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4 years ago

In root locus technique, angle between adjacent asymptote is (a) 180°/(m+ n) (b) 360°/(m + n) (c) 360°/(m-n) (d) 180°/(m-n)

Engineering

1 answer:

Inessa [10]4 years ago

4 0

Answer:

Option (d)

Explanation:

In the root locus, method, the angle of asymptote or the angle between adjacent asymptote is calculated by using the given formula:

Angle of asymptote, $\angle A = \frac{180^{\circ}(2p + 1)}{m - n}$

If p = 0 (say)

then the above eqn becomes:

$\angle A = \frac{180^{\circ})}{m - n}$

where

m = no. of poles

n = no. of zeroes

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katrin2010 [14]

Answer:

Explanation:

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3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a T-s diagram?

gayaneshka [121]

Answer:

Heat

Explanation:

Carnot cycle:

Carnot cycle is the ideal cycle for all working engine .Carnot cycle all processes are reversible.It have fore process Out of two are constant temperature process and other two are isentropic process(reversible adiabatic).

We know that area under T-s diagram represents the heat.

So $Q=\int Tds$

From cycle we can say that

$q_{in}=T_2\left ( s_2-s_1 \right )$

$q_{out}=T_1\left ( s_2-s_1 \right )$

4 0

4 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D

slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

6 0

3 years ago

Why is it important to keep hand tools free of oil and grease?

Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0

4 years ago