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Thermodynamics Chapter Four Lecture: Noor M. Jasim THERMODYNAMICS Chapter Four Work and Energy Noor M. Jasim 2009 Thermodynamics Chapter Four Lecture: Noor M. Jasim Chapter Four Work and Energy Example (3-1) Unit of mass of fluid at a pressure of 3 bar, and specific volume of 0.18 m3/kg expands reversibly in the piston cylinder assembly according to the processes (a) constant pressure process until the volume is doubled, (b) according to the relation PV=c until the final volume will be 0.36 kg/m3, and (c) according to the relation pv2=c until the final volume reaches to 0.36 m3/kg. Determine the work done in each case. Solution According to Fig. (3.4) a) for constant pressure process P=c , then kN W P 2 V 2 V1 ( m 3 ) 1 2 m m3 m3 3 1.013 10 2 ( 1kg 0.36 1kg 0.18 ) = 54.702 kJ kg kg p (a) p=c 1 2 (b) T=C (c) 2\ pvn = C 1 2\\ v 0.18 0.36 Fig. (3. 4) b) for constant temperature process T = c , then V2 V kN m3 W mRT n mp1V1 n 2 1( kg ) 3 10 2 1.013 2 01.8 kg n 2 37.9( kJ ) 1 2 V1 V1 m c) For pvn = constant where n=2 Thermodynamics Chapter Four Lecture: Noor M. Jasim p 2V 2 p 1V1 W m 1 2 1n n since p 2 V1 and then p2 p1 1 75.97(kN / m 2 ) , Hence 2 p1 V2 2 75 .97 0.36 3 101 .3 0.18 W 1kg 13 .7kJ 1 2 2 Example (3.2) Piston cylinder assembly contains 1 kg of gas at initial pressure of 40 bar. The gas expands reversibly according to a law (PV1.3 = c) until the volume is doubled. The gas is then cooled reversibly at constant pressure until the piston firmly locked in position until the pressure rises to the original value of 40 bar. Calculate the net work done by the gas if the initial volume equals 0.02 m3 Solution According to figure (3-5) p 1 p1 = 40x102 kN/m2 Pv1.3 = 0 Net work = 17.5 kJ p2 = 16.25x102 kN/m2 P 2 3 0.02 0.04 v Fig. (3.5) Since, p 1V 11.3 p 2V 21.3 , therefore 1.3 1.3 V 1 p 2 p 1 1 40 16.25bar , hence V 2 2 P V P1V1 2 16.25 0.04 40 0.02 W m 2 2 ( 10 ) 50kJ and 1 2 1 n 0.3 W p 2 V3 V2 , since V3 = V1, hence 2 3 W 16.25 10 2 0.02 0.04 32.5 kJ , and W 0 23 3 1 the net work done = W W W W = 50 – 32.5 + 0 = 17.5 kJ 1 2 2 3 34 Thermodynamics Chapter Four Lecture: Noor M. Jasim Example: (3.3) Consider, the main shaft of the gear box transmitted power as a system. If the torque applied is 1kJ and the shaft rotates at speed of 2400 rpm. Determine the power transmitted by the shaft. Solution 2400 W shaft 2n ( kW ) 2 1( kJ / s ) 251.2kW 60 Example (3.4) A volume of 0.02m3 of air is contained in piston-cylinder assembly with pressure of 100kPa. At this position a linear spring that has K=100kN/m is touching the piston but exerting no force on it. If a heat is transferred to the gas, the gas is then expanded and the piston rises and compresses the spring until the volume inside the cylinder is doubled. If the cross section area of the piston is 0.2m2. Calculate (a) the final pressure inside the cylinder, (b) the total work done by the mass, and (c) the fraction of this work done against the spring to compress it. Solution Given: piston cylinder device contains gas, spring connected between the piston and the cylinder externally. Find: p2 , w, Wspring Schematic and Given data: schematic diagram and p-V diagrams for given data is shown in Figure (3.8). k = 100 kN/m p 2 p2 = 150 kPa pm = 125 kPa Wspring I A = 0.2 m2 Wwith no p1 = 100 kPa 1 P P1 = 100 kPa spring action II V1 = 0.02 m3 v 0.02 0.04 Fig. (3.8) Assumptions: The gas is a closed system. Analysis: The enclosed volume at the final state is V2 2V1 2 0.02 0.04m 3 hence, the displacement of the piston (and the spring) becomes ( V 2 V1 ) 0.04 0.02m 3 x 0.1m A 0.2 m 2 Thermodynamics Chapter Four Lecture: Noor M. Jasim The force applied by the linear spring at the final state is determined using equation (3.12), then F kx 100kN / m 0.1m 10kN The additional pressure added by the spring on the gas at this state is F 10kN P 50kN / m 2 A 0.2m 2 Then the final pressure of gas inside the cylinder rises by the additional pressure that created from the effect of the spring, then the final pressure is: Pf = Pi + Pdue to spring action ( kPa ) = 100 + 50 = 150 kPa 2 b) W pdV 1 2 1 There is no exact relation between P and V given directly from the given data, but we know that the relation between the force exerted on the spring and its elongation or compression is linear within the elastic limits then the mean pressure is pi p f 100 150 pm 125kPa 2 2 then the W can be determined as follows 1 2 W pm V f Vi 1250.04 0.02 2.5 kJ 1 2 c) The spring work can be determined as: W spring 1 2 2 1 k x 2 x 1 100 ( 0.1 ) 2 0 0.5 kJ 2 2 or The spring work equal to the triangle area on p-V diagram Figure (3.8) W spring 1 125 100kPa0.04 0.02 05kJ 2 Example (3.6) A car has a mass of 1ton need to be accelerated from a velocity of 40 km/h to 90km/h is 5 minutes on a level road. Determine the power required for acceleration. Solution 1 The acceleration work = kinetic energy = m(V22 V12 ) 2 2 40 10 3 2 90 10 3 KE 1000 1 250.8 10 3 kJ 2 3600 3600 The power required is determined as: 250.8 10 3 Power = ( k .E ) / time 836kW 5 60 Thermodynamics Chapter Four Lecture: Noor M. Jasim Thermodynamics Chapter Four Lecture: Noor M. Jasim

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posted: | 10/22/2011 |

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