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Aleksandr-060686 [28]
3 years ago
14

Using any of the bilinear transform, matched pole-zero, or impulse invariance techniques in converting a continuous-time system

to a discrete-time system will always preserve the system stability.
a. True
b. False
Engineering
2 answers:
leonid [27]3 years ago
4 0

Answer:

A. True

The bilinear transform is employed in digital signal processing and discrete-time control theory which helps in transforming continuous-time system representations to discrete-time

Natasha2012 [34]3 years ago
3 0

Answer:

A - TRUE

Explanation:

The bilinear transform (also known as Tustin's method) is usually applied in digital signal processing and discrete-time control theory. It is used to transform continuous-time system representations into discrete-time and vice versa.

This transform usually preserves stability and also maps every point of the frequency response of the continuous-time filter.

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What is Not considered as metric system ? a. Length b. Mass c. Time d. Volume e. Temperature
vampirchik [111]

Answer:

D.) Volume

Explanation:

7 0
3 years ago
Read 2 more answers
I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal
tigry1 [53]

Answer:

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>

<em></em>

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

where, Nₐ₀-Nₐ = is the amount in moles of A used up

            Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if  3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

5 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
3 years ago
(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500
lubasha [3.4K]

Answer:

You can look it up

Explanation: if you don't know what it is look it up on .

6 0
3 years ago
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