topof door

Bottomof door

ydy

Which after integrating, and inserting our unique values gives

= 110.1 kN

ii) Centreof P

. ressure=MomentsForce

= MF

Moments= ρ gw ∫Topof door

Bottomof door

y2 dy

= 643.8 kN m

ForceOnDoor= 110.1 k N Already calculated

= 5.85m

_______________________________________________________________________________________________WATS 2. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Q2 i) As before but now with a different fluid and the integration is carried out from the free surface to the bottom of the door. i.e. some of the door is above the fluids free surface.

From the data given the bottom of the door is at a location 4.8m – 0.52 m = 4.28m below the free surface.

= 40.0 kN

See how this force is lower than the force previously calculated (Q1i). This makes sense since a) there is less fluid hence all of the door is not exposed to the fluids force. In this case, this reduction in force, due to the depth reduction, is not offset by the slight increase in density. It is wise to take a few minutes to see if your answers appear to make sense.

ii) Following this ‘make sense’ approach through I would expect the line of action to be less that that calculated in part ii) of Q1. Lets see…

Centreof P. ressure=Moments

Force= M

F

= 114.1 kN m

So far so good. i.e. the moments are less.

= 2.85m

Which again is lower so at least it seems to make sense.

iii) In this case the force is the same as previously calculated i.e. 39984 N (40.0 kN) but the doors width and height have now been transposed. i.e. –

door height = 0.5mdoor width = 4.8m.

ForceOnDoor=ρ gw ∫

X

Doorheight+ X

ydy

Where x, in the limits of integration, is the depth of fluid from the free surface to the top of the door.

Integrating gives

_______________________________________________________________________________________________WATS 2. Mark Russell (2005)Student number51 School of Aerospace, Automotive and Design Engineering

University of Hertfordshire

Since force on door is already known, i.e. 39984 N we can write

Expanding the brackets and tidying gives

From which, X, the additional fluid depth above the door, can be found. In this case it is calculated as 1.658 m

iv) Assuming now that the door lies in the floor of the tank.

Force=Pressure∗Area

= 2.4m2

Force already calculated as 39984 N hence

= 16660 Pa.

Since Pbase= ρ gh rearranging for h gives .

Therefore depth of tank = 1.91m.

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

University of Hertfordshire