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3 years ago

Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:

Engineering

1 answer:

Pavlova-9 [17]3 years ago

3 0

Answer:

a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

$a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100$

No roots not in the LHP

$b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480$

2 roots not in the LHP

$c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8$

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

$d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78$

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

$e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25$

2 roots not in LHP

check:

$a (s) = 0$ ⇒

$s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_- 0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\$

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A two-stroke CI. engine delivers 5000 kWwhile using 1000 kW to overcome friction losses. It consumes 2300 kg of fuel per hour at

Lady_Fox [76]

Answer:

(a) Indicating power(IP)=6000 KW

(b) $\eta_{mech}$ =0.833

(d) $\eta_{BPth}$ =0.1865

Explanation:

Break power(BP) =5000 KW

Friction power(FP)=1000 KW

Consumption of fuel per hour=2300 kg/hr

CV=42000 KJ/kg

We know that

Indicating power(IP)=Break power(BP)+Friction power(FP)

⇒IP=5000+1000 KW

IP=6000 KW

(a)

Indicating power(IP)=6000 KW

(b)

Mechanical efficiency $\eta_{mech}=\dfrac{BP}{IP}$

$\eta_{mech}=\dfrac{5000}{6000}$

$\eta_{mech}$ =0.833

(c)

Air fuel ratio= $\dfrac{mass \ of \ air}{mass \ of \ fuel}$

consumption of air per hour=20 $\times$ 2300 kg/hr

So consumption of air per hour =46000 kg/hr

(d)

Break thermal efficiency $\eta_{BPth}=\dfrac{IP}{\dot{m_f}\times CV}$

$\dot{m_f}=\dfrac{2300}{3600}$

=0.638 kg/s

$\eta_{BPth}=\dfrac{5000}{{0.638}\times 42000}$

$\eta_{BPth}$ =0.1865

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4 years ago

Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How

pogonyaev

Answer:

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) The kilograms of total ferrite and is 0.8311 kg

The kilograms of total cementite is 0.1689 kg

c) The kilograms of total cementite is 0.9343 kg

Explanation:

Given:

1 kg of austenite

1.15 wt% C

Cooled to below 727°C

Questions:

a) What is the proeutectoid phase?

b) How many kilograms each of total ferrite and cementite form, Wf = ?, Wc = ?

c) How many kilograms each of pearlite and the proeutectoid phase form, Wp = ?

d) Schematically sketch and label the resulting microstructure

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) To get the mass of the total ferrite form:

$W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }$

Here,

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = composition of alloy = 1.15 wt%

Substituting values:

$W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg$

To get the mass of the total cementite:

$W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg$

c) The mass of pearlite:

$W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg$

d) In the diagram you can see the different compositions (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

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3 years ago

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th

Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

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solution

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3 years ago

In the High Low Logic Index low levels are bearish and high levels are bullish, generally True False

Irina-Kira [14]

Answer:

True

Explanation:

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3 years ago

Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ; B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

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3 years ago