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sladkih [1.3K]
3 years ago
7

In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivale

nt of this circuit using phasors. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.

Engineering
1 answer:
sashaice [31]3 years ago
8 0

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivalent of this circuit using phasors.

a. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.

b. What is the value of the angle associated with the phasor Vth, in degrees?

c. Now, calculate the Thevenin impedance, Zth. What is the magnitude of this phasor?

d. What is the angle associated with the phasor Zth, in degrees?

Answer:

Vth = 6 < 45° V

Zth = 1.414 < 45°

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

Explanation:

The given voltage is

V(t)=12cos(2000t+45)

In phasor form,

V(t) = 12 < 45° V

So the magnitude of voltage is 12 V and the phase angle is 45°

Also the frequency ω = 2000

then the inductance is

L₁ = L₂ = L₃ = jωL = j×2000×0.003 = j6 Ω

and the capacitance is

C₁ = 1/jωC = 1/(j×2000×250x10⁻⁶) = -j2 Ω

and the resistance is

R₁ = R₂ = 2 Ω

Thevenin voltage:

The Thevenin voltage is the voltage that appears across the open-circuited terminals a-b (after removing L₃)

The Thevenin voltage is given by

Vth = V(t) × [ (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ]

Please note that there is no current flow in the capacitor due to open-circuited terminal a-b

Vth = 12 < 45°  × [ (2 + j6) / (2 + j6) + (2 + j6) ]

Vth = 12 < 45°  × [ (2 + j6) / (4 + j12) ]

Vth = 4.24264 + j4.24264 V

In phasor form,

Vth = 6 < 45° V

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

Thevenin Impedance:

The Thevenin Impedance is the impedance of the circuit calculated when looking from the terminal a-b

Zth = [ (R₁ + L₁) × (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ] + (-j2)

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (-32 + j24) / (4 + j12) ] -j2

Zth = [ (1 +j3) ] - j2

Zth = 1 + j Ω

In phasor form,

Zth = 1.414 < 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

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Answer:

0.31

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Explanation:

Given:-

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- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

   

                           w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

                           

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

             

- Compute the quality of the mixture at condenser inlet by the following relation:

                           x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947

- Determine the isentropic ( h4s ) at this state as follows:

                          h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31

Answer: The thermal efficiency of the cycle is 0.31

       

   

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