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.
What is “Hardenability” of steels? How it is measured? Why it is important in applications such as Axil rod of cars.
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
Saturated water vapor undergoes a throttling process from 1bar to a 0.35bar. What is the change in temperature for this process? O -4.2C O -11.3C CA-17.7C O No change in temperature for a throttling process
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 .
A workpiece of 2000 mm length and 300 mm width was machined by a planning operation with the feed set at 0.3 mm/stroke. If the machine tool executes 10 double strokes/min, the planning time for a single pass will be?
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.
Which of the following is not a fuel? a)- RP-1 b)- Nitrogen Tetroxide c)- Liquid Hydrogen d)- Methane
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 .
An air conditioner unit uses an electrical power input of 100W to drive the system and rejects 440W of heat to the kitchen air. Calculate the air conditioner's cooling rate and its coefficient of performance β.??
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]
Glass-ceramic is a fine-grained crystalline ceramic formed by the controlled crystallization of a ceramic. ( True , False )
Which of the following is/are NOT an alloy (mark all that apply)? a. Type M tool steel b. Stainless steel c. Titanium d. Brass e. Inconel
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.
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 420 K, 350 kPa, and velocity of 3 m/s. At the exit, the temperature is 300 K and the velocity is 460 m/s. Using the ideal gas model for air with constant ep=1.011 k/kg. K, determine: (a) the area at the inlet, in m2 (b) the heat transfer to the nozzle from its surroundings, in kW.
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]
A nozzle in a horizontal orientation is designed to have steady flowing steam exit it with a velocity of 250 m/s. If the outlet specific enthalpy of the steam is 1,986 kJ/kg, what is the required inlet specific enthalpy? Assume that heat transfer to the surroundings and the inlet steam velocity are negligible.
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
In determining liquid propellant performance a combination of A. High chemical energy & high molecular weight B. High chemical energy & low molecular weight C. Low chemical energy & low molecular weight D. Great taste & less filling
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.
A bronze statue weighing 4 tonnes with a base of area 0.8 m2 is placed on a granite museum floor. The yield strength of the bronze is 240 MPa. What is the true area of contact, between the base and the floor?
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
Name one aluminium alloy used in low pressure die casting and one in high pressure die casting? Explain the major reasons why one is different to the other?
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.
The speed of sound in air is proportional to the square root of the absolute temperature. If the speed of sound is 349 m/s when the air temperature is 20 °C, what is the temperature of the air when the speed of sound is 340 m/s? Give your answer in °C, K, °F, and R (Rankine).
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
Define the difference between elastic and plastic deformation in terms of the effect on the crystal lattice structure.
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.
A Carnot engine whose low-temperature reservoir is at 19.1°C has an efficiency of 30.7%. By how much should the Celsius temperature of the high-temperature reservoir be increased to increase the efficiency to 52.0%?
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]
What components determines the direction of moment?
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.
Describe the grain structure of a metal ingot that was produced by slow-cooling the metal in a stationary open mold.
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.
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat transfer surface is held at 20°C. What percentage change does the inclusion of the sensible heat correction term make to the estimated heat transfer condensing film coefficient?
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 %.
Amorphous material is characterized by by a) organized crystalline structure; b) high hardness and ductility c)the chaotic arrangement of atoms or high hardness; d) excellent magnetic, electrical properties, atomic chaotic layout, high hardness.
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.
Carbon dioxide at 20°C flows in a pipe at a rate of 0.005 kg/s. Determine the minimum diameter required if the flow is laminar (answer in m).
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
Which of the following is not an example of heat generation? a)- Exothermic chemical reaction in a solid b)- Endothermic Chemical Reactions in a solid c)- Nuclear reaction in nuclear fuel rods d)- Electric resistance heater
Answer:
b) Endothermic Chemical Reactions in a solid
Explanation:
Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.
A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.
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
The following yield criteria are dependent on hydrostatic stress (a) Maximum distortion energy and maximum normal stress (b) Tresca and Mohr-Coulomb (c) Tresca and von-Mises (d) Maximum normal stress and Mohr-Coulomb
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.
Which of the following is least likely to affect the convection heat transfer coefficient? a)- Thermal conductivity of the fluid b)-Geometry of the solid body c)-The roughness of the solid surface d)-Type of fluid motion (laminar or turbulent) e)- Fluid velocity f)- Density of the solid body g)-Dynamic viscosity of the fluid
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.
How does a 2.5 MW wind turbine costing $ 4 million compare to a 5-kw wind turbine $3 /W? a) Same $/w b) Smaller $/w c) Larger $/w
Different between boring and turning?
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:
A(n)______ is a device used to ensure positive position of a valve or damper actuator A. calibrator B. positioner C. actuator D. characteristic cam
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.
For tool A, Taylor's tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is (a) 26.7 (b) 42.5 (c) 80.7 (d) 142.9
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]
A cubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30 upward incline. The horizontal velocity of the box is 20 ft/s. Determine how far up the incline the box will travel (report center of mass distance along the inclined surface, not vertical distance)
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
_____The coefficients, i.e. a and b of van der Waals equation can be determined by (A) critical condition of the gas, (B) curve fit of p-v-t experimental data points, (C) statistical analysis
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.