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3 years ago

14

Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______

1 answer:

Alina [70]3 years ago

5 0

Answer:

Multiplying impulse response by t ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

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Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

andriy [413]

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$

Explanation:

Given data:

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from steady flow enerfy equation

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

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$s_1 = 1.708 Btu/lbm -R$

for P1 = 40 lbf/in^2

$H_1 = 1193.5 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

$v_2 = 2016.80 ft/s$

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$\Delta s = s_2 - s_1$

$s_1 = 1.708 Btu/lbm -R$

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$s_2 is 1.714 Btu/lbm -R$

$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

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3 years ago

An inductor of ????=6.95 H with negligible resistance is placed in series with a ℰ=12.5 V battery, a ????=3.00 Ω resistor, and a

elena-14-01-66 [18.8K]

Answer:

a) I=0 b) 4.17V c) 0.354 A d) 14.5s

Explanation:

a) consider circuit in the attachment

i(t)= E/R (1- e^(-t/RL))

i(0)= 12.5/3×(1-e^(0/RL))

i(0)=0

b) at t⇒∞

i(∞)= 12.5/3× (1- e^(-∞/RL))

= 4.17V

c) 1/RL= 1/(6.95×3)= 0.0479616

i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)

= 0.354A

d) I/2= I (1- e^(-t/RL))

t= - RL ln0.5

t= - 3×6.95 × (-0.693)

t= 14.5 s

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4 years ago