The mechanical dynamometer is an instrument used to measure forces or to calculate the weight of objects. The traditional dynamometer, invented by Isaac Newton, bases its operation on the stretching of a spring that follows the law of elasticity of Hooke in the measurement range. Like a scale with elastic spring, it is a spring scale, but it should not be confused with a scale of saucers (instrument used to compare masses).
These instruments consist of a spring, generally contained in a cylinder that in turn can be inserted into another cylinder. The device has two hooks or rings, one at each end. The dynamometers have a scale marked on the hollow cylinder that surrounds the spring. When hanging weights or exerting a force on the outer hook, the cursor of that end moves on the external scale, indicating the value of the force.
The dynamometer works thanks to a spring or spiral that has inside, which can be lengthened when a force is applied on it. A point or indicator usually shows, in parallel, the force.
For all substances, Cp>C. Why is that?
What is the name of the instrument that measures or senses the vibration and is commonly referred to as a pickup or sensor.
Answer Explanation :
ACCELEROMETERS: Accelerometers are used for sensing the vibration. basically accelerometers are used for measuring high as well as low frequency. when a transducer is used in addition with another device to measure vibration is called pickups. Seismic instruments are commonly used vibration pickups
PROXMITY PROBE: Proximity probe is used for the measurements of vibration the vibration sensitivity is highest around 8 to 16 hertz
The Reynolds number is the major parameter that relates fluid flow momentum to friction forces. How is the Reynolds number defined? How does the boundary layer thickness vary with Reynolds number?
Answer:
Reynolds number determines whether a flow is laminar or turbulent flow.
Explanation:
Reynolds number is defined as ratio of inertia force to the viscous force. it is a dimension less number. Reynolds number is used to describe the type of flow in a fluid whether it is laminar flow or turbulent flow. Reynolds number is denoted by Re.
When Reynolds number is in the range of 0 to 2000, the flow is considered to be laminar.
When Reynolds number is in the range of 2000 to 4000, the flow is considered to be transition.
And when Reynolds number is more than 4000, the flow is turbulent flow.
The boundary layer thickness for a fluid is given by
δ = [tex]\frac{5\times x}{\sqrt{Re}}[/tex]
where δ is boundary layer thickness
x is distance from the leading edge
Re is Reynolds number
Thus from the above boundary layer thickness equation, we can see that the boundary layer thickness varies inversely to square root of reynolds number.
Typical metals show elastic-plastic behaviour in tension and shear but not in compression. a)True b)- False
Answer: True
Explanation:Typical metals have a property of ductility and malleability that is metals can be drawn into wires or any other shape by beating or stretching the metal by putting the tensile strength or shear strength that pulls them apart . But while compression the metals are squeezed together which affects the hardness of a metal and they are not able to bear the compression force well and thus cannot show elastic-plastic behavior while compression .Therefore the statement given is true typical metals show elastic-plastic behavior in tension and shear but not in compression.
A particle moves along a straight line such that its acceleration is a=(4t^2-2) m/s, where t is in seconds. When t = 0, the particle is located 2 m to the left of the origin, and when t = 2, it is 20 m to the left of the origin. Determine the position of the particle when t=4s.
Answer with Explanations:
We are given:
a(t)=4*t^2-2............................(1)
where t= time in seconds, and a(t) = acceleration as a function of time.
and
x(0)=-2 .................................(2)
x(2) = -20 ............................(3)
where x(t) = distance travelled as a function of time.
Need to find x(4).
Solution:
From (1), we express x(t) by integrating, twice.
velocity = v(t) = integral of (1) with respect to t
v(t) = 4t^3/3 - 2t + k1 ................(4)
where k1 is a constant, to be determined.
Integrate (4) to find the displacement x(t) = integral of (4).
x(t) = integral of v(t) with respect to t
= (t^4)/3 - t^2 + (k1)t + k2 .............(5) where k2 is another constant to be determined.
from (2) and (3)
we set up a system of two equations, with k1 and k2 as unknowns.
x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 ......................(6)
substitute (6) in (3)
x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20
16/3 -4 + 2k1 -2 = -20
2k1 = -20-16/3 +4 +2 = -58/3
=>
k1 = -29/3 ....................................(7)
Thus substituting (6) and (7) in (5), we get
x(t) = (t^4)/3 - t^2 - 29t/3 + 2 ..............(8)
which, by putting t=4 in (8)
x(4) = (4^4)/3 - (4^2 - 29*4/3 +2
= 86/3, or
= 28 2/3, or
= 28.67 (to two places of decimal)
The answer is "28.87 m" and the further calculation can be defined as follows:
Given:
[tex]\to a=(4t^2-2)\ \frac{m}{s}\\\\ [/tex]
When
[tex] \to t=0 \ \ \ \ \ \ \ v= 2\ m\\\\ \to t=2 \ \ \ \ \ \ \ v= 20\ m\\\\ \to t=4\ \ \ \ \ \ \ v=?[/tex]
To find:
value=?
Solution:
[tex]V(t)=\int a(t)\ dt=\int (4t^2-2)\ dt=\frac{4}{3}t^3-2t+C_1\\\\ x(t)=\int V(t)\ dt =\int (\frac{4}{3}t^3-2t+C_1)\ dt=\frac{1}{3}t^4-t^2+C_1t+C_2\\\\ x(0)=-2=\frac{1}{3} 0^4-0^2+C_1(0)+C_2\to C_2=-2\ m\\\\ x(2)=-20=\frac{1}{3} 2^4-2^2+C_1(2)-2\to C_1=-\frac{29}{3}\ \frac{m}{s}\\\\ x(4) = \frac{1}{3}4^4-4^2 -\frac{29}{3}4-2 = 28.87 \ m \\\\[/tex]
The particle will be at 28.87 m at the right of the origin.
Learn more about velocity:
brainly.com/question/24376522
Define the isentropic efficiency for each of the following 3. a. i. Adiabatic turbine ii. Adiabatic compressor iii. Adiabatic nozzle
Answer:
a)[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
Explanation:
a)
Adiabatic turbine
Adiabatic turbine means turbine can not reject or take heat from surrounding.
Isentropic efficiency of turbine can be define as the ratio of actual work out put to the Ideal or isentropic work out put.Ideal means when turbine will give maximum work and there is no any friction losses we can say when process is isentropic.
Isentropic efficiency of turbine=(Actual work output)/(Ideal work output)
[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)
Adiabatic compressor
Isentropic efficiency of compressor can be define as the ratio of ideal or isentropic work in put to the actual work in put.
Isentropic efficiency of turbine=(Ideal work input)/(actual work input)
[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)Adiabatic nozzle
We know that nozzle is device which used to accelerate the fluid.
Basically it covert pressure energy to kinetic energy.
Isentropic efficiency of nozzle can be define as the ratio of actual enthalpy drop put to the Ideal or isentropic enthalpy drop.
[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
A mass of 2 kg is suspended from a vertical spring of stiffness 15 kN/m and subject to viscous damping of 5 Ns/m. What is the amplitude of the forced oscillations produced when a periodic force of amplitude 25 N and angular frequency 100 rad/s acts on the mass? What is the maximum force transmitted to the support of the spring?
Answer:
Amplitude of A is 4.975 mm and total force is 94.3 N
Explanation:
given data in question
mass (m) = 2 kg
stiffness (k) = 15 kN/m
viscous damping (c) = 5Ns/m
amplitude (F) = 25 N
angular frequency (ω) = 100 rad/s
to find out
amplitude of the forced and maximum force transmitted
Solution
static force for transmitted is mg i.e 2 × 9.81 = 19.6 N .............. 1
we know the amplitude formula i.e.
Amplitude of A = amplitude / [tex]\sqrt{c^{2}\omega^{2} + (k - m \omega^{2})^{2}[/tex]
now put the value c k m and ω and we find amplitude
Amplitude of A = 25 / [tex]\sqrt{5^{2} * 100^{2} + (15000 - 2 * 100^{2})^{2}[/tex]
Amplitude of A = 4.975 mm
now in next part we know the maximum force value when amplitude is equal displacement i.e.
maximum force = amplitude of A [tex]\sqrt{k^{2}+c^{2}\omega^{2}}[/tex]
now put all these value c , ω k and amplitude and we get
maximum force = 4.975 [tex]\sqrt{15000^{2}+5^{2} * 100^{2}}[/tex]
maximum force = 74.7 N .......................2
total force is combine equation 1 and 2 we get
total force 19.6 + 74.7 = 94.3 N
An example of Ferrous alloy is Brass a)-True b)-False
Answer: False
Explanation: No, brass is not a ferrous alloy.
Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.
_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares
e)-consumables
Answer:
a)-Bin stock items free issue
Explanation:
Bin stock items free issue items are similar to the free issue items, but their access is limited.
Bin stock items free issue items are similar to the free issue items, but their access is limited.
A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is 1.5 in-kip. Find the torque capacity using the uniform-pressure assumption.
Answer:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × [tex]\pi[/tex] × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × [tex]\pi[/tex] × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
How does the thermal efficiency of an ideal cycle, in general, compare to that of a Carnot cycle operating between the same temperature limits?
Answer:
Rankine cycle less efficient as compare to Carnot cycle operating betwwen same temperature limit.
Explanation:
We know that Carnot's cycle is an ideal cycle for all heat engine which operating between same temperature.It is a reversible cycle which have all process reversible that is why it have maximum efficiency.
On the other hand Rankine cycle is a practical working cycle so it is impossible to make all process reversible .In practical there will be always loss due to this any process can not make 100 % reversible.That is why Rankine cycle have low efficiency as compare to Carnot cycle operating between same temperature limits.
Define the work Envelope for a Robot
Heat transfer by is the fastest mode of heat transfer that requires no intervening medium. a)-conduction b)-convection c)-radiation d)-conduction and convection
Answer:
C. Radiation
Explanation:
Heat transfer through Conduction is a slow process and both Conduction and convection require material medium
Answer:
C radiation
Explanation:
Assume that light of wavelength 6000A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Answer:
θ=0.0288 radian
Explanation:
resolution limit is the minimum angular separation of two sources that can be viewed distinctly by telescope
[tex]\theta =\frac{1.22\times \lambda}{D}[/tex]
[tex]\lambda=6000\times 10^{-8} cm=6 \times 10^{-5} cm[/tex]
[tex]d=100 inch=100\times 2.54=254cm[/tex]
[tex]\theta = \frac{1.22 \times 6 \times 10^{-5}}{254}[/tex]
θ=0.0288 radian
Convert 10.25 degrees into radians; and π, π/2 and π/3 radians into degrees.
Answer:
0.1788 ,180°,90°,60°
Explanation:
CONVERSION FROM DEGREE TO RADIANS: For converting degree to radian we have to multiply with [tex]\frac{\pi}{180}[/tex]
using this concept 10.25°=10.25×[tex]\frac{\pi}{180}[/tex]=0.1788
CONVERSION FROM RADIAN TO DEGREE: For converting radian to degree we have to multiply with[tex]\frac{180}{\pi}[/tex]
using this concept π=π×[tex]\frac{180}{\pi}[/tex]
=180°
[tex]\frac{\pi}{2}[/tex]= [tex]\frac{\pi}{2}[/tex×[tex]\frac{180}{\pi}[/tex]
=90°
[tex]\frac{\pi}{3}[/tex]= [tex]\frac{\pi}{3}[/tex]×[tex]\frac{180}{\pi}[/tex]
=60°
Answer:
10.25° = 0.1790 radians
π radians = 180°
π/2 radians = 90°
π/3 radians = 60°
Explanation:
The conversion of degree into radians is shown below:
1° = π/180 radians
So,
10.25° = (π/180)*10.25 radians
Also, π = 22/7
So,
[tex]10.25^0=\frac{22\times10.25}{7\times180}radians[/tex]
Solving it we get,
10.25° = 0.1790 radians
The conversion of radians into degree is shown below:
1 radian = 180/π°
(a)
π radians = (180/π)*π°
Thus,
π radians = 180°
(b)
π/2 radians = (180/π)*(π/2)°
[tex]\frac {\pi }{2} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{2}^0[/tex]
π/2 radians = 90°
(c)
π/3 radians = (180/π)*(π/3)°
[tex]\frac {\pi }{3} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{3}^0[/tex]
π/3 radians = 60°
What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d). 74%
Answer:
(b) 56%
Explanation:
the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature
here we have given T₁ (Higher temperature)= 600+273=873
lower temperature T₂=110+273=383
Efficiency of power cycle is given by =1-[tex]\frac{T2}{T1}[/tex]
=1-[tex]\frac{383}{873}[/tex]
=1-0.43871
=.56
=56%
Work done by a system during a process can be considered as a property of the system. a)True b) False
Answer:
b) False
Explanation:
Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.
Explain the difference between thermal equilibrium and thermodynamic equilibriurm.
Answer:
A body is said to be in thermal equilibrium where all the system temperatures are the same.
A body is said to be in thermodynamic equilibrium if all three requirements of equilibrium are met
Explanation:
A body is said to be in thermal equilibrium where all the system temperatures are the same. In this case, there will be no temperature gradient between the system and the environment.
while, A body is said to be in thermodynamic equilibrium if all three requirements of equilibrium are met, i.e. mechanical equilibrium, chemical equilibrium and thermal equilibrium.
Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)? b. What would be the equivalent reading of a Mercury barometer (mm of Hg)?
Answer:
(a) water height =408.66 in.
(b) mercury height=30.04 in.
Explanation:
Given: P=14.769 psi ( 1 psi= 6894.76 [tex]\frac{N}{m^2}[/tex] )
we know that [tex]P=\rho\times g\timesh[/tex]
where [tex]\rho =Density,g=9.81\frac{m}{s^2}[/tex]
h=height.
Given that P=14.769 psi ⇒P= 101828.6 7[tex]\dfrac{N}{m^2}[/tex]
(a) [tex]P=\rho_{w}\times g\times h_{w}[/tex]
[tex]\rho_{w}=1000\frac{Kg}{m^3}[/tex]
⇒101828.67=[tex]1000\times 9.81\times h_{w}[/tex]
[tex]h_{w}[/tex]=10.38 m
So water barometer will read 408.66 in. (1 m=39.37 in)
(b) [tex]P=\rho_{hg}\times g\times h_{hg}[/tex]
[tex]\rho_{hg}[/tex]=13600
So 101828.67=[tex]13600\times 9.81\times h_{hg}[/tex]
[tex]h_{hg}[/tex]=0.763 m
So mercury barometer will read 30.04 in.
Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive machining tool.
Answer:
Explained
Explanation:
Cutting tools:
1. Cutting tools can either be single point or multi point.
2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.
3.Cutting tools have definite shapes and geometry.
Abrasive machining tools
1. Abrasive tools are always multi point tools.
2. Abrasive tools composed of abrasives bounded in medium of resin or metal.
3. They do not have definite geometry of shape
Cutting tools engage materials with a sharp edge for aggressive material removal, while abrasive machining tools employ hard particles to wear away material for high precision and smoothness in finishing operations.
Explanation:The differences between cutting tools and abrasive machining tools are fundamental in the processes they are used for and their operational principles. Cutting tools are typically used in operations like turning, milling, and drilling where the tool itself engages the material to be cut with a sharp edge, removing materials in the form of chips. These tools are often made of high-speed steel, carbide, ceramics, or other hard materials and are precisely shaped according to the specific cutting operation.
On the other hand, abrasive machining tools, which include grinding wheels, sandpaper, and abrasive belts, remove material through the action of hard, abrasive particles that are either bonded to the tool's surface or are used as loose grains. Abrasive machining is used for finishing surfaces to a high degree of smoothness, precision, and complex shapes that cutting tools cannot achieve. These tools are made of materials like aluminum oxide, silicon carbide, diamond, or cubic boron nitride.
To summarize, cutting tools use a sharp edge to remove material in a defined shape, while abrasive tools use hard particles to wear away material for finishing and shaping. Moreover, cutting tools are applied in more aggressive material removal processes and quicker operations like shaping or roughing out material, whereas abrasive machining is associated with finishing operations that require high precision and smoothness.
How is entropy defined on a differential basis? (use an equation)?
Answer:
Differential entropy differ's from the normal entropy as in case of differential entropy random variable needs not to be discrete.
The differential concept of entropy is as follows
Let X be a random variable which is continuous
Let[tex]f(x)[/tex] be it's probability distribution function
We have differential entropy h(X) defined as
[tex]h(X)=-\int _{X}f(X)log[f(X)]dx[/tex]
The random variable 'X' need not to be discrete and it is continuous.
An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates waste heat to a sink at 0 °C. He also claims that this engine produces 3.3 kW while rejecting heat at a rate of 4.4 kW. Is this claim valid?
Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=[tex]\frac{Work produced}{Heat supplied}[/tex]
and
Heat Supplied [tex]{\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )[/tex]
[tex]{Q_s}=3.3+4.4=7.7KW[/tex]
η=[tex]\frac{3.3}{7.7}[/tex]
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-[tex]\frac{T_ {sink}}{T_ {source}}[/tex]
η=1-[tex]\frac{273}{1023}[/tex]
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.
What is the quality of a two-phase liquid-vapor mixture of H20 at 20 °C with a specific volume of 10 m^3/kg?
Answer:
Quality of vapor is equal to 17.3%.
Explanation:
We know that if we know only one property in side the wet region then we will find the other property by using steam property table.
So pressure at saturation temperature 20°C
[tex]v_f= 0.001\frac{m^3}{Kg} ,v_g= 57.76\frac{m^3}{Kg}[/tex]
So specific volume v
[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]
Where x is quality of mixture
Now putting the values ,given that [tex]v=10m^3/kg[/tex]
[tex]10=0.001+x(57.76-0.001)\frac{m^3}{Kg}[/tex]
x=0.173
So quality of vapor is equal to 17.3%.
In solid rocket motor terminology, a sliver is: A. A basic geometric shape B. Leftover propellant C. A test article for propellant characterization D. A type of igniter
Answer:
In the question given out of the four options
option B. Leftover propellant
is correct.
Explanation:
Solid rocket motors have rocket engines that uses solid fuels or propellants such as HTPB and PBAN as the most commonly used binders.
These rockets always have the propellant in adequate amount can be powered for long enough without much fuel degradation and therefore are reliable and mostly used in military applications.
Since these have fuel or propellant storage and therefore to indicate it, solid rocket motor identifies the left over propellant as ' a silver' .
Explain why different types of equipment are required for proper conditioning of air
Answer:
Different types of equipment are required for proper conditioning of air because every air conditional space faces some geometrical and environmental issues or problems. There are some different types of equipment used for conditioning of air that are air system, water system and air-water system. In many cases the air conditioning of the system varies with size of the equipment.
A system that is not influence anyway by the surroundings is called a)- control mass system b)- Isothermal system c)-- isolated system d)- open system
Answer:
Isolated system
Explanation:
By definition of a closed system it means that a system that does not interact with it's surroundings in any manner
The other options are explained as under:
Isothermal system : It is a system that does not allow it's temperature to change
Control Mass system : It is a system whose mass remains conserved which means the mass entering the system equals the mass leaving the system
Open system: It is a system that allows transfer of mass and energy across it's boundary without any opposition i.e freely.
Why/how is a paperclip able to float on water?
Answer:
Water surface tension
Explanation:
The water around the paperclip forms a kind of elastic surface, deforming, in which the clip can stay afloat.
This is because the molecules that are on the surface of the water, already in contact with the air, try to cling to those that are next to them and those that are immediately below.
If we added even if it were just a drop of soap in the water, the clip would go to the bottom, because the soap has the ability to decrease the surface tension of the water.
Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)- occurs at the inner surface of a shaft.
Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that
[tex]\frac{T}{J}= \frac{\tau }{r}[/tex]
where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that
[tex]\tau = \frac{T.r}{J}[/tex]
Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.
Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow
Answer:
[tex]\omega_y,\omega_x,\omega_Z[/tex] all are zero.
Explanation:
We know that if flow is possible then it will satisfy the below equation
[tex]\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0[/tex]
Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.
And velocity potential function [tex]\phi[/tex] given as follows
[tex]u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}[/tex]
Rotationality of fluid is given by [tex]\omega[/tex]
[tex]\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z[/tex]
[tex]\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x[/tex]
[tex]\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y[/tex]
So now putting value in the above equations ,we will find
[tex]\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},[/tex]
[tex]\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}[/tex]
So [tex]\omega_y[/tex]=0
Like this all [tex]\omega_y,\omega_x,\omega_Z[/tex] all are zero.
That is why velocity potential flow is irroational flow.
An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal efficiency of 70%. Comment on the validity of this claim.
Answer Explanation:
the efficiency of the the engine is given by=1-[tex]\frac{T_2}{T_1}[/tex]
where T₂= lower temperature
T₁= Higher temperature
we have given efficiency =70%
lower temperature T₂=27°C=273+27=300K
higher temperature T₁=627°C=273+627=900K
efficiency=1-[tex]\frac{T_2}{T_1}[/tex]
=1-[tex]\frac{300}{900}[/tex]
=1-0.3333
=0.6666
=66%
66% is less than 70% so so inventor claim is wrong