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4 years ago

Given the unity feedback system

Engineering

1 answer:

LenaWriter [7]4 years ago

3 0

Answer:

A.) 0 > K > 9.6

B.) K = 9.6

C.) w = +/- 2 sqrt (3)

Explanation:

G(s)= K(s+4)/s(s+1.2)(s+2)

For a closed loop stability, we can analyse by using Routh - Horwitz analysis.

To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.

Please find the attached files for the solution

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The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de

garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = $\frac{\pi d^{2} }{4}$

d = 1.5

$\frac{\pi *1.5^{2} }{4}$

= 1.767 inches²

<h3>Solve for inertia</h3>

$\frac{\pi *0.75^4}{4}$

= 0.2485inches⁴

Solve for the tensile force from here

$\frac{F}{A} +\frac{Mc}{I}$

30x10³ = $\frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\$

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0

2 years ago

Ayden read 84 pages in 2 hours. At that rate, how many pages can he read in 5 hours

Papessa [141]

Answer:210

Explanation:

Divide 84 by 2 and then multiply by 5 by that number

3 0

4 years ago

Read 2 more answers

A 10 kHz sinusoidal is to be transmitted using FM in the presence of additive white Gaussian noise. The SNR improvement at the d

Mashutka [201]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

4 0

3 years ago

A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of

-Dominant- [34]

Answer:

a. $\tau=51.55 MPa$

b. $\tau=42.95MPa$

c. $\theta=7.67\times 10^{-3}$ rad.

Explanation:

Given: $D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa$

We know that

$\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}$

J for hollow shaft $J=\dfrac{\pi (D_o^4-D_i^4)}{64}$

(a)

Maximum shear stress $\tau =\dfrac{16T}{\pi Do^3(1-K^4)}$

$K=\dfrac{D_i}{D_o}$ ⇒K=0.83

$\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}$

$\tau=51.55 MPa$

(b)

We know that $\tau \alpha r$

So $\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}$

$\dfrac{51.55}{\tau}=\dfrac{210}{175}$

$\tau=42.95MPa$

(c)

$\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}$

$\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}$

$\theta=7.67\times 10^{-3}$ rad.

8 0

4 years ago

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

ivolga24 [154]

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg, s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291 - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ = 1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683 + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

$y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284$

The turbine control volume is given as follows;

$\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2} \right ) + \left (1 - y \right )\left (h_{2} - h_{3} \right )$

= (3507.41 - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

$\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6} \right ) + \left (1 - y \right )\left (h_{5} - h_{4} \right )$

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 - 151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

$\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7} \right )$

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

$\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}$

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, $\dot{m_{1}}$ , into the first turbine stage is given as follows;

$\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}$

$\dot{m_{1}}$ = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

$\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}$

$\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5} \right ]$

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

4 0

3 years ago