Solution:
Given:
[tex]T_{H}[/tex] = 1200 K
[tex]T_{L}[/tex] = 600 K
Q = 100 kJ
The Entropy change of the two reservoirs is given by the sum of entropy change of each reservoir system and is given by the formula:
[tex]\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}[/tex]
[tex]\Delta s = \frac{Q(T_{L}-T_{_{H}})}{T_{H}T_{L}}[/tex]
[tex]\Delta s = \frac{-100(600-1200)}{1200\times 600}[/tex]
[tex]\Delta s = 0.0833kJ/K
Since, the change in entropy is positive and according to the Increase in entropy principle, for any process the total change in entropy of a system is always greater than or equal to zero (with its enclosing adiabatic surrounding).
Therefore, the entropy principle is satisfied.
A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T. What is the magnitude of the maximum torque that can be applied to the coil by the magnetic field?
Answer:
Explanation:
it is given that diameter = 8.6 cm
[tex]radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m[/tex]
current =2.7 ampere
number of turns = 15
[tex]area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}[/tex]
magnetic field =0.56 T
maximum torque= BINASINΘ for maximum torque sinΘ=1
so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm
What is hot tear in casting? How can we avoid it?
Answer:
Hot tear casting is the discontinuity or failures occurs during cooling or solidification phase in casting process. It is also known as hot crack
Explanation:
In case when the casting is of weak metal, when it is hot and casted, during the process residual stresses cause failures or discontinuity in casting as it cools resulting in hot tears i,e., material casted is partly liquid and partly solid.
Preventive measures:
Ensure proper mold designA no. of gates should be used to even out temperature gradient. Over heated mold parts should not be usedProper choice of mold materialsWhich 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.
In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetnic
Answer:
The correct answer is C) Trimetric
Explanation:
The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.
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
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.
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
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.
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
Silicon is an intrinsic semiconductor. Adding a small amount of phosphorus provides extra electrons. As a resuit, phosphorus is an p-type dopant. (True , False )
It's an n-type dopants, as it makes the silicone n-type after the doping. so, False
(true/false) Moment thickness, is an index that is proportional to an increment in momentum flow due to the presence of the boundary layer. if false explain why?
Answer:
True
Explanation:
Moment thickness, is an index that is proportional to an increment in momentum flow due to the presence of the boundary layer.
Solid rockets can experience significant 2 phase flow. a) True b) False
Answer:
the answer is false solid rockets can experience significant 2 phase flow
Which of the following statements are correct? (a) A substance will emit radiation at a particular wavelength only (b) All substances emit radiation (c) Only some substances emit radiation (d) Bodies black in colour are known as black bodies
Answer: b)All substances emit radiation
Explanation: This is a universal statement that all bodies emit radiations . The radiations are usually in the form of electromagnetic radiations that are being emitted in accordance with temperature that varies from body to body.Even human bodies along with any other body emit radiations These radiations are absorbed by the black body.Therefore the correct option is option(b).
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
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.
_____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.
At the mid-plane of a plate in pure bending the stresses are minimum. a)True b)- False
Answer:
I'm pretty sure its true
Explanation:
The point where all three phases coexist on a P-T diagram.
Answer:
The point where all three phases coexist on a P-T diagram is called triple point.
Explanation:
A phase diagram of a substance is a graphical representation of the transition of its physical state under various temperature and pressure conditions.
A typical phase diagram usually consists of:
(A) Fusion curve - The curve that shows that the solid and the liquid state in in equilibrium with each other.
(B) Vaporization curve - The curve that shows that the liquid and the gaseous state in in equilibrium with each other.
(C) Sublimation curve - The curve that shows that the solid and the gaseous state in in equilibrium with each other.
(D) Triple point - This is a point in the phase diagram in which all the three state exists in equilibrium.
For example, Considering the phase diagram of water, The triple point occurs at a pressure of 0.6 kPa and 0.01⁰C.
The phase diagram is shown in image below:
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:
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.
In an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling water for every Kilogram of air. Calculate the work input for the compression stroke per kilogram of air?
Answer:
work is 50 kj
Explanation:
Given data
heat (Q) = 50 kj
To find out
work input for the compression stroke per kilogram of air
Solution
we will apply here "first law of thermodynamics" i.e.
The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.
ΔU = Q – W ................1
here ΔU is change in internal energy, Q is heat and W is work done
here U = 0 because air compressor the compression takes place at a constant internal energy in question
so that by equation 1
Q = W
and Q = 50
so work will be 50 kj
Stated as an equation, what is the Clausius Inequality?
Answer:
Clausius inequality is defined as, it applies in the cycle of real engine and there is negative entropy change. When the entropy given in the cycle during the environment is larger than entropy transferred into heat engine from hot reservoir. As, entropy of reversible system are zero.
Clausius Inequality is defined as:
[tex]R_2-R_1> \oint_{T_1}^{T_2}\frac{dQ}{T}[/tex]
Where R1 and R2 are not be equal.
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.
Oxygen enters an insulated 14.2-cm-diameter pipe with a velocity of 60 m/s. At the pipe entrance, the oxygen is at 240 kPa and 20°C; and, at the exit, it is at 200 kPa and 18°C Calculate the rate at which entropy is generated in the pipe.
Answer:
Entropy generation==0.12 KW/K
Explanation:
[tex]s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}[/tex]
[tex]s_2-s_1=0.891\ln \frac{291}{293}-0.2598\ln \frac{200}{240}[/tex]
[tex]s_2-s_1=0.0412\frac{KJ}{kg-K}[/tex]
Mass flow rate= [tex]\rho\times\dfrac{\pi}{4}d^2V[/tex]
[tex]\rho_1=\dfrac {P_1}{RT_1}[/tex]
[tex]\rho_1=\dfrac {240}{0.2598\times 293}[/tex]
[tex]\rho_1=3.51\frac{kg}{m^3}[/tex]
mass flow rate=[tex]\rho_1A_1V_1[/tex]
So by putting the values
Mass flow rate=2.97 kg/s
So entropy generation=(2.97)(0.0412)
=0.12 KW/K
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]
Glass-ceramic is a fine-grained crystalline ceramic formed by the controlled crystallization of a ceramic. ( True , False )
Segmented solid rockets require what special component? A. Igniters B. Liners C. O-rings D. Nozzle gimbals
Answer: D) Nozzle gimbals
Explanation: Igniters are present for the ignition power in the segmented solid rocket, liners are the basic requirement in the engine of the rocket,O-rings are also a general part of the segmented solid rocket .So, these are the general purpose parts of the segmented solid rocket but nozzle gimbals are the special units of the rocket as they provide improved thrust in various direction in accordance with the rocket.So, option (d) is the correct option.
Answer:
D. Nozzle gimbals
Explanation:
Segmented solid rockets require the special component, nozzle gimbals.
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
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]