Answer:
t =253.8s
Explanation:
Chvorinov's Rule can be written as:
[tex]t=B(\frac{V}{A} )^{n}[/tex]
where t is the solidification time,
V is the volume of the casting,
A is the surface area of the casting that contacts the mold,
n is a constant
B is the mold constant
The S.I. units of the mold constant B are s/m2.
According to Askeland, the constant n is usually 2.
[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3
[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2
[tex]V/A=9.198[/tex]mm
[tex]t = 3.0*9.198^2[/tex] =253.8s
Answer:
Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method:
Minimum time required at constant n = 1.5 : t = 1.408751434 minutesMaximum time required at constant n = 2.0 : t = 4.286694102 minutesExplanation:
Data:
Aluminum disc
Diameter (D) = 500 mm
Thickness = Height (h) = 20 mm
Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]
Required:
Solidification time (t) in minutes = ?
Formula:
The solidification time can be found by using the Chvorinov's Rule:
[tex]t = C (\frac {V}{A})^{n}[/tex]
Where;
t = solidification time
C = mold constant
V = Volume of disc
A = Surface area of disc
n = constant
Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.
Solution:
First, we will find the volume of the disc
disc = cylinder
therefore, Volume of cylinder is given by:
[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]
Where:
V = Volume of Cylinder
H = Height of disc
D = Diameter of disc
Now, putting dimensional values in above equation
[tex]V = \frac{\pi }{4} * 500^{2} *20[/tex]
V = 3926990.817 [tex]mm^{3}[/tex]
Second, we will find the surface area of the disc
Therefore, surface area of cylinder is given by:
[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]
Where:
A = Surface area of disc
D = Diameter of disc
H = Height of disc
Now, putting dimensional values in above equation
[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]
A = 424115.0082 [tex]mm^{2}[/tex]
Finally, Moving towards the final solution
Chvorinov's Rule with Askeland Method n = 2:Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method n = 1.5 (Minimum Solidification Time)Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]
t = 84.52508604 seconds
Converting to minutes
t = 1.408751434 minutes
Chvorinov's Rule with Degarmo Method n = 2.0 (Maximum Solidification Time)
Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change in kinetic energy of the object, in kJ?
Answer:2058.992KJ
Explanation:
Given data
Mass of object[tex]\left ( m\right )[/tex]=521kg
initial velocity[tex]\left ( v_0\right )[/tex]=90m/s
Final velocity[tex]\left ( v\right )[/tex]=14m/s
kinetic energy of body is given by=[tex]\frac{1}{2}[/tex][tex]m[/tex][tex]v^{2}[/tex]
change in kinectic energy is given by substracting final kinetic energy from initial kinetic energy of body.
Change in kinetic energy=[tex]\frac{1}{2}\times m[/tex][tex]\left ( V_0^{2}-V^2\right )[/tex]
Change in kinetic energy=[tex]\frac{1}{2}\times521[/tex][tex]\left ( 90^{2}-14^2\right )[/tex]
Change in kinetic energy=2058.992KJ
What is the no-slip condition? What causes it?
The no-slip condition is the viscosity of a fluid.This is most likely brought on by stress or stressful situations.
Answer/Explanation:
The no-slip condition for viscous fluids assumes that at a solid boundary, the fluid will have zero velocity relative to the boundary. Along with a flow when adhesion is stronger than cohesion. ... Basically the molecules of the fluid crash into the molecules of the wall and get stopped.
The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False
Answer:
(B) FALSE
Explanation:
view factor [tex]F_{ij}[/tex] depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by [tex]F_{ij}[/tex]
The use of zeroes after a decimal point are an indicator of accuracy. a)True b)- False
Answer:
True.
Explanation:
Yes, zeroes indicates the precision after the decimal point.
For smallest of the calculation in to get the precise value is very important for that the calculation can have very minute changes in decimal point as more accurate the calculation is more zeroes will be in the decimal value .
Most of the instrument calculating weights is said to precise by how much decimal they can calculate.
A spherical steel container 3 feet in diameter is buried in a land fill. The container is filled with a chemical that keeps the outer surface of the container at 100°F, whereas the earth's surface is at 50°F. Determine the heat transfer from the container if it is buried under 3 feet of earth.
Answer:
Q = 378.247 Bt/hr
Explanation:
given data:
diameter of container = 3 m
so r = 1.5 m
T1 = 50°C
T2 = 100°C
depth y = 3 ft
Heat transfer is given as Q
[tex]Q = SK\Delta T[/tex]
Where
S = Shape factor for the object
[tex]S = \frac{4\pi r}{1-\frac{r}{2y}}[/tex]
[tex]S = \frac{4\pi *1.5}{1-\frac{1.5}{2*3}}[/tex]
S = 25.132 ft
[tex]Q = SK\Delta T[/tex]
Q = 25.132*0.301 *(100-50)
Q = 378.247 Bt/hr
A mass of 7 kg undergoes a process during which there is heat transler frorn the mass at a rate of 2 kJ per kg, an elevation decrease of 40 m, and an increase in velocity from 13 m/s to 23 m/s. The specific internal energy decreases by 4 kJ/kg and the acceleration of gravity is constant at 9.7 m/s2. Determine the work for the process, in k.J
Answer:44.61 KJ
Explanation:
Let h_1,V_1,Z_1 be the initial specific enthalpy,velocity&elevation of the system
and h_2,V_2,Z_2 be the Final specific enthalpy,velocity&elevation of the system
mass(m)=7kg
Applying Steady Flow Energy Equation
[tex]m\left [ h_1+\frac{v_1^2}{2g}+gZ_1\right ][/tex]+Q=[tex]\left [ h_2+\frac{v_2^2}{2g}+gZ_2\right ][/tex]+W
[tex]h_1-h_2=4 KJ/kg[/tex]
[tex]V_1=13m/s[/tex]
[tex]V_2=23m/s[/tex]
[tex]Z_1-Z_2=40m[/tex]
substituting values
[tex]7\left [ h_1+\frac{13^{2}}{2g}+gZ_1\right ]+7\times2[/tex] = [tex]\left [ h_2+\frac{23^2}{2g}+gZ_2\right ][/tex]+W
W=[tex]7\left [h_1-h_2+\frac{V_1^2-V_2^2}{2000g}+g\frac{\left (Z_1-Z_2 \right )}{1000}\right ][/tex]+Q
W=[tex]7\left [4+\frac{13^2-23^2}{2000g}+g\frac{\left (40 \right )}{1000}[\right ][/tex]+[tex]2\times 7[/tex]
W=44.61KJ
The equation of motion is not valid without the assumption of an inertial frame. a) True b)- false
Answer:
True
Explanation:
when we write equation of motion we have to assume frame of reference as because frame of reference is the property that in frame of reference the body is not accelerated and net force acting on the body is zero.
when body is assumed to be any frame of reference we can assume the body is at rest and moving with constant speed.
Answer:
true
Explanation:
hope this helped , God bless
List fabrication methods of composite Materials.
Answer:
Fabrication method of composite materials varies for one product of material to other product. It is basically developed to meet the product requirement.
The fabrication of composite parts are depends upon different factors that are:
Depend on the Characteristics of strengthening and matrices. The details of the product and the shape and size also.Application or end uses.Types of fabrication methodologies are :
Press moldingCompression moldingContact moldOpen molding Tube rolling
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:
a)COP=5.01
b)[tex]W_{in}=2.998[/tex] KW
c)COP=6.01
d)[tex]Q_R=17.99 KW[/tex]
Explanation:
Given
[tex]T_L[/tex]= -12°C,[tex]T_H[/tex]=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
[tex]COP=\dfrac{T_L}{T_H-T_L}[/tex] ,T in Kelvin.
[tex] COP=\dfrac{261}{313-261}[/tex]
a)COP=5.01
Given that refrigeration effect= 15 KW
We know that [tex]COP=\dfrac{RE}{W_{in}}[/tex]
RE is the refrigeration effect
So
5.01=[tex]\dfrac{15}{W_{in}}[/tex]
b)[tex]W_{in}=2.998[/tex] KW
For heat pump
So COP of heat pump is given as follows
[tex]COP=\dfrac{T_h}{T_H-T_L}[/tex] ,T in Kelvin.
[tex] COP=\dfrac{313}{313-261}[/tex]
c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put
[tex]Q_R=Q_A+W_{in}[/tex]
Given that [tex]Q_A=15[/tex]KW
We know that [tex]COP=\dfrac{Q_R}{W_{in}}[/tex]
[tex]COP=\dfrac{Q_R}{Q_R-Q_A}[/tex]
[tex]6.01=\dfrac{Q_R}{Q_R-15}[/tex]
d)[tex]Q_R=17.99 KW[/tex]
Modified Reynolds Analogy is often used in the Heat Exchanger industry. a) True b) False
Answer: True
Explanation: Modified Reynolds Analogy is based on heat transfer process in the Plate Heat Exachanger's(PHE) which is done in the phases that have complex form plates shape by being changed into sheet metal. This analogy helps in the increment of the reliability and operation to be carried out easily in the industries.Thus the given statement is true.
If a point is positioned 16 inches from the origin on the x-y plane, and it is 12 inches above the x axis, what is its position in terms of the unit vectors i and j?
Answer:
[tex]p=12\hat{i}+10.58\hat{j}[/tex]
Explanation:
distance of point from the origin is 16 inch
x-axis distance =12 inch
[tex]p=\sqrt{x^2+y^2[/tex]
[tex]y=\sqrt{p^2-x^2} \\y=\sqrt{16^2-12^2}\\ y=\sqrt{112}[/tex]
y=10.58 inch
[tex]p=12\hat{i}+10.58\hat{j}[/tex]
What is the advantage to use a multistage compression refrigeration system over a single stage compression system?
Answer:
the advantages of using multistage compression refrigeration system over a single stage compression system are as follows:
Explanation:
It results in increased volumetric efficiency of compressor due to decrease in pressure ratio in each stage.Cost of operation is comparatively lowUniformity in torque is achieved thus reducing the size of the flywheel.Reduced size of condensor as a result of heat removal during condensationLower temperature at the end of compression resulting in effective lubrication and increased compressor life.List irreversibilities
Answer:
Some of the irreversibilities are listed below:
Plastic deformation of solidsTransfer of heat over finite difference of temperatureWhen two fluids are mixed together the process is irreversibleCombustion of a gasCurrent flowing through a finite resistor Diffusion and free compression or expansion of gasRelative motion of body with force of frictionProcesses involving chemical reactions(spontaneous)Cylinder cushions at the ends of the cylinder speeds the piston up at the end of the stroke. a)- True b) False
The turbine blade tip speed v for r=1m and 5 rev/sec is a) 5 m/s b) 0.25 m/s c) 0.20 m/s
Answer:
(a) 5 m/sec
Explanation:
we have given r=1 meter
angular velocity ω =5 revolution/sec
we have to find the velocity of turbine blade tip
the velocity of turbine blade tip is given by v =rω
where v = velocity of turbine blade tip
r = radius
ω = angular velocity
so v =5×1= 5 meter/sec
so the option (a) will be the correct option as in option (a) velocity is given as 5 meter/sec
What is the difference between a refrigeration cycle and a heat pump cycle?
Answer:
In refrigeration cycle heat transfer from inside refrigeration
In heat pump cycle heat transfer from environment
Explanation:
heat cycle is mechanical process use for cool the temperature but
In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator
A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's velocity is such that the average heat transfer coefficient is 230 W/m'K and the efficiency of the transmission is n=0.95 (i.e. the remainder is heat), calculate the surface temperature if the ambient temperature is 15°C. Assume that heat transfer only occurs on one face of the cube. NOTE: 1 hp=745 W
Answer:
The surface temperature is 921.95°C .
Explanation:
Given:
a=25 cm ,P=350 hp⇒P=260750 W
Power transmitted [tex]0.95\times 260750[/tex]W and remaining will lost in the form of heat.This heat transmitted to air by the convection.
h=230[tex]\frac{W}{m^2-K},\eta =0.95[/tex]
Actually heat will be transmit by the convection.
In convection Q=hA[tex]\Delta T[/tex]
So [tex]P=\Delta T\times Q[/tex]
[tex]0.05\times 260750=230\times0.25^2\(T-15)[/tex]
T=921.95°C
So the surface temperature is 921.95°C .
The area under the moment diagram is shear force. a)-True b)-False
Answer:
False
Explanation:
as we know that [tex]V(x)=\frac{dM}{dx}\\ \\=> M(x) = \int\limits^x_o {V(x)} \, dx \\\\[/tex]
=> Area under shear diagram gives the moment at any point but the reverse cannot be established from the same relation
Air is compressed in an isentropic process from an initial pressure and temperature of P1 = 90 kPa and T1=22°C to a final pressure of P2=900 kPa. Determine: a)- The final temperature of the air. b)-The work done per kg of air during the process.
Answer:
a) [tex]T_2=569.35 K[/tex]
b)Work done per kg of air=196.84 KJ/Kg
Explanation:
Given: [tex]\gamma =1.4[/tex] for air.
[tex]P_1=90 KPa ,T_=22^\circ C,P_2=900 KPa[/tex]
We know that
[tex]\dfrac{T_2}{T_1}=\left (\frac{P_2}{P_1}\right )^{\dfrac{{\gamma-1}}{\gamma}}[/tex]
So [tex]\dfrac{T_2}{295}=\left (\frac{900}{90}\right )^{\dfrac{{1.4-1}}{1.4}}[/tex]
[tex]T_2=569.35 K[/tex]
(a) [tex]T_2=569.35 K[/tex]
(b)Work for adiabatic process
W=[tex]\frac{P_1V_1-P_2V_2}{\gamma -1}[/tex]
We know that PV=mRT for ideal gas.
W=[tex]mR\frac{T_1-T_2}{\gamma -1}[/tex]
Now by putting values
work per kg of air=[tex]0.287\times \frac{295-569.35}{1.4 -1}[/tex]
Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)
So work done per kg of air=196.84 KJ/Kg
The Poisson effect does not apply to shear strains. a)True b)- False
Answer:
true
Explanation:
Shear strains and direct strains are independent components in a strain tensor at a point. Also the poisson ratio is defined as
μ = - [tex]\frac{lateral strain}{longitudinal strain}[/tex]
Which is independent of shear strains Thus this affect does not apply on shear strains
The shear force diagram is always the slope of the bending moment diagram. a)True b)- False
Answer:
True
Explanation:
Shear force diagram is a diagram which is drawn by calculating the shear force either to the right of the section or to the left of the section .
shear force is also be define as the change of moment w.r.t to distance
V=[tex]\frac{\mathrm{d}M }{\mathrm{d} x}[/tex]
where V is shear force and M is bending moment.
from the equation we can clearly say that shear force diagram is slope of bending moment diagram.
As the car gets older, will its engine's compression ratio change, if yes how.
Answer:
yes
Explanation:
yes, it is correct as car gets older the engine's compression ratio changes .
compression ratio is the volume in maximum compression chamber to the volume in the piston at full compression .
More and more engine will be used if the ring wears compression goes down
or there can be one condition if carbon build up on the piston then this can alter the compression ratio.
What does STP and NTP stands for in temperature measurement?
STP stands for standard temperature pressure and NTP stands for normal temperature pressure
Describe the basic types of chips produced in metal-cutting operations.
Answer:
Chips are of three types --
1. Continuous chip
2. Discontinuous or segmental chip
3. Continuous chip with built up edge
Explanation:
Conventional machining process always removes some excess part of the metal in the form of Chips. Every machinist should be well aware of the type of chip formed as it gives the knowledge of the machining process. The chips forms give the knowledge of --
1. Dimension of tool
2. feed rate
3. cutting speed
4. nature of tool
5. Friction between tool and work piece
the different types of chips are :
1. Continuous chips :
Continuous chips are long ribbon like coil that are bonded together. The continuous chips undergoes plastic deformation continuously. This is the most desirable form of chip produced. When such chips are formed, the cutting is smooth with good surface finish. Mostly ductile material forms continuous chips.
2. Discontinuous chips :
Discontinuous chips are formed when metals are machined and the material gets deformed easily. Brittle materials forms discontinuous chips. Discontinuous chips are in the form of loose broken chips that are not continuous. Discontinuous chips are formed when depth of cut and feed is large and cutting sped is low.
3. Continuous chips with built up edge :
These chips are similar to the continuous chips where surface finish is not smooth. When ductile materials are machined at low cutting speed, a portion of work material tends tends to stick at the rake face of the tool due to the friction between the tools and the chip. This is known as built up edge.
A student checks her car's tyre air pressure at a petrol station and finds it is 31 psi. Re-express this as an absolute pressure in kpa. Note any assumptions you make.
Answer:
Absolute Pressure=315.06256 kPa
Explanation:
Gauge pressure= 31 psi
Atmospheric Pressure at Sea level= 1 atm=101.325 kPa
[tex]1\ psi=6.89476\ kPa\\\Rightarrow 31\ psi=31\times 6.89476\\\Rightarrow 31\ psi=213.73756\ kPa=Gauge\ Pressure\\Absolute\ Pressure=Gauge\ Pressure+Atmospheric\ Pressure\\\Rightarrow Absolute\ Pressure=213.73756+101.325\\\therefore Absolute\ Pressure=315.06256\ kPa[/tex]
A free particle has kinetic energy equal to 35eV a)- What is the velocity of the particle? b)- If this velocity is known to within 0.2% accuracy, what is r of the position that particle? Assume that the mass of the particle is 2 x 10^-26 kg. Use h and give the answer in nm.
Given:
kinetic energy of free particle, KE = 35ev
1eV = [tex]1.6\times 10^{-19}[/tex] J
mass of the particle, m = [tex]2\times 10^{-26}[/tex] Kg
accuracy in velocity= 0.2%= 0.002
Solution:
a) We know that
KE = [tex]\frac{1}{2}mv^{2}[/tex]
v = [tex]\sqrt{\frac{2KE}{m}}[/tex]
⇒ v = [tex]\sqrt{\frac{2\times 35\times1.6\times 10^{-19} }{2\times 10^{-26}}}[/tex]
v = [tex]2.36\times10^{4}[/tex] m/s
b) From Heisenberg's uncertainity principle:
[tex]\Delta x\Delta p = \frac{h}{4\pi}[/tex]
[tex]\Delta x.(mv) = \frac{h}{4\pi}[/tex]
[tex]\Delta x = \frac{h}{4\pi\times 2\times 10^{-26}\times2.36 \times 10^{4}\times 0.002}[/tex]
[tex]\Delta x = 0.56nm[/tex]
If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K
Answer:
0.25 J/K
Explanation:
Given data in given question
heat (Q) = 100 J
temperature (T) = 400 K
to find out
the change in entropy of the given system
Solution
we use the entropy change equation here i.e
ΔS = ΔQ / T ...................a
Now we put the value of heat (Q) and Temperature (T) in equation a
ΔS is the entropy change, Q is heat and T is the temperature,
so that
ΔS = 100/400 J/K
ΔS = 0.25 J/K
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R. Are these measurements reasonable? Why?
The maximum efficiency must be greater than the actual efficiency of the heat engine. These measurements are not reasonable.
What is Efficiency?The efficiency is defined as the work done by the engine divided by the heat supplied.
So, maximum efficiency η = 1 - T₁/T₂ = W/Qs
Where T₁ is the lower temperature and T₂ is the higher temperature.
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R
Substitute the value into the expression , we get
η = 1 -(540 / 900) x 100 %
η = 40 %
η = Work done/heat supplied
Substitute the value into the expression , we get
η = 160/300 x 100 %
η = 53 %
So, the maximum efficiency must be greater than the actual efficiency of the heat engine.
Thus, these measurements are not reasonable.
Learn more about efficiency.
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The experimentalist's claim that the heat engine converts more work than allowed by the Carnot efficiency is not reasonable because it exceeds the theoretical maximum efficiency, violating the second law of thermodynamics.
Explanation:The experimentalist's claim that a heat engine receives 300 Btu of heat from a source at 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R can be assessed using the second law of thermodynamics and the concept of Carnot efficiency. The Carnot efficiency can be calculated using the temperatures of the hot and cold reservoirs (Th and Tc):
Carnot Efficiency (EffC) = 1 - (Tc/Th)
Assuming the temperatures are in degrees Rankine (where 0 R is absolute zero), we can calculate the maximum theoretical efficiency for a perfectly reversible engine:
EffC = 1 - (540 R / 900 R) = 1 - 0.6 = 0.4 (or 40%)
However, the experimentalist's claim states that 160 Btu is converted to work out of the 300 Btu received, which represents an actual efficiency (EffA) of:
EffA = Work done / Heat received = 160 Btu / 300 Btu = 0.5333 (or 53.33%)
Since the actual efficiency (53.33%) claimed by the experimentalist exceeds the maximum possible Carnot efficiency (40%) for a heat engine operating between these temperatures, the experimentalist's measurements are not reasonable and violate the second law of thermodynamics.
Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity
Explanation:
(a)
The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity. For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat. The SI unit of heat capacity is W/m.K.
The expression for thermal conductivity is:
[tex]q=-\kappa \bigtriangledown T[/tex]
Where,
q is the heat flux
[tex]\kappa[/tex] is the thermal conductivity
[tex] \bigtriangledown T[/tex] is the temperature gradient.
(b)
Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.
The expression for Heat capacity is:
[tex]C=\frac{E}{\Delta T}[/tex]
Where,
C is the Heat capacity
E is the energy absorbed/released
[tex]\Delta T[/tex] is the change in temperature
(c)
Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.
The expression for thermal diffusivity is:
[tex]\alpha=\frac{\kappa}{C_p \times \rho}[/tex]
Where,
[tex]\alpha[/tex] is thermal diffusivity
[tex]\kappa[/tex] is the thermal conductivity
[tex]C_p[/tex] is specific heat capacity at constant pressure
[tex]\rho[/tex] is density
What is the function of the following: 1- Oil rings 2- Flywheel 3- Timing gears
Answer:
Explanation:
The functions of the following are as follows:
1). Oil Rings:
Regulates oil within cylinder walls.Helps to keep cylinder walls lubricatedPrevent Heat transferReduce friction between piston and cylinder2). Fly Wheel:
It is used to store rotational energyIt resist any change in rotational speed due to its moment of inertiaIt helps to control the orientation of the mechanical system3). Timing gears:
It provides synchronization in the rotation of crankshaft and the camshaft so as to provide proper valve opening and closing time during each cylinder's stroke(intake and exhaust).