A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this refrigeration cycle b)-If the cycle is absorbing 15 kW at the -12C temperature, how much power is required? c)-If a Carnot heat pump operates between the same temperatures as the above refrigeration cycle, determine the COP of heat pump? d)-What is the rate of heat rejection at the 40°C temperature if the heat pump absorbs 15 kw at the -12 C temperature?

Answer:

the required diameter is 0.344 m

Explanation:

given data:

flow is laminar

flow of carbon dioxide Q = 0.005 Kg/s

for  flow to be laminar,  Reynold's number must be less than 2300 for pipe flow and it is given as

[tex]\frac{\rho VD}{\mu }<2300[/tex]

arrange above equation for diameter

\frac{\rho Q D}{\mu A }<2300

dynamic density of carbon dioxide = 1.47×[tex]10^{-5}[/tex] Pa sec

density of carbon dioxide is 1.83 kg/m³

[tex]\frac{1.83\times 0.0056\times D}{1.47\times 10^{-5}\times \frac{\pi}{4} \times D^{2} }<2300[/tex]

[tex]\frac{1.83\times 0.0056}{1.47\times 10^{-5}\times \frac{\pi}{4} \times 2300}= D[/tex]

D = 0.344 m

Answer:

[tex]h_{1}[/tex] = 2017.25 kJ/kg

Explanation:

GIVEN DATA:

Exit velocity [tex]v_{2}[/tex] = 250 m/s

outlet enthalpy [tex]h_{2}[/tex]= 1986 kJ/kg

inlet velocity  [tex]v_{1}[/tex]= 0

heat transfer Q = 0

from steady flow energy equation(SFEE) between inlet and exit point

[tex]h_{1}+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2}+ Q[/tex]

[tex]h_{1}=h_{2}+\frac{v_{2}^{2}}{2}[/tex]

[tex]h_{1}[/tex]=2017.25 kJ/kg

Answer: C) actuator

Explanation:

Actuator is the device that used to provides the power and manipulate the motion of the moving parts of the valve and damper is used to control the flow of the fluid. Actuator is the device or the mechanism which are used to control valve automatically and valve is a device which is used to control and regulate the fluid by rotating the flow.

Answer:

The difference between them lies in the area of the workpiece from which the material is removed. Turning is designed to remove material from the external surface of a workpiece, whereas boring is designed to remove material from the internal surface of a workpiece.

Explanation:

Answer:

c). Tresca and von-Mises

Explanation:

Tresca yield criteria states that when maximum shear stress becomes greater than the yield strength, the materials starts to yield.

Von -Mises is also known as Distortion energy theory. This theory states that failure occurs when a body is acted upon to a bi axial stresses or tri axial stresses when at any point the strain energy of distortion by unit volume of the body  equal to the specimen of the strain energy of distortion by unit volume when yielding starts in tension test.

Thus most successful and commonly used yield criteria are the Von-Mises criteria and Tresca criteria.

Answer: B- Nitrogen Tetroxide

Explanation: Except for the nitrogen tetroxide , other given all options are fuel .Nitrogen Tetroxide is a chemical compound having brownish-red color which is in liquid form having a unpleasant smell, therefore it does not belong to the category of fuel because it cannot be used as a substance for production of heat or power .

Answer:

true area of contact is 1.7 * [tex]10^{-4}[/tex] m²

Explanation:

Given data

mass (m) = 4000 tonnes

yield strength = 240 MPa i.e. = 240 * [tex]10^{6}[/tex]

base area = 0.8 m²

To find out

the true area of contact

Solution

we have given yield strength and weight

so with we can find contact area directly we know that

area is equal to weight / yield strength

so we will put weight and yield strength value in this formula

and weight = mass * 9.81 = 4 * 9.81 = 39.24 tonnes = 39240 N

area = weight /  yield strength

area = 39240 / 240 * [tex]10^{6}[/tex]

true area of contact = 1.7 * [tex]10^{-4}[/tex] m²

Answer:

The correct option is : c. Titanium

Explanation:

1. Type M tool: It belongs to the high- speed group of tool steels and used as a cutting tool material. It is a multi component alloy system Fe–C–X, where X is molybdenum.

2. Stainless steel: It is also called inox steel. It is a steel alloy with about 10.5% chromium and 1.2% carbon by mass.

3. Titanium: It is a chemical element, having atomic number 22 and mass 47.867 u. It is silver lustrous transition metal with a symbol Ti.

4. Brass: It is an alloy of zinc and copper metal.

5. Inconel: It is a family of austenitic nickel-chromium based superalloys.

Therefore, Titanium is not an alloy.

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 [tex]h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}[/tex]

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

[tex]Total\ heat\ =C_p\Delta T+\Delta h[/tex]

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  [tex]h_1=\dfrac{\Delta h}{\Delta T}[/tex]

  [tex]h_1=\dfrac{2406}{20}[/tex]

[tex]h_1=120.3\frac{KJ}{kg-m^2K}[/tex]

Now film coefficient after inclusion of sensible heat

 [tex]h_2=\dfrac{total\ heat}{\Delta T}[/tex]

 [tex]h_2=\dfrac{2,544}{20}[/tex]

[tex]h_2=127.2\frac{KJ}{kg-m^2K}[/tex]

[tex]So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100[/tex]

             [tex]=\dfrac{127.2-120.3}{120.3}\times 100[/tex]

                   =5.75 %

So Percentage change 5.75 %.

Answer and Explanation :

HARDENABILITY OF STEEL:  Hardenability of steel is related to the its ability to form martensite when it is quenched. Hardenability is the measurement of capacity that how hard would be the steel when  it is quenched.

The hadenability of steel can be measured as maximum diameter of rod which will have 50% martensite

its application in axial rods of car because it is very hard  

Answer:

26.667

Explanation:

Given Data

For Tool A

Life exponent [tex]{\ n_1}[/tex]=0.45

Constant [tex]{C_1}[/tex]=90

For tool B

Life exponent [tex]{n_2}[/tex]=0.3

Constant [tex]{C_2}[/tex]=60

and tool life equation is

[tex]VT^{n}=c[/tex]

[tex]VT_{A}^{0.45}=90[/tex]

[tex]T_{A}^{0.45}=\frac{90}{V}[/tex]

[tex]T_{A}=\frac{90}{V}^{\frac{1}{0.45}}[/tex]

[tex]For Tool B[/tex]

[tex]VT_{A}^{0.3}=60[/tex]

[tex]T_{B}^{0.3}=\frac{60}{V}[/tex]

[tex]T_{B}=\frac{60}{V}^{\frac{1}{0.3}}[/tex]

[tex]T_{A}>T_{B}[/tex]

[tex]\frac{90}{V}^{\frac{1}{0.45}}>\frac{60}{V}^{\frac{1}{0.3}}[/tex]

[tex]V>26.667[/tex]

Answer:

The temperature of the high-temperature source must increase in 12.23ºC.

Explanation:

For a Carnot engine, the efficiency is defined as:

[tex]n = 1- T2/T1 [/tex]

Where T2 and T1 are the low and the high-temperature sources respectively. Therefore for the value of T2 of 19.1ºC and the n equal to 30.7% (0.307), the T1 Temperature can be calculated as:

[tex]n = T1/T1 - T2/T1[/tex]

[tex]n = (T1-T2)/T1[/tex]

[tex]T1.n = (T1-T2)[/tex]

[tex] T1-T1.n =T2[\tex]

[tex] T1(1-n) =T2[/tex]

[tex] T1 =T2/(1-n)[/tex]

[tex] T1 = 19.1\ºC /(1-0.307)[/tex]

[tex] T1 = 27.56\ºC[/tex]

Then for the new effciencie n' of 52% (0.52) the new temeperature T1' will be:

[tex] T1' =T2/(1-n')[/tex]

[tex] T1' = 19.1\ºC /(1-0.52[/tex]

[tex] T1' = 39.79\º C[/tex]

Finally the increment of temperature is:

[tex] AT1 =T1'-T1[/tex]

[tex] AT1 =39.79\º C-27.56\º C[/tex]

[tex] AT1 =12.23\º C[/tex]

[tex] AT1 =12.23\º C[/tex]

Answer:

(a)[tex]A_1=0.26 m^2[/tex]

(b)Q= -35.69 KW

Explanation:

Given:

[tex]P_1=350 KPa,T_1=420 K,V_1=3 m/s,T_2=300 K,V_2=460 m/s[/tex]

We know that foe air [tex]C_p=1.011\frac{KJ}{kg-k}[/tex]

Mass flow rate for air =2.3 kg/s

(a)

By mass balancing [tex]\dot{m}=\dot{m_1}\dot{m_2}[/tex]

[tex]\dot{m}=\rho AV[/tex]

[tex]\rho_1A_1V_1=\rho_2A_2V_2[/tex]

[tex]\rho_1 =\dfrac {P_1}{RT_1},R=0.287\frac{KJ}{kg-K}[/tex]

[tex]\rho_1 =\dfrac {350}{0.287\times 420}[/tex]

[tex]\rho_1=2.9\frac{kg}{m^3}[/tex]

[tex]\dot{m}=\rho_1 A_1V_1[/tex]

[tex]2.3=2.9\times A_1\times 3[/tex]

[tex]A_1=0.26 m^2[/tex]

(b)

Now from first law for open(nozzle) system

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}[/tex]

Δh=[tex]C_p(T_2-T_1)\frac{KJ}{kg}[/tex]

[tex]1.011\times 420+\dfrac{3^2}{2000}+Q=1.011\times 300+\dfrac{460^2}{2000}[/tex]

Q=-15.52 KJ/s

⇒[tex]Q= -15.52\times 2.3[/tex] KW

  Q= -35.69 KW

If heat will loss from the system then we will take negative and if heat will incoming to the system we will take as positive.

Answer:

A) A1 ==0.2829 m^2

B) [tex]\frac{dQ}{dt} = -105.5 kW[/tex]

Explanation:

A) we know from continuity equation

[tex]\frac{dm}{dt} = \frac{A_1 v_1}{V_1}[/tex]

solving for A1

[tex]A_1 = \frac{\frac{dm}{dt} V_1}{v_1}[/tex]

we know V = \frac{RT}{P}  as per ideal gas equation, so we have

[tex]A_1 = = \frac{\frac{dm}{dt} \frac{RT_1}{P_1}}{v_1}[/tex]

       [tex]= \frac{2.3 \frac{0.287 \times 450}{350}}{3}[/tex]

        =0.2829 m^2

b) the energy balanced equation is

[tex]\frac{dQ}{dt} = \frac{dm}{dt} ( Cp(T_2 -T_1) + \frac{V_2^2 - V_1^2}{2})[/tex]

[tex]= 2.3 ( 1.011(300 - 450) + [\frac{460^2+3^2}{2}])[/tex]

[tex]\frac{dQ}{dt} = -105.5 kW[/tex]

Given:

Let the speed of sound be represented by 'v' then

v ∝ [tex]\sqrt{T}[/tex]              (1)

[tex]v_{1}[/tex] = 349 m/s

[tex]v_{2}[/tex] = 340 m/s

[tex]T_{1}[/tex] = 20°C = 273+20 = 293 K

Formulae used:

1)  °C = K + 273

2) K = °C - 273

3) °F = 1.8°C + 32

4) °R = °F + 459.67

Solution:

From eqn (1),

[tex]\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}[/tex]

[tex]T_{2}[/tex] = [tex]T_{2} = (\frac{v_{2}}{v_{1}})^{2}T_{1}[/tex]

[tex]T_{2} = (\frac{340}{349})^{2}{293}[/tex] = 278.08 K

Now, Usinf formula (1), (2), (3) and (4) respectively, we get

1) T = 293 K

2) T = 293 -278.8 = 5.08°C

3) T = 1.8(5.08) + 32=41.14°F

4) T = 41.14 + 459.67 = 500.81°R

Answer:

The position vector of the  point and the direction of the force define direction of the moment a force generates.

Explanation:

Moment generated by force about any point 'o' is defined by

[tex]\overrightarrow{dM}=\overrightarrow{dr}\times \overrightarrow{dF}[/tex]

The above expression being a cross product of vectors [tex]\overrightarrow{dr}[/tex] and[tex]\overrightarrow{dF}[/tex] the moment at point 'o' will depend on direction of both these vectors.

Answer:

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

Given :

Initial properties of piston  cylinder assemblies

Temperature, [tex]T_{1}[/tex] = -20°C

Quality, x = 70%

           = 0.7

Final properties of piston  cylinder assemblies

Temperature, [tex]T_{2}[/tex] = 180°C

Pressure, [tex]P_{2}[/tex] = 6 bar

From saturated ammonia tables at [tex]T_{1}[/tex] = -20°C  we get

[tex]P_{1}[/tex] = [tex]P_{sat}[/tex] = 1.9019 bar

[tex]v_{f}[/tex] = 0.001504 [tex]m^{3}[/tex] / kg

[tex]v_{g}[/tex] = 0.62334 [tex]m^{3}[/tex] / kg

[tex]u_{f}[/tex] = 88.76 kJ/kg

[tex]u_{g}[/tex] = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

[tex]v_{1}[/tex] = [tex]v_{f}[/tex]+x ([tex]v_{g}[/tex]-[tex]v_{f}[/tex]

                       = 0.001504+0.7(0.62334-0.001504)

                       = 0.43678 [tex]m^{3}[/tex] / kg

[tex]u_{1}[/tex] = [tex]u_{f}[/tex]+x ([tex]u_{g}[/tex]-[tex]u_{f}[/tex]

                       = 88.76+0.7(1299.5-88.76)

                       =936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

[tex]v_{2}[/tex] = 0.3639 [tex]m^{3}[/tex] / kg

[tex]u_{2}[/tex] = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

           [tex]w = \left (\frac{P_{1}+P_{2}}{2}  \right )\left ( v_{2}-v_{1} \right )[/tex]

           [tex]w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )[/tex]

           [tex]w = -28794.52[/tex] J/kg

                       = -28.8 kJ/kg

Now for heat transfer

[tex]q = (u_{2}-u_{1})+w[/tex]

[tex]q = (1688.2-936.27)-28.8[/tex]

          = 723.13 kJ/kg

Answer:

(B) Curve fit of p-v-t experimental data points

Explanation:

The constants a and b have positive values and are characteristic of the individual gas. The van der Waals state equation approximates the ideal gas law PV = nRT as the value of these constants approaches zero. The constant a provides a correction for intermolecular forces. The constant b is a correction for finite molecular size and its value is the volume of one mole of atoms or molecules.

Answer:

-25.63°C.

Explanation:

We know that throttling is a constant enthalpy process

      [tex]h_1=h_2[/tex]

From steal table

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

  Temperature at saturation pressure 1 bar is 99.63°C and  Temperature at saturation pressure 0.35 bar is about 74°C .

So from above we can say that change in temperature is -25.63°C.

But there is no any option for that .

Answer:

Cooling Rate=340 W

Coefficient of Performance β=3.4

Explanation:

[tex]Desired\ effect= Cooling\ Rate=Q_L= 440-100=340\ W\\ W_{net,in}=Work\ in=100\ W[/tex]

[tex]Coefficient\ of\ performance (\beta) =\frac {Desired\ Out}{Required\ In}=\frac {Cooling\ {Effect}}{Work\ In}=\frac {Q_L}{W_{net,in}}[/tex]

[tex]Coefficient\ of\ performance (\beta) =\frac {440-100}{100}=3.4[/tex]

Answer:

Explanation:

Low pressure die casting -

Also called the cold chamber die casting .

Example is -

A380 - having the composition , Al ( > 80% ) , Cu( 3 - 4% ) , Si ( 7.5 - 9.5% )

High Pressure die casting -

Also called hot chamber die casting .

Example is -  

ZAMAK 2 - having composition , Al ( 3.5 - 4.3% ) , Cu ( 2.5 - 3.5% ) , Zn( > 90% )

Low pressure die casting -

This type of die casting is perfect for the metals with  high melting point , for example aluminium . during this process , the metal is liquefied by very high temperature in the furnace  and then loaded in to the cold chamber to be injected to the die.

High Pressure die casting -

The metal is melted in a container and then a piston injects the liquid metal under high pressure into the die . low melting point metals that don not chemically attack are ideal for this die casting , example Zinc.

Answer:

Explained

Explanation:

In case case of slow cooling of a metal ingot, the micro structure is more like coarse. At the surface due to high heating rate ( surface will to exposed to higher temperature than the inner part and for longer time) the grains are small as grains will get lesser time to cool. Where as we go inside the grains will be gradually elongated. At the center we will find equiaxed grains.  

Answer: B) High chemical energy and low molecular weight

Explanation: Liquid propellant is a a single chemical compound or mix of other chemicals as well. It is used in the rocket that uses them as a major part for the fuel. A liquid propellant is supposed to have high chemical energy and low molecular weight so that is can be ignited with ease. High chemical energy can release good amount of heat in a chemical reaction and thus is good igniting compound for the liquid propellant rocket.

Answer:

The planning time of the planner machine is 100 minute

Explanation:

Planning machine

A planning machine is a metal working machine that gives a flat surface to the work piece. Here the work-piece reciprocates and the feed is given to the tool.  A planning machine is used for heavy duty work and are often large in size.

 The machining time or the planning time of a planning machine is given by,

[tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]

where, [tex]L_{w}[/tex] is the total length of travel of job

                                    = width of the job

                                    = 300 mm

            [tex]N_{s}[/tex] is number of strokes per min

                                     = 10 double strokes per min

            f is feed of the tool, mm per stroke

                                      = 0.3 mm per stroke

Therefore,  [tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]

                                            = [tex]\frac{300}{10\times 0.3}[/tex]

                                           = 100 min

Therefore the planning time is 100 minute.

Explanation  & answer:

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

Answer:

f)Density of the solid body

Explanation:

We know that heat transfer due to convection is given by

            Q=hAΔT

Where

A is the area ,which represent the geometry of solid body or surface.

h is the heat transfer coefficient which depends on

                             1.Thermal conductivity of fluid

                             2.Motion of fluid

                             3.Type of fluid flow(Laminar or Turbulent)

                             4.Viscosity of fluid

                             5.Surface condition

So from the above parameters ,we can say that heat transfer due to convection does not depends on density of the solid body.

Elastic deformation is a temporary, reversible change in a material's crystal lattice under stress, following Hooke's law, and is depicted as a linear response on a stress-strain graph. Plastic deformation results in permanent, irreversible changes in the crystal structure, typically involving dislocations and is characterized by the yield point and yield stress. Factors such as temperature and rate of stress application influence a rock's response to stress.

Deformation in materials can be categorized into two types: elastic deformation and plastic deformation. Elastic deformation refers to temporary changes in the crystal lattice that are reversible when the applied stress is removed. It follows Hooke's law, where the force is proportional to the displacement, and is shown as a linear region on a stress-strain graph. On the other hand, plastic deformation results in permanent changes to the lattice structure. Dislocations play a significant role in this process, where planes of atoms slip past one another. The moment when deformation transitions from elastic to plastic is known as the yield point, with associated yield stress. Beyond this point, deformation is irrevocable, and with continued stress, the material will eventually fracture.

At a microscopic level, plastic deformation involves the movement of dislocations and the introduction of an extra plane of atoms in the crystal structure, which allows atoms to move more easily under stress. This results in a permanently altered lattice configuration. Elastic deformation, by contrast, can be envisioned as if atoms were connected by springs that return to their original positions after the removal of stress.

The ability of rocks to deform elastically or plastically before breaking depends on several factors including temperature, water content in clay-bearing rocks, the rate at which stress is applied, and the inherent strength of the rock. A fundamental understanding of these principles is essential in geology and materials science.

Elastic deformation is reversible, maintaining lattice structure. Plastic deformation is irreversible, causing permanent lattice rearrangement.

Elastic and plastic deformation are two different responses of materials to applied stress, and they affect the crystal lattice structure differently:

1. Elastic Deformation :

  - Elastic deformation occurs when a material is subjected to stress, but it returns to its original shape and size once the stress is removed.

  - In elastic deformation, the atomic or molecular bonds within the crystal lattice are stretched or compressed, causing the material to temporarily change shape.

  - Within the elastic limit, the crystal lattice structure remains intact, and the atoms or molecules maintain their relative positions.

  - The deformation is reversible, meaning the material returns to its original state when the applied stress is released.

2. Plastic Deformation :

  - Plastic deformation occurs when a material is subjected to stress beyond its elastic limit, causing permanent changes in shape or size even after the stress is removed.

  - In plastic deformation, the atomic or molecular bonds within the crystal lattice undergo significant rearrangement or sliding.

  - Plastic deformation leads to the permanent displacement of atoms or molecules within the lattice structure, resulting in the material maintaining a new shape or size.

  - The material undergoes irreversible changes in its crystal lattice structure due to dislocation movement, grain boundary sliding, or other mechanisms.

  - Plastic deformation is characteristic of materials undergoing permanent deformation, such as metals being shaped or formed through processes like forging, rolling, or extrusion.

In summary, the difference between elastic and plastic deformation lies in the extent of the changes to the crystal lattice structure and whether the deformation is reversible or permanent. Elastic deformation involves temporary changes within the elastic limit, whereas plastic deformation involves permanent changes beyond the elastic limit.

Answer:

C.The chaotic arrangement of atoms or high hardness

Explanation:

We know that atomic arrangement in Solids are of two types

 1)Crystalline

 2)Amorphous

Crystalline arrangement have periodic arrangement where as Amorphous arrangement have random arrangement.

Generally all metal have Crystalline arrangement and material like wood ,glass have  random arrangement ,that is why wood and glass is called Amorphous.We know that wood act as a insulator for conductivity and glass is a brittle and hard material.

So from above we can say that Amorphous material have chaotic arrangement of atoms or have high harness,so our option c is right.

Answer:

b) Endothermic Chemical Reactions in a solid

Explanation:

Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.