A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)

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:

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:

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:

Otto cycle for 4 stroke engine:

Assumptions:

1.Air is a working fluid it will behave like ideal gas.

2.Mass of air is constant(close system)

3.All process is reversible process.

4.Specific heat of air does not depends on temperature.

4 stroke engine is an internal combustion engine.It works on 4 processes like intake ,compression,power and heat exhaust.To complete one cycle ,piston moves from top dead center to bottom dead center two times.

From the Otto cycle

Process 1-2 is isentropic  compression.

Process 2-3 is heat addition.

Process 3-4 is isentropic expansion.

Process 4-1 is heat rejection.

Petrol engine works on Otto cycle.  

Efficiency of cycle [tex]\eta[/tex]

[tex]\eta=\dfrac{W_{net}}{Q_{supply}}[/tex]

The Otto cycle is a four-step thermodynamic cycle used in four-stroke internal combustion engines to convert heat into work. It consists of the intake, compression, power, and exhaust strokes. The cycle helps in understanding engine efficiency and performance.

The Otto cycle is a thermodynamic cycle that is used in four-stroke internal combustion engines. It consists of four processes: intake, compression, power, and exhaust.

The intake stroke is when the mixture of fuel and air is drawn into the combustion chamber. The compression stroke compresses the mixture adiabatically, increasing its temperature and pressure. During the power stroke, the mixture is ignited, creating a rapid increase in pressure that pushes the piston down. Finally, the exhaust stroke expels the burnt gases from the combustion chamber.

The Otto cycle is an idealized representation of the processes that occur in an engine and it describes the thermodynamic changes that are involved in converting heat into work. It helps in understanding the efficiency and performance of four-stroke engines.

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:

C. Radiation

Explanation:

Heat transfer through Conduction is a slow process and both Conduction and convection require material medium

Answer:

C radiation

Explanation:

Answer:

Differential entropy differ's from the normal entropy as in case of differential entropy random variable needs not to be discrete.

The differential concept of entropy is as follows

Let X be a random variable which is continuous

Let[tex]f(x)[/tex] be it's probability distribution function

We have differential entropy h(X) defined as

[tex]h(X)=-\int _{X}f(X)log[f(X)]dx[/tex]

The random variable 'X' need not to be discrete and it is continuous.

Answer:

Quality of vapor is equal to 17.3%.

Explanation:

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

So pressure at saturation temperature 20°C

[tex]v_f= 0.001\frac{m^3}{Kg} ,v_g= 57.76\frac{m^3}{Kg}[/tex]

So specific volume v

[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]  

Where x is quality of mixture

Now putting the values ,given that [tex]v=10m^3/kg[/tex]

[tex]10=0.001+x(57.76-0.001)\frac{m^3}{Kg}[/tex]  

x=0.173

So quality of vapor is equal to 17.3%.

Answer:

A. True

Explanation:

Answer:

θ=0.0288 radian

Explanation:

resolution limit is the minimum angular separation of  two sources that can be viewed  distinctly    by telescope

[tex]\theta =\frac{1.22\times \lambda}{D}[/tex]

[tex]\lambda=6000\times 10^{-8} cm=6 \times 10^{-5} cm[/tex]

[tex]d=100 inch=100\times 2.54=254cm[/tex]

[tex]\theta = \frac{1.22 \times 6 \times 10^{-5}}{254}[/tex]

θ=0.0288 radian

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

[tex]\frac{T}{J}= \frac{\tau }{r}[/tex]

where, T = torque

            J = polar moment of inertia of shaft

            τ = torsional shear stress

             r = raduis of the shaft

Therefore from the above relation we see that

            [tex]\tau = \frac{T.r}{J}[/tex]

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × [tex]\pi[/tex] × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × [tex]\pi[/tex] × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque =  30316.369 lb-in

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:

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:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=[tex]\frac{Work produced}{Heat supplied}[/tex]

and

Heat Supplied [tex]{\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )[/tex]

[tex]{Q_s}=3.3+4.4=7.7KW[/tex]

η=[tex]\frac{3.3}{7.7}[/tex]

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-[tex]\frac{T_ {sink}}{T_ {source}}[/tex]

η=1-[tex]\frac{273}{1023}[/tex]

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

Answer:

In the question given out of the four options

option B. Leftover propellant

is correct.

Explanation:

Solid rocket motors have rocket engines that uses solid fuels or propellants such as HTPB and PBAN as the most commonly used binders.

These rockets always have the propellant in adequate amount can be powered for long enough without much fuel degradation and therefore are reliable and mostly used in military applications.

Since these have fuel or propellant storage and therefore to indicate it, solid rocket motor identifies the left over propellant as ' a silver' .

Answer:

Reynolds number determines whether a flow is laminar or turbulent flow.

Explanation:

Reynolds number is defined as ratio of inertia force to the viscous force. it is a dimension less  number. Reynolds number is used to describe the type of flow in a fluid whether it is laminar flow or turbulent flow. Reynolds number is denoted by Re.

When Reynolds number is in the range of 0 to 2000, the flow is considered to be laminar.

When Reynolds number is in the range of 2000 to 4000, the flow is considered to be transition.

And when Reynolds number is more than 4000, the flow is turbulent flow.

                     The boundary layer thickness for a fluid is given by

                                      δ = [tex]\frac{5\times x}{\sqrt{Re}}[/tex]

where δ is boundary layer thickness

           x is distance from the leading edge

           Re is Reynolds number

Thus from the above boundary layer thickness equation, we can see that the boundary layer thickness varies inversely to square root of reynolds number.

Answer:

Water surface tension

Explanation:

The water around the paperclip forms a kind of elastic surface, deforming, in which the clip can stay afloat.

This is because the molecules that are on the surface of the water, already in contact with the air, try to cling to those that are next to them and those that are immediately below.

If we added even if it were just a drop of soap in the water, the clip would go to the bottom, because the soap has the ability to decrease the surface tension of the water.

Answer:

the interest rate that Sandy is getting is (C.) 15.10%

Explanation:

Given data

in 4 year cash pay (p1)  = $2000

in 8 year cash pay (p2)  = $3000

time period (t1) = 4 years

time period (t2) =  8 years

loan value = $2000

To find out

interest rate

solution

first we know amount will be paid in first 4 year is

$2000 [tex](1+r/100)^{12t}[/tex]

$2000 [tex](1+r/100)^{48}[/tex]                  ...................1

now we calculate the next payment will paid after 4 year i.e.

$2000 [tex](1+r/100)^{12t}[/tex] - $2000  

$2000 [tex](1+r/100)^{48}[/tex]   - $2000  ..................2

after full time period of payment total amount will be paid by equation 1 and 2  i.e.

$3000 = $2000 [tex](1+r/100)^{48}[/tex] ×$2000 [tex](1+r/100)^{48}[/tex] - $2000

$3000 = $2000 ( [tex](1+r/100)^{48}[/tex] ×  [tex](1+r/100)^{48}[/tex] - 1 )   .....3

now we have solve   [tex](1+r/100)^{48}[/tex] this eqution

so we consider  [tex](1+r/100)^{48}[/tex] = A

so new equation will be by equation 3

$3000/$2000 = ( A - 1 ) × A

3/2 = A² - A

solve this equation we get  2A² - 2 A - 3 = 0 so A = 1.823

now we compute A again in  [tex](1+r/100)^{48}[/tex] = A

[tex](1+r/100)^{48}[/tex] = 1.823

so rate (r) = 1.258 % / month

and rate yearly = 1.258 ×12

the interest rate that Sandy is getting yearly 15.10 %

Answer:

A body is said to be in thermal equilibrium where all the system temperatures are the same.

A body is said to be in thermodynamic equilibrium if all three requirements of equilibrium are met

Explanation:

A body is said to be in thermal equilibrium where all the system temperatures are the same. In this case, there will be no temperature gradient between the system and the environment.

while, A body is said to be in thermodynamic equilibrium if all three requirements of equilibrium are met, i.e. mechanical equilibrium, chemical equilibrium and thermal equilibrium.

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:

Explained

Explanation:

Cutting tools:

 1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

Cutting tools engage materials with a sharp edge for aggressive material removal, while abrasive machining tools employ hard particles to wear away material for high precision and smoothness in finishing operations.

The differences between cutting tools and abrasive machining tools are fundamental in the processes they are used for and their operational principles. Cutting tools are typically used in operations like turning, milling, and drilling where the tool itself engages the material to be cut with a sharp edge, removing materials in the form of chips. These tools are often made of high-speed steel, carbide, ceramics, or other hard materials and are precisely shaped according to the specific cutting operation.

On the other hand, abrasive machining tools, which include grinding wheels, sandpaper, and abrasive belts, remove material through the action of hard, abrasive particles that are either bonded to the tool's surface or are used as loose grains. Abrasive machining is used for finishing surfaces to a high degree of smoothness, precision, and complex shapes that cutting tools cannot achieve. These tools are made of materials like aluminum oxide, silicon carbide, diamond, or cubic boron nitride.

To summarize, cutting tools use a sharp edge to remove material in a defined shape, while abrasive tools use hard particles to wear away material for finishing and shaping. Moreover, cutting tools are applied in more aggressive material removal processes and quicker operations like shaping or roughing out material, whereas abrasive machining is associated with finishing operations that require high precision and smoothness.

Answer:  0.025 in = 0.065 mm

Explanation:  To convert the value in inches to mm we have to multiply the inches by the conversion factor 25.4.

So, 0.025 × 25.4 = 0.065 mm (millimeter)

Conversion formula for calculation in (inch) into mm is:

      Value in mm = Value in in × 25.4

Answer:

.025 inch = 0.635 mm

Explanation:

We know that 1 inch = 2.54 cm

also we know that 1 cm = 10 mm

Thus 1 inch = 2.54 x 10 mm  

=> 1 inch = 25.4 mm

Hence 0.025 inch = 25.4 x .025 mm

Thus .025 inch = 0.635 mm

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:

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:

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:

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:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi     ( 1 psi= 6894.76 [tex]\frac{N}{m^2}[/tex] )

we know that   [tex]P=\rho\times g\timesh[/tex]

where [tex]\rho =Density,g=9.81\frac{m}{s^2}[/tex]

     h=height.

Given that P=14.769 psi ⇒P= 101828.6 7[tex]\dfrac{N}{m^2}[/tex]

(a) [tex]P=\rho_{w}\times g\times h_{w}[/tex]  

     [tex]\rho_{w}=1000\frac{Kg}{m^3}[/tex]

⇒101828.67=[tex]1000\times 9.81\times h_{w}[/tex]

[tex]h_{w}[/tex]=10.38 m

So water barometer will read 408.66 in.            (1 m=39.37 in)

(b)  [tex]P=\rho_{hg}\times g\times h_{hg}[/tex]

     [tex]\rho_{hg}[/tex]=13600

So 101828.67=[tex]13600\times 9.81\times h_{hg}[/tex]

[tex]h_{hg}[/tex]=0.763 m

So mercury barometer will read 30.04 in.

Answer:

a)-Bin stock items free issue

Explanation:

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Bin stock items free issue items are similar to the free issue items, but their access is limited.

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.