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valina [46]
3 years ago
9

Harmonic excitation of motion is represent as

Engineering
1 answer:
Gennadij [26K]3 years ago
6 0

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

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What can be the main disadvantage of pulse amplitude modulation?​
Feliz [49]

Answer:

transmission bandwidth required is very large.

Explanation:

4 0
2 years ago
Water flows in a constant diameter pipe with the following conditions measured:
Burka [1]

Answer:

a) h_L=-3.331ft

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

p_a =31.1psi=4478.4\frac{lb}{ft^2}

p_b =27.3psi=3931.2\frac{lb}{ft^2}

For above conversions we use the conversion factor 1psi=144\frac{lb}{ft^2}

z_a =56.7ft

z_a =68.8ft

h_L =? head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L

3)Part a

And on this case we have all the values in order to replace and solve for h_L

\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L

h_L=(71.769+56.7-63-68.8)ft=-3.331ft

4)Part b

Analyzing the value obtained for \h_L is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

5 0
3 years ago
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
marysya [2.9K]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

4 0
2 years ago
Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m
Aloiza [94]

Explanation:

uuui ielts k oshru with the best of my life u

3 0
2 years ago
It is suggested that the kitchen in your house could be cooled in the summer by closing the kitched from the rest of the house a
jeka57 [31]

Answer:

False

Explanation:

When we open the gate of refrigerator then the temperature of kitchen get increase.

But we think how it is possible that temperature of kitchen get increases.Actually the heat rejecting coil(condenser) is also present inside the room that is why it reject more heat as compare to heat absorbed by evaporator coil.This is the reason to increases the temperature of kitchen.

If heat rejecting coil put outside the kitchen then it will produces cooling effect even door of refrigerator open.This concept use in air conditioning.

8 0
3 years ago
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