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Systems, Accounting and Modeling Approach Examples showing how students are taught to solve problems. Examples* Heat Engine Performance Monkeys on a Rope Colliding Train Cars Grain Conveyor Belt *Before reading the complete solution to each problem, you are encouraged to sketch out your own solution to the problem. Example - Carnot Cycle Revisited What is the maximum thermal efficiency of a steady-state heat engine that receives energy by heat transfer at a surface temperature TH and rejects energy by heat transfer at a surface temperature T ? Q!H,in TH W! net,out L TL Analysis What’s the system? What properties should we count? What is the time interval? What are the important interactions? Q!L,out dEsys 0,SS ! = Q net,in dt 0 = Q! W! H,in - Q! net,out L,out + W! = Q! H,in - W! net,in Closed System Q! H,in net,out TH - Q! L,out dSsys 0,SS Q! j = ∑T dt j + S!gen Q! L,out Q! Q! H,in - L,out 0= T T H L net,out TL j W! + S!gen T L + T S! Q! = Q! L,out H,in T L gen H T L + T S! W! = Q! - Q! net,out H,in H,in T L gen η= W! net,out = Q! H,in H T 1- L T H T S!gen - L Q! H,in Closed System Q! H,in TH η= W! ! T T S net,out = 1 - L - L gen T Q! Q! H,in H T = 1- L η T max H H,in W! net,out TL Q! L,out T S!gen L ≥O Q! H,in Example - Monkeys on a Rope Three monkeys A, B, and C with masses of 10, 12, and 8 kg, respectively, are climbing up and down the rope suspended from point D. At the instant shown in the figure, A is descending the rope with an acceleration of 1.6 m/s2, and C is pulling himself up with an acceleration of 0.9 m/s2. Monkey B is climbing up with a constant speed of 0.6 m/s. D A B Determine the tension T in the rope at D, in newtons. Analysis What’s the system? What properties should we count? What is the time interval? What are the important interactions? C D A B C Physical System T D A WA B WB C WC Physical System Free-body Diagram " dPsys = dt " ∑Fext,j Closed System T j " " " " " ∑Fext,j = W A +WB +WC +T WA " " " " Psys = mAVA +mBVB +mCVC WB j WC " " " " " " " d m V +m V +m V = W + W + W + T A B C dt A A B B C C " " " " dVA dV " " " dV C B = m g + m g + m g + T + m + m m B dt A B C dt C dt A Closed System " " " " dVA " " " dV dV C B T = m -g + mB -g + mC -g dt dt dt A x WA WB WC m m m T = 10 kg -1.6+9.81 2 + 12 kg 0+9.81 2 + 8 kg 0.9+9.81 2 s s s = + + 82.1N 117.7 N 85.6 N T T = 285.4 N Examples - Rail Cars on the Move A 45-Mg railroad car moving with a velocity of 3 km/h is to be coupled to a 25-Mg car which is at rest. Determine (a) the final velocity of the coupled cars (b) the average impulsive force acting on each car if the coupling is completed in 0.3 s. VA A VB = 0 B Part (a) - Final velocity after coupling System: Assume an open, moving system. Initially it contains car B only and finally it contains both cars. Property: x " " " " dPsys ! - ∑m ! eVe = ∑F + ∑mV ext,j i i dt j in out VA A VB = 0 B Conservation of Linear Momentum (x-direction) dPsys,x ! = mV i x,i dt t t 2 dPsys,x ! dt = mV ∫ dt ∫ i x,i dt t t 1 1 2 Integrate over time interval t1 to t2. 1 - Initial state 2 - Final state VB,1 = 0 P sys,x,2 m +m V A B -P sys,x,1 =mV A A,1 0 AB,2 -mV B B,1 mV V = A A,1 AB,2 m +m A B =m V A A,1 Part (b) Coupling Force System: Assume a closed system that only contains car B throughout the process. This system moves with car B. " 0 " " " dPsys ! - ∑m ! eVe = ∑F + ∑mV ext,j i i dt j in out Property: x B Fcoupling Conservation of Linear Momentum (x-direction) dPsys,x = Fx dt t VB,1 = 0 VB,2 = VAB,2 from part (a) t 2 dPsys,x ∫ dt dt = ∫ Fx dt t t 1 1 2 Integrate over time interval t1 to t2. 1 - Initial state 2 - Final state P sys,x,2 mV B B,2 -P sys,x,1 = Fx,avg #t 0 - mV B B,1 = Fx,avg #t mBVB,2 Fx,avg = #t Example - Conveyor Belt Grain falls from a hopper onto a conveyor belt at the rate of 200 kg/min. The conveyor belt carries the grain away at a constant velocity of 2 m/s. Determine the force on the belt required to keep the belt moving at a constant speed. Hopper Belt Grain Vbelt Hopper Belt Grain Vbelt Grain entering 1 2 Grain leaving 3 Belt entering Fbelt Steady-state open system Belt leaving x 1 3 2 Conservation of mass for this system. System Fbelt dmsys ! ! ! ! =m +m -m -m grain,1 belt,2 belt,3 grain,3 dt msys =m +m belt,sys 0, SS dmbelt,sys dt ! ! =m -m belt,2 belt,3 ! ! m =m belt,2 belt,3 grain,sys 0, SS dmgrain,sys dt ! ! =m -m grain,1 grain,3 ! ! m =m grain,1 grain,3 Conservation of Linear Momentum for this system. ( X-direction) 0, SS 1 2 x 3 System Fbelt dPx,sys ! ! eVx,e = ∑Fx,ext,j + ∑ mV m ∑ i x,i dt j in out ! ! 0 = Fbelt + m V +m V grain,1 x,grain,1 belt,2 x,belt,2 ! ! -m V -m V belt,3 x,belt,3 grain, x,grain,3 0 0 ! ! 0 = Fbelt + m V -V V + m V grain x,grain,1 x,grain,3 belt x,belt,2 x,belt,3 ! ! Fbelt = m V =m V grain x,grain,3 grain belt End of Examples For additional information about the RH Sophomore Engineering Curricula or the Systems, Accounting, and Modeling Approach contact --Don Richards Rose-Hulman Institute of Technology 5500 Wabash Ave. - CM 160,Terre Haute, IN 47803 Email: [email protected] URL: http://www.rose-hulman.edu/~richards Phone: 812-877-8477 Or check the Foundation Coalition Web Site at http://www.foundationcoalition.org