Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m =
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn =
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
This question is incomplete, the complete question as well as the missing diagram is uploaded below;
Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.
Take = 880 kg/m³ and
= 1260 kg/m³
Answer:
the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
Explanation:
Given that;
Inlet velocity of Glycerin, = 6 m/s
Inlet velocity of oil, = 3 m/s
Density velocity of glycerin, = 1260 kg/m³
Density velocity of glycerin, Take = 880 kg/m³
Volume of tank V = 4 m
from the diagram;
Diameter of glycerin pipe, = 100 mm = 0.1 m
Diameter of oil pipe, = 80 mm = 0.08 m
Diameter of outlet pipe = 120 mm = 0.12 m
Now, Appling the discharge flow equation;
π/4 × ()²
+ π/4 × (
)²
= π/4 × (
)²
we substitute
π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²
0.04712 + 0.0150796 = 0.0113097
0.0621996 = 0.0113097
= 0.0621996 / 0.0113097
= 5.5 m/s
Now we apply the mass flow rate condition
so we substitute
1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5
1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335
59.3712 + 13.27 = 0.06220335p
72.6412 = 0.06220335p
p = 72.6412 / 0.06220335
p = 1167.8 kg/m³
Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
