Ask question

Ask question

All categories

2 years ago

5

Q1-A Lossless transmission line is 80 cm long and operates at a frequency of 500 MHz. Two line parameters are L=0.15μH/m and C=9

0pf/m. Find the characteristic impedance the phase constant, the velocity on the line, and the input impedance for ZL =80 Ω

1 answer:

hjlf2 years ago

6 0

The characteristic impedance, the phase constant, the velocity on the line, and the input impedance are respectively;

A) Z_o = 40.82 Ω

B) β = 11.543 rad/m

C) v_p = 2.72 × 10^(8) m/s

D) Z_in = 115.91 Ω

We are given;

Inductance; L = 0.15 μH/m = 0.15 × 10^(-6) H/m

Capacitance; C = 90 pf/m = 90 × 10^(-12) f/m

Frequency; f = 500 MHz = 500 × 10^(6) Hz

Load impedance; Z_l = 80 Ω

Length of transmission line; l = 80cm = 0.8m

A) Formula for characteristic impedance is;

Z_o = √(L/C)

Thus;

Z_o = √((0.15 × 10^(-6))/(90 × 10^(-12)))

Z_o = 40.82 Ω

B) Formula for the phase constant is;

β = ω√(LC)

Where;

ω = 2πf

Thus;

β = (2π × 500 × 10^(6))√(0.15 × 10^(-6) × 90 × 10^(-12))

β = 11.543 rad/m

C) Formula for phase velocity is;

v_p = ω/β

v_p = (2π × 500 × 10^(6))/11.543

v_p = 2.72 × 10^(8) m/s

D) Formula for the input impedance is;

Z_in = Z_o[(Z_l + Z_o*tanβl)/(Z_o + Z_l*tanβl)]

Z_in = 40.82[(80 + 40.82*tan(11.543*0.8))/(40.82 + 80*tan(11.543*0.8))]

Z_in = 40.82(72.1335/25.4031)

Z_in = 115.91 Ω

Read more about impedance at; brainly.com/question/25153504

You might be interested in

Rubber bushings are used on suspensions to

Harlamova29_29 [7]

D. All of the above

4 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0

3 years ago

A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti

natima [27]

Answer:

Final mass of Argon= 2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

= (200 * 303 ) / (450 * 219 ) * 4

= 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0

3 years ago

2. The device that varies incoming line pressure using centrif-

Kryger [21]

Answer:

I believe reverse boost valve.

3 0

3 years ago