ENGINEERING COLLEGE

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following: a. The amount by which this specimen will elongate in the direction of the applied stress.
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer 1

Answer:

1)ΔL = 0.616 mm

2)Δd = 0.00194 mm

Explanation:

We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.

Thus;

Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

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A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase and the rest is in the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 20%. (a) Determine the initial temperature. (b) Determine the final temperature. (c) Determine the mass of liquid water when the piston first starts moving. (d) Determine the work done during this process in kJ.
Please define the specific heat of material?
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Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle b. the back work ratio c. the net power developed, in kW d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.

The larger the Bi number, the more accurate the lumped system analysis. a)-True b)- False

Answers

Answer:

b). False

Explanation:

Lumped body analysis :

Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.

Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.

The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.

In general it is assume that for a lumped body analysis, Biot number $\leq$ 0.1

Therefore, the smaller the Biot number, the more exact is the lumped system analysis.

Which property of real numbers is shown below?3 + ((-5) + 6) = (3 + (-5)) + 6
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication

Answers

The property of realnumbers is shown below is associative property of addition. The correct option is C.

What is associative property of addition?

According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.

According to the commutative property of addition, you can rearrange the addends without altering the result.

When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.

This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.

Thus, the correct option is C.

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Answer:

Explanation:

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

Answers

Answer:

$\dot W_(out) = 3374.289\,(BTU)/(s)$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_(out) + \dot m \cdot (h_(in) - h_(out)) = 0$

The turbine power output is:

$\dot W_(out) = \dot m\cdot (h_(in)-h_(out))$

The volumetric flow is:

$\dot V = (\pi)/(4) \cdot \left( (2)/(12)\,ft \right)^(2)\cdot (620\,(ft)/(s) )$

$\dot V \approx 13.526\,(ft^(3))/(s)$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,(ft^(3))/(lbm)$

The mass flow is:

$\dot m = (\dot V)/(\nu)$

$\dot m = (13.526\,(ft^(3))/(s) )/(1.33490\,(ft^(3))/(lbm) )$

$\dot m = 10.133\,(lbm)/(s)$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,(BTU)/(lbm)$

State 2 (Saturated Vapor)

$h = 1146.1\,(BTU)/(lbm)$

The turbine power output is:

$\dot W_(out) = (10.133\,(lbm)/(s) )\cdot (1479.1\,(BTU)/(lbm)-1146.1\,(BTU)/(lbm))$

$\dot W_(out) = 3374.289\,(BTU)/(s)$

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provideA. expansion joints
B. isolation joints
C. control joints
D. construction joints

Answers

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.

Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.

Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.

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A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.

Answers

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Will mark brainliest if correctWhen a tractor is driving on a road, it must have a SMV sign prominently displayed.

True
False

Answers

Answer: true

Explanation:

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