This paper Stiffness of System is fourth article about Micro Robot Control in laminar flow thesis. Here is the first one(https://tunavatansever.com/2021/05/09/micro-robot-control/). This time we are considering how to calculate stiffness of the micro robots in laminar flow. Have a nice readings…
In order to understand the stability of our robot, we had to find the stiffness of the robot. Stiffness formula is:
Stiffness:? /∆? (newton\m) (2.10)
We looking for y-axis stiffness. In order to find the value of the stiffness value on the y-axis, we need to calculate the force on the y-axis acting on the robot. The force that affects the robot on the y axis is the drag force. Drag force formula is:
FD =1/2(ρ.V^2.CD) (2.11)
Velocity formula in drag force formula (2.11) is:
? = ? /?
The variables and values in the above equation are given in the table below.
Table 2.3: Quantity of Stiffness Equations
|FD||Drag Force||3.85 (Newton)|
|ρ||Density of Fluid||998.206 (kg\?3)|
|? ?||Fluid Velocity||0,027 (m\s)|
|Q||Flow Rate||35 (ml\min)|
|A||Area of Channel||21 (??2)|
When the data given in the table is written in the equation and the units are equalized,
FD = 1/2( 998,206 . (35/21)^2) . 0,005 =3.85 (Newton) (2.13)
drag force is calculated as 3.85 Newton.
The system has a horizontal length of 0.06 meters. This causes a drop in pressure. This decrease in pressure also causes a reduction in the force acting on the microrobot. The pressure drop mentioned here, the formula for rectangular pipe is:
∆? = ?h? = ??h? (2.14)
h? = 2*? (? *?^2)/ (?h *?) (2.15)
? = 16/Re (in laminar flow) (2.16)
Re = (v∗?h∗ρ )/ μ (for rectangular duct) (2.17)
The units and values of the variables in these equations are given in the table below.
Table 2.4: Quantity of Pressure Drop Equation
|∆?||Pressure Drop||1.542 (pascal)|
|h?||Frictional Head Losses||0.0001577 (m)|
|L||Length of Pipe||0,003 (m)|
|? ?||Velocity of Fluid||0,027 (m\s)|
|g||Gravitational Acceleration||9.80665 (m/?2)|
|ρ||Density of Fluids||998.206 (kg\?3)|
|μ||Dynamic Viscosity||0.00105 (N*s)|
|?h||Hydraulic Diameter||0.0042 (m)|
When the data given in the table is written instead of the equation and the units are equalized pressure drop is calculated as:
∆? = 1.542 (pascal) (2.18)
The change in pressure reduced the force on the y-axis acting on our micro robot.
? = ?/? (2.19)
Calculated from equation (2.19).
? = ?.? (2.20)
? = 1.542*0.09817=0.1513 ?? (2.21)
Stiffness calculated from equation (2.10).
Stiffness: ?/ ∆? = 3.85−0.0000001513/1 ≈3.85 (newton\mm) (2.22)
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