Mais conteúdo relacionado Mais de Mark Russell (10) WATS 6 Fluid Mechanics and Thermodynamics- Master And Solution1. Fluid Mechanics and Thermodynamics<br />Weekly Assessed Tutorial Sheets <br />Tutor Sheets: WATS 6.<br />The WATS form a collection of weekly homework type problems in the form of out-of-class tutorial sheets. <br />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.<br />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.<br />FURTHER INFORMATION<br />Please see http://tinyurl.com/2wf2lfh to access the WATS Random Factor Generating Wizard. <br />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.<br />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.<br /> <br />Fluid Mechanics and Thermodynamics<br />Weekly Assessed Tutorial Sheet 6 (WATS 6)<br />TUTOR SHEET – Data used in the Worked Solution<br />Q1. Consider the pipe and tank layout shown in figure 1. Assuming a fluid with a relative density of 1.06 flows through a 68 mm diameter pipe from the large tank to the small tank - calculate - <br />i)the velocity of the fluid flowing through the pipe (m/s) (3 marks)<br />ii)the Reynolds Number of the flow (1 mark).<br />iii)the likely nature of the flow regime i.e. laminar, transitional or turbulent(1 mark).<br />iv)the mass flow rate of fluid flowing through the pipe system (kg/s) (1 mark)<br />v)the volume flow rate of fluid flowing through the pipe system (m3/s) (1 mark).<br />Assume now that the velocity for part i) has been calculated to be 2.960 m/s calculate <br />vi)the head loss associated with the pipe line only (m) (1 mark)<br />vii)the pressure loss associated with the pipe line only (Pa) (1 mark)<br />viii)the head loss due to all the minor losses (m) (2 mark)<br />ix)the pressure loss due to all the minor losses (Pa) (1 mark)<br />x)the loss coefficient of the valve and (2 mark)<br />xi)the ratio, as a percentage, of the minor to the pipe losses.(%) (1 mark)<br />You may assume the following :<br />The friction factor associated with the interaction of the fluid and the pipe surface is 0.00600.<br />The fluids kinematic viscosity is 1.13 x 10-6 m2/s<br />The loss coefficients associated with the fluid as it leaves and enters the tanks are 0.78 and 1.02 respectively. <br />Figure 1. Drawing for Q1.<br />14.20 mPipe length 229 m1.70 mValve.Pressure loss = 33 Pa <br /> WATS 6 <br />Worked solution<br />This sheet is solved using the TUTOR data set. <br />Q1 i)the velocity of the fluid flowing through the pipe <br />The actual total head loss must equal the available head. For this case the available head is 14.2 – 1.7m = 12.5m. It is assumed that the flow of water somehow does not drain the large tank nor does it change the height of the fluid of the low tank. <br />For the system shown the actual total head loss is the sum of the head losses due to –<br />a)The exit from the large tank in-to the pipe line.<br />b)The pipe itself.<br />c)The valve and <br />d)The inlet from the pipeline into the small tank.<br />For fittings, the head loss is usually calculated via <br /> (m) <br />Where-as for straight pipe the head loss is usually calculated via<br /> (m) <br />Collecting all the terms together allows us to write<br />Since the pressure loss of the valve is given we need to re-write this as a head loss. i.e. <br /> Writing this for the valve and collecting the student specific data gives.<br /> which is <br /> Hence <br /> = 1.72 m/s<br />ii)the Reynolds Number of the flow.<br /> In this case you are given kinematic viscosity () and not dynamic viscosity () hence <br /> because <br />Application of student specific data gives. <br /> = 103504<br />iii)the likely nature of the flow regime i.e. laminar, transitional or turbulent.<br />For flow in pipes turbulence is likely when Re > 4000. In this case, therefore, the flow is likely to be turbulent. Note we do not say the fluid is turbulent but the flow is turbulent.<br />iv)the mass flow rate of fluid flowing through the pipe system<br /> and therefore i.e. for this case <br /> = 6.62 kg/s<br />v)the volume flow rate of fluid flowing through the pipe system<br />Volume flow rate therefore for this case<br /> = 0.00625 m3/s<br />vi)the head loss associated with the pipe line only<br /> <br /> = 36.09 m<br />vii) Pressure loss associated with the pipe only <br /> = 374931 Pa<br />viii) Head losses due to all the minor losses.<br />In this case the minor losses are derived from the exit from the tank into the pipe, the valve and the exit from the pipe into the small tank. Remembering <br />For tank to pipe and pipe to tank losses.<br /> = 0.804m<br />For the valve<br />= 0.0032m<br />Therefore total minor losses = 0.804 + 0.003 = 0.807 m<br />ix) Pressure losses due to all the minor losses.<br /> therefore <br /> = 8344 Pa<br />x)the loss coefficient of the valve.<br />Recall the minor head losses, (i.e. fittings etc), are found via - <br /> or in terms of pressure loss is <br /> writing for the loss coefficient () gives <br /> Which, using the data, gives, <br /> = 0.0071<br />xi)The ratio, as a percentage, of the minor to the pipe losses.<br />The minor losses = 8344 Pa<br />The pipe losses = 374931Pa<br />Hence the minor losses to the pipe losses = = 0.0225 which is 2.25 %<br />Hope this helps.<br />If you see any errors or can offer any suggestions for improvements then please <br />e-mail me at m.b.russell@herts.ac.uk<br />Credits<br />This 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. 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