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WATS 10 Fluid Mechanics and Thermodynamics- Master And Solution
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WATS 10 Fluid Mechanics and Thermodynamics- Master And Solution

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The WATS form a collection of weekly homework type problems in the form of out-of-class tutorial sheets. …

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.

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  • 1. Fluid Mechanics and Thermodynamics Weekly Assessed Tutorial Sheets Tutor Sheets: WATS 10. 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.
  • 2. Fluid Mechanics and Thermodynamics Weekly Assessed Tutorial Sheet 10 (WATS 10) TUTOR SHEET – Data used in the Worked Solution Q1). A cylinder of diameter 0.48 m, fitted with a gas tight, frictionless piston contains dry saturated steam at 8 bars. Heat is added until the final temperature is 316 ° C. Determine i) the work done (kJ) [3 dp] (2 marks) ii) the heat supplied (kJ) [2 dp] (1 mark) iii) the change in internal energy (kJ). [2 dp] (2 marks) You may assume that the piston is free to rise as a consequence of any expansion and at the start of the process the distance between the piston and the bottom of the cylinder is 0.34 m. Q2) 11.10 kg/s of air enters a turbine at 923 °C with a velocity of 90 m/s. The air expands adiabatically as it passes through the turbine and leaves with a velocity of 114 m/s. It then enters a diffuser where the velocity is reduced to a negligible value. If the turbine produces 2.25 MW calculate i) The temperature of the air at exit from the turbine (°C) [1 dp] (3 marks) ii) The temperature of the air at exit from the diffuser (°C) [1 dp] (3 marks) You may assume that the Cp for air is constant and has a value of 1.005 kJ/kg K. _______________________________________________________________________________________________ WATS 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 3. WATS 10 Worked solution This sheet is solved using the TUTOR data set. Q1 The first thing to note about this question is that it is a closed system. The cylinder within the piston will move and no frictional forces are included. Since heat is added to the cylinder the gases will expand forcing the piston to rise. The staring point for a closed, (non-flowing), first law problem is Q1− 2 − W1− 2 = ∆U i) The work done 2 Work _ done = ∫ Pdv 1 i.e. you now need to find the relationship between pressure and volume. In this case, however, there is no relationship because pressure, ( P ), is constant, so the integral becomes 2 Work _ done = P ∫ dv After integration this becomes 1 Work _ done = P [ V ] 1 2 On application of the ‘limits of integration’ the result is simply Work _ done = P [ V2 − V1 ] Hence we now need to find V1 and V2 Using the fact that at the start of the process the fluid is saturated steam will allow the initial specific volume to be found. For this data this is v1 = 0.2403 m3/kg And using the fact that the final temperature and pressure are known, i.e. 316°C at 8 bars - Don’t forget for this problem the pressure remains constant allows the final specific volume to be found. – For this data this is v2 = 0.334 m3/kg Note you may have to interpolate from the tables. _______________________________________________________________________________________________ WATS 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 4. To calculate the actual work done and not the work pr unit mass you now have to calculate the mass of the fluid in the cylinder. This can be found via A* L π *0.482 *0.34 mass = which is mass = = 0.256 kg v1 4*0.2403 1 Hint: Don’t forget =ρ v This now allows the work done to be calculated. Work _ done = (8*105 ) * 0.256*(0.334 − 0.2403) = 19190 J = 19.2 kJ ii) The heat supplied. The heat supplied can be found via Q1− 2 = ∆U + W1− 2 The above has just found the work done, (W1−2 ) , but since there isn’t data on the change in internal energy the above cannot yet be solved. Hence the following will answer part iii) first and then come back with the remaining data. iii) The change in internal energy. Using the steam tables at the conditions noted for part i) above gives u1 = 2577 kJ/kg and u2 = 2824 kJ/kg. Again there may have been a need to interpolate from the tables. The change in specific internal energy, ∆u , is therefore ∆u = u2 − u1 = 2824 – 2577 = 247 kJ/kg. But since the question asks for ‘change in internal energy’ and not ‘change in specific internal energy’ this value needs to be multiplied by the mass. i.e. by definition ∆U = m * ∆u = 0.256 * 247 = 63.2 kJ This information can now be fed back to help answer part ii) i.e Q1− 2 = ∆U + W1− 2 = 63.2 + 19.2 = 82.4 kJ Q2. This question is different from that above due to the fact that the fluid is flowing through a series of ‘components’ thus making it an open system. Since no mass is stored it is reasonable to assume that the system is in steady-state: in this instance the Steady Flow Energy Equation (SFEE) applies. i.e. _______________________________________________________________________________________________ WATS 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 5. C12 C2 Q1− 2 + m(h1 + + gZ1 ) = W1− 2 + m( h2 + 2 + gZ 2 ) 2 2 Figure 1. Sketch to support Q2. 1 2 3 4 Turbine Diffuser Work Although the question states that the fluid flows through the two components you need to tackle them as individual entities. i.e. solve for the turbine first and then solve for the diffuser. _______________________________________________________________________________________________ WATS 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 6. For the turbine. A read of the question will show that the potential energy at the inlet is the same as the potential energy at the outlet. By definition, since it is a turbine we are dealing with, there will be some work-done. That is the job of a turbine !; and for this student ‘data set’ the work-done is given as 2.25 MW. Further, since no data is given regarding heat transfer we will assume that it is adiabatic. Adding this data to a sketch of the problem gives Heat = 0 kW 1 2 Turbine Inlet conditions Outlet conditions Temp = 923 Deg C Temp = ? Deg C C = 90 m/s C = 114 m/s Mass flow rate = 11.1 kg/s Work Mass flow rate = 11.1 kg/s 2.25 MW In this case, therefore, the SFEE reduces to C12 C2 m( h1 + ) = W1− 2 + m(h2 + 2 ) This can be re-arranged to give 2 2 C2 − C12 2 −W1− 2 = m(h2 − h1 + ) 2 And since the question asks for the temperature at the outlet of the turbine it is worth remembering the relationship (h2 − h1 ) = Cp (T2 − T1 ) . Substituting this into the above gives  (C 2 − C12 )  −W1− 2 = m Cp (T2 − T1 ) + 2   2  Inserting the student unique data into the above gives  (1142 − 902 )  −2.25*106 = 11.1* (1.005*1000)*(T2 − [923 + 273]) +   2  From which T2 = 992 K Which is 992 – 273 = 719°C. For the diffuser. Noting that the steam flows from the turbine to the diffuser allows us to obtain more data. i.e. the inlet velocity and the inlet temperature to the diffuser is clearly the exit velocity and exit temperature from the turbine. In a diffuser no work is done and also assuming that the diffuser behaves adiabatically allows a sketch of this to be shown. _______________________________________________________________________________________________ WATS 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 7. Heat = 0 kW 3 4 Diffuser Inlet conditions Outlet conditions Temp = 719 Deg C Temp = ? Deg C C = 114 m/s C = 0 m/s Mass flow rate = 11.1 kg/s Mass flow rate = 11.1 kg/s Work 0 MW The SFEE for the diffuser can now be stated as  (C 2 − C32 )  0 = m Cp (T4 − T3 ) + 4   2  Which upon substitution of the student unique vales gives  (02 − 1142 )  0 = 11.1* (1.005*1000)*(T4 − [719 + 273]) + 2   2  From which T4 can be found. T4 = 998 K Which is 998 – 273 = 725°C. 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 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire
  • 8. Credits 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. 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 10. Mark Russell (2005) Student number 51 School of Aerospace, Automotive and Design Engineering University of Hertfordshire