Which of the following is not an example of heat generation? a)- Exothermic chemical reaction in a solid b)- Endothermic Chemical Reactions in a solid c)- Nuclear reaction in nuclear fuel rods d)- Electric resistance heater

Answer:d

Explanation:

Melt spinning is the economical process to manufacture synthetic fiber.

Major limitations of Melt spinning is

Although it is very efficient method as it does not create any environment Pollution

Answer:

Re=100,000⇒Q=275.25 [tex]\frac{W}{m^2}[/tex]

Re=500,000⇒Q=1,757.77[tex]\frac{W}{m^2}[/tex]

Re=1,000,000⇒Q=3060.36 [tex]\frac{W}{m^2}[/tex]

Explanation:

Given:

For air      [tex]T_∞[/tex]=25°C  ,V=8 m/s

  For surface [tex]T_s[/tex]=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

[tex]Re<30\times10^5[/tex]⇒Laminar flow

[tex]Re>30\times10^5[/tex]⇒Turbulent flow

Take Re=100,000:

 So this is case of laminar flow

  [tex]Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}[/tex]

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24[tex]\times 10^{-3}[/tex]

So    [tex]Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}[/tex]

Nu=187.32   ([tex]\dfrac{hL}{K_{air}}[/tex])

187.32=[tex]\dfrac{h\times2.75}{26.24\times 10^{-3}}[/tex]

     ⇒h=1.78[tex]\frac{W}{m^2-K}[/tex]

heat transfer rate =h[tex](T_∞-T_s)[/tex]

                           =275.25 [tex]\frac{W}{m^2}[/tex]

Take Re=500,000:

So this is case of turbulent flow

  [tex]Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}[/tex]

[tex]Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}[/tex]

Nu=1196.18  ⇒h=11.14 [tex]\frac{W}{m^2-K}[/tex]

heat transfer rate =h[tex](T_∞-T_s)[/tex]

                             =11.14(179-25)

                           = 1,757.77[tex]\frac{W}{m^2}[/tex]

Take Re=1,000,000:

So this is case of turbulent flow

  [tex]Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}[/tex]

[tex]Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}[/tex]

Nu=2082.6  ⇒h=19.87 [tex]\frac{W}{m^2-K}[/tex]

heat transfer rate =h[tex](T_∞-T_s)[/tex]

                             =19.87(179-25)

                           = 3060.36 [tex]\frac{W}{m^2}[/tex]

Answer:

traditional casting --  sand casting

non traditional casting  - investment casting

Explanation:

Traditional casting is related to sand casting in which liquid is poured into mold cavity. once the poured material cool and solidify the casting is removed.

traditional casting is cheaper than non traditional casting . Mold use in traditional casting is very cheap .

 non traditional casting include investment casting, die casting , wax casting.  In this casting  fix dimension is provided unlike sand casting.  

Answer:

N = 38546.82 rpm

Explanation:

[tex]D_{1}[/tex] = 150 mm

[tex]A_{1}= \frac{\pi }{4}\times 150^{2}[/tex]

              = 17671.45 [tex]mm^{2}[/tex]

[tex]D_{2}[/tex] = 250 mm

[tex]A_{2}= \frac{\pi }{4}\times 250^{2}[/tex]

              = 49087.78 [tex]mm^{2}[/tex]

The centrifugal force acting on the flywheel is fiven by

F = M ( [tex]R_{2}[/tex] - [tex]R_{1}[/tex] ) x [tex]w^{2}[/tex] ------------(1)

Here F = ( -UTS x [tex]A_{1}[/tex] + UCS x [tex]A_{2}[/tex] )

Since density, [tex]\rho = \frac{M}{V}[/tex]

                        [tex]\rho = \frac{M}{A\times t}[/tex]

                        [tex]M = \rho \times A\times t[/tex][tex]M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t[/tex]

                        [tex]M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37[/tex]

                        [tex]M = 8252963901[/tex]

∴ [tex]R_{2}[/tex] - [tex]R_{1}[/tex] = 50 mm

∴ F = [tex]763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}[/tex]

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

[tex]33618968.38 = 8252963901\times 50\times \omega ^{2}[/tex]

∴ ω = 4036.61

We know

[tex]\omega = \frac{2\pi N}{60}[/tex]

[tex]4036.61 = \frac{2\pi N}{60}[/tex]

∴ N = 38546.82 rpm

Answer:

Thermosetting polymers are polymers that becomes soft and pliable when heated is false

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

Answer: False

Explanation: Horizontal axis wind turbines are usually used for generation of the electric power on the off-shore. The generation of horizontal-axis wind turbine works well when it is installed away from the shore because it supports large sized wind turbines so that they can generate high amount of electricity.They are usually not preferred for the on-shore wind farms because they can have small sized wind turbines only.Therefore the statement given is false.

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × [tex]10^{-3}[/tex] A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × [tex]10^{-3}[/tex]

R2 = 1500Ω

Answer:

A good quality refrigerant should be eco friendly.

Explanation:

A refrigerant is a substance that can extract and transfer heat from body to another body or medium.

The desirable qualities required for a refrigerant are :

1. A refrigerant should not deplete ozone layer.

2. A good quality refrigerant should have a low boiling point.

3. It should also have a low melting point.

4. Thermal conductivity of the refrigerant should be high for fast heat transfer.

5. It should have low specific heat.

6. It should have high latent heat.

7. It should have low vapour density.

8. Refrigerant should have high critical pressure and temperature.

9. It should have high enthalpy of vapourization for maximum heat absorption.

10. Refrigerants should not be toxic in nature and non flammable.

11. It should have high coefficient of performance for the working temperature range.

12. It should be easily available and cheap.

Flash Chamber :

A flash chamber in the refrigeration system is also know as the mixing chamber. It is normally used in multistage refrigeration system and is placed in between the expansion valve and the evaporator.

The flash chamber sends only the liquid refrigerant to the evaporator by seperating the liquid from the vapour refrigerant in order to increase the efficiency.

Answer:

275 Kelvin

Explanation:

Coefficient of Performance=11

[tex]T_H=\text {Absolute Temperature of high temperature reservoir=300 K}[/tex]

[tex]T_L=\text {Absolute Temperature of low temperature reservoir}[/tex]

[tex]\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}[/tex]

Given:

max displacement, A = 3.2 m

f= 50 Hz

at t = 0, displacement, d = 150 cm = 1.5 m

Solution:

Displacement in the general form is represented by:

d = Asin(ωt ± α)

d = 3.2sin(2πft ± α)

d = 3.2sin(100πt ± α)                    

where,

A = 3.2 m,            

ω = 2πf = 100π

Now,

at t = 0,

1.5 = 3.2sin(100π(0) ± α )

1.5 = 3.2sinα

sin α = [tex]\frac{1.5}{3.2}[/tex] = 0.4687

α = [tex]sin^{-1}(0.46875)[/tex] = 27.95° = 0.488 radian

Now, we can express displacement in the form of 'Asin(wt + α)' as:

d = 3.2sin(100πt ± 0.488 )

Answer:

[tex]e^{ix} = cosx + i sinx[/tex]

Explanation:

In mathematics, Euler's formula is an equation in complex analysis, that gives a relationship between an exponential factor and the trigonometric functions.

The Euler equation is:

[tex]e^{ix} = cosx + i sinx[/tex]

Here,

e - base of the natural logarithm

i - imaginary unit

x - argument given in radians

sin , cos - trigonometric functions sine and cosine respectively.

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=[tex]\frac{30}{15}[/tex] ft/s

velocity=2 ft/s

power = force[tex]\times[/tex] velocity

power=[tex]{4500}\times2[/tex]

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= [tex]\frac{9000}{550} =16.36[/tex] hp

In vibration analysis, damping cannot always be disregarded. This is especially the case when the system is excited near the resonance frequency.

Answer:

The application of force is the main difference between truss and frame.

Explanation:

Truss:

Truss is a collection of beams,which use to handle the tensile and   compression loads . That collection of beams creates rigid structure.

The load on the truss will be acting always at the  at the hinge. Truss is     widely used in the construction areas.                

Frame:

       Like truss, it is also a combination of beams and used to handle the loads. The main difference between truss and frame is the application of load. In the frame load can apply at the any point of the member of frame along  with hinge.

Truss are connected by pin joint and can not transfer moment ,on the other hand frames are connected by rigid joint like welding so frame can transfer moment.

Truss and frame both forms a rigid structure and is used in the construction areas.  

Answer:

The correct answer is c. object snap  

Explanation:

In Autocad, the object snap is defined as a drawing aid that is used together with other different commands to help to draw accurately.  It also allows snapping onto a specific object location when there is a picking point. and thus, it helps to select the center of a line.

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

Answer:

Power consume by compressor=113,726.87 KW

Explanation:

Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]

 [tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]

Actually compressor is an open system, so here we will use first law of thermodynamics for open system .

We know that first law of thermodynamics for steady flow

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

We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]and we take the air as ideal gas.

System is in steady state then mass flow rate in =mass flow rate out

Mass flow rate= [tex]density\times area\times velocity[/tex]

So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex]     ,[tex]\rho =\frac{P}{RT}[/tex]

                                   =1.23×200×2 Kg/s

                                  =541.17 Kg/s

[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]

[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex]  ,[tex]\rho =\frac{P}{RT}[/tex]

[tex]V_{2}[/tex]=80.07 m/s

Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]

So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]

             [tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]

Now by putting the values

[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]

Here Q=0 because heat transfer is zero here.

W= -210.15 KJ/kg

So power consume by compressor=541.17×210.15

                                                          =113,726.87 KW

Answer: 12

Explanation: In FCC metal lattice there are total four octahedral slip plane and six direction where each are common to two octahedral plane   and so that gives total of three slip direction . So the total slip system is multiplication of the slip plane and the slip direction that is twelve.

Slip system=Slip plane×Slip direction

Slip system =4×3=12

Therefore there are total 12 slip system in the FCC metal lattice.

Answer:

Failure Rate

Explanation:

Failure rate is the frequency with which an engineered system or component fails, expressed for example in failures per hour.

It is often denoted by the Greek letter λ (lambda) and is important in reliability theory.

Failure rate is usually time dependent, and an intuitive corollary is that the rate changes over time versus the expected life cycle of a system

Answer:

[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]

Explanation:

Now from first law for open system

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]

Here given  Q=0 ,w=0

So [tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]

[tex]V_1=4 m/s,V_2=2 m/s[/tex]

[tex]h_1+\dfrac{4^2}{2000}=h_2+\dfrac{2^2}{2000}[/tex]

[tex]h_2-h_1=6\times 10^{-3}[/tex]

So increase in specific enthalpy

[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]

Answer:

5000J

Explanation:

Given in the question that

Heat added to the coffee cup is, ΔS = 20 J/K

The temperature of the coffee, T = 250 K

Now, using the formula for the entropy change

[tex]\bigtriangleup S=-\frac{\bigtriangleup H}{T}[/tex] ...........(1)

Where,

ΔS is the entropy change

ΔH is the enthalpy change

T is the temperature of the system

substituting the values in the equation (1)

we get

[tex]20=-\frac{\bigtriangleup H}{250}[/tex]

ΔH=250×20

ΔH=5000 J

Answer:

The importance of ferrite and austenite stabilizing elements in steels .

Explanation:

Alloying -

The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .

The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .

Stabilizing austenite -

The process by which temperature is increased , in which Austenite exists .

Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .

Hence, the elements are -

Cobalt , Nickel , Manganese , Copper.

The examples of the Austenitic steels are -

Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.

Stabilizing ferrite –

The process by which temperature is decreased , in which austenite exists .

Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.

Hence, the elements are -  

Aluminium , Silicon , Tungsten , Chromium , Molybdenum , Vanadium

The examples of the Ferritic steels are -

F-Cr alloys , transformer sheets steel ( 3% Si ).

Answer:

Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.

Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc.  Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.

The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.

Answer:

Diesel engine's working is based on internal combustion. Diesel engine can be categorized as two stroke diesel engines and 4 stroke diesel engines.

Let us look into the working of a 4 stroke diesel engine:

In a four stroke diesel engine the operation continues by rotating in cycle of 4 strokes or stages. In these 4 stages the piston moves up and down the crankcase twice or the crankshaft rotates twice. the four strokes are listed below:

1) Inlet: In this, the air mixture is made to enter through the inlet valve as the piston moves downwards.

2) Compression: Inlet valve is closed and the air mixture is compressed and the piston moves up. Fuel is injected into the heated air mixture through central injection valve and ignition takes place without the need of spark plug.

3) Expansion or Power: After ignition, mixture burns and due to expansion of gas, piston moves down, it drives the crankshaft to send power to the wheel.

4) Exhaust: This is the last stroke which results in the piston moving up to let out the exhaust gases

Moreover, in 2 stroke diesel engine, there is only one rotation of crankshaft and one stroke includes inlet and compression and the other includes expansion and exhaust

Answer and explanation :

Fluid distribution is a new technique to produce and to transmit power from one place to other its play a major role in power distribution it is a process of using fluid (any type of fluid as oil or water ) under pressure to generate to control or to transmit  

fluid power system is divided into two types

Answer:

1 ton refrigeration =3.517 kJ/s = 3.517 kW

Explanation:

Refrigeration capacity is defined at the  measure of the effective cooling capacity of a refrigerator which is  expressed in Btu per hour or in tons.

1 ton capacity is a unit of air conditioning and refrigeration which  measure the capacity of air conditioning and refrigeration unit.

One ton  is equal to removal of 3025kcal heat per hour

1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP

Answer:

A. True

Explanation:

Answer:

B. 3 m

Explanation: For plastic injection moulded the thickness is generally between 2 mm to 3 mm

the wall is not too thick because during cooling process there should be defects so thickness of wall is no too high and there is also a problem if we use thicker wall that we need more material for moulding process so the thickness should be in between 2 to 3 mm which is in option B so option B will be the correct option