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3 years ago

9

There is a black-box system (i.e we do not know the transfer function of the system). When we fed a step input into the system,

we got a ramp out. What do you expect the output to be if we feed in a ramp input

1 answer:

xxMikexx [17]3 years ago

8 0

I need the point brodbhshdhdhd

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Write the 5 important of profession education in a point

tensa zangetsu [6.8K]

did it help ? I was going to put the same things so

3 0

3 years ago

An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci

Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0

3 years ago

Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the

VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

$V_2=I_2R_2=V_{ab}$

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

$V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}$

And

$V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A$

$V_{ab}'=I_1'R_{2||3}=3V=V_{2}'$

In the second case we can use an equivalent resistance between R2 and (R1+R4):

$V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}$

And

$V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A$

$V_{ab}''=I_3'R_{2||1-4}=2V$

If we consider both batteries:

$V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V$

7 0

4 years ago

A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure meth

GuDViN [60]

Answer:

mass of LNG: 129501.3388 kg

quality: 0.005048662

Explanation:

Volume occupied by liquid:

400 m^3*0.9 = 360 m^3

Volume occupied by vapor

400 m^3*0.1 = 40 m^3

A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present

For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg

From specific volume definition:

vf = liquid volume/liquid mass

liquid mass = liquid volume/vf

liquid mass = 360 m^3/0.002794 m^3/kg

liquid mass = 128847.5304 kg

vg = vapor volume/vapor mass

vapor mass = liquid volume/vg

vapor mass = 40 m^3/0.06118 m^3/kg

vapor mass = 653.8084341 kg

total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg

Quality is defined as the ratio between vapor mass and total mass

quality = 653.8084341 kg/129501.3388 kg = 0.005048662

4 0

3 years ago

A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat

Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

$Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ$

b) The coefficient of performance is:

$COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6$

c) The coefficient of performance of a reversible heat pump is:

$COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l} }$

Th=20+273=293 K

Tl=10+273=283K

Replacing:

$COP_{reversible}=\frac{293}{293-283}=29.3$

4 0

3 years ago