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2 years ago

50.38

Engineering

1 answer:

klemol [59]2 years ago

6 0

Answer:

International Building Code (IBC)

Explanation:

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A controller for a satellite attitude control with transfer function G = 1 s 2 has been designed with a unity feedback structure

S_A_V [24]

Answer:

type 2, k = 4

Explanation:

(a) The transfer function of the controller for a satellite attitude control is

$G = \frac{1}{s^2}$

The transfer function of unity feedback structure is

$D(s) = \frac{10(s+2)}{s+5}$

To determine system type for reference tracking, identify the number of poles at origin in the open-loop transfer function.

For unity feedback system, the open-bop transfer function

$G(s)D_c(s)=\frac{1}{s^2}\frac{10(s+2)}{s+5}$

$=\frac{10(s+2)}{s^2(s+5)}$

Determine the poles in G(s)4(s).

s = 0,0,-5

Type of he system is decided by the number of poles at origin in the open loop transfer function.

Since, there are two poles at origin, the type of the system will be 2.

Therefore, the system type is

Type 2

check the attached file for the concluding part of the solution

5 0

3 years ago

1. You do not need to remove the lead weights inside tires before recycling them.

makvit [3.9K]

It true -.-.-.-.-.-.-.-.-.-.-.-.-.- EZ

7 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz

ra1l [238]

Answer:

Flow rate is $1.82\times 10^{-8} m^{3}/s$

Explanation:

Given information

Density of oil, $\rho_{oil}= 850 Kg/m^{3}$

kinematic viscosity, $v= 0.00062 m^{2} /s$

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

$\bar v=\frac {\triangle P\pi D^{4}}{128\mu L}$ where $\mu$ is dynamic viscosity and $\triangle P$ is pressure drop

At the bottom of the tank, pressure is given by

$P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}$

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still $25015.5 N/m^{2}$

Dynamic viscosity, $\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s$

Now the volume flow rate will be

$\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s$