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Tom [10]
3 years ago
15

Write out simple definitions in words and equations for the following:

Engineering
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = S_{21} a_{1} + S_{22} a_{2}

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = \frac{V1^-}{V1^+} |v2^+=0

d) S12 : this is the gross voltage gain

S12 = \frac{V1^-}{V2^+}| v1 ^+

e) S21 : This is the forward voltage gain

    S21 = \frac{V2^-}{V1^+} | v2^+

f) S22 : output port voltage reflection coefficient

   S22 = \frac{v2^-}{v2^+} | v1^+ = 0\frac{v2^-}{v2^+} | v1^+ = 0

Explanation:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = S_{21} a_{1} + S_{22} a_{2}

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = \frac{V1^-}{V1^+} |v2^+=0

d) S12 : this is the gross voltage gain

S12 = \frac{V1^-}{V2^+}| v1 ^+

e) S21 : This is the forward voltage gain

    S21 = \frac{V2^-}{V1^+} | v2^+

f) S22 : output port voltage reflection coefficient

   S22 = \frac{v2^-}{v2^+} | v1^+ = 0\frac{v2^-}{v2^+} | v1^+ = 0

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In alternating current, how often does the current alternate between negative and positive?
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Answer:

AC voltages alternate/cycle, at a rate of 60 times each second.

7 0
3 years ago
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Consider a Pitot static tube mounted on the nose of an experimental airplane. A Pitot tube measures the total pressure at the ti
kotykmax [81]

Answer:

M∞ = 0.53

M∞ = 1.5

M∞ = 3.1

Explanation:

Find: For each case the free stream Mach number.

-Pitot pressure=1.22×10^5N/m2 , static pressure=1.01 × 10^5N/m2 .

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

            P = 1.01 × 10^5 .. static pressure

            Po = 1.22×10^5   ... pitot pressure ( hydrodynamic )

- Take the ratio:

            P / Po = (1.01 × 10^5) / (1.22×10^5) = 0.8264.

- Use Table A.13 and look up the ratio P/Po = 0.8264 for Mach number M∞.

            M∞ = 0.53

Find:

-Pitot pressure=7222 lb/ft^2 , static pressure=2116 lb/ft^2

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

            P = 2116 .. static pressure

            Po = 7222   ... pitot pressure ( hydrodynamic )

- Take the ratio:

            P / Po = (2116) / (7222) = 0.2930.

- However, since this is supersonic, a normal shock sits in front of the Pitot tube.  Hence, Po is now the total pressure behind a normal shock wave. Thus, we have  to use Table A.14.

            P1 = 2116 .. static pressure

            Po2 = 7222   ... pitot pressure ( hydrodynamic )

- Take the ratio:

            Po2 / P1 = (7222) / (2116) = 3.412.

- Use Table A.14 and look up the ratio Po2/P1 = 3.412 for Mach number M∞.

            M∞ = 1.5

Find:

-Pitot pressure=13197 lb/f^t2 , static pressure=1020 lb/ft^2

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

            P = 1020 .. static pressure

            Po = 13197   ... pitot pressure ( hydrodynamic )

- Take the ratio:

            P / Po = (1020) / (13197) = 0.0772.

- Again, since this is supersonic, a normal shock sits in front of the Pitot tube.  Hence, Po is now the total pressure behind a normal shock wave. Thus, we have  to use Table A.14.

            P1 = 1020 .. static pressure

            Po2 = 13197   ... pitot pressure ( hydrodynamic )

- Take the ratio:

            Po2 / P1 = (13197) / (1020) = 12.85.

- Use Table A.14 and look up the ratio Po2/P1 = 12.85 for Mach number M∞.

            M∞ = 3.1

3 0
3 years ago
An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 0
4 years ago
A developer has requested permission to build a large retail store at a location adjacent to the intersection of an undivided fo
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Answer:

Detailed solution is given below:

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3 years ago
A weight-lifting athlete raises a mass of 160 kg through a vertical distance of 1.4 m. What force did
Over [174]

Answer:

1568N

2195.2J

Explanation:

Given parameters:

Mass of the weight = 160kg

Distance  = 1.4m

Unknown:

Force applied to lift the weight = ?

Energy expended  = ?

Solution:

The force applied in moving a body with a given mass through a distance is the weight;

     Force applied  = mg

Where m is the mass

           g is the acceleration due to gravity

i.  Applied force = 160 x 9.8  = 1568N

ii. The energy used to lift the weight is given as;

     Energy  = mgh

h is the vertical distance

     Energy  = 1568 x 1.4  = 2195.2J

8 0
3 years ago
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