Explain the Otto cycle of a 4 stroke engine.

Answer: Covalent bond

Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms  and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .

Answer:

Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.

Answer:

The given statement is False.

Explanation:

This is because at high temperature the creep rate depends on grain boundary area, increasing with an increase in grain boundary area thus decreasing the grain size. Thus at higher temperatures, grain size has opposite effect , the bigger the grain size the slower the slower is the creep rate, the larger the grain boundary area.

Answer:

a

Explanation:

a

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

           C=capacitance

           V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C=[tex]\frac{Q}{V}[/tex]

=500×[tex]10^{-12}[/tex]×[tex]\frac{1}{250}[/tex]

=2×[tex]10^{-12}[/tex]

=2 pF

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

[tex] A=\pi*D^2/4 [/tex]

Replacing the diameter the area results:

[tex] A= 17.76 in^2 [/tex]

Therefore the the stress results:

[tex] σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi [/tex]

Answer:  b) FCOR( Face-centered orthorhombic)

Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face  with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.

Answer:

The correct option is (A) TRUE

Explanation:

A composite material is the material having characteristics that are improved and different from the constituting materials. A composite material is produced by combining two or more materials. The constituting materials have significantly different physical or chemical properties.

Some of the composite materials include-Ceramic matrix composites, Metal matrix composites etc.

A ceramic matrix composite is formed by embedding ceramic fibers in a  ceramic matrix.

Also, a tungsten carbide- cobalt is a metal matrix composite, in which the tungsten carbide and cobalt metal matrix are fused together.

Therefore, the given statement: A composite material is a mix of two different materials that are fused together on the atomic level to form a new substance with improved properties is TRUE.

Answer:

The correct answer is C. World Coordinate System

Explanation:

The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.

Answer:

18931.4

Explanation:

Given : velocity of the electron = 2.0 [tex]\times[/tex]10

            mass of the electron = 9.19[tex]\times[/tex] 103

we know that reduced planks constant, h = 6.5821[tex]\times[/tex] [tex]10^{-16}[/tex] eV s

We know from uncertainity principle,

[tex]\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}[/tex]

[tex]\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}[/tex][tex]\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}[/tex]

[tex]\Delta \textup{x}[/tex] = 18931.4 m

Hence, uncertainty in position of the electron is 18931.4

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   [tex]Pr=\dfrac{\nu }{\alpha }[/tex]

  Where [tex]\nu[/tex] is the molecular diffusivity of momentum

             [tex]\alpha[/tex] is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    [tex]Pr=\left (\dfrac{\delta }{\delta _t}\right )^3[/tex]

Where [tex]\delta[/tex] is the hydrodynamic boundary layer thickness and [tex]\delta_t[/tex] is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

Answer:

0.424

Explanation:

The electrical energy is the electrical power times time:

E = Pt

E = (IV)t

E = (1.5 A) (110 V) (300 s)

E = 49,500 J

The heat absorbed by the cookie dough is the mass times specific heat capacity times increase in temperature:

Q = mCΔT

Q = (1 kg) (4200 J/kg/K) (5 K)

Q = 21,000 J

So the fraction of electrical energy converted to internal energy is:

Q / E = 21,000 / 49,500

Q / E = 0.424

The remainder of the electrical energy is used to do work.

To find the average number of styrene and butadiene units in a poly(styrene-butadiene) copolymer with Mn of 1,350,000 g/mol, add the molecular weights of styrene and butadiene to get the molecular weight of the repeat unit (158.24 g/mol) and divide Mn by this number to get approximately 8530 repeat units, which includes about 4265 of each monomer type.

To calculate the average number of styrene and butadiene repeat units per molecule for a poly(styrene-butadiene) copolymer with a number-average molecular weight (Mn) of 1,350,000 g/mol, we need the molecular weights of the monomer units. The molecular weight of styrene (C8H8) is approximately 104.15 g/mol, and the molecular weight of butadiene (C4H6) is approximately 54.09 g/mol. Since the copolymer is alternating, each repeat unit consists of one styrene and one butadiene unit.

To find the total molecular weight of the repeat unit, we add the molecular weights of styrene and butadiene: 104.15 g/mol + 54.09 g/mol = 158.24 g/mol. Dividing the number-average molecular weight of the copolymer by the molecular weight of the repeat unit gives us the average number of repeat units per molecule: 1,350,000 g/mol \/ 158.24 g/mol \approximately 8530 repeat units\.

The average number of styrene units per molecule will be approximately 4265, and the same for butadiene, since there is one of each in every repeat unit in an alternating copolymer structure.

Answer: True

Explanation: Injector orifice is the factor which describes the size of the opening of the injector .There are different pattern and size of the opening for the injector which affects the mixture of the chemical substance that is used for the production of the energy that is known as propellant.

The pattern and size of the orifice will define the variation in the amount of energy that could be produced.Thus the statement given is true.

Answer: 5.35m

Explanation:

By using energy equation:

[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g}  =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]

[tex]\gamma=specific weight[/tex]

[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]

[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]

[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81}  +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]

[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]

exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m

head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]

[tex]h_L=K\times V_e[/tex]

head loss as a function of exit velocity head =5.35m

Answer:

 Is required a 0.8 inches diameter steel shaft.

Explanation:

With the power P and the rotating speed n (RPM), we can find the torque applied:

T = P/N

Before calculating the torque, we convert the power and rotating speed units:

[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]

[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]

Replacing the values, the torque obtained is:

[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]

Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:

[tex]Smax =\ \frac{T.R}{J}[/tex]

Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:

[tex]J =\frac{\pi.D^4}{32}[/tex]

Where D is shaft's diameter. Replacing the expression of J in

[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]

As the radius is half of the diameter:

[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]

For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:

[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]

Solving for D:

[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]

Answer:

A reheat cycle is used to increase the overall efficiency of the power plant.                            

Explanation:

The Reheat cycle is used --

1. to increase the efficiency of the turbine

2. to maintain the quality of the steam

3. to provide higher pressure ratio

The reheat cycle is used to to increase the net work output of the power plant.In reheat cycle, two turbines , one low pressure turbine and one high pressure turbine is used to increase the efficiency. The main purpose of the reheat cycle is to maintain the quality of the steam to 0.85 at the exit of the turbine.

               The figure given below shows a reheat rankine cycle.

In the cycle,

Process 1-2 is high pressure turbine

Process 2-3 is the reheater

Process 3-4 is the low pressure turbine

Process 4-5 is condenser

Process 5-6 is pump

Process 6-1 is boiler

A reheat rankine cycle is 30 -40% efficient than a simple rankine cycle.  

Answer and Explanation:

The increase in entropy principle is defined as the process in which the total change in entropy of system with its adiabatic surroundings is always positive or equal to zero. The increase in entropy mostly takes place when a solid becomes liquid because randomness is increases when solid becomes liquid so entropy is also increases.

example of increase of entropy is when solid burns and become ash, ice melting  

Answer: False

Explanation:

Pure silicon is a good semiconductor, which conducts electricity when is mixed with some other component and can also work as an insulator.It is found abundantly on the earth's crust area. It is widely used in mixed form in the advanced electronic technologies but not in pure silicon form . So silicon is the important material in the advanced electronic technologies but not the pure silicon form.

Answer:

Explanation:

The deatailed diagram of VCRS is given below such

1-2=Isentropic compression in which temperature increases at constant entropy

2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)

3-4=Irreversible expansion or throttling in which enthalpy remains constant

4-1=Isobaric heat addition(Evaporation)

Answer: d) Prandtl number

Explanation: Prandtl number is basically defined as the ratio between the fluid's viscosity to the thermal conductivity.It doesn't have any sort of dimension. The fluids which are discovered with the small Prandtl numbers are considered as good fluids as they have a smooth rate of flow and as the number increases the fluid are not considered as reliable. Thus,option (d) is the correct option.

Answer:

t =253.8s

Explanation:

Chvorinov's Rule can be written as:

[tex]t=B(\frac{V}{A} )^{n}[/tex]

where t is the solidification time,

V is the volume of the casting,

A is the surface area of the casting that contacts the mold,

n is a constant

B is the mold constant

The S.I. units of the mold constant B are s/m2.

According to Askeland, the constant n is usually 2.

[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3

[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2

[tex]V/A=9.198[/tex]mm

[tex]t = 3.0*9.198^2[/tex] =253.8s

Answer:

Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes

Chvorinov's Rule with Degarmo Method:

Explanation:

Data:

Aluminum disc

Diameter (D) = 500 mm

Thickness = Height (h) = 20 mm

Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]

Required:

Solidification time (t) in minutes = ?

Formula:

The solidification time can be found by using the Chvorinov's Rule:

[tex]t = C (\frac {V}{A})^{n}[/tex]

Where;

t = solidification time

C = mold constant

V = Volume of disc

A = Surface area of disc

n = constant

Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.

Solution:

First, we will find the volume of the disc

disc = cylinder

therefore, Volume of cylinder is given by:

[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]

Where:

V = Volume of Cylinder

H = Height of disc

D = Diameter of disc

Now, putting dimensional values in above equation

[tex]V = \frac{\pi }{4} * 500^{2}  *20[/tex]

V = 3926990.817 [tex]mm^{3}[/tex]

Second, we will find the surface area of the disc

Therefore, surface area of cylinder is given by:

[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]

Where:

A = Surface area of disc

D = Diameter of disc

H = Height of disc

Now, putting dimensional values in above equation

[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]

A = 424115.0082 [tex]mm^{2}[/tex]

Finally, Moving towards the final solution

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]

t = 257.2016461 seconds

Converting to minutes

t = 4.286694102 minutes

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]

t = 84.52508604 seconds

Converting to minutes

t = 1.408751434 minutes

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]

t = 257.2016461 seconds

Converting to minutes

t = 4.286694102 minutes

Answer:

(b)     Irreversible cycle.

Explanation:

 Given;

 [tex]T_2=540R ,Q_2= 500 Btu/s ,T_1=240 R ,Q_1= 200 Btu/s [/tex]

To find the validity of cycle

      [tex]\oint _R\frac{dQ}{T}\leq0[/tex]

If it is zero then cycle will be reversible cycle and if it is less than zero then cycle will be irreversible cycle.These are possible cycle.

If it is greater than zero ,then cycle will be impossible .

Now find

[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}=\dfrac{200}{240}-\dfrac{500}{540}[/tex]

[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}[/tex]= -0.09

It means that this  cycle is a irreversible cycle.

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

Answer:

True

Explanation:

By definition of steady flow we have

[tex]\frac{\partial f(x,y,z,t) }{\partial t}=0[/tex]

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

Answer:

D Is the answer babe....

Answer:

d)- ON/OFF.

Explanation:

ON/OFF components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated.

Answer: c)-Yes, spheroidized steel are considered as composite.the dispersed phase is  iron carbide.

Explanation: Spheroidized steel are the alloy that have iron as the basic part that have been heat treated to increase their ductility and malleability property .They are considered as composite because they are made up of  iron alloys.  The heat treatment is usually for the carbon steel and so the dispersed phase that is obtained is iron carbide.

Answer: Mollier diagram is the diagram or graph representing the relation established between enthalpy ,air, moisture and temperature.  

Explanation: Mollier graph represents the basic thermodynamic properties along with terms of enthalpy and entropy. The factors like air, moisture , temperature etc. are plotted on the graph which makes it easy to understand. It is used in the field building designers and engineer.They are usually plotted for the gases with purity with general pressure and temperature.

Answer Explanation :

ADVANTAGES OF ISOTHERMAL PROCESS

DISADVANTAGES OF ISOTHERMAL PROCESS :

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

Explanation:

a). Operational Performance

     It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.

b).  Salvage externalities

   Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.

c). Value Vs Risk

   When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.

d). Initial expenditure

   Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.

e). Maintenance implications

   It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.