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babymother [125]

3 years ago

9

GG(ss) = 1 ss2 + 3 We want to design a feedback control system in a unity feedback structure by adding a controller DD(ss) = ss+

1 ss+2 . (a) Using the Routh Stability criterion, determine the stability of the feedback system. (b) Determine the system type and the corresponding steady-state error.

1 answer:

saw5 [17]3 years ago

3 0

Answer:

Explanation:

The explanation is given in the picture that is shown below

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Determine the number of flipflops required to build a binary counter that count from 0 to 2043

Pie

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

3 0

3 years ago

In fully developed laminar flow in a circular pipe the velocity at R/2 (mid-way between the wall surface and the centerline) is

Butoxors [25]

Answer:

$u_{max} = 17.334\,\frac{m}{s}$

Explanation:

Let consider that velocity profile inside the circular pipe is:

$u(r) = 2\cdot U_{avg} \cdot \left(1 - \frac{r^{2}}{R^{2}} \right)$

The average speed at $r = \frac{1}{2} \cdot R$ is:

$U_{avg} = \frac{13\,\frac{m}{s} }{2\cdot \left(1-\frac{1}{4} \right)}$

$U_{avg} = 8.667\,\frac{m}{s}$

The velocity at the center of the pipe is:

$u_{max} = 2\cdot U_{avg}$

$u_{max} = 17.334\,\frac{m}{s}$

6 0

2 years ago

Read 2 more answers

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

IceJOKER [234]

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

$q=-\kappa \bigtriangledown T$

Where,

q is the heat flux

$\kappa$ is the thermal conductivity

$\bigtriangledown T$ is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

$C=\frac{E}{\Delta T}$

Where,

C is the Heat capacity

E is the energy absorbed/released

$\Delta T$ is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

$\alpha=\frac{\kappa}{C_p \times \rho}$

Where,

$\alpha$ is thermal diffusivity

$\kappa$ is the thermal conductivity

$C_p$ is specific heat capacity at constant pressure

$\rho$ is density

6 0

3 years ago

Which of the following is NOT true concerning the color of minerals? A. Some minerals have a consistent color, but many have a r

mixer [17]

Answer:

Option D

A mineral’s color reflects the wavelengths of light that are absorbed by the mineral.

Explanation:

Color is one of the physical properties of minerals. Many minerals have a wide range of colors but there are some minerals with one consistent color and such minerals are referred as monochromatic minerals for example azurite. Normally, the streak color tends to be less variable than the color of the whole mineral and impurities or minor chemical components in a mineral react and often control the display color of resultant mineral. Option D is incorrect since mineral's color don't reflect wavelengths of light absorbed by such minerals.

5 0

3 years ago

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

$\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}$

using the energy equation for entry and exit value :

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}$

where

$\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}$

$= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\$

$= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\$

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\$

$\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})$

L.H.S = R.H.S

7 0

2 years ago