Chemical Engineering - Apllied Mechanics

Mechanical systems with transmissions

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1. 1 DOF MECHANICAL SYSTEMS Let us consider a mechanical system composed of an engine, E, a mechanical transmission, T, and motor-driven system, MDS (Figure 1.1). The engine provides a drive torque M d while the angular velocity of the drive (input) shaft is ω i. Conversely, the angular velocity of the driven (output) shaft of the transmission is denoted ω o. The transmission is characterised by a transmission ratio oi τ=ωω =and an efficiency coefficient η. In the end, M r is the resistance torque exerted by the motor-driven system on the output shaft of the transmission. Figure 1.1 = The engine can be composed of some=translating, rotating or roto-translating rigid bodies. The angular velocity of the j-th mechanical component, subjected to a rotational motion, is ij ω , while J jd=is=the= corresponding mass moment of inertia evaluated with respect to the barycentre=(for the sake of simplicity, let= us assume that any component is subjected to a planar motion).=Vector= Gik v is=the absolute=velocity of the centre of gravity, G ik, of the k-th mechanical component while kdm is the corresponding mass. Therefore, the drive part of the system (Figure 1.2) can be modelled considering the drive torque, M d, provided by the engine, the drive shaft and a “virtual” fly-wheel, mounted on the drive shaft, whose mass moment of inertia is 'Jd. Besides, it is necessary to take into account the equivalent resistance torque '1r M exerted by the remaining system on the drive shaft. For a given time t, and a corresponding angular velocity iω of the drive shaft (or input shaft), the kinetic energy of the fly-wheel must be equal to the global kinetic energy of the drive part of the real system. That is: 12 ' i ii iGi Gi 11 11 1JJ 22 2 j j jkk k nn dd djk m == ×= × + × ∑∑ vv ωω ω ω (1.1) The mass moment of inertia of the equivalent fly-wheel can be constant or it may depend on the drive shaft angular position, that is it may depend on time t. 'Jd const = or ( ) ( ) '''J J () Jd dd d tt ϕ = = (1.2) Figure 1.2 Drive part of the system Figure 1.3 Driven part of the system , including the transmission The remaining part of the system is shown in Figure 1.3. The resistance torque exerted by the motor-driven system on the transmission output shaft is M r. The motor-driven system can be composed of some translating, rotating or roto-translating rigid bodies. The angular velocity of the j–th mechanical component, subjected to a rotational motion, is oj ω , while Jjr is the corresponding mass moment of inertia evaluated with respect to the barycentre (once again, for the sake of simplicity, let us assume that any component is subjected to a planar motion). Vector Gok v =is=the absolute velocity of the centre of gravity, G ok, of the k-th mechanical component, while krm is the corresponding mass. Therefore, this sub-system (Figure=1.3) can be= modelled considering the=resistance torque, M r, exerted by the motor-driven system, MDS, on the output shaft, the transmission, T, the output shaft and a “virtual” fly-wheel, mounted on the output shaft, whose mass moment of inertia is 'Jr. Besides, it is necessary to take into account the equivalent drive torque '1d M exerted on the input shaft of the transmission. For a given time t, and a corresponding angular velocity oω of the output shaft, the kinetic energy of the fly-wheel must be equal to the global kinetic energy of the motor- driven part of the real system. That is: 34 ' o oo oGo Go 11 11 1 JJ 22 2 jj jkk k nn rr r jk m == ×= × + × ∑∑ vv ωω ω ω (1.4) The mass moment of inertia of the equivalent fly-wheel, 'Jr, can be constant or it may depend on the output shaft angular position, that is it may depend on time t. 'Jr const = or ( ) ( ) ''' o J J () Jrrr tt ϕ = = (1.5) The power balance of the sub-system illustrated in Figure 1.2 is: () '' i i i 1iJ0 dd r×− ×+×= MM ω ωω ω (1.6) The power of the equivalent resistance torque '1r M must be equal to the sum of the power of any force and torque acting on the remaining part of the global system. That is: ( )( ) '''1ioi i i oo 1 JJ r rdd r η  × = × −− × − × − ×  MM M  ω ωω ωω ωω (1.7) The eq.(1.6) can be rewritten as: '' 1i 1i 0 dr× + ×= MMωω (1.8) where: () '' 1ii i i J d dd× = ×− × MM ω ω ωω (1.9) Being the equivalent resistance torque '1r M opposite to the input shaft angular velocity, we obtain: ''1i 1i M M0 drω − ω= (1.10) Now, let us consider the sub-system illustrated in Figure 1.3. The power balance of this sub-system is: ( ) '''1i o1i oo 1 J0 d rdr η  ×+ ×−− ×− × =  MM M  ω ωω ωω (1.11) The power of the equivalent drive torque '1d M must be equal to the sum of the power of any force and torque acting on the remaining part of the global system. That is: ( ) ''1ii i i J d dd× = ×− × MM  ω ω ωω (1.12) The eq.(1.11) can be rewritten as: ( ) '''1io1i oo 1J r rdr η  × = × −− × − ×  MM M  ω ωω ωω = (1.13)= = Now, let=us consider the=isolated sub-system illustrated in Figure=1.4. The equivalent torque exerted by the remaining part of the complete system on the transmission output shaft is= '2r M . The power balance of this sub-system is:= () () () ''' i i i1i 2 oJ10 ddd r η × − × −− × + × = MMM  ω ωωω ω = (1.14)= That is:= () () ''' 1i1i 2 o 10 ddr η × −− × + × = M MMω ωω = (1.15)= The power of the equivalent resistance torque '2r M =must be equal to the sum of the power of any force and= torque acting on the remaining part of the global system.=That is:= '' 2 oo oo J r rr  × = ×− ×  MM  ω ω ωω =(1.16)= = = = Figure= 1.4 Drive part of the system =including the =transmission = Figure 1.5 Driven =part of the system = = Now, let us consider the=isolated sub-system illustrated in Figure=1.5. The equivalent torque exerted by the= remaining part of the complete system on the transmission output shaft is= '2 d M . The power balance of this sub-system is:= ''2 o o oo J0 d rr × + ×− × = MM  ω ω ωω = (1.17)= The power of the equivalent drive torque= '2d M =must be equal to the sum of the power of any force and= torque acting on the remaining part of the global system. That is:= ( ) ( )( ) '''2oi iii ii J1 J d dddd η × = ×− × −− ×− × MMM  ω ω ωωω ωω == (1.18)= = = In the end, let us consider the=isolated transmission=illustrated in Figure=1.6. The power balance of this sub- system is:= () () ''' 1i1i 2 o 10 ddr η × −− × + × = M MMω ωω = (1.19)= Where the equivalent torques= '2r M =and= '1d M =are given by eqs.(1.16) and (1.12), respectively.= Besides, it is: ''2o 2odr ×= × MM ωω with ''2 o 1idd η ×= × MM ωω = (1.20)= = = Figure= 1.6 Transmission = = = =