Tuesday, June 2, 2020

Department of Computing Engineering and Technology Coursework - 825 Words

Department of Computing Engineering and Technology (Coursework Sample) Content: DEPARTMENT OF COMPUTING, ENGINEERING AND TECHNOLGYEAT106 à ¢Ã¢â€š ¬ THERMODYNAMICS AND FLUID MECHANICSREFERRED COURSEWORK 2013-2014NAME:DATE:Question 1Water at 50 degrees Celsius flows at a mass flow rate of 20 kg/s in a 200 mm diameter pipeline. * Find the density and dynamic viscosity of the water at this temperatureletà ¯Ã¢â€š ¬Ãƒ ¯Ã‚ Ãƒ ¯Ã¢â€š ¬be the dynamic viscosityà ¯Ã‚ (kg/ms)= {[2.414 x 10-5)] x[10(247k/(temp-140k))]}={[2.414 x 10^-5)] x [10(247k/(50 + 273 -140))]}=0.00002414 x 101.4=0.000605914kg/msà ¯Ã‚  =6. 091 x 10^ -4 kg/msDensity=988.1kg/m3(Source Engineeringtoolbox.com, 2014) * Calculate the velocity of the water.Q=AVV=Q/AVolume flow rate= {(mass low rate)/ density}Q=20/988.1= 0.0202m3/sA=à ¯Ã‚ r2=à ¯Ã‚  x 0.1 x 0.1=0.031m2V=0.0202/0.031=0.652m/s * Hence, calculate the Reynolds number Re.Re=density x velocity x diameter)(Dynamic viscosity)Re=à ¯Ã‚ Ã‚ ²Ãƒ ¯Ã¢â€š ¬xVx Dà ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢ ‚ ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã¢â€š ¬Ãƒ ¯Ã‚ =988.1x 0.652 x 0.26. 091 x 10^ -4Re=21255.94 * Is the flow laminar or turbulent?The flow is turbulentBecause the Reynolds number is greater than 2000Question 2Water flows in a circular pipe. At one section the diameter of the pipe is 0.3 m, the static pressure is 260 kPa, the velocity is 3 m/s, and the elevation is 10 m above ground level. The elevation at a section downstream is at ground level and the pipe diameter is 0.15 m.Find the gauge pressure at the downstream section.Let upward section be section 1Pressure at 1= 260Kpa =260 x 103 N/m2Area 1 = (à ¯Ã‚ /4) x (D2)= (à ¯Ã‚ /4) x (0.3 x 0.3)=0.07071m2Let downstream section be section 2Pressure at 2= PArea 2= (à ¯Ã‚ /4) x (0.15 x 0.15)=0.0176Equation of flow rate, Q= AVQ is constantAV1=AV23 x 0.07071= V2 x 0.0176V2=12.052m/sBernoulli equation(Source Princeton.edu, 2014){[260 x 103)/ 999 x 9.81]+ [(32)/ 2 x 9.81] + 10}={ [P/999 x 9.81]+ [12.052x12.052)/ 2x9.81] + 0.075}26.5+0.45 9+10=7.478+ P/999 x 9.81P=29.179 x 999 x 9.81P=285, 959.74 N/m2P= 285.959 KpaQuestion 3A 40 m long horizontal pipe is connected to a water tank at one end and discharges freely into the atmosphere at the other. For the first 25 m of its length from the tank, the pipe is 150 mm diameter. Its diameter is then suddenly enlarged to 300 mm, as shown. The water level in the tank is 8 m above the centre of the pipe. If the Darcy-Weisbach friction factor in both pipes is 0.04, calculate the exit velocity V2 of the water and the volumetric flow rate in m3/s.(Source Nptel.ac.in, 2014)Using line XX as the datumUsing pipe1 and the tankBernoulli equationP1=P2=0Vt for tank=0Pt/à ¯Ã‚ g + V2t/2g + Zt=P1/2g + {V12/2 x g} +Z1+hf(The crossed values mean they are zero)8+0.075=V21 /2x9.81 +0.15+ (0.04x25xV2)/ (0.15x2x9.81)