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UkoKoshka [18]
3 years ago
13

Using Circuit 5.1, answer the following questions. Calculate the voltage across the capacitor, assuming that the circuit has bee

n left for a significant amount of time after the power supplies are connected to the other components (this is the initial voltage, vc(0- )). 2. At t = 0, the 10V source is turned off (V=0). Determine how the voltage across the capacitor changes as a function of time. That is, determine an equation for vc (t). a. Find vc(0+) b. Find vc([infinity]) c. Find  d. Write an equation for vc (t) e. Calculate the voltage for the following times: t = , t = 2, t = 3, t = 4, t = 5, t = 6, t = 10,

Engineering
1 answer:
PSYCHO15rus [73]3 years ago
8 0

Complete Question: The complete Question is attached as a file below and the circuit of consideration is contained therein.

Answer:

1) vc(0-) = 6.7 v

2a) vc(0+) = 6.7 v

2b) vc(∞) = 0.825 v

2c) τ = 0.0238775 sec

2d) vc (t) = 0.825 + 5.875 exp(-41.88t) v

e)

vc (τ) = 2.9863 v

vc (2τ) = 1.62 v

vc (3τ) = 1.1175 v

vc (4τ) = 0.9326 v

vc (5τ) = 0.8646 v

vc (6τ) = 0.8396 v

vc (10τ) = 0.8253 v

Explanation:

The circuit analysis and the calculations involved in this question are handwritten and contained in the files attached below.

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Pumped-storage hydroelectricity is a type of hydroelectric energy storage used by electric power systems for load balancing. The
NikAS [45]

Answer:

A) energy loss E = pgQtH

Where p = density in kg/m3

g = gravity acceleration in m/s2

Q = flow rate in m3/s

t = time taken for flow in sec

H = height of flow in m

B) power required to run pump;

P = pgQH

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
What forces are not present in space
ss7ja [257]

Answer:

C.) Weight and distance I believe

Explanation:

7 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
3 years ago
How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?
nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

\frac{5400cu-yd}{1-0.25} =7200cu-yd

The volume at loose is:

V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd

If the Herrywampus has a capacity of 30 cubic yard:

\frac{8640cu-yd}{30cu-yd/trip} =288trip

b) By weight

The swell factor in terms of percent swell is equal to:

pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }

pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd

The weight of backfill is:

8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton

The Herrywampus has a capacity of 40 ton:

\frac{10800}{40ton/trip} =270trip

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0
3 years ago
Using the characteristics equation, determine the dynamic behavior of a PI controller with τI = 4 applied to a second order proc
Sladkaya [172]

Answer:

The values of Kc that render this closed-loop process unstable are in the interval

(Kc < 0)

Explanation:

The transfer function of a PI controller is given as

Gc = Kc {1 + (1/sτI)}

τI = 4

Gc = Kc {1 + (1/4s)}

Gc = Kc {(4s+1)/(4s)}

Divide numerator and denominator by 4

Gc = Kc {(s+0.25)/(s)}

For a second order process, the general transfer function is given by

Gp = Kp {1/(τn²s² + 2ζτns + 1)}

Kp = 2, τn = 5 and ζ = 1.5

Gp = 2/(25s² + 15s + 1)

Divide numerator and denominator by 25

Gp = 0.08/(s² + 0.6s + 0.04)

Ga = 1

Gs = 1

We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0

The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.

Gp*Ga*Gc*Gs + 1 = 0

Becomes

[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0

[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1

0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s

0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s

s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.

For the substitution,

We put s = jw into the equation. (frequency analysis)

Note that j = √(-1)

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

we then collect terms with j and terms without.

(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0

Meaning,

0.08Kcw + 0.04w - w³ = 0 (eqn 1)

0.02Kc - 0.6w² = 0 (eqn 2)

0.02 Kc = 0.6 w²

Kc = 15w²

Substituting this into eqn 1

0.08Kcw + 0.04w - w³ = 0

Kc = 15w²

0.08(15w²)w + 0.04w - w³ = 0

1.2w³ + 0.04w - w³ = 0

0.2w³ + 0.04w = 0

w = 0 or 0.2w² + 0.04 = 0

0.2w² = -0.04

w² = -0.2

w = ± √(-0.2)

w = ± 0.4472j or w = 0

Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0

The turning points for the curve of the closed loop transfer function occur when

Kc = 0 or Kc = -3

To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.

Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0

Note that, one positive characteristic root or pole is enough to make the system unstable.

We pick a value for Kc in that interval and evaluate the closed loop transfer function.

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

- First of, let Kc = - 4 (Kc < -3)

s³ + 0.6s² - 0.28s - 0.08 = 0

Solving the polynomial

s = (-0.22002), 0.44223, (-0.82221)

One positive pole means the closed loop transfer function is unstable in this region

Let Kc = -3

s³ + 0.6s² - 0.20s - 0.06 = 0

s = 0.37183, (-0.21251) or (-0.75933)

One positive pole still means that the closed loop transfer function is still unstable.

Then the next interval

Let Kc = -1

s³ + 0.6s² - 0.04s - 0.02 = 0

Solving this polynomial,

s = 0.18686, (-0.1749) or (-0.61196)

The function is unstable in the region being investigated.

Let Kc = 0

s³ + 0.6s² + 0.04s = 0

s = 0, -0.0769, -0.5236

One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.

Let Kc = 1, Kc > 0

s³ + 0.6s² + 0.12s + 0.02 = 0

s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)

All the real negative parts of the poles are all negative, this indicates stability.

Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)

Hope this Helps!!!

8 0
3 years ago
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