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Harman [31]
3 years ago
8

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

parameter, K, in terms of the original parameters in each case. Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s).
a) s + (1/τ) = 0 versus parameter τ
b) s2 + cs + c + 1 = 0 versus parameter c
c) (s + c)3 + A(Ts + 1) = 0
i. versus parameter A
ii. versus parameter T
iii. versus the parameter c, if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all.
d) 1 + (kp + k1/s + kDs/Ts + 1)G(s) = 0. Assume that G(s) = A c(s)/d(s), where c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s).
i. versus kp
ii. versus kI
iii. versus kD
iv. versus τ

Engineering
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

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Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0
3 years ago
500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether
BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that   X follows a  Binomial distribution with (n= 500, p = 0.05)

The  probabilities  are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to  15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to  29)  

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)  

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0
3 years ago
Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci
Mademuasel [1]

Answer:

please find attached.

Explanation:

4 0
3 years ago
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp
cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

6 0
3 years ago
When moving cylinders always remove and make
Karolina [17]

Unless cylinders are firmly secured on a special carrier intended for this purpose, regulators shall be removed and valve protection caps put in place before cylinders are moved. A suitable cylinder truck, chain, or other steadying device shall be used to keep cylinders from being knocked over while in use.

5 0
3 years ago
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