Calculation of Piston Diameter for Oil Displacement

Q1)
Volume of oil displaced by the piston = volume of oil remove by the 4 holes
Volume = area * velocity
V* A = 4*v*a
A = π∗D2
4 = ?
a = π∗d2
4 = π∗6∗10−3
4 =4.71239∗10−3 m2
Forming of equation in relation to the piston and the holes
Given that v = d2 ∆ p
32 μL
V*A = 4*a* d2 ∆ p
32 μL
Multiply both side [A – 4a]
[A – 4a]*V*A = d2 ∆ p
32 μL *[A – 4a]
[A – 4a]*V*A = a* d2 ∆ p
8 μL
Diving both side with V
∆ P=force=F
[A – 4a]* A =
1
V ∗d2 F
8 μL
= a* d2 C
8 μL = a2 C
2 μπL
A2 – 4Aa - a2 C
2 μπL =0

Substituting for a and C
A2 – 18.84956*10-3A - 2.4182=0
Solving by use of quadratic method the above equation
A = 1.55599032
π D2
4 =1.55599032
D2 = 1.981148407
D = 1.4075 m
Q2)
Maximum acceleration
Amax = A*ω2
Where, A = amplitude
And ω2= k
m
Since, F*t = m*Vmax
Vmax = A*ω
amax = (Vmax)ω

amax = F∗t
m ∗ √ k
m
amax = 400∗0.003
4.7 ∗
√ 2000
4.7
amax = 5.267 m/s2
b) deflection of in the string
amax = xm * ω2
5.267 = xm * ( √ 2000
4.7 )2
Xm = 0.01238 m
Q3)

Two blade aircraft propeller can be considered as Bifillar suspension system
Tension in wire A and B
TA = TB = T = w/2
And resistivity torque due to small displacement
T = TA * ∅∗¿r1 + TB * ∅∗¿r2
Torque = 2*T¿ ∅∗¿r1 = W*∅∗¿r .......(1)
From the geometry ∅= rθ
l
From equation 1
Torque = W∗r2 θ
l
We know that
T = I*∝
I = w
g ∗k2 ...2
∝= T
I =
w r2 θ
l
w
g ∗k2
....3
K = radius of gyration
W2 = ∝
θ = g r 2
lW 2