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Mechanical Engineering
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1/11
Lab-2: Prelab
1. The given EOM is

 

  


 

 


At steady state, the pump speed is
and water height
.

Linearizing the above non-linear EOM,
 

 






Obtaining the value of various term individually,






The linearized equation is:












2. Open-loop dynamics: In order to obtain the open-loop dynamics of the system under study,
the above obtained linearized time domain equation is transformed to frequency domain
through Laplace transform.
Replacing, 
by and 
by








 
Laplace transform of the above equation is (after zeroing the initial condition)





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








 





 





 





 



 
where,


and,

3. The closed-loop transfer function with PI control block is shown below:
G(s)C(s)
Y(s)
U(s)
E(s)
R(s)
The closed loop transfer function is given by:


  

where,

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Lab-2: Prelab 1. The given EOM is 𝐴𝑐 ℎ1̇ = 𝐾𝑝 (𝑣𝑝 − 𝑣𝑝𝑚𝑖𝑛 ) − 𝐶𝑑 𝐴0 √2𝑔ℎ1 𝐾𝑝 𝐶𝑑 𝐴0 ⇒ ℎ1̇ = (𝑣𝑝 − 𝑣𝑝𝑚𝑖𝑛 ) − √2𝑔ℎ1 𝐴𝑐 𝐴𝑐 ⇒ ℎ1̇ = 𝑓(ℎ1 , 𝑣𝑝 ) At steady state, the pump speed is ̅̅̅ 𝑣𝑝 and water height ̅̅̅ ℎ1 . ̅̅̅ ̅̅̅1 , ̅̅̅) ℎ1̇ = 𝑓(ℎ 𝑣𝑝 = 0 Linearizing the above non-linear EOM, ̅̅̅1 , ̅̅̅) ℎ1̇ = ̅̅̅ ℎ1̇ + ∆ℎ1̇ = 𝑓(ℎ 𝑣𝑝 + 𝑑𝑓 𝑑𝑓 | ∆ℎ1̇ + | ∆𝑣 ̇ 𝑑ℎ1 (ℎ̅̅̅̅1,𝑣̅̅̅̅) 𝑑𝑣𝑝 (ℎ̅̅̅̅,𝑣̅̅̅̅) 𝑝 𝑝 1 𝑝 Obtaining the value of various term individually, 𝑑𝑓 𝐶𝑑 𝐴0 2𝑔 | =− √ 𝑑ℎ1 (ℎ̅̅̅̅1 ,𝑣̅̅̅̅) 2𝐴𝑐 ̅̅̅ ℎ1 𝑝 𝐾𝑝 𝑑𝑓 | = 𝑑𝑣𝑝 (ℎ̅̅̅̅,𝑣̅̅̅̅) 𝐴𝑐 1 𝑝 The linearized equation is: ∆ℎ1̇ = 𝑑𝑓 𝑑𝑓 | ∆ℎ1̇ + | ∆𝑣 ̇ 𝑑ℎ1 (ℎ̅̅̅̅,𝑣̅̅̅̅) 𝑑𝑣𝑝 (ℎ̅̅̅̅ ̅̅̅̅) 𝑝 1 𝑝 ⇒ ∆ℎ1̇ = − 1 ,𝑣𝑝 𝐾𝑝 𝐶𝑑 𝐴0 2𝑔 ∆𝑣 √ ∆ℎ1 + 2𝐴𝑐 ̅̅̅ 𝐴𝑐 𝑝 ℎ1 2. Open-loop dynamics: In order to obtain the open-loop dynamics of the system under study, the above obtained linearized time domain equation is transformed to frequency domain through Laplace transform. Replacing, ∆ℎ1 by 𝑦 and ∆𝑣𝑝 by 𝑢 ∆ℎ1̇ = − 𝐾𝑝 𝐶𝑑 𝐴0 2𝑔 ? ...
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