Concept explainers
The power available from a wind turbine is calculated by the following equation:
where P 5 power [watts], A = sweep area (circular) of the blades [square meters], ρ = air density [kilograms per cubic meter], and v = velocity [meters per second]. The world’s largest sweep area wind turbine generator in Spain has a blade diameter of 420 feet [ft]. The specific gravity of air is 0.00123. Assuming a velocity of 30 miles per hour [mi/h] and the power produced is 5 megawatts [MW], determine the efficiency of this turbine.
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- The Ideal gas law shows the relationship between some common properties of ideal gases and is written as PV = nRT. P = pressure, V = volume, n = number of moles of ideal Gas, R = General gas constant = 8.314 kJ/(kmol.K) and T = absolute temperature. 5 mol ideal gas at 22°C is placed in a cube with edge lengths of 0.750 meters. Calculate the pressure of the Ideal gas as [Pa].arrow_forwardThe title of our study is "Design and Fabrication of an Agricultural Spraying Attachment for an Autonomous Rover" we need help in the machine design part of our study, specifically Pump selection, battery selection, and center of gravity computation I already sent you the image of the rover Rover weight capacity 35 kilo grams (payload) Rover measurements: Wheel distance - 14.5 inches and 8.5 inches Rover platform size (width & length) 14.4 inches x 10.4 inches Height- 15.8 inches. Water tank selection More or less 25 liters *For tank selection we need to justify the size with the center of gravity, proving that is the maximum dimension or liters of the tank because the center of gravity will be higher making the rover prone to toppling Pump selection *Our sprayer design has two nozzles which splits using a tee splitter, we need to find the size of the pump to obtain the same discharge rate of a normal sprayer. A normal sprayer has a 12V pump and a pressure from 0.2 to 0.45 mpa…arrow_forwardSuppose you're converting a bill of materials for machining feedstock from the British gravitational (U.S. system) of units to International System (SI) units. Your goal is to have all of the material quantities in kilograms. Your bill of materials lists values for required weights of brass rod in pounds. To convert to these values to kilograms you first you divide by gravitational acceleration in feet per second squared to get a mass in slugs, then convert slugs to kilograms using standard conversion factors. Group of answer choices True Falsearrow_forward
- 1. Energy consumption in a country from 2001 to 2011 increased from 1041.3 million to 1526.1 million BOE (barrels of oil equivalent). If the high calorific value (HHV) of the oil is 141,000 kJ/gallon what is the increase in energy consumption over 10 years in Joule units. What is the rate of energy consumption per year? 2. Based on data on the rate of energy consumption per year from question no. 1, if the growth rate energy 2.5% per year, calculate the doubling time (td) and what is the approximate rate of energy growth in 2021.arrow_forwardProcess Fluid Mechanics for Chemical Engineers , please help in this Question , thanks.arrow_forward#01: Explain the following terms in your own words. a) Aerodynamics. b) Mach Number. c) Sub-Sonic. d) Angle of Attack. e) Thrust. f) Chord Line. g) Drag Coefficient. h) Pitch Moment. i) Wind Tunnel. j) Skin Drag.arrow_forward
- The differential change in pressure p close to the surface of a static fluid is given by the following expression:dp/dy = -3Ap2,where A is a constant, with units of 1/(atm•m), and p is the pressure in atm. The pressure at the surface of the fluid is p(0) = 1 atm, and the coordinate y here is positive upwards with origin at the surface. An absulute pressure gauge is placed at a depth 0.19m in the fluid. What would be the reading of the pressure gauge in units in atm? you can take the constant A=1(atm.m)^-1arrow_forwardFor the Following question Graph all 4 : [I just need all 4 graphs and please explain and make clean solution] Position vs time Velocity vs time Acceleration vs time Force vs time [For your convenience, I have solved the numerical solutions for the problem] (Please Look at the picture since it is much cleaner) Question : A 550 kilogram mass initially at rest acted upon by a force of F(t) = 50et Newtons. What are the acceleration, speed, and displacement of the mass at t = 4 second ? a =(50 e^t)/(550 ) [N/kg] v = ∫_0^t▒(50 e^t )dt/(550 )= v_0 +(50 e^t-50)/550=((e^t- 1))/11 x = ∫_0^t▒(e^t- 1)dt/(11 )= x_0 +(e^t- t - 1)/(11 ) a(4s)=(50*54.6)/550= 4.96[m/s^2 ] v(4s)=((e^4-1))/11= 4.87[m/s] x(4s)=((e^4- 4 - 1))/11= 4.51 [m]arrow_forwardPi constant to be used: 3.14159 Capillary tube: radius = 0.02 cm; length = 9 cm; pressure = 0.7 mmHg Liquid sample: density = 4 ml; time of flow = 4.5 seconds Determine (a) pressure in MKS unit, (b) pressure in CGS unit, (c) the viscosity of the liquid, (d) If the weight was given as 8.5 grams, what will be the density of the liquid? (e) the kinematic viscosity of the liquid? (f) and the fluidity of the liquid?arrow_forward
- Chapter 7, Problem 63, there is an 'h' value that is solved when expressing half the width of the Rankine Oval that I am not quite understanding the process by which the value is solved for. Could you break down the process by which h=0.02423 m? It's been a while since I was last in school, so the mathematical method by which 'h' was solved is a little unclear to me.arrow_forwardCompletely solve and box the final answer. Please show the units to show the unit cancellations. Write legibly 2. During a steady flow process, the pressure of the working substance drops from 1,380 kpa to 138 kpa, the speed increases from 61 m/s to 305 m/s, the internal energy of the open system decreases 58.1 KJ/kg, and the specific volume increases from 0.0625 m3/kg to 0.5 m3/kg. No heat is transferred, determine the work in KW if the mass flow at the rate of 272.15 kg/hr.arrow_forwardPLEASE ANSWER THEM ALL FOR AN UPVOTE. THANK YOU. Given: Pi constant to be used: 3.14159 Capillary tube: radius = 0.02 cm; length = 9 cm; pressure = 0.7 mmHg Liquid sample: density = 4 ml; time of flow = 4.5 seconds Ans. the ff questions. •What is the pressure in MKS unit? •What is the pressure in CGS unit? •What is the viscosity of the liquid? • If the weight was given as 8.5 grams, what will be the density of the liquid? •What is the kinematic viscosity of the liquid? •What is the fluidity of the liquid?arrow_forward
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