Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
1 answer:
Answer:
β = = 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = =
next we determine the damping condition using the damping formula
β = = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
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Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top = 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π/4) ×
= √(4Q/π
)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828
Answer:
Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.
Note: kindly find an attached diagram to the complete question given below.
Sources: The diagram/image was researched and taken from Slader website.
Explanation:
Solution
Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:
Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A
= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A
Here
The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.
Thus
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q = T∞₁ -T₁/1/h₁A
T₁ = T∞₁ - Q/h₁A
Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A
Now
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q= T₂ -T∞₂/1/h₂A
T₂ = T∞₂ + Q/h₂A
Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A
