Provide main reasons for the short shot during the injection molding.

Answer:

a) 4365 kJ

Explanation:

In any thermodynamic system, any heat change is accompanied by the change in temperature.  The relation between heat released/gained in a system and the temperature is:

Q=mcΔT

where,

Q is the amount of heat absorbed or released

m is the mass

ΔT is the change in temperature

c is called the specific heat.  

Specific heat is defined as heat gained by 1 unit mass of any sample to raise the temperature of the sample by 1 °Celsius.

Thus, from the question:

Mass of aluminum =5 kg

Final temperature = 1000°C

Initial temperature = 30°C

ΔT = (1000 -30)°C = 970°C

Specific heat of aluminum = 900 J/kg°C

Thus, Amount of heat required:

Q = 5 kg×900 J/kg°C×970°C = 4365000 J

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,  Heat gained by aluminum =4365000 ×10⁻³ J = 4365 kJ

Answer:

Probaility

Explanation:

The reliability of a component, system or unit can be defined as the probability that that component can operate for a certain period of time (mission time) without losing its function.

Answer:

0.740833917 ton/hr

Explanation:

Given:

Cooling load, 8890.007 Btu/hr = 2.605 kW

Room size = 180 [tex]ft^{2}[/tex]

According to the thumb rule

1 ton of refrigerant = 12000Btu

Hence for 8890.007 Btu/hr,

the mass flow rate of the refrigerant is =8890.007 / 12000

                                                                = 0.740833917 ton per hr

Hence, mass flow rate is 0.740833917 ton/hr

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

Understanding analytical methods is essential in systems engineering for problem-solving and system comprehension, with real-world applications in areas like environmental monitoring.

The importance of analytical methods in systems engineering management is crucial for problem-solving and comprehending the physics of the situation. Analytical methods help in selecting the right system and developing solutions efficiently. For instance, in the field of environmental monitoring programs, the use of analytical methods aids in assessing system health and trends accurately.

Answer:

Explanation:

Ionic bonding are formed when two opposite ion attract each other. Ionic bond is generally between metal and non metal. Example NaCl

Covalent bonding is the bonding which is formed by the sharing of electron pair covalent bond is generally between two non metal.Example H₂O

Ionic bond is generally electrostatic in nature which lead to attraction of positive and negative ion

ionic bond does not have definite  shape where covalent has definite shape

melting point of covalent bond is low where as ionic bond have high melting point.

Given:

[tex]T_{1}[/tex] = 50°C = 273+50 =323 K

[tex]T_{2}[/tex] = 300°C = 273+300 =573 K

Solution:

We know that:

specific heat for gold, c = 0.129 J/g°C

Also, change in entropy, ΔS is given by:

[tex]\Delta S = cln\frac{T_{f}}{T_{i}}[/tex]

After the bars brought in contact with each other,

final temperature, [tex]T_{f}[/tex] = [tex]\frac{T_{1}+T_{2}}{2}[/tex]

final temperature, [tex]T_{f}[/tex] = [tex]\frac{323+573}{2}[/tex] = 448K

Now, entropy for first gold bar, using eqn-1

[tex]\Delta S = cln\frac{T_{f}}{T_{1}}[/tex]

[tex]\Delta S_{1} = 0.129ln\frac{448}{323}[/tex] =0.042 J/K

[tex]\Delta S_{2}[/tex] = 0.129ln [tex]\frac{448}{573}[/tex] = - 0.032 J/K    

Total entropy generation,

[tex]\Delta S_{1}[/tex] = [tex]\Delta S_{1}[/tex] + [tex]\Delta S_{2}[/tex]

[tex]\Delta S[/tex] = 0.042 + (- 0.032) = 0.010 J/K

Answer:

(b) vpltage

Explanation:

we know the expression for voltage across the inductor V=L[tex]\frac{di}{dt}[/tex] which clearly shows voltage across the inductor is directly proportional to rate of change of current similarly current across the capacitor I=C[tex]\frac{dv}{dt}[/tex] from the expression we can see that current across the capacitor is directly proportional to rate of change of voltage. so from above discussion it is clear that response of an iductor to current is similar to response of capacitor to voltage

Answer: d) property tests

Explanation: Product service life can be referred as the life that define the service that can be provided by the product manufactured.The service life contains the testing and calculation of the product's quality, reliability, maintenance factor etc. These factors are known as the property of the product and so is calculated by the property test. Therefore option (d) is the correct option because other option does not define the factors for defining the product service life.

Answer:

Equilibrium of a system is defined as when the concentration of reactants and products of a chemical reaction are in equilibrium and their ratio does not vary. In another words we can say that, when the forward reaction rate is similar to the reverse reaction rate.

Thermodynamics equilibrium is defined as in an isolated system when there is no change in energy and entropy. It is determined by the its intensive properties like volume and pressure.  

Answer:

a)initial quality of water 0.724.

b)mass of water =2.45 kg

c)T=70°F

Explanation:

h=732.5 Btu/lbm

 h=763.33 KJ/kg      (1 Btu=1.05 kj)

T=70°F⇒T=21.1°C

[tex]V=150 in^3=0.002458m^3[/tex]

(a)From steam table

Properties of saturated steam at 21.1°C  

 [tex]h_f= 88.56\frac{KJ}{Kg} ,h_g= 2539.49\frac{KJ}{Kg}[/tex]

To find dryness fraction

[tex]h=h_f+x(h_g-h_f)\frac{KJ}{Kg}[/tex]

[tex]763.33=88.56+x(2539.49-88.56)\frac{KJ}{Kg}[/tex]

x=0.27

So initial quality of water 0.724.

(b)

[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]

where v is specific volume

From steam table at 21.1°C  

[tex]v_f= 0.001\frac{m^3}{Kg} ,v_g= 54.13\frac{m^3}{Kg}[/tex]

V=[tex]m_f\times v_f[/tex]

0.002458=[tex]m_f\times 0.001[/tex]

So mass of water =2.45 kg

(c)

Actually water will take latent heat ,it means that heating of will take place at constant temperature and constant pressure.So we can say that final temperature of water will remain same (T=70°F).

Answer:

96.1%

Explanation:

We know that lift force

[tex]F_L=\dfrac{1}{2}C_L\rho AV^2[/tex]

                                                                    ------------(1)

Where [tex]C_L[/tex] is the lift force coefficient .

          ρ is the density of fluid.

         A is the area.

        V is the velocity.

Now when speed is increased by 2 % and all other parameter remains constant except [tex]C_L[/tex] .

Let;s take new value of lift force coefficient is [tex]C_L'[/tex] .

[tex]F_L=\dfrac{1}{2}C_L'\rho A(1.02V)^2[/tex]

                                                                         -----------(2)

Now from equation 1 and 2

[tex]C_L\times V^2=C_L'\times1.0404 V^2[/tex]

⇒[tex]C_L'=0.961C_L[/tex]

So we can say that revised value of  lift force coefficient is 96.1% of original value.

Answer:  Exploration includes plethora of activities and depend upon the kind  of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

Answer:

Yes ,at low temperature ductile material behaves like brittle.

Explanation:

Yes,as temperature decreases a ductile material can become brittle.

In metals ductile to brittle transition temperature is around 0.1 to 0.2 Tm(melting point temperature) and in ceramics  ductile to brittle transition temperature is around 0.5 to 0.7 Tm(melting point temperature) .

We can easily see that from graph between fracture toughness and  temperature.In the graph when temperature is low then the ductile material is behaving like brittle material.But when temperature is above a particular value then material behaves like ductile.

Answer:

From the multiple choices provided for the action of capacitor, option

(d) stores electrical energy

is correct

Explanation:

A capacitor is basically a two terminal device that stores electrical energy in the electric field in its vicinity. It is apassive element.

The property of a capacitor to store electrical energy or the effect of a capacitor is known as its capacitance. The capacitance of a capacitor is given by:

Q = CV or C = [tex]\frac{Q}{V}[/tex]

It is originally known as condensor and it can be said that a capacitor adds capacitance to the circuit.

Answer:

(b) kJ/kg

Explanation:

The ratio of amount of energy required to change the temperature of the substance by certain magnitude and this magnitude of temperature change is known as heat capacity of the substance.

The expression for Heat capacity is:

C=E/ΔT

Where,

C is the Heat capacity

E is the energy absorbed/released

ΔT is the change in temperature

The SI unit of heat capacity is J/K.

(a) Tons represents the unit of mass (1000 kg)

(c) kW represents the unit of power (1000 W)

(d) Btu represents the unit of heat (1055 J)

The units from the options that can be a unit of heat capacity is (b) kJ/kg.

Answer:

11.34 g/cm3

Explanation:

At room temperature, where it is in a solid state, it is 11.34 [tex]\frac{g}{cm^{3}}[/tex]. While at melting temperature, at 327.5 ° C, it is 10.66 [tex]\frac{g}{cm^{3}}[/tex]

Answer:

a) True

Explanation:

Roller can provide reaction for push support but it can not provide reaction for pull support.  

From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.

On the other hand hinge support can provide reaction in both the direction.

So we can say that roller can not proved reaction for pull support.

Given:

Outer Diameter, OD = 90 mm

thickness, t = 0.1656 mm

Poisson's ratio, [tex]\mu[/tex] = 0.334

Strength = 320 MPa

Pressure, P =3.5 MPa

Formula Used:

1).  [tex]Axial Stress_{max}[/tex] = [tex]P[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}][/tex]

2). factor of safety, m = [tex]\frac{strength}{stress_{max}}[/tex]

Explanation:

Now, for Inner Diameter of cylinder, ID = OD - 2t

ID = 90 - 2(1.65) = 86.7 mm

Outer radius,  [tex]r_{o}[/tex] = 45mm

Inner radius,  [tex]r_{i}[/tex] = 43.35 mm

Now by using the given formula (1)

  [tex]Axial Stress_{max}[/tex] = [tex]3.5[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}][/tex]

  [tex]Axial Stress_{max}[/tex] = [tex]3.5\times 26.78[/tex] =93.74 MPa

Now, Using formula (2)

factor of safety, m = \frac{320}{93.74} = 3.414

Answer:

C.30°

Explanation:

Given that compound swivel base is set on 60° at the lathe center line index.

We need to find reading on cross slide index

We know that relationship between center line index and cross slide index in angle 2∝=β

Where ∝ Angle of swivel and β is the reading on cross line index.

So by using above  relationship between center line index and cross slide index  

2∝=60°⇒∝=30°

So our option is C.

Answer:

yes it is possible

Explanation:

dislocation are if two type edge and screw dislocations

edge dislocation is a defect where an extra half plane is inside the lattice.

and screw dislocation is one in which can be assumed as the first half of the crystal slips over another.

These dislocation can coexist together where the line direction and burger vectors are  neither parallel nor perpendicular then at that condition both dislocation screw and edge will coexist

Answer:

Tensile stress is 14.15 N/[tex]mm^{2}[/tex]

Explanation:

When any object is subjected to an external force, the body offers a resisting force which is equal and opposite to the external load. This resisting force is called stress. Thus stress is defined as the force acting perpendicular to the given cross sectional area of the object.

Mathematically,  stress , σ = [tex]\frac{force }{area}[/tex]

Given : Tensile force, F = 400 N

            Diameter of the rod, d = 6 mm

            Area of the rod is given by, A = [tex]\frac{\pi }{4}\times d^{2}[/tex]

                                                              = [tex]\frac{\pi }{4}\times 6^{2}[/tex]

                                                              =28.27 [tex]mm^{2}[/tex]

Therefore, the tensile tress is,   σ = [tex]\frac{force }{area}[/tex]

                                                        = [tex]\frac{400 }{28.27}[/tex]

                                                       = 14.149 N/[tex]mm^{2}[/tex]

                                                       [tex]\simeq[/tex] 14.15 N/[tex]mm^{2}[/tex]

Thus, tensile stress experieced by the rod is 14.15 N/[tex]mm^{2}[/tex]

Answer:

T = 1.17 x [tex]10^{-3}[/tex] N-m

Explanation:

Given :

Gap between the two plates , dy = 1 mm

                                                  dy = 1 x [tex]10^{-3}[/tex] m

Angular velocity of the top plate , ω = 5 rad/s

Diameter of the plate, D = 20 cm

Radius of the plate, R = 10 cm

                                    = 0.1 m

Temperature of the kerosene = 40°C

Viscosity of kerosene at 40°C = 0.0015 Pa-s

Now let us take a small elemental ring of thickness dr at a radius r.

Therefore, area of this elemental ring of dr = 2πrdr

Now linear velocity at radius r = ω x r

                                                   5r m/s

Now applying Newtons law of viscosity we get,

Shear stress, τ = μ.[tex]\frac{du}{dy}[/tex]

   [tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{du}{dy}[/tex]

  [tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{5r-0}{1\times 10^{-3}}[/tex]

  [tex]\Rightarrow \frac{F_{s}}{A}=\mu \times 10^{3}\times 5r[/tex]

  [tex]F_{s}=\mu \times 10^{3}\times 5r\times 2\pi rdr[/tex]

  [tex]F_{s}=5 \times 10^{3}\times \mu \times r\times 2\pi rdr[/tex]

  [tex]F_{s}=\frac{18849}{400}\times r^{2}dr[/tex]

Now we know torque due to small strip,

 dT = [tex]F_{s}[/tex] x r

 dT = [tex]\frac{18849}{400}\times r^{3}dr[/tex]

Therefore total torque for r=0 to r=R can be calculated. So by integrating,

[tex]\int dT=\int_{0}^{R}\frac{18849}{400}\times r^{3}dr[/tex]

[tex]T = \frac{18849}{400}\times \frac{R^{4}}{4}[/tex]

[tex]T = 47.1225\times \frac{0.1^{4}}{4}[/tex]

T = 1.17 x [tex]10^{-3}[/tex] N-m

Answer:690.21 mm of Hg

Explanation:

Given data

depth of lake[tex]\left (h\right )[/tex]=237.3m

density of lake water[tex]\left ( \rho\right )[/tex]=[tex]989kg/m^3[/tex]

acceleration due to gravity[tex]\left ( g\right )[/tex]=[tex]9.806m/s^2[/tex]

air  above water has a barometric  pressure of 29.83in of Hg=711.2 mm of Hg

Altitude correct for pressure=1.87in of Hg=47.5 mm of Hg

Absolute pressure above water =711.2-47.5mm of Hg=663.7 mm of Hg

Absolute pressure at bottom of Lake=Pressure at surface level +[tex]\left ( \rho \times g\times h\right )[/tex]

Absolute pressure at bottom of Lake=663.7+[tex]\frac{989\times 9.81\273.3}{10^{5}}[/tex]

Absolute pressure at bottom of Lake=690.21 mm of Hg

Answer:

Explanation:

Molten metal is forced through a small orifice and is shatter by a jet of compressed air,inert gas .

In Atomization, the particles shape is analysed  by the rate of solidification and varies from spherical to highly irregular shape.

oxide of metals are transformed to pure metal powder when undefended to under melting point gases results in a product of spongy material.

It is used for Iron,copper,tungsten,Nickel etc.

Answer:

(1)Atomizing process

 (2)Gaseous reduction

Explanation:

The first step in powder metallurgy is the production of powder,because the property of the final product depends on the powder.

The methods for the production of powder are as follow

                    (1)Atomizing process

                    (2)Gaseous reduction

(1)Atomizing process:

     In the Atomizing process the molten metal is passing through an orifice into a stream of inert gas.Due to this rapid cooling of metal occurs and then it will in very fine particle .

(2)Gaseous reduction:

In this process powder is producing by grinding of metallic oxide to a fine state,after that reducing it by carbon mono oxide.

Answer Explanation:

CASCADE REFRIGERATION SYSTEMS :

cascade refrigeration systems are commonly used in the liquefaction of natural gas and some other gases

examples of cascade refrigeration system: LNG (Liquefied natural gas ) plants mostly used cascade refrigeration system

ADVANTAGES OF CASCADE REFRIGERATION SYSTEMS:

A cascade refrigeration system utilizes multiple linked refrigeration cycles to achieve ultra-low temperatures not possible with a single cycle. It features a cascade control mechanism for precise temperature management, using heat exchangers to interconnect and sequentially cool each stage.

A cascade refrigeration system is a complex refrigeration strategy that employs multiple refrigeration cycles linked together to achieve cooling. Typically, these systems are utilized in environments requiring very low temperatures, which cannot be efficiently or cost-effectively achieved through a single refrigeration cycle. The essence of a cascade system lies in its deployment of two or more refrigeration units in series, each referred to as a "stage," with each successive stage operating at a lower temperature range than its predecessor.

The interconnection between these stages involves a heat exchanger, where a refrigerant from one cycle cools the condenser of the next lower temperature cycle. This process is crucial for achieving the desired low temperatures. One common application is in industrial refrigeration, where maintaining precise low temperatures is critical for process efficiency and product quality.

A key feature of cascade refrigeration systems is the use of cascade control, which enhances system responsiveness to temperature changes and maintains tight control over the process. This control mechanism consists of a master-slave relationship, where the output of one control loop (master) serves as the setpoint for the next (slave), creating a highly integrated control environment. Such setup enables rapid adjustment to changing operating conditions, ensuring the system swiftly responds to disturbances while maintaining the desired temperature levels.

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

[tex]P_{1}[/tex] = 5 Mpa

[tex]T_{1}[/tex] = 1623°C

                       = 1896 K

[tex]V_{1}[/tex] = 0.05 [tex]m^{3}[/tex]

Also given [tex]\frac{V_{2}}{V_{1}} = 20[/tex]

Therefore, [tex]V_{2}[/tex] = 1  [tex]m^{3}[/tex]

R = 0.27 kJ / kg-K

[tex]C_{V}[/tex] = 0.8 kJ / kg-K

Also given : [tex]P_{1}V_{1}^{1.25}=C[/tex]

   Therefore, [tex]P_{1}V_{1}^{1.25}[/tex] = [tex]P_{2}V_{2}^{1.25}[/tex]

                     [tex]5\times 0.05^{1.25}=P_{2}\times 1^{1.25}[/tex]

                     [tex]P_{2}[/tex] = 0.1182 MPa

a). Work transfer, δW = [tex]\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]

                                  [tex]\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}[/tex]

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = [tex]\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]

  =[tex]\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W[/tex]

  =[tex]\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200[/tex]

  = 197.7 kJ

Answer:

The correct answer is option B i.e. 86 f

Explanation:

The freezing point of water is 32°f and boiling point is 212 °f.  Thus difference in boiling and freezing point is exactly come out to be 180°. therefore fahrenheit scale is 1/180 interval of both scale ( freezing and boiling point)

given data:

absolute temperature is 303 K

We know by general formula of temperature in fahrenheit

°F = [tex]\frac{9}{5}[/tex] × (K-273) +32

°F = [tex]\frac{9}{5}[/tex] × (303-273) +32

°F =  86 F

Answer:

Heat loss 67%.

Explanation:

We know that heat transfer

    Q=AUΔT

Where U is the overall heat transfer coefficient ,A is the area and ΔT is the temperature difference.

Now heat transfer in terms of U-factor

[tex] Q_1=AU_1ΔT[/tex]

[tex] Q_2=AU_2ΔT[/tex]

Given that temperature difference is same in both condition so

[tex]\dfrac{Q_1}{U_1}=\dfrac{Q_2}{U_2}[/tex]

[tex]\dfrac{Q_1}{Q_2}=\dfrac{U_1}{U_2}[/tex]

[tex] heat\ loss=\dfrac{Q_2-Q_1}{Q_1}[/tex]

[tex] heat\ loss=\dfrac{U_2-U_1}{U_1}[/tex]

Given that[tex]U_2=0.33U_1[/tex]

[tex] heat\ loss=\dfrac{0.33U_1-U_1}{U_1}[/tex]

Heat loss 67%.

Resistance zero meaning superconductor, so True.