Race Car Aerodynamics Gregor Seljak April 8‚ 2008 1 Introduction First racing cars were primarily designed to achieve high top speeds and the main goal was to minimize the air drag. But at high speeds‚ cars developed lift forces‚ which affected their stability. In order to improve their stability and handling‚ engineers mounted inverted wings profiles1 generating negative lift. First such cars were Opel’s rocket powered RAK1 and RAK2 in 1928. However‚ in Formula‚ wings were not used
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BMEDME613 Aerodymamics Assignment Aerodynamic Analysis on Winglet By: V.Devanandh‚ 114009037‚ B.Tech-Mechanical Engineering(III year)‚ SASTRA University. Prof.Dhananjay‚ SASTRA University. Submitted to: April 12‚ 2013 Contents 1 Introduction 2 Scope of the Assignment 3 Review of Basic concepts 3.1 3.2 Wingtip vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Induced drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Date and Observations: I’ll send you the picture by text message: (561-542-5608) Calculations: 1. Write the balanced equation for the reaction conducted in this lab‚ including appropriate phase symbols. Mg(s) + 2HCl(aq) --> H2(g) + MgCl2(aq) 2. Determine the partial pressure of the hydrogen gas collected in the gas collection tube. The partial pressure of the hydrogen gas is 1.07 atm 3. Calculate the moles of hydrogen gas collected. pv=mrt ; n= .0013mol of hydrogen gas 4. If magnesium was
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Using this online NPV Calculation Tool http://finance.thinkanddone.com/online-n… we get the following NPV at 15% Net Cash Flows CF0 = -3000000 CF1 = 1100000 CF2 = 1450000 CF3 = 1300000 CF4 = 950000 Discounted Net Cash Flows DCF1 = 1100000/(1+0.15)^1 = 1100000/1.15 = 956521.74 DCF2 = 1450000/(1+0.15)^2 = 1450000/1.3225 = 1096408.32 DCF3 = 1300000/(1+0.15)^3 = 1300000/1.52087 = 854771.1 DCF4 = 950000/(1+0.15)^4 = 950000/1.74901 = 543165.58 NPV Calculation NPV = 956521.74
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the airplane generally from nose to tail. b. It consists of the main airplane structure like frames‚ stringers‚ longerons‚ keel beam and skin. 2. Wings a. This is the component or section of the airplane that is responsible for aerodynamic lift. b. Lift is the force that raises the airplane up. 3. Empennage a. This is the airplane’s tail section which consists of the vertical stabilizer (or vertical fin) and horizontal stabilizer. On most large airplanes the empennage
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CULINARY CALCULATIONS The Standardized Recipe The standardized recipe is the hallmark of the foodservice industry today. The information contained in the standardized recipe ensures that a consistent product is always served to the guest. A consistent product means the look‚ taste‚ texture‚ and portion size of the menu item is the same each time the item is prepared and served‚ regardless of who is in the kitchen on a given day. Each standardized recipe has a specific yield‚ which can be increased
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Many nurses are weak with drug calculations of all sorts. This article will help to review the major concepts related to drug calculations‚ help walk you through a few exercises‚ and provide a few exercises you can perform on your own to check your skills. There are many reference books available to review basic math skills‚ if you find that you have difficulty with even the basic conversion exercises. Common Conversions: 1 Liter = 1000 Milliliters 1 Gram = 1000 Milligrams 1 Milligram = 1000
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Yr: 2014 Yr: 2013 Remarks 1.1 Profitibality Ratio 1.1.1 ROE= x 100 = 10.71% x100 = 10.30% Higher the better 1.1.2 ROTA= x100 = 18.27% x100 = 19.45% Higher the better 1.1.3 GPM= x100 = 71.67% x100 =70.35% Higher the better 1.1.4 OPM= x100 =28.72% x100 =29.79% Higher the better 1.1.5 NPM= x100 =43.48% x100 =39.46% Higher the better 1.1.6 NPM= x100 =38.74% x100 =35.04% Higher the better 1.2 Asset Utilisation 1.2.1 TA TURN= x100 =18.27% x100 =19.45% Higher the better 1.2.2
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Sample Calculations I-Beam (S8x18.4) Dimensions: D= 8 in; h= 7.148 in; bf= 4.001 in; tw= 0.271in; tf= 0.426in; L (length of the beam) =18.4 in I= (bf*D3 – h3 (bf – tw))/12= 57.6 in4; E (Referenced value of Young’s modulus) = 29X106 psi Theoretical Strain: ε= σ/E= (M*y)/(E*I) P = load a = distance from support to the applied load (48 in) y = distance from neutral axis to the extreme element in y-direction The sing in the theoretical strain (±) determines if the strain is in compression
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CALCULATIONS Determining the amount Limiting Reagent used. nlimiting reagent = Molarity x Volume or Mass / Molar Mass Example: Limiting reagent is 5mL of 1.0 M HCl nlimiting reagent = Molarity x Volume nlimiting reagent = (1.0 [mol/L]) x 0.005 [L]) = 0.005 mol Determining the qrxn and qcal. qrxn + qcal = 0 -qrxn = qcal qrxn = ΔHrxn x nlimiting reagent qcal = Ccal ΔT qrxn = - Ccal ΔT + mcsolid ΔT (note: only if there is a precipitate formed in the reaction)
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