A gas contained within a piston-cylinder undergoes the follow change in states: Process 1: Constant volume from p1 = 1 bar V1 = 2.6 m3 to state 2 with p2 = 2.7 bar Process 2: Compression to V3 = 1.5 m3, which the pressure-volume relationship is pV = constant. Process 3: Constant pressure to state 4, where V4 = 0.5 m3. Sketch the processes on p-V graph and evaluate the work for each process in kJ.

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

It's an n-type dopants, as it makes the silicone n-type after the doping. so, False

Answer: B) High chemical energy and low molecular weight

Explanation: Liquid propellant is a a single chemical compound or mix of other chemicals as well. It is used in the rocket that uses them as a major part for the fuel. A liquid propellant is supposed to have high chemical energy and low molecular weight so that is can be ignited with ease. High chemical energy can release good amount of heat in a chemical reaction and thus is good igniting compound for the liquid propellant rocket.

Answer:

The point where all three phases coexist on a P-T diagram is called triple point.

Explanation:

A phase diagram of a substance is a graphical representation of the transition of its physical state under various temperature and pressure conditions.

A typical phase diagram usually consists of:

(A) Fusion curve - The curve that shows that the solid and the liquid state in in equilibrium with each other.

(B) Vaporization curve - The curve that shows that the liquid and the gaseous state in in equilibrium with each other.

(C) Sublimation curve - The curve that shows that the solid and the gaseous state in in equilibrium with each other.

(D) Triple point - This is a point in the phase diagram in which all the three state exists in equilibrium.

For example, Considering the phase diagram of water, The triple point occurs at a pressure of 0.6 kPa and 0.01⁰C.

The phase diagram is shown in image below:

Answer:

-25.63°C.

Explanation:

We know that throttling is a constant enthalpy process

      [tex]h_1=h_2[/tex]

From steal table

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

  Temperature at saturation pressure 1 bar is 99.63°C and  Temperature at saturation pressure 0.35 bar is about 74°C .

So from above we can say that change in temperature is -25.63°C.

But there is no any option for that .

Answer:

The temperature of the high-temperature source must increase in 12.23ºC.

Explanation:

For a Carnot engine, the efficiency is defined as:

[tex]n = 1- T2/T1 [/tex]

Where T2 and T1 are the low and the high-temperature sources respectively. Therefore for the value of T2 of 19.1ºC and the n equal to 30.7% (0.307), the T1 Temperature can be calculated as:

[tex]n = T1/T1 - T2/T1[/tex]

[tex]n = (T1-T2)/T1[/tex]

[tex]T1.n = (T1-T2)[/tex]

[tex] T1-T1.n =T2[\tex]

[tex] T1(1-n) =T2[/tex]

[tex] T1 =T2/(1-n)[/tex]

[tex] T1 = 19.1\ºC /(1-0.307)[/tex]

[tex] T1 = 27.56\ºC[/tex]

Then for the new effciencie n' of 52% (0.52) the new temeperature T1' will be:

[tex] T1' =T2/(1-n')[/tex]

[tex] T1' = 19.1\ºC /(1-0.52[/tex]

[tex] T1' = 39.79\º C[/tex]

Finally the increment of temperature is:

[tex] AT1 =T1'-T1[/tex]

[tex] AT1 =39.79\º C-27.56\º C[/tex]

[tex] AT1 =12.23\º C[/tex]

[tex] AT1 =12.23\º C[/tex]

Answer:

The position vector of the  point and the direction of the force define direction of the moment a force generates.

Explanation:

Moment generated by force about any point 'o' is defined by

[tex]\overrightarrow{dM}=\overrightarrow{dr}\times \overrightarrow{dF}[/tex]

The above expression being a cross product of vectors [tex]\overrightarrow{dr}[/tex] and[tex]\overrightarrow{dF}[/tex] the moment at point 'o' will depend on direction of both these vectors.

Answer:

(B) Curve fit of p-v-t experimental data points

Explanation:

The constants a and b have positive values and are characteristic of the individual gas. The van der Waals state equation approximates the ideal gas law PV = nRT as the value of these constants approaches zero. The constant a provides a correction for intermolecular forces. The constant b is a correction for finite molecular size and its value is the volume of one mole of atoms or molecules.

Answer:

f)Density of the solid body

Explanation:

We know that heat transfer due to convection is given by

            Q=hAΔT

Where

A is the area ,which represent the geometry of solid body or surface.

h is the heat transfer coefficient which depends on

                             1.Thermal conductivity of fluid

                             2.Motion of fluid

                             3.Type of fluid flow(Laminar or Turbulent)

                             4.Viscosity of fluid

                             5.Surface condition

So from the above parameters ,we can say that heat transfer due to convection does not depends on density of the solid body.

Answer:

The correct option is : c. Titanium

Explanation:

1. Type M tool: It belongs to the high- speed group of tool steels and used as a cutting tool material. It is a multi component alloy system Fe–C–X, where X is molybdenum.

2. Stainless steel: It is also called inox steel. It is a steel alloy with about 10.5% chromium and 1.2% carbon by mass.

3. Titanium: It is a chemical element, having atomic number 22 and mass 47.867 u. It is silver lustrous transition metal with a symbol Ti.

4. Brass: It is an alloy of zinc and copper metal.

5. Inconel: It is a family of austenitic nickel-chromium based superalloys.

Therefore, Titanium is not an alloy.

Answer:

c). Tresca and von-Mises

Explanation:

Tresca yield criteria states that when maximum shear stress becomes greater than the yield strength, the materials starts to yield.

Von -Mises is also known as Distortion energy theory. This theory states that failure occurs when a body is acted upon to a bi axial stresses or tri axial stresses when at any point the strain energy of distortion by unit volume of the body  equal to the specimen of the strain energy of distortion by unit volume when yielding starts in tension test.

Thus most successful and commonly used yield criteria are the Von-Mises criteria and Tresca criteria.

Explanation  & answer:

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

Answer: B- Nitrogen Tetroxide

Explanation: Except for the nitrogen tetroxide , other given all options are fuel .Nitrogen Tetroxide is a chemical compound having brownish-red color which is in liquid form having a unpleasant smell, therefore it does not belong to the category of fuel because it cannot be used as a substance for production of heat or power .

Answer and Explanation :

HARDENABILITY OF STEEL:  Hardenability of steel is related to the its ability to form martensite when it is quenched. Hardenability is the measurement of capacity that how hard would be the steel when  it is quenched.

The hadenability of steel can be measured as maximum diameter of rod which will have 50% martensite

its application in axial rods of car because it is very hard  

Answer:

The difference between them lies in the area of the workpiece from which the material is removed. Turning is designed to remove material from the external surface of a workpiece, whereas boring is designed to remove material from the internal surface of a workpiece.

Explanation:

Answer:

[tex]h_{1}[/tex] = 2017.25 kJ/kg

Explanation:

GIVEN DATA:

Exit velocity [tex]v_{2}[/tex] = 250 m/s

outlet enthalpy [tex]h_{2}[/tex]= 1986 kJ/kg

inlet velocity  [tex]v_{1}[/tex]= 0

heat transfer Q = 0

from steady flow energy equation(SFEE) between inlet and exit point

[tex]h_{1}+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2}+ Q[/tex]

[tex]h_{1}=h_{2}+\frac{v_{2}^{2}}{2}[/tex]

[tex]h_{1}[/tex]=2017.25 kJ/kg

Answer:

true area of contact is 1.7 * [tex]10^{-4}[/tex] m²

Explanation:

Given data

mass (m) = 4000 tonnes

yield strength = 240 MPa i.e. = 240 * [tex]10^{6}[/tex]

base area = 0.8 m²

To find out

the true area of contact

Solution

we have given yield strength and weight

so with we can find contact area directly we know that

area is equal to weight / yield strength

so we will put weight and yield strength value in this formula

and weight = mass * 9.81 = 4 * 9.81 = 39.24 tonnes = 39240 N

area = weight /  yield strength

area = 39240 / 240 * [tex]10^{6}[/tex]

true area of contact = 1.7 * [tex]10^{-4}[/tex] m²

Answer: D) Nozzle gimbals

Explanation: Igniters are present for the ignition power in the segmented solid rocket, liners are the basic requirement in the engine of the rocket,O-rings are also a general part of the segmented solid rocket .So, these are the general purpose parts of the segmented solid rocket but nozzle gimbals are the special units of the rocket as they provide improved thrust in various direction in accordance with the rocket.So, option (d) is the correct option.

Answer:

D. Nozzle gimbals

Explanation:

Segmented solid rockets require the special component, nozzle gimbals.

Answer:

Clausius inequality is defined as, it applies in the cycle of real engine and there is negative entropy change. When the entropy given in the cycle during the environment is larger than entropy transferred into heat engine from hot reservoir. As, entropy of reversible system are zero.

Clausius Inequality is defined as:

[tex]R_2-R_1> \oint_{T_1}^{T_2}\frac{dQ}{T}[/tex]

Where R1 and R2 are not be equal.

Answer:

26.667

Explanation:

Given Data

For Tool A

Life exponent [tex]{\ n_1}[/tex]=0.45

Constant [tex]{C_1}[/tex]=90

For tool B

Life exponent [tex]{n_2}[/tex]=0.3

Constant [tex]{C_2}[/tex]=60

and tool life equation is

[tex]VT^{n}=c[/tex]

[tex]VT_{A}^{0.45}=90[/tex]

[tex]T_{A}^{0.45}=\frac{90}{V}[/tex]

[tex]T_{A}=\frac{90}{V}^{\frac{1}{0.45}}[/tex]

[tex]For Tool B[/tex]

[tex]VT_{A}^{0.3}=60[/tex]

[tex]T_{B}^{0.3}=\frac{60}{V}[/tex]

[tex]T_{B}=\frac{60}{V}^{\frac{1}{0.3}}[/tex]

[tex]T_{A}>T_{B}[/tex]

[tex]\frac{90}{V}^{\frac{1}{0.45}}>\frac{60}{V}^{\frac{1}{0.3}}[/tex]

[tex]V>26.667[/tex]

Answer:

I'm pretty sure its true

Explanation:

Answer:

the required diameter is 0.344 m

Explanation:

given data:

flow is laminar

flow of carbon dioxide Q = 0.005 Kg/s

for  flow to be laminar,  Reynold's number must be less than 2300 for pipe flow and it is given as

[tex]\frac{\rho VD}{\mu }<2300[/tex]

arrange above equation for diameter

\frac{\rho Q D}{\mu A }<2300

dynamic density of carbon dioxide = 1.47×[tex]10^{-5}[/tex] Pa sec

density of carbon dioxide is 1.83 kg/m³

[tex]\frac{1.83\times 0.0056\times D}{1.47\times 10^{-5}\times \frac{\pi}{4} \times D^{2} }<2300[/tex]

[tex]\frac{1.83\times 0.0056}{1.47\times 10^{-5}\times \frac{\pi}{4} \times 2300}= D[/tex]

D = 0.344 m

Answer:

Explanation:

it is given that diameter = 8.6 cm

[tex]radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m[/tex]

current =2.7 ampere

number of turns = 15

[tex]area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}[/tex]

magnetic field =0.56 T

maximum torque= BINASINΘ  for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm

Answer:

work is 50 kj

Explanation:

Given  data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another  location. i.e.

ΔU = Q – W                        ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

Given:

Let the speed of sound be represented by 'v' then

v ∝ [tex]\sqrt{T}[/tex]              (1)

[tex]v_{1}[/tex] = 349 m/s

[tex]v_{2}[/tex] = 340 m/s

[tex]T_{1}[/tex] = 20°C = 273+20 = 293 K

Formulae used:

1)  °C = K + 273

2) K = °C - 273

3) °F = 1.8°C + 32

4) °R = °F + 459.67

Solution:

From eqn (1),

[tex]\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}[/tex]

[tex]T_{2}[/tex] = [tex]T_{2} = (\frac{v_{2}}{v_{1}})^{2}T_{1}[/tex]

[tex]T_{2} = (\frac{340}{349})^{2}{293}[/tex] = 278.08 K

Now, Usinf formula (1), (2), (3) and (4) respectively, we get

1) T = 293 K

2) T = 293 -278.8 = 5.08°C

3) T = 1.8(5.08) + 32=41.14°F

4) T = 41.14 + 459.67 = 500.81°R

Answer:

Hot tear casting is the discontinuity or failures occurs during cooling or solidification phase in casting process. It is also known as hot crack

Explanation:

In case when the casting is of weak metal, when it is hot and casted, during the process residual stresses cause failures or discontinuity in casting as it cools resulting in hot tears i,e., material casted is partly liquid and partly solid.

Preventive measures:

Answer:

Explained

Explanation:

In case case of slow cooling of a metal ingot, the micro structure is more like coarse. At the surface due to high heating rate ( surface will to exposed to higher temperature than the inner part and for longer time) the grains are small as grains will get lesser time to cool. Where as we go inside the grains will be gradually elongated. At the center we will find equiaxed grains.  

Answer: b)All substances emit radiation

Explanation: This is a universal statement that all bodies emit radiations . The radiations are usually in the form of electromagnetic radiations that are being emitted in accordance with temperature that varies from body to body.Even human bodies along with any other body emit radiations These radiations are absorbed by the black body.Therefore the correct option is option(b).

Answer:

The planning time of the planner machine is 100 minute

Explanation:

Planning machine

A planning machine is a metal working machine that gives a flat surface to the work piece. Here the work-piece reciprocates and the feed is given to the tool.  A planning machine is used for heavy duty work and are often large in size.

 The machining time or the planning time of a planning machine is given by,

[tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]

where, [tex]L_{w}[/tex] is the total length of travel of job

                                    = width of the job

                                    = 300 mm

            [tex]N_{s}[/tex] is number of strokes per min

                                     = 10 double strokes per min

            f is feed of the tool, mm per stroke

                                      = 0.3 mm per stroke

Therefore,  [tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]

                                            = [tex]\frac{300}{10\times 0.3}[/tex]

                                           = 100 min

Therefore the planning time is 100 minute.