Define full and partial well abandonment.

Answer:

true.

Explanation:

but i am not 100% sure

Answer:

Critical Reynolds number is a number or the threshold or the limit at which the laminar flow changes to turbulent flow.

Explanation:

It is basically the ratio of inertial forces to viscous forces and helps to predict if the flow is laminar or turbulent.

from the experiments, for flow in a pipe of diameter D, critical Reynolds number=2300

For Laminar flow: Reynolds number is less than 2000

For transitional flow: Reynolds number is in between 2000-4000, the flow is unstable

For Turbulent flow: Reynolds number is greater than 3500

Reynolds number is different for different geometries

Answer:

a) reciprocating pump

b) rotary pump

c) linear pump

Explanation:

Positive displacement pump has enlarged  cavity on the suction side and decreasing cavity on the discharge side and in  positive displacement pump

the amount of fluid captured and inside and then discharged is same.

There are three type of positive displacement pump

a) reciprocating pump is the pump which two valve during suction outlet valve is closed and  for delivery the inlet valve is closed and pump is used for discharge

b) rotary pump fluid move with the help of rotary the rotation of rotary displaces water.

c) linear pump is the pump in which displacement is linear rope and chain pump is example of this type of pump.

Answer: True

Explanation: It is correct that during a cycle the energy of the system remains unchanged. Total energy consist of the initial energy and the final energy in a system and both remain equal in a system whether they are in cycle or not  ans also according to the law of conservation of energy ,total energy cannot be destroyed or created. Thus the while a cycle goes on energy will remain the same.

Answer:32.4m/[tex]s^2[/tex]

Explanation:

Given data

[tex]radius\left ( r\right )[/tex]=0.4m

Intial angular velocity[tex]\left ( \omega_0\right )[/tex]=4rad/s

angular acceleration[tex]\left ( \alpha\right )[/tex]=5rad/[tex]s^2[/tex]

angular velocity after 1 sec

[tex]\omega[/tex]=[tex]\omega_0 [/tex]+[tex]\alpha\times\t[/tex]

[tex]\omega[/tex]=4+5[tex]\left ( 1\right )[/tex]

[tex]\omega[/tex]=9rad/s

Velocity of point on the outer surface of disc[tex]\left ( v\right )[/tex]=[tex]\omega_0\timesr[/tex]

v=[tex]9\times0.4[/tex] m/s=3.6m/s

Normal component of acceleration[tex]\left ( a_c\right )[/tex]=[tex]\frac{v^2}{r}[/tex]

[tex]a_c[/tex]=[tex]\frac{3.6\times3.6}{0.4}[/tex]=32.4m/[tex]s^2[/tex]

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

[tex]Aa.Va=Ab.Vb=Q[/tex]

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

[tex]Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3[/tex]

[tex]Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3[/tex]

Using the volume rate:

[tex]Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s[/tex]

[tex]Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s[/tex]

Assuming no losses, the energy equation for fluids can be written as:

[tex]Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb[/tex]

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

[tex]Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za[/tex]

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

[tex]Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m[/tex]

[tex]Pb = 70000\ Pa-27303\ Pa - 14715\ Pa[/tex]

[tex]Pb = 27,996\ Pa = 28\ kPa[/tex]

Answer:

The answer is a) True  

Explanation:

"Statically Indeterminate" means the number of unknowns you have exceed the number of available equations you can get from the static analysis, then the Static (equilibrium) analysis isn't enough to solve the problem.  

Tracing constitutive models help to know where a load or reaction is located, also what kind of support settlements are used. This way you will know how the deflection behaves and also the momentum and torques location depending on the method you decide to use to solve the Statically Indeterminate system.

Answer:

Enthalpy almost doubles.

Explanation:

Argon

Cp = Specific heat at constant volume = 0.520 kJ/kgK

T₁ = Initial temperature = 75°C

T₂ = Final temperature = 35°C

Enthalpy

Δh = CpΔT

⇒Δh = Cp(T₂-T₁)

⇒Δh = 0.520×(35-75)

⇒Δh = -20.8 kJ/kg

Neon

Cp = Specific heat at constant volume = 1.03 kJ/kgK

T₁ = Initial temperature = 75°C

T₂ = Final temperature = 35°C

Δh = Cp(T₂-T₁)

⇒Δh = 1.03×(35-75)

⇒Δh = -41.2 kJ/kg

Enthalpy will change because Cp value is differrent.

Enthalpy almost doubles.

Answer:227.56 mm

Explanation:

Given data

Elastic modulus[tex]\left ( E\right )[/tex]= 117 GPa

Diameter[tex]\left ( d\right )[/tex]=3.6mm

force applied[tex]\left ( F\right )[/tex]=2460N

Area of cross-section[tex]\left ( A\right )[/tex]=[tex]\frac{\pi}{4}\times d^{2}[/tex]=10.18[tex]mm^{2}[/tex]

and change in length is given by

[tex]\Delta L[/tex]=[tex]\frac{FL}{AE}[/tex]

[tex]\Delta {0.47\times 10^{-3}}[/tex]=[tex]\frac{2460\times L}{10.180\times 117\times 10^{3}}[/tex]

L=[tex]0.47\times 10^{-3}\times 10.18\times 117\times 10^{3][/tex]

L=227.56 mm

Answer:

d). a and c

Explanation:

According to Pascal's law, when force is applied to a fluid, the pressure will increase equally in all direction of the container. Pascal law says that the pressure is transmitted undiminised in all direction inside a closed container. And the pressure inside the closed container is constant when measured at the same height above the datum.

Hence option d is correct.

Answer:

1). Entropy can be defined as a measure of sytem's thermal energy per unit temperature unavailable for ding useful work.

In other words, we can say that its a measure of the degree of randomness of a defined system.

Entropy-Environment relation:

Entropy is the fundamental concept that applies to the environment, humans or the entire universe.

The Second law  of thermodynamics explains the environmental impact of entropy. Reversing a process requires work, so reversing environmental process takes energy. Environmental impacts are higher at higher entropies and harder to reverse. Entropic flows and entropy measures can be used to prioritize the impacts which need action.

Increase of Entropy Principle:

This principle states that the total change in entropy of a system with its enclosed adiabatic surrounding is always positive i.e., greater than or equal to zero.

Given:

[tex]\sigma _{s} = 2.8\times 10^{4} \Omega-m[/tex]

electron concentration, n = [tex]2.9\times 10^{22} m^{-3}[/tex]

[tex]\mu _{h} = 0.14[/tex]

[tex]\mu _{e} = 0.023[/tex]

Solution:

Let holes concentration be 'p'

[tex]\sigma _{s}[/tex] = ne[tex]\mu _{e}[/tex] +pe[tex]\mu _{h}[/tex]     (1)

substituting all given values in eqn (1):

[tex]2.8\times 10^{4} = 2.9\times 10^{22}\times 1.6 \times 10^{-19}\times 0.14 + p\times1.6 \times 10^{-19}\times 0.023[/tex]

The cocentration of holes is:

[tex]p = 7.432\times 10^{24} m^{-3}[/tex]

Answer:para comprar terreno por ejemplo

to buy land for example

Explanation: meter, centimeter

Answer:

d = 13 mm

d =  15.68 mm

Explanation:

Given data

force = 20000 N

stress = 150 N/mm²

strain = 0.0005

E = 207 GPa

Solution

we know stress = force / area

so 0.0005 = 20000 / area

area = [tex]\pi[/tex]/4 × d²

put the area in stress equation and find out d

d² = 4×force / [tex]\pi[/tex] ×stress

d² = 4× 20000 / [tex]\pi[/tex] ×150

d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×150}[/tex]

d = 13 mm

and now we know starin = stress / E

same like stress we find d here

d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×0.0005×207×10³ }[/tex]

so d =  15.68 mm

Answer:

a). d = 13 mm

b). d = 16 mm

Explanation:

a). Given :

  Force = 20000 N

  Maximum stress, σ = 150 N/[tex]mm^{2}[/tex]

Therefore, we know that that

σ = [tex]\frac{Force}{area}[/tex]

150 = \frac{Force}{\frac{pi}{4}\times d^{2}}

150 = \frac{20000}{\frac{pi}{4}\times d^{2}}

[tex]d^{2}[/tex] = 169.76

d = 13.02 mm

d [tex]\simeq[/tex] 13 mm

b). Given :

   Strain, ε =  0.0005

   Young Modulus, E =  207 GPa

                                   = 207[tex]\times[/tex][tex]10^{3}[/tex] MPa

Therefore we know that, Stress σ = E[tex]\times[/tex]ε

                                                         = 207[tex]\times[/tex][tex]10^{3}[/tex][tex]\times[/tex]0.0005

                                                         = 103.5 N/[tex]mm^{2}[/tex]

We know that  

σ = [tex]\frac{Force}{Area}[/tex]

103.5 = [tex]\frac{Force}{\frac{pi}{4}\times d^{2}}[/tex]

[tex]d^{2}[/tex] = 246.27

d = 15.69 mm

d [tex]\simeq[/tex]16 mm

Answer Explanation : The general principles for design for assembly (DFA) are,

STEPS TO MINIMIZE THE NUMBER OF PARTS

Answer:

Volumetric flow rate = 0.4773 m³/s

Mass flow rate = 477.3 kg/s

It will take 286.38 seconds to fill a tank with measurements 5 m x 6 m x 20 m

Explanation:

Given:

Diameter of the pipe through which the water is flowing = 450 mm

Radius = Diameter/2

Thus, Radius of the pipe = 225 mm

The conversion of mm into m is shown below:

1 mm = 10⁻³ m

Radius of the pipe = 225×10⁻³ m

The area of the cross-section = π×r²

So, Area of the pipe = π×/(225×10⁻³)² m² = 0.1591 m²

Also, Given : The water flowing rate = 3 m/s

Volumetric flow rate is defined as the amount of flow of the fluid in 1 sec.

[tex]Volumetric\ flow= \frac {Volume\ passed}{Time taken}[/tex]

This, can be written as Velocity of the fluid from the cross-section area of the pipe.

Q = A×v

Where,

Q is Volumetric flow rate

A is are though which the fluid is flowing

v is the velocity of the fluid

So,

Q = 0.1591 m²×3 m/s = 0.4773 m³/s

Mass flow rate is defined as the mass of the fluid passes per unit time.

[tex]\dot {m}= \frac {Mass\ passed}{Time taken}[/tex]

The formula in terms of density can be written as:

[tex]Density=\frac{Mass}{Volume}[/tex]  

So, Mass:

[tex]Mass= Density \times {Volume}[/tex]

Dividing both side by time, we get:

[tex]\dot {m}= Density \times {Q}[/tex]

Where,

[tex]\dot {m}[/tex] is the mass flow rate

Q is Volumetric flow rate

Density of water = 1000 kg/m³

Thus, Mass flow rate:

[tex]\dot {m}= 1000 \times {0.4773} Kgs^{-1}[/tex]

Mass flow rate = 477.3 kg/s

The time taken to fill the volume of measurement 5 m× 6 m× 20 m can be calculated from the formula of volumetric flow rate as:

t= Q×V

So,

Volume of Cuboid = 600 m³

Time = 0.4773 m³/s × 600 m³ = 286.38 s

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe[tex]\left ( D\right )=0.15m[/tex]

Velocity of water in pipe[tex]\left ( V\right )=15cm/s[/tex]

We know viscosity of water is[tex]\left (\mu\right )=8.90\times10^{-4}pa-s[/tex]

Pressure drop is given by hagen poiseuille equation

[tex]\Delta P=\frac{128\mu \L Q}{\pi D^4}[/tex]

We have asked pressure Drop per unit length i.e.

[tex]\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}[/tex]

Substituting Values

[tex]\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}[/tex]

[tex]\frac{\Delta P}{L}[/tex]=0.1898 Pa/m

Answer:

[tex]K=C+273.15[/tex]

Explanation:

Kelvin's climbing represents the absolute temperature. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.

The Celsius scale has an artificial zero, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.

Answer:

The given statement is true.

Explanation:

Answer  and Explanation:

The two principles or corollaries of Carnot Theorem are listed below:

1). The efficiencies of all the reversible heat engines between any two thermal reservoirs working between the same temperatures will be equal to each other.

2). For every Carnot engine working between any two thermal reservoirs will have the same efficiency independent of the operating conditions and the nature of working substance. It only depends on the temperature of the thermal energy reservoirs.

Answer:

Explanation:

We know that Drag force[tex]F_D[/tex]

  [tex]F_D=\dfrac{1}{2}C_D\rho AV^2[/tex]

Where

             [tex]C_D[/tex] is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

 When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

Answer:

The correct option is : d) b and c.

Explanation:

A superalloy, also known as a high performance alloy, is a class of alloys that shows the following characteristics: high mechanical strength, surface stability, resistance to thermal creep and resistance to oxidation or corrosion at high temperatures.

Therefore, a nickel base superalloy is commonly used for the high temperature gas turbine blades because of high thermal creep strength and high maximum working temperature due to its stability at high temperatures.

Answer:

Explanation:

k_max = 26.9 w/mk

  k_min  =  22.33 w/mk

Explanation:

a) the maximum thermal conductivity is given as

K_MAX = k_m v_m + k_p v_p

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.15

v_p = proportion of carbide in the cement = 0.85

[tex]K_MAX = k_m v_m + k_p v_p[/tex]

            = 66*0.15 + 20*0.85

           k_max = 26.9 w/mk

b) the minimum thermal conductivity is given as

[tex]k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}[/tex]

          = \frac{20*66}{20*0.15 +66*0.85}

        k_min  = 22.33 w/mk

Answer:

Four rules that will control or eliminate pump cavitation are:

Explanation:

With reference to the above respective rules:

Answer:

Harmful effect associated with extraction of Aluminum, Gold and Copper are:

During the melting of aluminium there is a released of per fluorocarbon are more harmful than carbon dioxide in the environment as they increased the level of green house gases and cause global warming. The process of transforming raw material into the aluminium are much energy intensive.

Gold mining industries destroyed land scopes and increased the amount of toxic level in the environment and they also dump there toxic waste in the natural water bodies, which increased the level of water pollution in the environment.

Copper mining causes the health problems like asthma and problem in respiratory system because of the inhalation of silica dust. It also increased the level of sulfur diode in the environment which cause acid rain and destroyed various trees and buildings in the nature.

Answer:

Stress is a tensor quantity and tensor quantities require 2 sub scripts for their complete definition. 1 subscript defines force and the other area.

Explanation:

Stress at a point is defined as follows

[tex]\sigma =\lim_{\Delta a\rightarrow 0}\frac{\overrightarrow{\Delta F}}{\Delta A}[/tex]

In the expression above in addition to the direction of force F we also need to define orientation of the area that we choose in defining the limit of the quantity on right hand side of the expression above hence we get a dobule subscript notation in stress at a point. 1 subscript defines force and the other area.

Answer:

δ₂ = 15.07 mm

Explanation:

Given :

When the leading edge, [tex]x_{1}[/tex] is 2.2 m, then boundary layer thickness,δ₁ = 10 mm = 0.01 m

[tex]x_{2}[/tex] = 5 m

Now we know that for a laminar flow, the boundary layer thickness is

δ = [tex]\frac{5.x}{\sqrt{Re_{x}}}[/tex] -------(1)

and Reyonlds number, Re is

               [tex]Re = \frac{\rho .v.x}{\mu }[/tex]------(2)

where ρ is density

           v is velocity

           x is distance from the leading edge

           μ is dynamic viscosity

from (1) and (2), we get

δ∝[tex]x^{1/2}[/tex]

Therefore,

[tex]\frac{\delta _{1}}{x_{1}^{1/2}}= \frac{\delta _{2}}{x_{2}^{1/2}}[/tex]

[tex]\frac{10}{2.2^{1/2}}= \frac{\delta _{2}}{5^{1/2}}[/tex]

δ₂ = 15.07 mm

Therefore, boundary layer thickness is 15.07 mm when the leading edge is 5 m.

Answer: Metals are good thermal conductors because they have close packaged metal ions in their lattice structure but in polymers they have discontinuous structure which make them poor as a conductor.

Explanation: Metals are considered as good thermal conductors because of their lattice structure which has tightly packed ions in it. In the outer shell of atoms the electrons are free to move and thus conduct the electricity but for the polymers, they have a different structure as compared to metals thus it makes it difficult for a polymer to conduct electricity due to high number of discontinuous particle chains. Therefore metals are good conductors , but polymers are not.

The answer is : True

Answer:

a. true

Explanation: