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Energy, work and Power Energy, work and Power Work is defined as using energy Work and energy are both measured in joules Energy, work and Power • Work done is defined as Force x distance moved in the direction of the force • The SI unit is the newton metre (Nm) • 1Nm is 1Joule Energy, work and Power A 10 kg mass has a weight of 98.1 Newtons (weight is a force and is found by multiplying mass x the gravitational field strength 9.81N/kg) 2metres 10kg If the mass is lifted vertically against the force by a distance of 2 metres then 98.1 x 2 (f x d) = 196.2 joules of energy are needed. 196.2 joules of work has been done Energy, work and Power 10kg 2metres The mass has now gained 196.2 joules of potential energy Power • Power is the rate at which energy is used (or work is done) • Power = energy/time • Measured in Joules/seconds • I Watt = I Joule/sec Power (question) If the power of the kettle is 2400W and there are assumed to be no losses, calculate how much energy will be used if the kettle is on for I minute . Power (question) - answer Power = energy ÷ time ( in seconds) Energy = power x time ( in seconds) Energy = 2400 x 60 = 144000 joules 1.44 x 105 joules Relationship between work, power and velocity Work done = Force x distance Velocity = distance / time Work done/time = ( Force x distance )/ time Work done /time = power and distance/ time = velocity Power = Force x velocity Relationship between power and velocity (question) A vehicle has a maximum speed of 50 m/s and a power output of 120 kW at this speed. Calculate the driving force generated by the engine. Relationship between work, power and velocity (question) - answer Power = Force x velocity Force = Power/velocity Force = 120000 watts/50 m/s = 2400 Newtons Relationship between work, power and velocity (question) • The brakes exert a total force of 8 kN and the car stops from 50 m/s in 200m. Calculate the work done by the brakes. Relationship between work, power and velocity (question) - answer Work done = force x distance Work done = 8kN x 200m 1600 kJ or 1.6MJ Specific Heat Capacity The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. Specific Heat Capacity Q = c.m.∆T Q = heat added C = specific heat capacity m = mass and ∆T = change in temperature Specific Heat Capacity (question) • The brake discs on the car in the last question have a total mass of 8 kg and are made from a material with a specific heat capacity of 1180 J/kgK. Neglecting any cooling effects, what will be the temperature rise of the brake discs during the stopping manoeuvre? Specific Heat Capacity (question) Q/c.m =∆T ∆T = 1600000/ 1180 x8 169K (or 169oC) Specific Heat Capacity (question) The specific heat capacity of water is 4200 J/Kg oC How much heat is required to raise the temperature of 10 Kg of water by 20oC Specific Heat Capacity (question) - answer Q = c.m.∆T Q = 4200 x 10 x 20 = 840000 joules or 8.4 x 105 joules Enthalpy Enthalpy is a measure of the total energy of a thermodynamic system Enthalpy (question) • If the specific enthalpy of water at 10oC is 43 kJ/kg and the specific enthalpy of steam at 100oC is 2675 kJ/kg, calculate the energy required to convert 1.5 kg of water at 10oC to steam at 100oC. Enthalpy (question) -answer • Q = m(H – h) = 1.5( 2675 – 43) = 3948 kJ Where H = enthalpy of steam, h = enthalpy of water, Q = quantity of energy and m = mass Gas Laws • Volume is inversely proportional to pressure • V = 1/P • Pressure x volume is constant. If pressure increases volume decreases Gas Laws Pressure x volume is constant p1 V1 = p2 V2 Where: p1 is the starting pressure V1 is the starting volume p2 is the finishing pressure V2 is the finishing volume This is known as Boyle’s Law Gas Laws (question) • An air tight syringe contains a volume of 8 x10-6 m3 of air at 1 × 105 Pa. pressure. The plunger is pushed until the volume of trapped air is 4 x 10-6 m3. If there is no change in temperature what is the new pressure of the gas? Gas Laws (question) - answer p1 V1 = p2 V2 p2 = (p1 V1) ÷ V2 = (1 x 105 Pa x 8 x10-6 m3) ÷ 4 x 10-6 m3 = 2 x 105 Pa Gas Laws Volume is directly proportional to temperature • volume ÷ temperature is constant. If temperature increases volume increases Gas Laws Pressure is directly proportional to temperature • pressure ÷ temperature is constant. If temperature increases pressure increases Gas Laws Combining all three laws we get (pressure x volume) ÷ volume is constant Gas Laws This equates to PV/T = constant (P1V1)/T1 = (P2V2)/T2 = constant Gas Laws • This means that in any system, if any of the quantities change then the others will change in proportion • Temperature has to be measured in Kelvin • (degrees Celsius + 273) • - 273oC = absolute zero Gas Laws (Question) Steam at a pressure of 20 x 105 Pa and a temperature of 127oC is compressed to 1/2 of its original volume, after which the temperature recorded is 327oC. Calculate the pressure required to achieve this. . Gas Laws (Question) - answer • (P1V1)/T1 = (P2V2)/T2 • P1 = 20 x 105 Pa P2 = ? • V2 = 1/2 of V1 so V1/ V2 = 2/1 • T1 = 400 Kelvin (K) and T2 = 600 Kelvin(K) Gas Laws (Question) - answer P2 = (P1V1 T2)/(V2T1) (20 x 105 Pa x 600 x 2)/400 (V1/ V2 = 2) = 60 x 105 Pa or 6 x 106 Pa Gas Laws ( Question) Diesel fuel is injected into a cylinder at an inlet pressure of 1.2 bar and a temperature of 24oC. The compression ratio of the engine is 18 : 1 and the minimum ignition temperature of diesel fuel is 400oC. Calculate the minimum cylinder pressure in bar required to achieve this (1 bar is approximately atmospheric pressure) Gas Laws (Question) - answer P2 = (P1V1 T2)/(V2T1) P2 = (1.2 x18 x 673)/ (1 x 297) = 48.9 bar Kinetic energy (movement) The equation for calculating kinetic energy is ½(mv2) Where m = mass and v = velocity Kinetic energy (question) • A car of mass 1500 kg is moving at 20m/s. Calculate its kinetic energy. Kinetic energy (question) - answer kinetic energy = ½(mv2) Mass = 1500 kg Velocity = 20m/s Ke = 0.5 x 1500 x202 300000 Joules (300kJ) Relationship between Gravitational potential energy and kinetic energy Gravitational potential energy is the energy an object has due to it’s perpendicular height Gravitational potential energy = mass x gravitational field strength x height GPE = mgh Relationship between Gravitational potential energy and kinetic energy (question) A 20kg mass is suspended 20 metres above the ground. Calculate the gravitational energy of the mass Relationship between Gravitational potential energy and kinetic energy (question) - answer GPE = mgh m = mass (kg), g = gravitational field strength 9.81 (N/kg) h = height (metres) Relationship between Gravitational potential energy and kinetic energy (question) - answer GPE = mgh GPE = 20 x 9.81 x 20 = 3924 Joules Relationship between Gravitational potential energy and kinetic energy If the 20kg mass falls to the ground the potential energy converts to kinetic energy. When the mass hits the ground all the energy is kinetic energy Relationship between Gravitational potential energy and kinetic energy (question) Calculate the velocity of the 20kg mass as it hits the ground Relationship between Gravitational potential energy and kinetic energy (question) - answer GPE = 3924 Joules Kinetic of the mass when it hits the ground = 3924 Joules Relationship between Gravitational potential energy and kinetic energy (question) - answer Kinetic energy = ½(mv2) = 3924 Joules v2 = (2 x 3924)/20 v2 = 392.4 v = √392.4 velocity = 19.8m/s 50o 500g A pendulum of length 3.0m is pulled aside to an angle of 50o with the vertical. Calculate the gain in potential energy of the 500g bob. (g = 9.81m/s2) A 50o B 500g To calculate the length AB Cosine 50o = adjacent/hypotenuse = AB/3 AB = 3 x Cosine 50o 1.9 m Increase in vertical height of the 500g bob = 3 – 1.9 = 1.1m A Increase in GPE = 0.5kg x 9.81 x 1.1m 5.4 Joules 50o B 500g A Calculate the velocity of the bob at the bottom of its swing. 50o B 500g Ke = ½(mv2) = 5.4J v2 = (2 x 5.4)/0.5 = 21.6 V = √21.6 V = 4.6 m/s