The Cv factor for a valve is 48. Compute the head loss when 30 GPM of water passes through the valve.

Answer:

(a) Shear plane angle will be 21.9°

(b) Shear train will be 2.7913 radian

Explanation:

We have given rake angle [tex]\alpha =5^{\circ}[/tex]

Thickness before cut [tex]t_1=0.01inch[/tex]

Thickness after cutting [tex]t_2=0.027inch[/tex]

Now ratio of thickness before and after cutting [tex]r=\frac{t_1}{t_2}=\frac{0.01}{0.027}=0.3703[/tex]

(a) Shear plane angle is given by [tex]tan\Phi =\frac{rcos\alpha }{1-rsin\alpha }[/tex], here [tex]\alpha[/tex] is rake angle.

So [tex]tan\Phi =\frac{0.3703\times cos5^{\circ}}{1-sin5^{\circ}}=0.4040[/tex]

[tex]\Phi =tan^{-1}0.4040[/tex]

[tex]\Phi =21.9^{\circ}[/tex]

(b) Shear strain is given by [tex]\gamma =tan(\Phi -\alpha )+cot\Phi[/tex]

So [tex]\gamma =tan(21.9 -5 )+cot21.9=2.7913radian[/tex]

(a) The shear plane angle is approximately [tex]\( 20.9^\circ \)[/tex].  

(b) The shear strain for the operation is approximately 2.90.

In an orthogonal cutting operation, we need to calculate the shear plane angle and the shear strain using the given parameters.

Given:

- Tool width [tex]\( w = 0.250 \)[/tex] in

- Rake angle [tex]\( \alpha = 5^\circ \)[/tex]

- Uncut chip thickness [tex]\( t_1 = 0.010 \)[/tex] in

- Deformed chip thickness [tex]\( t_2 = 0.027 \)[/tex] in

(a) Shear Plane Angle

The shear plane angle [tex]\( \phi \)[/tex] can be calculated using the chip thickness ratio  r and the rake angle [tex]\( \alpha \)[/tex].

1. Calculate the chip thickness ratio r :

  [tex]\[ r = \frac{t_1}{t_2} = \frac{0.010 \text{ in}}{0.027 \text{ in}} = 0.3704 \][/tex]

2. Use the relationship for shear plane angle [tex]\( \phi \)[/tex] :

  [tex]\[ \tan(\phi) = \frac{r \cos(\alpha)}{1 - r \sin(\alpha)} \][/tex]

  Substituting r and [tex]\( \alpha = 5^\circ \)[/tex] :

  [tex]\[ \tan(\phi) = \frac{0.3704 \cos(5^\circ)}{1 - 0.3704 \sin(5^\circ)} \][/tex]

3. Calculate [tex]\( \cos(5^\circ) \)[/tex] and [tex]\( \sin(5^\circ) \)[/tex] :

  [tex]\[ \cos(5^\circ) \approx 0.9962, \quad \sin(5^\circ) \approx 0.0872 \][/tex]

4. Substitute and calculate [tex]\( \tan(\phi) \)[/tex] :

  [tex]\[ \tan(\phi) = \frac{0.3704 \times 0.9962}{1 - 0.3704 \times 0.0872} \approx \frac{0.3690}{0.9677} \approx 0.3816 \][/tex]

5. Find [tex]\( \phi \)[/tex] :

  [tex]\[ \phi = \tan^{-1}(0.3816) \approx 20.9^\circ \][/tex]

(b) Shear Strain

The shear strain [tex]\( \gamma \)[/tex] in the shear plane can be calculated using:

  [tex]\[ \gamma = \cot(\phi) + \tan(\phi - \alpha) \][/tex]

1. Calculate [tex]\( \cot(\phi) \)[/tex] :

  [tex]\[ \cot(\phi) = \frac{1}{\tan(\phi)} = \frac{1}{0.3816} \approx 2.62 \][/tex]

2. Calculate [tex]\( \phi - \alpha \)[/tex] :

  [tex]\[ \phi - \alpha = 20.9^\circ - 5^\circ = 15.9^\circ \][/tex]

3. Calculate [tex]\( \tan(15.9^\circ) \)[/tex] :

  [tex]\[ \tan(15.9^\circ) \approx 0.2844 \][/tex]

4. Calculate [tex]\( \gamma \)[/tex] :

  [tex]\[ \gamma = 2.62 + 0.2844 \approx 2.9044 \][/tex]

The shear plane angle and shear strain are crucial in determining the efficiency of the cutting operation. The shear plane angle is derived from the relationship between the undeformed and deformed chip thicknesses and the rake angle. The shear strain reflects the material deformation occurring during the cutting process, combining the geometric factors and material properties.

Answer:

P= 45.384 kW

Explanation:

given data:

m = 1000 kg

u = 0

v = 100 km/hr = 250/9  m/s

t = 10 sec

g =9.81 m/s2

5% gradient

from figure we can have

[tex]tan\theta = \frac{0.05x}{x}[/tex]

[tex]\theta =  tan^{-1}0.05 [/tex]

[tex]\theta = 2.86[/tex] from equation of motion we have

v = u + at

[tex]\frac{250}{9} = 0 + 9*10[/tex]

[tex]a = \frac{25}{9} m/s^2[/tex]

distance covered  in 10 sec

from equation of motion

[tex]s = ut + \frac{1}{2}at^2[/tex]

[tex]s = 0*10+ \frac{1}{2}*\frac{25}{9}*10^2[/tex]

s = 138.8 m

from newton's 2nd law of motion along inclined position

[tex]F -mgsin\theta  = ma[/tex]

solving for f  

[tex]f = mgsin\theta +ma[/tex]

[tex]F = 1000*9.81*SIN2.8624 +1000*\frac{25}{9}[/tex]

F = 3267.67 N

work done is given as W

[tex]W = F* s[/tex]

and power [tex]P = \frac{W}{t}[/tex]

[tex]P = \frac{F*s}{t}[/tex]

[tex]P = \frac{3267.67*\frac{1250}{9}}{10}[/tex]

P = 45384.30 W

P= 45.384 kW

Answer:

Force will be equal to 0.08522 horsepower

Explanation:

We given weight of elevator = 5000 lb

Height of the elevator = 30 ft

Time t = 320 sec

We have to calculate the power in horsepower

For let first find power in lb-ft/sec

So power in lb-ft/sec [tex]=\frac{5000lb\times 30ft}{320sec}=46.875lb-ft/sec[/tex]

Now we know that 1 horsepower = 550 lb-ft/sec

So 46.785 lb-ft/sec = [tex]\frac{46.875}{550}=0.08522horsepower[/tex]

Answer with Explanation:

1) The advantages of fission energy are:

a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.

b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c

2) The disadvantages of fission energy are:

a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.

b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.

Answer:

[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

[tex]Re=\dfrac{F_i}{F_v}[/tex]

For plate can be given as

[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

[tex]Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.[/tex]

[tex]Reynolds\ number\is \ <\ 5 \times 10 ^5\ then\ flow\ will\ be\ laminar.[/tex]

Calculate the elongation of a cylindrical nickel wire under a specific load by using the formula for elastic deformation and provided values.

To calculate the elongation of the cylindrical nickel wire under a load, we can use the formula for elastic deformation: elongation = (F * L) / (A * E), where F is the load, L is the length, A is the cross-sectional area, and E is the Young's modulus.

Substitute the given values: diameter = 1.8 mm (radius = 0.9 mm), load = 290 N, length = 2.6 × 10^4 mm, and Young's modulus for nickel = 207 × 10^9 N/m^2. Solve for the elongation to find the answer.

The elongation of the cylindrical nickel wire under the given load is found to be **insert answer here**.

The elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.

Step 1

To calculate the elongation of the nickel wire under the applied load, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] of a material is directly proportional to the applied force ([tex]\( F \)[/tex]) and the material's elastic modulus ([tex]\( E \)[/tex]), and inversely proportional to its cross-sectional area ([tex]\( A \)[/tex]) and original length ([tex]\( L_0 \)[/tex]). Mathematically, it's expressed as:

[tex]\[ \Delta L = \frac{F \cdot L_0}{A \cdot E} \][/tex]

Where:

- F is the applied force (290 N)

- [tex]\( L_0 \)[/tex] is the original length of the wire (2.6 × [tex]10^4[/tex] mm = 26,000 mm)

- A is the cross-sectional area of the wire

- E is the elastic modulus of nickel (207 × [tex]10^9[/tex] [tex]N/m^2[/tex])

Step 2

First, let's calculate the cross-sectional area (A) of the wire using its diameter (d ):

[tex]\[ A = \frac{\pi d^2}{4} \][/tex]

Given that the diameter [tex](\( d \))[/tex] is 1.8 mm, we have:

[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \][/tex]

Now, let's calculate the elongation ([tex]\( \Delta L \)[/tex]) using Hooke's Law:

[tex]\[ \Delta L = \frac{290 \times 26,000}{A \times 207 \times 10^9} \][/tex]

Step 3

Finally, we can substitute the values and solve for [tex]\( \Delta L \).[/tex] Let's do the calculations.

First, let's calculate the cross-sectional area A:

[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \]\[ A = \frac{\pi \times 3.24 \times 10^{-6}}{4} \]\[ A = 2.55 \times 10^{-6} \, \text{m}^2 \][/tex]

Step 4

Now, let's calculate the elongation [tex](\( \Delta L \))[/tex] using Hooke's Law:

[tex]\[ \Delta L = \frac{290 \times 26,000}{2.55 \times 10^{-6} \times 207 \times 10^9} \]\[ \Delta L = \frac{7,540,000}{528.15} \]\[ \Delta L = 14.295 \, \text{m} \][/tex]

So, the elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.

Answer and Explanation:

Pyroelectric material

Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.

The potential difference generated is for very less time

The generation of potential difference is due to change in position of atoms after heating or cooling.

Answer:

Heat supplied = 208.82 BTU/s

Heat rejected  =  66.82 BTU/s

Carnot thermal efficiency = 0.68

Explanation:

Data  

Hot temperature,[tex] T_H [/tex] = 1200 F + 459.67 = 1659.67 R

Cold temperature,[tex] T_C [/tex] = 70 F + 459.67 = 529.67 R

Engine power, [tex] \dot{W} = 200 hp \times 0.71\frac{BTU/s}{hp} = 142 \frac{BTU}{s} [/tex]  

Carnot thermal efficiency is computed by

[tex] \eta = 1 - \frac{T_C}{T_H} [/tex]

[tex] \eta = 1 - \frac{529.67 R}{1659.67 R} [/tex]  

[tex] \eta = 0.68 [/tex]  

Efficiency is by definition

[tex] \eta = \frac{\dot{W}}{\dot{Q_{in}}} [/tex]

[tex] \dot{Q_{in}} = \frac{\dot{W}}{\eta} [/tex]

[tex] \dot{Q_{in}} = \frac{142 \frac{BTU}{s}}{0.68} [/tex]

[tex] \dot{Q_{in}} = 208.82 \frac{BTU}{s} [/tex]

where [tex] \dot{Q_{in}} [/tex] is the heat supplied

Energy balance in the engine  

[tex] \dot{Q_{in}} = \dot{W} + \dot{Q_{out}} [/tex]

[tex] \dot{Q_{out}} = \dot{Q_{in}} - \dot{W} [/tex]

[tex] \dot{Q_{out}} = 208.82 \frac{BTU}{s} - 142 \frac{BTU}{s} [/tex]

[tex] \dot{Q_{out}} = 66.82 \frac{BTU}{s} [/tex]

where [tex] \dot{Q_{out}} [/tex] is the heat rejected

Answer with Explanation:

For a linearly responsive system principle of superposition states that

"The cumulative response of the given system to forces of different magnitudes is the sum of the individual responses of the system to the individual forces"

In a less formal manner principle of superposition states that the effect of various forces acting together on any body is the sum of all the effects on the body produced when each force acts individually.

Now we know that

[tex]Stress=\frac{Force}{Area}[/tex]

Since stress at any point is in linear relation with the force hence we can conclude that the state of stress at any point due to different forces acting together is the sum of the individual stresses due to individual forces alone.

Mathematically

Let the stress due to a force [tex]F_i[/tex] be [tex]\sigma _i[/tex]

and the stress due to combined forces be [tex]\sigma _f[/tex]

thus according to principle of superposition we have

[tex]\sigma_f=\sum_{i=1}^{n}(\sigma_i)[/tex]

Answer:[tex]0.01 N/m^2[/tex]

Explanation:

Given

Distance between Plates(dy)=1 cm

Relative Velocity(du)=10 cm/s

Dynamic viscosity[tex]\left ( \mu \right )=0.001 Ns/m^2[/tex]

We know shear stress is given by [tex]\tau [/tex]

[tex]\tau =\frac{\mu du}{dy}[/tex]

where du=relative Velocity

dy=Distance between Plates

[tex]\tau =\frac{0.001\times 10}{1}=0.01 N/m^2[/tex]

Answer:

Area under the strain-stress curve up to fracture gives the toughness of the material.

Explanation:

When a material is loaded by external forces stresses are developed in the material which produce strains in the material.

The amount of strain that a given stress produces depends upon the Modulus of Elasticity of the material.

Toughness of a material is defined as the energy absorbed by the material when it is loaded until fracture. Hence a more tough material absorbs more energy until fracture and thus is excellent choice in machine parts that are loaded by large loads such as springs of trains, suspension of cars.

The toughness of a material is quantitatively obtained by finding the area under it's stress-strain curve until fracture.

Answer:

In Rankine 524.07°R

In kelvin 291 K

In Fahrenheit 64.4°F  

Explanation:

We have given temperature 18°C

We have to convert this into Rankine R

From Celsius to Rankine we know that  [tex]T(R)=(T_{C}+273.15)\frac{9}{5}[/tex]

We have to convert 18°C

So [tex]T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R[/tex]

Conversion from Celsius to kelvin

[tex]T(K)=(T_{C}+273)[/tex]

We have to convert 18°C

[tex]T(K)=(18+273)=291K[/tex]

Conversion of Celsius to Fahrenheit

[tex]T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F[/tex]

Answer:

The specific volume of the super heated steam is 0.2139 m³/kg.

Explanation:

Super heated steam is the condition where only compressed vapor of water present. This is not containing any liquid form of water. Super heated steam has very high pressure depending upon temperature. If heat supplied increases after saturation vapor condition of the water, the water fuses into steam completely. Properties of super heated steam are taken from the super heated steam table.

Given:  

Temperature of superheated steam is 300°C.  

Pressure of the super heated steam is 1.2 Mpa.

Calculation:

Step1  

Value of specific volume of the super heated steam is taken from superheated steam table at 300°C and 1.2 Mpa as follows:

Specific volume is 0.2139 m³/kg.

Step2

All other properties are also taken from the table as:

Internal energy is 2789.7 kj/kg.

Enthalpy is 3046.3 kj/kg.

Entropy is 7.034 kj/kgK.

The part of table at the temperature 300°C and pressure 1.2 Mpa is shown below:

 Thus, the specific volume of the super heated steam is 0.2139 m³/kg.

In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.  

Answer:

 Rage pressure:

The rage pressure in terms of engineering is the protected shield hard plastic shell and made up of the hard material that is basically used in the protection.

It is the effective material and used in the sewing machine so the pressure developed due to the hard material is known as rage pressure and it has strong elasticity.  

 Absolute pressure:

The pressure which is relative to the perfect vacuum is known as absolute pressure. The absolute pressure is basically measured against the atmospheric pressure.

The absolute pressure is defined as the total pressure in the fluid at the point is equal to the sum of the atmospheric pressure and gauge.  

Answer:

1.Molding

a.Cold molding

b.Hot molding

2.Slab stock

Explanation:

Polymer foam:

 When solid and gas phases are mixed then they produce polymer foam.The gas used during forming of polymer foam is know as blowing agent and can be physical or chemical.Physical agent will not react and act as inert while chemical agent take part in reaction.

Blowing agent-Carbon diaoxide and Hydrochlorofluorocarbons.

Example of polymer foams-Polyurethane ,Starch etc.

Polymer foam can be produce by following process

1.Molding

a.Cold molding

b.Hot molding

2.Slab stock

Answer:

The volume flow rate of air is [tex]Q=A\times V[/tex]

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

[tex]Volume=Area\times V=A\times V[/tex]

Hence the volume flow rate is [tex]Q=V\times A[/tex]

Given : Sample size : n= 5

The proportion nonconforming : p= 0.10

Binomial probability formula :-

[tex]P(x)=^nC_x p^{x}(1-p)^{n-x}[/tex]

The probability of zero nonconforming unit in the sample :-

[tex]P(0)=^5C_0 (0.10)^{0}(1-0.1)^{5}\\\\=(1)(0.9)^5\ \ [ \because\ ^nC_0=1]=0.59049[/tex]

∴ The probability of zero nonconforming unit in the sample= 0.59049

The probability of one nonconforming unit in the sample :-

[tex]P(1)=^5C_1 (0.10)^{1}(0.9)^{4}\\\\=(5)(0.1)(0.9)^5\ \ [ \because\ ^nC_1=n]=0.295245[/tex]

∴ The probability of one nonconforming unit in the sample=0.295245

The  probability of 2 or more nonconforming units in the sample :-

[tex]P(X\geq2)=1-(P(0)+P(1))=1-(0.59049+0.295245)\\\\=1-0.885735=0.114265[/tex]

∴ The  probability of 2 or more nonconforming units in the sample=0.114265

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = [tex]\sqrt{300^2+50^2}[/tex]

or

The ground speed = [tex]\sqrt{92500}[/tex]

or

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

Answer:

Time period will be 1.26 sec

Explanation:

We have given amplitude A = 0.1 M

Speed [tex]\frac{dx}{dt}=0.5m/sec[/tex]

The displacement equation of simple harmonic motion is given by

[tex]x(t)=Asin\omega t[/tex]

Differentiating both side

[tex]\frac{dx}{dt}=A\omega cos\omega t[/tex]

In question it is given that at t=0, x=0 and [tex]\frac{dx}{dt}=0.5m/sec[/tex]

So [tex]0.5=0.1\omega cos0[/tex]

[tex]\omega =5sec^{-1}[/tex]

Now period of oscillation [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{5}=1.26sec[/tex]

The magnitude and direction of the resultant force from two applied forces, P and Q, at a point can be determined graphically using the triangle rule by forming a vector triangle, measuring the resultant vector, and calculating its angle.

To determine the magnitude and direction of the resultant force using the triangle rule when two forces, P and Q, are applied at a point, you can follow a graphical method. Picture two vectors representing these forces originating from the same point. Since P = 60 lb and Q = 25 lb, you draw them to scale, with the tail of Q starting at the head of P to form a triangle. By connecting the tail of P to the head of Q, you form the resultant vector.

To find the magnitude of the resultant force, measure the length of the resultant vector using the same scale and apply trigonometry or the Pythagorean theorem if the angle is known. The direction of the resultant force is given by the angle it makes with either P or Q, which can be measured with a protractor in reference to a baseline, such as the horizontal.

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

Answer:

What is the difference Plastic vs elastic deformation

Explanation:

The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.

The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.

Answer:

The pump operates satisfactorily.

Explanation:

According to the NPSH available definition:

[tex]NPSHa =  \frac{P_{a} }{density*g} + \frac{V^{2} }{2g} - \frac{P_{v}}{density*g}[/tex]

Where:

[tex]P_{a} absolute pressure at the inlet of the pump [/tex]

[tex]V velocity at the inlet of te pump = 4m/s[/tex]

[tex]g gravity acceleration = 9,8m/s^{2}[/tex]

[tex]P_{v} vapor pressure of the liquid, for water at 65°C = 25042 Pa[/tex]

The absolute pressure is the barometric pressure Pb minus the losses: Suction Lift PLift and pipe friction loss Ploss:

To convert the losses in head to pressure:

[tex]P = density*g*H [/tex]

So:

[tex]P_{b}  = 760 mmHg = 101325 Pa[/tex]

[tex]P_{lift}  = 33634,58 Pa[/tex]

[tex]P_{loss}  = 8648,89 Pa[/tex]

The absolute pressure:

[tex]P_{a} = P_{b} - P_{lift} - P_{loss} = 59044,53 Pa[/tex]

replacing on the NPSH available equiation:

[tex]NPSHa =  6,14 m + 0,816 m - 2,6 m = 4,356 m [/tex]

As the NPSH availiable is higher than de required the pump should operate satisfactorily.

Answer:

For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.

Explanation:

Factor of safety:

  When materials are subjected to stress then we have to prevent it from a failure so we multiple stress by a factor and that factor is called factor of safety.

Factor of safety can be given as

[tex]FOS=\dfrac{Maximum\ strength}{Applied\ stress}[/tex]

Factor of safety is not a fixed quantity is varies according to the situation.

For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.

We know that brittle material did not shows any yield point and gets break without showing a indication but  ductile materials shows a yield point and gives indication before fracture.

Answer:

Pressure- temperature diagram of the fluid is the phase lines that separate all the phases.

Explanation:

Step1

Pressure temperature diagram is the diagram that represents the all the phases of the fluid by separating a line. There is no phase change region in the pressure temperature diagram out of 15 possible diagrams. There are three lines that separate the phase of the fluid. These three lines are fusion, vaporization and sublimation.

Step2

The intersecting point of these lines is triple point of fluid. Out of 15 possible phase diagram, only pressure temperature diagram has triple point as a point. In other diagrams phase change region is present and triple point is not a point. Critical point is the point in all possible property diagrams.

Pressure temperature diagram is shown below:

Answer: 10.29 sec.

Explanation:

Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.

If the car starts from rest, this means the following:

ΔK = 1/2 m*vf ²

As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:

100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec

Now, we calculate the change in energy:

ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

[tex]\sigma=\frac{\textup{PD}}{\textup{2t}}[/tex]

where, t is the thickness

on substituting the respective values, we get

[tex]100=\frac{\textup{2\times800}}{\textup{2t}}[/tex]

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

Answer with Explanation:

From newton's second law the acceleration produced by a force on a mass 'm' is given by

[tex]Acceleration=\frac{Force}{Mass}[/tex]

Applying the given values in the above equation we get

[tex]Acceleration=\frac{kvx}{m}[/tex]

Also we know that acceelration of a particle can ve mathem,atically written as

[tex]a=\frac{v\cdot dv}{dx}[/tex]

Applying the given values in the above equation we get

[tex]\frac{kvx}{m}=\frac{v\cdot dv}{dx}\\\\\Rightarrow {kx}\cdot dx=m\cdot dv\\\\\int kxdx=\int mdv\\\\\frac{kx^2}{2}=mv-c[/tex]

'c' is the constant of integration

whose value is found that at x =0 v= [tex]v_o[/tex]

Thus

[tex]c={mv_o}[/tex]

Thus the velocity as a function of position is

[tex]v=\frac{1}{m}(\frac{kx^2}{2}+c)[/tex]

Now  by definition of velocity we have

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

Using the function of velocity in the above relation we get

[tex]\frac{dx}{kx^{2}+\sqrt{2c}}=\frac{dt}{2m}\\\\\int \frac{dx}{(\sqrt{k})^2x^{2}+(\sqrt{2c})^2}=\int \frac{dt}{2m}\\\\\frac{1}{\sqrt{2kc}}\cdot tan^{-1}(\frac{(\sqrt{k})x}{\sqrt{2c}})=\frac{t}{2m}+\phi \\\\[/tex]

where

[tex]\phi [/tex] is constant of integration

Now it is given that at t = 0 ,x = 0

thus from the above equation of position and time we get [tex]\phi =0[/tex]

Thus the position as a function of time is

[tex]x(t)=\sqrt{\frac{2c}{k}}\cdot tan(\frac{kct}{\sqrt{2}m})[/tex]

where c=[tex]mv_o[/tex]

Answer:

a)True

Explanation:

While machining of ductile material and high feed and low cutting speed welding action take place between tool material and chip material this welding action is called built up edge. Built up edge action takes place due to high temperature.

To decrease the built up edge action

1. Increase the rake angle.

2. Increase the cutting speed

3. Decrease the feed rate

4. Use cutting fluid