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tatiyna
3 years ago
12

6msection of 150lossless line is driven by a source with vg(t) = 5 cos(8π × 107t − 30◦ ) (V) and Zg = 150 . If the line, which h

as a relative permittivity r = 2.25, is terminated in a load ZL = (150 − j50) , determine: (a) λ on the line. ∗ (b) The reflection coefficient at the load. (c) The input impedance. (d) The input voltage Vi. (e) The time-domain input voltage vi(t).

Engineering
1 answer:
Rainbow [258]3 years ago
7 0

Answer:

a. 5m

b. r = 0.16 e^-80.5◦

c. Zpn = (115.7 + j27.4) ohms

d. Vi = 2.2e^-j22.56◦ volts

e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts

Explanation:

In this question, we are tasked with calculating a series of terms.

Please check attachment for complete solution and step by step explanation

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A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at
Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}C

Temperature of air, T_{\infinity}=20^{o}C

Velocity of air flow V=5.5m/s

The film temperature is determined as:

T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}C

Thermal conductivity, K = 0.03443 W/m^oC; kinematic viscosity v=2.860 \times 10^{-5} m^2/s; Prandtl number Pr=0.70275

The reynolds number for the flow is determined as:

Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92

sice the obtained reynolds number is less than 2\times10^5, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}

Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}

Rewrite and solve for h

h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/m

6 0
3 years ago
20 points and brainliest A, B, C, D
Annette [7]
B !! Is the correct answer
5 0
3 years ago
Read 2 more answers
Why is a universal joint useful?
zubka84 [21]

Answer:

the are useful because they transmit a rotary motion they are used in shafts like the driveshaft in a car

Explanation:

3 0
4 years ago
Quadrature encoders operate with two photodetectors offset by _______ degrees. (a) 45 (b) 90 (c) 180 (d) 270
Lemur [1.5K]

Answer:

i think it should be 90 degree

5 0
3 years ago
What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?
Brut [27]
2.31 ft/psig

1.98 psig would equalize feet of head 4’7”

4.583/2.31 = 1.98. .583 is 7”/12”


So round off to 2.0 psig to get steady stream of bubbles
4 0
3 years ago
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