Fluid Mechanics and ThermodynamicsWeekly Assessed Tutorial Sheets

Tutor Sheets: WATS 11.

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 11 (WATS 11)

TUTOR SHEET – Data used in the Worked Solution

Q1). The performance characteristics of a turbo-compressor are being estimated on a test-rig. The shaft power driving the compressor is 2.00 kW. Air enters the compressor at atmospheric pressure of 1 bar and a temperature of 15C and is compressed to a pressure of 2.2 bar and temperature of 106C been estimated. The rate of heat loss from the compressor casing is approximately 200 W and the exit velocity of the compressed air is 120 m/s. Calculate -

i) the mass flow rate (kg/s) of the air flowing through the compressor. You may assume the inlet velocity to be negligible. [6 dp] (7 marks)ii) the cross-sectional area (mm2) of the exit from the compressor [3 dp] (5 marks)

The velocity of the air leaving the compressor has to be reduced to 10 m/s by the use of an adiabatic diffuser. Assuming the pressure at exit is 1 bar determine -

iii) the temperature (C) of the air at the diffuser exit [3 dp] (5 marks)iv) the cross-sectional area (mm2) of the diffuser exit [2 dp] (3 marks)

It has been decided to cool the air leaving the diffuser back to the initial conditions of 15C, by means of a water-cooled heat exchanger, determine -

v) the water mass flow rate(kg/s) required in the heat exchanger if the water inlet temperature is 5C and the water exit temperature cannot exceed 10C. [6 dp] (5 marks)

You may assume the following: Cp water = 4.2 kJ/kg K, Cp air = 1.005 kJ/kg K.

Q2) A pipe system carries water from a reservoir and discharges it as a free jet. The entrance to the pipe is via a bell mouth at a depth of X from the reservoir surface. From the entrance the pipe extends 60m, in the horizontal plane before a pipe bend tuning the pipe vertically. The pipe then rises for 20m before a second bend turning it back to the horizontal where it extends for another 60m before the exit. For a volume flow rate of 0.10m3/s through a 200mm diameter pipe calculate

i) the velocity of the fluid flowing through the pipeline [6 dp] (6 marks)ii) the Reynolds Number of the flow [0 dp] (3 marks)iii) the head loss (m) of the pipe-work (exc. fittings) [6 dp] (3 marks)iv) the pressure loss (Pa) of the pipe-work (exc. fittings) [0 dp] (2 marks)v) the head loss (m) of the fittings work (exc. pipe-work) [6 dp] (3 marks)vi) the pressure loss (Pa) of the fittings (exc. pipe-work) [0 dp] (2 marks)vii) the required depth X of the bell mouth entrance to the pipe [3 dp] (6 marks)

You may take the loss factor of each bend to be 0.40, the bell mouth entrance to be 0.05 and the exit 1.0. The friction factor is to be taken as 0.005. For the purpose of this exercise these values are assumed to be constant and independent of velocity. Further the dynamic viscosity and the fluid density can be assumed to be independent of temperature and have the values of 1.0300 x 10-3 kg m/s and 1000 kg/m3 respectively .

_______________________________________________________________________________________________WATS 11.

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

University of Hertfordshire

WATS 11 Worked solution

This sheet is solved using the TUTOR data set.

Q1 . This is an open system combining two components / devices (compressor and diffuser). Since the two devices are installed in series we will tackle them individually.

For reference we will assume the following

Point 1 entry to compressorPoint 2 exit from compressor

Point 3 entry to diffuserPoint 4 exit from diffuser

For the compressor

i) Mass flow rateAssuming the compressor is horizontal removes any changes in potential energy for the systems hence the following applies

remembering also that

gives

Substituting known values, and remembering the sign convention for work and heat gives,

Written for mass flow rate gives = 0.018kg/s

_______________________________________________________________________________________________WATS 11.

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

University of Hertfordshire

ii) Cross sectional area of compressor exit.

By continuity

Or i.e. using specific volume in lieu of density.

We now have two unknowns with one equation - insoluble. Recourse to the ideal gas laws will yield information on the specific volume i.e.

rearranged for specific volume gives

Substituting known values gives = 0.494 m3/kg

Using this data allows the continuity equation to be solved and hence calculate the exit cross sectional area

= = 0.0000741 m2 = 74.1 mm2

For the diffuser

iii) Temperature of the air at the exit of the diffuser

Re-writing the SFEE for the diffuser gives.

Note the rate of heat transfer = 0W, (adiabatic) and the rate of doing work = 0W.

Therefore

=

From which T4 = 386.1K = 113.1C

iv) Cross sectional area of diffuser exit.

This follows the same ideas as given in the answer to part ii). In this case the specific volume at the outlet of the diffuser = 1.108 m3/kg.

_______________________________________________________________________________________________WATS 11.

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

University of Hertfordshire

hence

= = 0.00019944 m2 = 199.44 mm2

v) mass flow rate of water

This simple requires you to construct an energy balance for the air and the water i.e.

rearranged for mass flow rate of water gives

= 0.085 kg/s

_______________________________________________________________________________________________WATS 11.

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

University of Hertfordshire

Q2.

i) velocity of flow through pipe-line

By continuity

0.03142 m2

= 3.183 m/s

ii) Reynolds number of flow

= 618058

iii) head loss of pipe excluding fittings

= 6.796 m

iv) Pressure loss of pipe excluding fittings

= 66669 Pa

v) head loss of pipe fittings excluding pipe-work

= 0.955 m

vi) Fittings pressure loss excluding pipe-work

= 9368 Pa

vii) Required depth

Constructing an energy balance using Bernoulli's equation gives

Making some assumptions and stating what we know i.e. _______________________________________________________________________________________________WATS 11.

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

University of Hertfordshire

therefore cancels

Tank is large compared to pipe therefore C1 = 0

h1 = x-20h2 = 0

This gives

i..e. x is the distance to the bell mouth and in this case '20' is the rise of the pipe from the bell mouthed entry. Substituting known values, and cancelling common density gives

Diving through by gravity and making x the subject gives

from which x = 20+0.5164+7.736 = 28.25m

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 11.

Mark Russell (2005)Student number 51 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 11.

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

University of Hertfordshire