Answer:
a)f=2.25 Hz
b)Time period T=.144 s
c)tex]V_{max}[/tex]=0.42 m/s
d)Phase angle Ф=87.3°
e) [tex]a_{max}=6.0041 [tex]\frac{m}{s^2}[/tex]
Explanation:
a)
Natural frequency
[tex]\omega _n=\sqrt {\dfrac{K}{m}}[/tex]
[tex]\omega _n=\sqrt {\dfrac{150}{0.75}}[/tex]
[tex]\omega _n[/tex]=14.14 rad/s
w=2πf
⇒f=2.25 Hz
b) Time period
[tex]=\dfrac{2π}{\omega _n}[/tex]
T=[tex]\frac{1}{f}[/tex]
Time period T=.144 s
c)Displacement equation
[tex]x=Acos\omega _nt+Bsin\omega _nt[/tex]
Boundary condition
t=o,x=0.03 m
t=0,v=.02m/s , V=[tex]\frac{dx}{dt}[/tex]
Now by using these above conditions
A=0.03,B=0.0014
x=0.03 cos14.14 t+0.0014 sin14.14 t
⇒x=0.03003sin(14.14t+87.3)
[tex]V_{max}=\omega_n X_{max}[/tex]
[tex]V_{max}=14.14\times 0.03003[/tex]=0.42 m/s
d)
Phase angle Ф=87.3°
e)
Maximum acceleration
[tex]a_{max}=(\omega _n )^2X_{max}[/tex]
[tex]a_{max}=(14.14)^20.03003[/tex]=6.0041 [tex]\frac{m}{s^2}[/tex]
Answer:
A. 2.249 hz
B. 0.45 s
C. 0.424 m/s
D. 66⁰
E. 6 m/s^2
Explanation:
Step 1: identify the given parameters
mass of the block (m)= 0.75kg
stiffness constant (k) = 150N/m
Amplitude (A) = 3cm = 0.03m
upward velocity (v) = 2cm/s
Step 2: calculate the natural frequency (F)by applying relevant formula in S.H.M
[tex]f=\frac{1}{2\pi } \sqrt \frac{k}{m}[/tex]
[tex]f=\frac{1}{2\pi } \sqrt \frac{150}{0.75}[/tex]
f = 2.249 hz
Step 3: calculate the period of the oscillation (T)
[tex]period (T) = \frac{1}{frequency}[/tex]
[tex]T = \frac{1}{2.249} (s)[/tex]
T = 0.45 s
Step 4: calculate the maximum velocity,[tex]V_{max}[/tex]
[tex]V_{max} = A\sqrt{\frac{k}{m} }[/tex]
A is the amplitude of the oscilation
[tex]V_{max} = 0.03\sqrt{\frac{150}{0.75} }[/tex]
[tex]V_{max} = 0.424(\frac{m}{s})[/tex]
Step 5: calculate the phase angle, by applying equation in S.H.M
[tex]X = Acos(\omega{t} +\phi)[/tex]
where X is the displacement; calculated below
Displacement = upward velocity X period of oscillation
[tex]displacement (X) = vt (cm)[/tex]
X = (2cm/s) X (0.45 s)
X = 0.9 cm = 0.009m
where [tex]\omega[/tex] is omega; calculated below
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{150}{0.75} }[/tex]
[tex]\omega= 14.142[/tex]
[tex]\phi = phase angle[/tex]
Applying displacement equation in S.H.M
[tex]X = Acos(\omega{t}+\phi)[/tex]
[tex]0.009 = 0.03cos(14.142 X 0.45+\phi)[/tex]
[tex]cos(6.364+\phi) = \frac{0.009}{0.03}[/tex]
[tex]cos(6.364+\phi) = 0.3[/tex]
[tex](6.364+\phi) = cos^{-1}(0.3)[/tex]
[tex](6.364+\phi)= 72.5⁰[/tex]
[tex]6.364+\phi =72.5⁰[/tex]
[tex]\phi[/tex] =72.5 -6.364
[tex]\phi[/tex] =66.1⁰
Phase angle, [tex]\phi[/tex] ≅66⁰
Step 6: calculate the maximum acceleration, [tex]a_{max}[/tex]
[tex]a_{max} = \omega^{2}A[/tex]
[tex]a_{max}[/tex] = 14.142 X 14.142 X 0.03
[tex]a_{max}[/tex] = 5.999 [tex](\frac{m}{s^{2} })[/tex]
[tex]a_{max}[/tex] ≅ 6 [tex](\frac{m}{s^{2} })[/tex]
The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 10^4 (Ω-m)1. The electron concentration is known to be 2.9x 10^22 m^-3. Given that the electron and hole mobilities are 0.14 and 0.023 m^2/N-s, respectively, calculate the hole concentration (in m^-3)
Given:
[tex]\sigma _{s} = 2.8\times 10^{4} \Omega-m[/tex]
electron concentration, n = [tex]2.9\times 10^{22} m^{-3}[/tex]
[tex]\mu _{h} = 0.14[/tex]
[tex]\mu _{e} = 0.023[/tex]
Solution:
Let holes concentration be 'p'
[tex]\sigma _{s}[/tex] = ne[tex]\mu _{e}[/tex] +pe[tex]\mu _{h}[/tex] (1)
substituting all given values in eqn (1):
[tex]2.8\times 10^{4} = 2.9\times 10^{22}\times 1.6 \times 10^{-19}\times 0.14 + p\times1.6 \times 10^{-19}\times 0.023[/tex]
The cocentration of holes is:
[tex]p = 7.432\times 10^{24} m^{-3}[/tex]
What is the approximate theoretical maximum efficiency of a heat engine receiving heat at 627°C and rejecting heat to 27°c? a)-23.2 b)-0.96 c)-0.04 d)-0.33 e)-0.66
Answer:
The correct option is (e) η = 0.66
Explanation:
given data:
temperature at entry is 627 degree C
temperature at exit is 27 degree C
the efficiency of a engine is given as η
[tex]\eta = (1-\frac{T_{L}}{T_{H}})*100[/tex]
where [tex]T_{L}[/tex] is temperature in kelvin at exit point
[tex]T_{H}[/tex] is temperature in kelvin at entry point
[tex]\eta = (1-\frac{27+273}{627+273})*100[/tex]
η = 66.66 %
Calculate the change in the enthalpy of argon, in kJ/kg, when it is cooled from 75 to 35°C. If neon had under-gone this same change of temperature, would its enthalpy change have been any different?
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 are the general principles of DFA? What are the steps to minimize the number of parts for an assembly?
Answer Explanation : The general principles for design for assembly (DFA) are,
MINIMIZE NUMBER OF COMPONENTUSE STANDARD COMMERCIALLY AVAILABLE COMPONENTS USE COMMON PARTS ACROSS PRODUCT LINESDESIGN FOR EASE OF PART FABRICATION DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITYMINIMIZE USE OF FLEXIBLE COMPONENTDESIGN FOR EASE OF ASSEMBLYUSE MODULAR DESIGNREDUCE ADJUSTMENT REQUIREDSTEPS TO MINIMIZE THE NUMBER OF PARTS
USE OF INCORPORATE HINGSUSE OF INTEGRAL SPRINGSUSE OF SNAP FITSUSE OF GUIDES BEARINGS USE OF COVERSThe slope of a moment diagram is the load. a)-True b)-False
Answer:
true.
Explanation:
but i am not 100% sure
The total energy of a system remains unchanged during a cycle. a)- True b)- False
Answer: True
Explanation: It is correct that during a cycle the energy of the system remains unchanged. Total energy consist of the initial energy and the final energy in a system and both remain equal in a system whether they are in cycle or not ans also according to the law of conservation of energy ,total energy cannot be destroyed or created. Thus the while a cycle goes on energy will remain the same.
Why metals are good thermal conductors, but polymers are not?
Answer: Metals are good thermal conductors because they have close packaged metal ions in their lattice structure but in polymers they have discontinuous structure which make them poor as a conductor.
Explanation: Metals are considered as good thermal conductors because of their lattice structure which has tightly packed ions in it. In the outer shell of atoms the electrons are free to move and thus conduct the electricity but for the polymers, they have a different structure as compared to metals thus it makes it difficult for a polymer to conduct electricity due to high number of discontinuous particle chains. Therefore metals are good conductors , but polymers are not.
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop per meter of pipe (in Pa/m)
Answer:0.1898 Pa/m
Explanation:
Given data
Diameter of Pipe[tex]\left ( D\right )=0.15m[/tex]
Velocity of water in pipe[tex]\left ( V\right )=15cm/s[/tex]
We know viscosity of water is[tex]\left (\mu\right )=8.90\times10^{-4}pa-s[/tex]
Pressure drop is given by hagen poiseuille equation
[tex]\Delta P=\frac{128\mu \L Q}{\pi D^4}[/tex]
We have asked pressure Drop per unit length i.e.
[tex]\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}[/tex]
Substituting Values
[tex]\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}[/tex]
[tex]\frac{\Delta P}{L}[/tex]=0.1898 Pa/m
The current that flows through series elements of a circuit is the same in each element. (a) True (b) False
Answer:
The given statement is true.
Explanation:
In a circuit, in which all circuit elements are connected in series, carries equal or same amount of current through each circuit element as in series circuit current have only one path through which it can flow and the voltage drop across each element depends on the values of resistances or reactances connected in the series circuit and vary depending on their respective respective resistances and reactances whereas the current flowing remain same throughout. Series circuit works on three rules:Current remains same throughout Total resistance equals the sum of individual resistances.Supply voltage equals the sum of voltage drops at each element.Describe harmful effect associated with extraction of Aluminum, Gold and Copper. Discuss each individually.
Answer:
Harmful effect associated with extraction of Aluminum, Gold and Copper are:
During the melting of aluminium there is a released of per fluorocarbon are more harmful than carbon dioxide in the environment as they increased the level of green house gases and cause global warming. The process of transforming raw material into the aluminium are much energy intensive.
Gold mining industries destroyed land scopes and increased the amount of toxic level in the environment and they also dump there toxic waste in the natural water bodies, which increased the level of water pollution in the environment.
Copper mining causes the health problems like asthma and problem in respiratory system because of the inhalation of silica dust. It also increased the level of sulfur diode in the environment which cause acid rain and destroyed various trees and buildings in the nature.
Why do we need the metering devices? List the basic types.
Answer:para comprar terreno por ejemplo
to buy land for example
Explanation: meter, centimeter
State the two Carnot corollaries (principles)
Answer and Explanation:
The two principles or corollaries of Carnot Theorem are listed below:
1). The efficiencies of all the reversible heat engines between any two thermal reservoirs working between the same temperatures will be equal to each other.
2). For every Carnot engine working between any two thermal reservoirs will have the same efficiency independent of the operating conditions and the nature of working substance. It only depends on the temperature of the thermal energy reservoirs.
Both equilibrium equations and constitutive models are needed to solve statically indeterminate problems. a)- True b)-False
Answer:
The answer is a) True
Explanation:
"Statically Indeterminate" means the number of unknowns you have exceed the number of available equations you can get from the static analysis, then the Static (equilibrium) analysis isn't enough to solve the problem.
Tracing constitutive models help to know where a load or reaction is located, also what kind of support settlements are used. This way you will know how the deflection behaves and also the momentum and torques location depending on the method you decide to use to solve the Statically Indeterminate system.
What is meant by the critical Reynolds number?
Answer:
Critical Reynolds number is a number or the threshold or the limit at which the laminar flow changes to turbulent flow.
Explanation:
It is basically the ratio of inertial forces to viscous forces and helps to predict if the flow is laminar or turbulent.
from the experiments, for flow in a pipe of diameter D, critical Reynolds number=2300
For Laminar flow: Reynolds number is less than 2000
For transitional flow: Reynolds number is in between 2000-4000, the flow is unstable
For Turbulent flow: Reynolds number is greater than 3500
Reynolds number is different for different geometries
Find the dimensions of a circular cross section steel bar subjected to tension by a force N 20000 N in two hypotheses: a) the maximum allowable stress is 150 N/mm2, b) the maximum acceptable strain is 0.0005. Take E 207GPa
Answer:
d = 13 mm
d = 15.68 mm
Explanation:
Given data
force = 20000 N
stress = 150 N/mm²
strain = 0.0005
E = 207 GPa
Solution
we know stress = force / area
so 0.0005 = 20000 / area
area = [tex]\pi[/tex]/4 × d²
put the area in stress equation and find out d
d² = 4×force / [tex]\pi[/tex] ×stress
d² = 4× 20000 / [tex]\pi[/tex] ×150
d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×150}[/tex]
d = 13 mm
and now we know starin = stress / E
same like stress we find d here
d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×0.0005×207×10³ }[/tex]
so d = 15.68 mm
Answer:
a). d = 13 mm
b). d = 16 mm
Explanation:
a). Given :
Force = 20000 N
Maximum stress, σ = 150 N/[tex]mm^{2}[/tex]
Therefore, we know that that
σ = [tex]\frac{Force}{area}[/tex]
150 = \frac{Force}{\frac{pi}{4}\times d^{2}}
150 = \frac{20000}{\frac{pi}{4}\times d^{2}}
[tex]d^{2}[/tex] = 169.76
d = 13.02 mm
d [tex]\simeq[/tex] 13 mm
b). Given :
Strain, ε = 0.0005
Young Modulus, E = 207 GPa
= 207[tex]\times[/tex][tex]10^{3}[/tex] MPa
Therefore we know that, Stress σ = E[tex]\times[/tex]ε
= 207[tex]\times[/tex][tex]10^{3}[/tex][tex]\times[/tex]0.0005
= 103.5 N/[tex]mm^{2}[/tex]
We know that
σ = [tex]\frac{Force}{Area}[/tex]
103.5 = [tex]\frac{Force}{\frac{pi}{4}\times d^{2}}[/tex]
[tex]d^{2}[/tex] = 246.27
d = 15.69 mm
d [tex]\simeq[/tex]16 mm
The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 rad/sec, what is the magnitudes of the normal component of acceleration of a point on the outer surface of the disk (0.4 m from the center) at the instant t=1 sec. a. 32. 4 m/sec^2 b. 81.8m/sec^2 c. 60.33 m/sec2 d. 4.33 m/sec2
Answer:32.4m/[tex]s^2[/tex]
Explanation:
Given data
[tex]radius\left ( r\right )[/tex]=0.4m
Intial angular velocity[tex]\left ( \omega_0\right )[/tex]=4rad/s
angular acceleration[tex]\left ( \alpha\right )[/tex]=5rad/[tex]s^2[/tex]
angular velocity after 1 sec
[tex]\omega[/tex]=[tex]\omega_0 [/tex]+[tex]\alpha\times\t[/tex]
[tex]\omega[/tex]=4+5[tex]\left ( 1\right )[/tex]
[tex]\omega[/tex]=9rad/s
Velocity of point on the outer surface of disc[tex]\left ( v\right )[/tex]=[tex]\omega_0\timesr[/tex]
v=[tex]9\times0.4[/tex] m/s=3.6m/s
Normal component of acceleration[tex]\left ( a_c\right )[/tex]=[tex]\frac{v^2}{r}[/tex]
[tex]a_c[/tex]=[tex]\frac{3.6\times3.6}{0.4}[/tex]=32.4m/[tex]s^2[/tex]
A cylindrical specimen of some metal alloy having an elastic modulus of 117 GPa and an original cross-sectional diameter of 3.6 mm will experience only elastic deformation when a tensile load of 2460 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.47 mm.
Answer:227.56 mm
Explanation:
Given data
Elastic modulus[tex]\left ( E\right )[/tex]= 117 GPa
Diameter[tex]\left ( d\right )[/tex]=3.6mm
force applied[tex]\left ( F\right )[/tex]=2460N
Area of cross-section[tex]\left ( A\right )[/tex]=[tex]\frac{\pi}{4}\times d^{2}[/tex]=10.18[tex]mm^{2}[/tex]
and change in length is given by
[tex]\Delta L[/tex]=[tex]\frac{FL}{AE}[/tex]
[tex]\Delta {0.47\times 10^{-3}}[/tex]=[tex]\frac{2460\times L}{10.180\times 117\times 10^{3}}[/tex]
L=[tex]0.47\times 10^{-3}\times 10.18\times 117\times 10^{3][/tex]
L=227.56 mm
Water flows in a pipe of diameter 450 mm at an average velocity of 3 m/s. Determine the volumetric flow rate and the mass flow rate. How long will it take to fill a tank with measurements 5 m x 6 m x 20 m?
Answer:
Volumetric flow rate = 0.4773 m³/s
Mass flow rate = 477.3 kg/s
It will take 286.38 seconds to fill a tank with measurements 5 m x 6 m x 20 m
Explanation:
Given:
Diameter of the pipe through which the water is flowing = 450 mm
Radius = Diameter/2
Thus, Radius of the pipe = 225 mm
The conversion of mm into m is shown below:
1 mm = 10⁻³ m
Radius of the pipe = 225×10⁻³ m
The area of the cross-section = π×r²
So, Area of the pipe = π×/(225×10⁻³)² m² = 0.1591 m²
Also, Given : The water flowing rate = 3 m/s
Volumetric flow rate is defined as the amount of flow of the fluid in 1 sec.
[tex]Volumetric\ flow= \frac {Volume\ passed}{Time taken}[/tex]
This, can be written as Velocity of the fluid from the cross-section area of the pipe.
Q = A×v
Where,
Q is Volumetric flow rate
A is are though which the fluid is flowing
v is the velocity of the fluid
So,
Q = 0.1591 m²×3 m/s = 0.4773 m³/s
Mass flow rate is defined as the mass of the fluid passes per unit time.
[tex]\dot {m}= \frac {Mass\ passed}{Time taken}[/tex]
The formula in terms of density can be written as:
[tex]Density=\frac{Mass}{Volume}[/tex]
So, Mass:
[tex]Mass= Density \times {Volume}[/tex]
Dividing both side by time, we get:
[tex]\dot {m}= Density \times {Q}[/tex]
Where,
[tex]\dot {m}[/tex] is the mass flow rate
Q is Volumetric flow rate
Density of water = 1000 kg/m³
Thus, Mass flow rate:
[tex]\dot {m}= 1000 \times {0.4773} Kgs^{-1}[/tex]
Mass flow rate = 477.3 kg/s
The time taken to fill the volume of measurement 5 m× 6 m× 20 m can be calculated from the formula of volumetric flow rate as:
t= Q×V
So,
Volume of Cuboid = 600 m³
Time = 0.4773 m³/s × 600 m³ = 286.38 s
Nickel superalloys are selected for high-temperature gas turbine blades because of: a)High modulus. b)High creep strength. c)High maximum working temperature. d) b and c.
Answer:
The correct option is : d) b and c.
Explanation:
A superalloy, also known as a high performance alloy, is a class of alloys that shows the following characteristics: high mechanical strength, surface stability, resistance to thermal creep and resistance to oxidation or corrosion at high temperatures.
Therefore, a nickel base superalloy is commonly used for the high temperature gas turbine blades because of high thermal creep strength and high maximum working temperature due to its stability at high temperatures.
Consider a Carnot cycle executed in a closed system with 0.0058 kg of air. The temperature limits of the cycle are300 K and 940 K, and the minimum and maximum pressures that occur during the cycle are20 kPa and 2,000 kPa. Assuming constant specific heats, determine the net work output per cycle.
Answer:0.646 KJ
Explanation:
Using First law for cycle
[tex]\sum Q=\sum W[/tex]
[tex]\sum Q=Q_{1-2}+Q_{3-4}[/tex]
For adiabatic process heat transfer is zero and for isothermal process
d(Q)=d(W)
[tex]Q_{1-2}=mRT_1\ln {\frac{P_1}{P_2}}[/tex]
Given [tex]P_1=2000KPa[/tex]
[tex]P_3=20KPa[/tex]
[tex]\left (\frac{T_2}{T_3}\right )^{\frac{\gamma }{\gamma -1}[/tex]=[tex]\left (\frac{P_2}{P_3}\right )}[/tex]
[tex]P_2=1089.06K[/tex]
[tex]Q_{1-2}=0.0058\dot 0.287\dot 940\ln \frac{2000}{1089.06}[/tex]=[tex]0.95KJ[/tex]
[tex]Q_{3-4}=mRT_2\ln {\frac{P_3}{P_4}}[/tex]
[tex]\left (\frac{T_1}{T_4}\right )^{\frac{\gamma }{\gamma -1}[/tex]=[tex]\left (\frac{P_1}{P_4}\right )}[/tex]
Now we have to find [tex]P_4=36.72KPa[/tex]
[tex]Q_{3-4}=0.0058\dot 0.287\dot 300\ln \frac{20}{36.72}[/tex]=[tex]-0.30341KJ[/tex]
[tex]Q_{net}=Q_{1-2}+Q_{3-4}[/tex]
[tex]Q_{net}=0.95-0.303=0.646KJ[/tex]
[tex]Q_{net}=W_{net}=0.646KJ[/tex]
A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False )
The answer is : True
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical example of the appropriate area to use?
Answer:
Explanation:
We know that Drag force[tex]F_D[/tex]
[tex]F_D=\dfrac{1}{2}C_D\rho AV^2[/tex]
Where
[tex]C_D[/tex] is the drag force constant.
A is the projected area.
V is the velocity.
ρ is the density of fluid.
Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.
Example:
When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.
Define full and partial well abandonment.
Answer:
Partial and Fully abandonment of well is a term used for oil well
Explanation:
Full Abandonment - This condition is depend on economics of the oil well. when production from oil well is come to that level when operating cost become higher than operating income, then it is need to shut down the well fully to prevent further loss.
Partial Abandonment - in this condition only the base part of the well is neglect permanently and from the upper portion of the well sidetracked well is build to acquire new target.
Pascal's law tells us that a. force per unit area is constant within a closed container completely filled with a fluid if measured at the same height above a datunm b. force is distributed equally on the walls of a vessel filled with a fluid. c, pressure is transmitted undiminished throughout a closed container d. a. and c e. a. and b
Answer:
d). a and c
Explanation:
According to Pascal's law, when force is applied to a fluid, the pressure will increase equally in all direction of the container. Pascal law says that the pressure is transmitted undiminised in all direction inside a closed container. And the pressure inside the closed container is constant when measured at the same height above the datum.
Hence option d is correct.
The relatonship between Kalven and celsius______
Answer:
[tex]K=C+273.15[/tex]
Explanation:
Kelvin's climbing represents the absolute temperature. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.
The Celsius scale has an artificial zero, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.
A fluid whose viscosity is unknown flow past a flare plate at a free stream velocity such that the the boundary layer thickness is 10mm at a distance of 2.2m from the leading edge. Assuming laminar flow throughout. Determine the boundary layer thickness at a distance 5m from the leading edge.
Answer:
δ₂ = 15.07 mm
Explanation:
Given :
When the leading edge, [tex]x_{1}[/tex] is 2.2 m, then boundary layer thickness,δ₁ = 10 mm = 0.01 m
[tex]x_{2}[/tex] = 5 m
Now we know that for a laminar flow, the boundary layer thickness is
δ = [tex]\frac{5.x}{\sqrt{Re_{x}}}[/tex] -------(1)
and Reyonlds number, Re is
[tex]Re = \frac{\rho .v.x}{\mu }[/tex]------(2)
where ρ is density
v is velocity
x is distance from the leading edge
μ is dynamic viscosity
from (1) and (2), we get
δ∝[tex]x^{1/2}[/tex]
Therefore,
[tex]\frac{\delta _{1}}{x_{1}^{1/2}}= \frac{\delta _{2}}{x_{2}^{1/2}}[/tex]
[tex]\frac{10}{2.2^{1/2}}= \frac{\delta _{2}}{5^{1/2}}[/tex]
δ₂ = 15.07 mm
Therefore, boundary layer thickness is 15.07 mm when the leading edge is 5 m.
What is "Double Subscript Notation for Stresses"?
Answer:
Stress is a tensor quantity and tensor quantities require 2 sub scripts for their complete definition. 1 subscript defines force and the other area.
Explanation:
Stress at a point is defined as follows
[tex]\sigma =\lim_{\Delta a\rightarrow 0}\frac{\overrightarrow{\Delta F}}{\Delta A}[/tex]
In the expression above in addition to the direction of force F we also need to define orientation of the area that we choose in defining the limit of the quantity on right hand side of the expression above hence we get a dobule subscript notation in stress at a point. 1 subscript defines force and the other area.
Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide particles in a metal matrix. Assume thermal conductivities of 20 and 66 W/m-K for the carbide and metal, respectively.
Answer:
Explanation:
k_max = 26.9 w/mk
k_min = 22.33 w/mk
Explanation:
a) the maximum thermal conductivity is given as
K_MAX = k_m v_m + k_p v_p
where k_m is thermal conductvitiy of metal
k_p is thermal conductvitiy of carbide
v_m = proportion of metal in the cement = 0.15
v_p = proportion of carbide in the cement = 0.85
[tex]K_MAX = k_m v_m + k_p v_p[/tex]
= 66*0.15 + 20*0.85
k_max = 26.9 w/mk
b) the minimum thermal conductivity is given as
[tex]k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}[/tex]
= \frac{20*66}{20*0.15 +66*0.85}
k_min = 22.33 w/mk
Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5 m above B. The pressure in the pipe at A is 70 kPa above atmospheric. Assuming no loss of energy between A and B, determine, a)- the velocity of A and B b)- Find pressure at B
Answer:
The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.
Explanation:
Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:
[tex]Aa.Va=Ab.Vb=Q[/tex]
Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:
[tex]Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3[/tex]
[tex]Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3[/tex]
Using the volume rate:
[tex]Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s[/tex]
[tex]Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s[/tex]
Assuming no losses, the energy equation for fluids can be written as:
[tex]Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb[/tex]
Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:
[tex]Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za[/tex]
Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:
[tex]Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m[/tex]
[tex]Pb = 70000\ Pa-27303\ Pa - 14715\ Pa[/tex]
[tex]Pb = 27,996\ Pa = 28\ kPa[/tex]
What are the three basic types of positive displacement pumps.
Answer:
a) reciprocating pump
b) rotary pump
c) linear pump
Explanation:
Positive displacement pump has enlarged cavity on the suction side and decreasing cavity on the discharge side and in positive displacement pump
the amount of fluid captured and inside and then discharged is same.
There are three type of positive displacement pump
a) reciprocating pump is the pump which two valve during suction outlet valve is closed and for delivery the inlet valve is closed and pump is used for discharge
b) rotary pump fluid move with the help of rotary the rotation of rotary displaces water.
c) linear pump is the pump in which displacement is linear rope and chain pump is example of this type of pump.