You want a potof water to boil at 105 celcius. How heavy a
lid should youput on the 15 cm diameter pot when Patm =
101kPa?

Answer:

10.41 kg

Explanation:

The gas state equation is:

p * V = n * R * T

For this equation we need every value to be in consistent units

1.5 bar = 150 kPa

5 C = 278 K

n = p * V / (R * T)

n = 150000 * 10 / (8.31 * 278) = 649 mol

Multiplying the amount of moles by the molecular weight of the gas we obtain the mass:

m = M * mol

m = 16.04 * 649 = 10410 g = 10.41 kg

Answer:

Air preheater:

 Air preaheater is a heat exchanger which take heat from flue gases and increase the temperature of the air before entering into the  combustion process.

As we know that ,flue gases contains high temperature and that temperature can be used by air preheater and it will reduce the amount of  heat addition in the combustion process and that leads to increase the efficiency of plant.When heat addition reduces for the same out put of power it means that less amount of fuel is required for the same out put of power as without using air preheater. The main purpose of air preheater is to increase the thermal efficiency.

By increasing the temperature of air it will reduce the amount of fuel reduce for the combustion.

Answer:

The given statement is true.

Explanation:

A chemical property of any material be it solid liquid or gas is defined as how it interacts chemically with other substances after an interaction takes place between them.  The interaction in language of chemistry is known as chemical reaction. Different materials in nature show different interactions with various other substances.The 2 substances can react with each other and form a different compound or may not react with each other and are termed as inert chemicals. Infact it is this interaction between the various chemicals that we can group different into classes based on their behavior with different chemicals. The interaction of different materials act as their signatures that help us in identifying them.

Answer:

[tex]\sigma _1=10.33MPa[/tex]

[tex]\sigma _2=5.16MPa[/tex]

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]

Longitudinal stress

[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]

Now by putting the values

[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]

[tex]\sigma _h=\dfrac{100\times 62}{2\times 0.3}[/tex]

[tex]\sigma _h=10.33MPa[/tex]

[tex]\sigma _l=5.16MPa[/tex]

So the principle stress are

[tex]\sigma _1=10.33MPa[/tex]

[tex]\sigma _2=5.16MPa[/tex]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = [tex]\frac{60+180}{2}[/tex]  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9  ) × g/4

R = ( 69  ) × 9.81/4  + ( 69 ) 9.81

R = 69  ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

Answer:

0.014 in

Explanation:

The Young's module of steel is of E = 210 GPa = 30*10^6 psi

The section of the bolt would be:

A = π/4 * D^2

The stiffness would be:

k = E * A / L

k = π * D^2 * E / (4 * L)

k = π * 0.25^2 * 30*10^6 / (4 * 10) = 147000 lb / in

If I apply a force of 2000 lb, I calculate the  stretching with Hooke's law:

Δx = f / k

Δx = 2000 / 147000 = 0.014 in

Explanation:

The two types of furnaces used in steel production are:

Basic oxygen furnace

In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.

Electric arc furnace

Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.

Answer:

ρ=962.16kg/m^3

Explanation:

The first thing we must do to solve is to find the mass of the specimen using the weight equation

w = mg

m=w/g

m=0.45/9.81=0.04587kg

To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).

We must bear in mind that the principle of archimedes indicates that the buoyant force is given by

F = ρgV

where V is the specimen volume and  ρ is the density of alcohol = 789kg / m ^ 3

considering the above we have the following equation

0.081=0.45-(789)(9.81m/s^2)V

solving for V

V=(0.081-0.45)/(-789x9.81)

V=4.7673x10^-5m^3

finally we found the density

ρ=m/v

ρ=0.04587kg/4.7673x10^-5m^3

ρ=962.16kg/m^3

Answer:

74.2 km from station A.

Explanation:

We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.

The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.

For this problem it is better to convert the speeds to km/min

45 km/h = 0.75 km/min

The equation for position under constant speed is:

X(t - t0) = X0 + v0 * (t - t0)

Since we know the time it will stop moving at this speed:

X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km

After it ran those 7.5 km it will keep running at 60 km/h.

60 km/h = 1 km/min

The position equation for it is now:

X(t - 10) = 7.5 + 1 * (t - 10)

The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.

75 km/h = 1.25 km/min

After those 5 minutes it will have traveled:

X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km

Then it travels at 105 km/h

105 km/h = 1.75 km/min

Its position equation is now:

X(t - 40) = 6.25 + 1.75 * (t - 40)

Equating both positions we find the time at which they would meet:

7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)

7.5 + t - 10 = 6.25 + 1.75*t - 70

t - 1.75*t = 6.25 - 70 +10 - 7.5

-0.75*t = -61.25

t = 61.25 / 0.75

t = 81.7 minutes

The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.

At that moment the freight train will be at:

X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is [tex]1.557\times10^{-7}[/tex]m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

[tex]Re=\frac{vd}{\nu}[/tex]

Here, v is velocity, [tex]\nu[/tex] is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

[tex]Re=\frac{vd}{\nu}[/tex]

[tex]Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}[/tex]

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

Answer:

The infiltration rate is of 0.042 m^3/s.

Explanation:

The total volume of the room is:

V = 3 * 10 * 20 = 600 m^3

If the air infiltration rate is of 1/4 volume per hour:

v = V/4 * 1 hour /3600 seconds

Replacing:

v = 600/4 * 1/3600 = 0.042 m^3/s

The infiltration rate would then be of 0.042 m^3/s.

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

[tex]v=\omega*r[/tex]

Where [tex]\omega[/tex] is the angular velocity and r the radius of the disk. This leads to:

[tex]v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s[/tex]

Answer:

0.2 m/s

Explanation:

Step 1: identify the given parameters

angular velocity, ω = 0.5 rad/s

radius of the disk, r = 0.4m

Note: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.

Step 2: calculate the velocity of the disk at the outer edge in m/s

Velocity = Angular velocity (rad/s) X radius (m)

Velocity = 0.5 rad/s X 0.4 m

Velocity = 0.2 m/s

Answer:

[tex]specific\ volume=0.00097\ m^3/kg[/tex]

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 [tex]kg/m^3[/tex]

Density of sea water = 1025  [tex]kg/m^3[/tex]

We know that

[tex]Density=\dfrac{mass}{Volume}[/tex]   ---1

Specific volume

[tex]specific\ volume=\dfrac{Volume}{mass}[/tex]    ---2

From equation 1 and 2

We can say that

[tex]specific\ volume=\dfrac{1}{density}\ m^3/kg[/tex]

[tex]specific\ volume=\dfrac{1}{1025}\ m^3/kg[/tex]

[tex]specific\ volume=0.00097\ m^3/kg[/tex]

Answer:

Time taken to reach 80 meters equals 4.4897 seconds.

Explanation:

We know that velocity is related to position as

[tex]v=\frac{ds}{dt}[/tex]

Now it is given that [tex]v=(11+0.2v[/tex]

Using the given velocity function in the above relation we get

[tex]\frac{ds}{dt}=(11+0.2s)\\\\\frac{ds}{(11+0.2s)}=dt\\\\\int \frac{ds}{(11+0.2s)}=\int dt\\\\[/tex]

Now since the limits are given as

1) at t = 0 , s=0

Using the given limits we get

[tex]\int_{0}^{80} \frac{ds}{(11+0.2s)}=\int_{o}^{t} dt\\\\\frac{1}{0.2}[ln(11+0.2s)]_{0}^{80}=(t-0)\\\\5\times (ln(11+0.2\times 80)-ln(11))=t\\\\\therefore t=4.4897seconds[/tex]

Answer:

The correct option is 'e': f = 0.05.

Explanation:

The head loss as given by Darcy Weisbach Equation is as

[tex]h_{l}=\frac{flv^{2}}{2gD}[/tex]

where

[tex]h_{l}[/tex] is head loss in the pipe

'f' is the friction factor

'l' is the length of pile

'v' is the velocity of flow in pipe

'D' is diameter of pipe

From equation of contuinity we have [tex]v=\frac{Q}{A}[/tex]

Thus using this in darcy's equation we get

[tex]h_{l}=\frac{flQ^{2}}{2gDA}[/tex]

where

'Q' is discharge in the pipe

'A' is area of the pipe [tex]A=\frac{\piD^2}{4}[/tex]

Applying the given values we get

[tex]h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}[/tex]

Solving for 'f' we get

[tex]f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05[/tex]

Answer:

Decrease

Explanation:

Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.

So we can say that when  with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.

Decrease

Answer:

Decreases

Explanation:

Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.

The modulus of elasticity is proportional to its tensile strength.

Answer:

  Diameter pitch is the parallel thread of the imaginary cylinder of the diameter that basically intersect in the surface of thread. It is also known as effective diameter. The diameter pitch is basically used to determine the threated parts.

The pitch diameter is the essential component for determining the basic compatibility in the externally and internally thread parts. The diameter pitch is the sensitive measuring tool for determine the specific measurements.

Answer:

(a) 10.76 [tex]ft^2/sec[/tex]

(b) 0.134 lbm

(c) 0.4308 Btu/lbm

Explanation:

We have to do conversion

(a) conversion of viscosity [tex]1m^2/sec\ to\  ft^2/sec[/tex]

We know that [tex]1m^2=10.76feet^2[/tex]

So [tex]1m^2/sec=10.76ft^2/sec[/tex]

(b) We have to convert power 100 W in hp

We know that 1 W = 0.00134 hp

So 100 W = 100×0.00134=0.134 hp

(c) We have to convert 1 KJ/kg to Btu/lbm

We know that 1 KJ = 0.94780 Btu

And 1 kg = 2.20 lbm

So [tex]1KJ/kg=\frac{0.9478Btu}{2.20lbm}=0.4308Btu/lbm[/tex]

Answer with Explanation:

The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:

1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.

2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.

3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity  the resistance it offer's to shear forces and the torques is more.

While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.

Answer:

5.9*10^-3 Pa*s

Explanation:

The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.

τ = μ * du/dr

u(r) = 0.4/r - 1000*r

The derivative is:

du/dr = -1/r^2 - 1000

On the radius of the inner cylinder this would be

u'(0.02) = -1/0.02^2 - 1000 = -3500

So:

τ = -3500 * μ

We don't care about the sign

τ = 3500 * μ

That is a tangential force per unit of area.

The area of the inner cylinder is:

A = h * π * D

And the torque is

T = F * r

T = τ * A * D/2

T = τ * h * π/2 * D^2

T = 3500 * μ * h * π/2 * D^2

Then:

μ = T / (3500 * h * π/2 * D^2)

μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s

Answer:

b)false

Explanation:

Specific cutting energy:

 Energy required to remove unit volume of material is called specific energy.In other words we can say that the ratio of energy to the volume removal rate is called specific cutting energy.

When cutting speed is increases then specific energy goes to decrease.As well as when depth of cut and feed of tool is increase then specific cutting energy will decrease.

Answer with Explanation:

Drag and lift are the forces that act on an object which moves in a fluid and are explained as under:

1)Drag: Drag is the force that opposes the motion of an object moving in a fluid and hence is analogous to fluid frictional force in a fluid. Power is needed to be spent by an object to overcome the frictional drag. The drag force is different from the classical friction as we know that the frictional force on a dry surface is independent of the velocity of the object but for drag the force is proportional to the square of the velocity of the object.

Mathematically

[tex]F_{Drag}=\frac{1}{2}C_{d}A\rho _{fluid}V^2[/tex]

where

[tex]C_d[/tex] is a constant known as coefficient of drag and A is the projected area of the object.

2) Lift: Lift force is a component of the force that a fluid exerts on an object moving in it that counters the weight of the object and hence has the tendency to lift the object and hence is known as lift force. The lift force is perpendicular to the incoming fluid. Lift force is useful as it allows the aeroplanes  to fly as the lift force that is generated by the air on the wings of the plane is used to overcome the force of gravity on the plane.

Mathematically

[tex]F_{Lift}=\frac{1}{2}C_{L}A\rho _{fluid}V^2[/tex]

where

[tex]C_L[/tex] is a constant known as coefficient of lift and A is the projected area of the object moving with velocity 'v'.

Answer:

The mass of the air is 4.645 kg.

The work done or heat transfer is 277.25 kj.

Explanation:

In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.

Given:  

Initial pressure of the gas is 2 bar.

Initial volume of the gas is 2 m³.

Initial temperature is 300 K.

Final pressure is 1 bar.

Calculation:  

Step1

Apply ideal gas equation for air as follows:

PV=mRT

[tex]2\times10^{5}\times2=m\times 287\times300[/tex]

m = 4.645 kg.

Thus, the mass of the air is 4.645 kg.

Step2  

For isothermal process work done is same as heat transfer.

Work done or the heat transfer is calculated as follows:

[tex]W=P_{i}V_{i}ln\frac{P_{i}}{P_{f}}[/tex]

[tex]W=2\times10^{5}\times2\times ln\frac{2}{1}[/tex]

[tex]W=2.7725\times10^{5}[/tex] j

Or,

W=277.25 kj.

Thus, the work done or heat transfer is 277.25 kj.  

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3

Rotor diameter, d = 100 m

or

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = [tex]\frac{1}{2}\rho V^3[/tex]

or

Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]

or

Rated power = 1345.66 W/m²

b) Mechanical power developed =  [tex]\frac{1}{2}\rho AV^3C_p[/tex]

Here, A is the area of the rotor

or

A = π × 50²

thus,

Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]

or

Mechanical power developed =  3169035.1875 W

Answer:

D/t>20

Explanation:

Lets take

D =Diameter of thin cylinder

t =Thickness of thin cylinder

So a cylinder is called thin cylinder if the ratio of diameter to the thickness is greater than 20 (D/t>20 ).

But on the other hand a cylinder is called thick cylinder is ratio of  thickness to the diameter is greater than 20 (t/D>20 ).

So the governing ratio of thin walled cylinder is 20.

Answer:

1.2253 kg/m3

Explanation:

from ideal gas equation we  know that

[tex]PV = n\timesRT[/tex]

[tex]\frac{n}{V} = \frac{P}{(RT)}[/tex]

where,

pressure P = 1.01325 x 10^5 N/m2 = 101325 Pascals  

gas constant R = 8.314 J/mol/K  

T = 288.15 K

[tex]\frac{n}{V} = \frac{101325}{(8.314*288.15K)}[/tex]

[tex]\frac{n}{V} = 42.2948967 Pa*mol/J[/tex]

we know that 1 Pa = 1 J/m3

so[tex]\frac{n}{V} = 42.2948967\  mol/m3[/tex]

as we know, Air consist of  21% oxygen and 79% nitrogen

therefore

Molar mass of air:

[tex]0.21 *(32 g/mol) + 0.79 *(28 g/mol) = 28.97 g/mol[/tex]

So  [tex](42.2948967\  mol/m3) \times 28.97\  g/mol[/tex]

= 1 225.27 g/m^3

= 1.22528316 kg/m^3 ~ 1.2253 kg/m3

Answer:

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Explanation:

Drag force:

 Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.

Drag force given as

[tex]F_D=\dfrac{1}{2}\rho\ A\ V^2[/tex]

So we can say that drag force depends on following properties

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Answer:

Yes

Explanation:

As we know that heat transfer take place from high temperature body to low temperature body.

In the given problem ,the temperature of the air is high as compare to the temperature  of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the  both body will become at the same  and this stage is called thermal equilibrium.

So an office worker claim is correct.

Answer:

Final temperature will be 16.57°C

Explanation:

We have given mass of nitrogen m = 6 kg

Initial temperature [tex]T_1=30^{\circ}[/tex]

Decrease in internal energy [tex]\Delta U=-60KJ[/tex]

Specific heat of nitrogen [tex]c_v[/tex]= 0.745 KJ/kg

Let final temperature is [tex]T_2[/tex]

Change in internal energy is given by [tex]\Delta U=mc_v\Delta T=mc_v(T_2-T_1)[/tex]

[tex]-60=6\times 0.745(T_2-30)[/tex]

[tex]T_2=16.57^{\circ}C[/tex]

So final temperature will be 16.57°C