ENGINEERING COLLEGE

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer 1

Answer:

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

Pressure ratio = $(4)/(5)$

= 0.8

From Isentropic Flow Tables

M = 0.58 A/A* = 1.213

b) 2.64 atm

Pressure ratio = $(2.64)/(5)$

= 0.528

From Isentropic Flow Tables

M = 1 A/A* = 1

c) 0.5 atm

Pressure ratio = $(0.5)/(5)$

= 0.1

From Isentropic Flow Tables

M =2.10 A/A* = 1.8369

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. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance would be needed for ???? = 0? 2. What feature in the time response of an RLC circuit distinguishes a critically damped response from an underdamped response? 3. Why must an op-amp be powered to be used in a circuit? 4. If you were handed a parts kit with an unknown op-amp, what information would you need to find prior to using it in a circuit?

Answers

Answer:

an attachment is below

Explanation:

1) the formula for damping coefficient id for RLC series circuit.

For \xi =0 you can make c=0 but inductor will still have some capacitance.

2) the responses of critically damped system and under damped system are shown with comments on their time response.

4) There can be many different answers to this question, but the 4 I have mentioned are the most important parameters we need to know about an unknown op-amp if we are to use it in our circuit.

Hope it answers all your questions.

If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential energy is unchanged, calculate its change in specificinternal energy if its mass is 5.4 kg. Report your answer in kJ/kg to one decimal place.

Answers

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

$\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U$ $15.5=-3.5+0+\bigtriangleup U$

$\bigtriangleup U=19$ kj.

Step3

Specific internal energy is calculated as follows:

$\bigtriangleup u=(\bigtriangleup U)/(m)$

$\bigtriangleup u=(19)/(5.4)$

$\bigtriangleup u=3.5$ kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

One kind of SS-3xX steel alloy has a melting point of 1450°c. Its specific heat = 0.46 J/g°C, and its heat of fusion 270 J/g. For a 200 kg block of this steel, determine how much heat is required to (a) raise its temperature from 25°C to its melting point and (b) transform it from solid to liquid phase.

Answers

Answer:

a)Q=131.1 MJ

b)Q=54 MJ

Explanation:

Given that

Mass ,m=200 kg

Specific heat Cp=0.46 J/g°C

Cp=0.46 KJ/kg°C

Heat of fusion = 270 J/g

Heat of fusion = 270 KJ/kg

Melting point temperature = 1450°C

Initial temperature = 25°C

Final temperature=1450°C

Heat required to rise temperature from 25°C to 1450°C.

Q= m CpΔT

Q=200 x 0.46 x (1450-25) KJ

Q=131,100 KJ

Q=131.1 MJ

Heat required to transform from solid phase to liquid phase

Q= Mass x heat of fusion

Q=200 x 27 KJ

Q=54,000 KJ

Q=54 MJ

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.

Answers

Answer:

critical stress required is 18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × $10^(-3)$ m

so we get here critical stress that is

$\sigma _c = \sqrt{(2E \gamma )/(\pi a)}$ ...............1

put here value we get

$\sigma _c$ = $\sqrt{(2* 225* 10^9 * 1 )/(\pi * 0.4* 10^(-3))}$

$\sigma _c$ = 18923493.9151 N/m²

$\sigma _c$ = 18.92 MPa

What character string does the binary ASCII code 1010100 1101000 1101001 1110011 0100000 1101001 1110011 0100000 1000101 1000001 1010011 1011001 0100001?

Answers

Answer: This is EASY!

Explanation:

To make it easy, you would convert those binary numbers and to denary. And this gives:

84 104 105 115 32 105 115 32 69 65 83 89 33

Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!

In a refrigerator, heat is transferred from a lower-temperature medium (the refrigerated space) to a higher-temperature one (the kitchen air). Is this a violation of the second law of thermodynamics

Answers

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