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zlopas [31]
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
Engineering
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

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Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

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h₁ = convection heat transfer coefficient = 4100 W/m² K

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Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

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Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

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3 years ago
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Answer:

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Explanation:

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
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Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

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1 + 8f² = (20+1)²

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f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

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c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

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