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 hydraulic cylinder has a piston diameter of 150mm and strokes at 300mm in 10 seconds. Calculate- A) The swept volume of the actuator in liters B) The pump displacement in Liters/minute
Answer:
(a)Volume in liters=5.3 liters.
(b)Volume in liters/minute=31.8 liters/minute.
Explanation:
Given:
Diameter of cylinder ,D=150 mm
Stroke,L=300 mm
Time ,t=10 sec
we know that swept volume of cylinder
[tex]V_{s}=\dfrac{\pi }{4}\times D^2\times L[/tex]
So [tex]V_{s}=\dfrac{\pi }{4}\times 0.15^2\times 0.3 m^{3}[/tex]
[tex]V_{s}=0.0053 m^3[/tex]
(a) Volume in liters =5.3 liters ( 1[tex]m^3[/tex]=1000 liters)
(b) When we divide swept volume by time(in minute) we will get liters/minute.
We know that 1 minute=60 sec
⇒10 sec=[tex]\frac{10}{60}[/tex] minute
So volume displace in liters/minute=31.8 liters/minute.
What is BCC unit cell and what is FCC unit cell? (10%) Sketch FCC and BCC unit cells by using a small dot to represent an atom.
Answer:
BCC is Body Centered Cubic structure and FCC is Face Centered Cubic structure.
Explanation:
BCC
In Body centered cubic structure, atoms exits at each corner of the cube and one atom is at the center of the cube. The total number of atoms that a BCC unit cell contains are 2 atoms and the coordination number of the BCC unit cell is 8, that is the number of neighbouring atoms. Packing factor which determines how loosely or closely a structure is packed by atoms is 0.68 for a BCC Unit cell. The packing efficiency is found to be 58%. Metals like tungsten, chromium, vanadium etc exhibit BCC structure.
A face Centered cubic structure consists of an atom at each corner of the cube and an atom at each face center of the cube. The total number of atoms that a BCC unit cell contains are 4 atoms per unit cell and the coordination number of FCC is 12. Packing factor of a FCC unit cell is 0.74%, thus FCC structure is closely packed than a BCC structure. Metals like aluminium, nickel, cooper, cadmium, gold exhibits FCC structure.
________What is the main role of dispersed phase in ceramic matrix composites? a)- to increases yield strength b)-to increase ultimate tensile strength c)-to increase plain strain fracture toughness d)-to reduce hardness
Answer: b) to increase the ultimate tensile strength
Explanation: Ceramic matrix composites(CMC) are the materials which consist of ceramic material as well as composites .There are two pats present in it - matrix part which is present for holding the material together and reinforcement part which provides the toughness to the material. When the ceramic matrix composites have the dispersed phase,it is basically due to the increasing of the toughness of the material and hence increasing the tensile strength .
(a) what is Linear equation (b) Why Laplace's equation is linear
Answer:
A) Linear Equation -
Linear equation has only one independent variable and when the linear equation plotted on a graph it forms a straight line. It is made up of two expressions equal to each other in a equation. Linear equation graph fits the Y= mx+a ( m=slope).
B) Laplace's equation is linear as it is a second order partial differential equation. So if we put dependent variable in differential equation it always show result in linear.
Most important question is "What is linear equation?". So, I will only answer that.
What is linear equation?
Linear equation is defined as one degree equation which make a straight line on a graph. Or one can say that when the highest degree of an equation is one than we call ot linear equation.
For example, 3x+4y= 8
In this equation there are three terms 3x, 4y and 8. Here in this equation "x" and "y" are variables while 8 is a constant, 3 is the coefficient of variable "x" and 4 is the coefficient of variable "y". The highest power of the variable "x" is one while the highest power of the variable "y" is also one. Hence, it is a one degree equation. Therefore, it is also known as linear equation.
Q: How to determine linear equation?
Ans: In order to determine a linear equation, look at the highest power of an equation. If the highest power of the equation is one, than it is a linear equation. If the highest power of the equation is not one or other than one like two or three, than it is not a linear equation.
For example, 2xy + 5 = 10
There are also three terms in this equation. The term "2xy" is considered as one term. The power of the variable "x" is one while the power of the variable "y" is also one hence, one plus one is equal to two. Thus the highest power of this equation is two. Therefore, it is neither one degree equation, nor a linear equation.
Q: How can we solve a linear equation in two variables?
Ans: There are different methods of solving linear equation in two variables, but we will solve a linear equation in two variables by adding equation one and equation two. After that, we will get the value of one variable and than we will put the value of that variable in the second equation also known as substitution method in order to get the value of second variable.
For example, x + y = 8 .... Equation One
X - y = 6 .... Equation Two
Now we will add equation one and equation two, as a result variable "y" will be cancelled out. We will get "2x = 14". So we will divide both sides by 2. Hence, we will get "x=7". Now we can put the value of "x" in either equation one or in equation two. For instance, we put the value of "x=7" in equation one. We get:
7 + y = 8
y = 8 - 7
y = 1
Thus, we successfully solve the linear equation in two variables where the value of variable "x=7" while the value of variable "y=1".
A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe has an inside diameter of 8.5 cm. The annulus between the cylinder ad the pipe contains stationary oil. The oil has a specific gravity of 0.92 and a kinematic viscosity of 5.57 x 10-4 m2/s. Most nearly, what is the force needed to move the cylinder along the pipe at a constant velocity of 1 m/s?
Answer :
The force needed to move the cylinder is 25.6 N
Further explanationGiven that,
Length of the cylinder, l = 0.5 m
Outer diameter of the cylinder, d = 8 cm = 0.08 m
Outer radius of the cylinder, [tex]r=0.04\ m[/tex]
Inside diameter of the pipe, d = 8.5 cm = 0.085 m
Inside radius of the pipe, [tex]r=0.0425\ m[/tex]
Specific gravity of the oil, [tex]\rho=0.92[/tex]
Density of oil, [tex]d=\rho\times \rho_w[/tex]
Kinematic viscosity of the oil, [tex]v=5.57\times 10^{-4}\ m^2/s[/tex]
Velocity of the cylinder, u = 1 m/s
We need to find the force needed to move the cylinder. Let the force is F.
Specific gravity is defined as the ratio of the density of the substance to the density of water.
Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.
Absolute viscosity is given by :
[tex]v=\dfrac{\mu}{d}[/tex]
Where, [tex]d[/tex] = density of oil
And [tex]d=\rho\times \rho_w[/tex] (density of oil = specific gravity × density of water )
[tex]d=0.92\times 10^3\ kg/m^3[/tex]
So,
[tex]\mu=v\times d[/tex]..............(1)
[tex]\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3[/tex]
[tex]\mu=0.512\ Pa-s[/tex]
The separation between the cylinder and pipe is given by :
[tex]dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m[/tex]
[tex]d_p\ and\ d_c[/tex] are diameter of pipe and cylinder respectively.
The mathematical expression for the Newton's law of viscosity can be written as:
[tex]\tau\propto\dfrac{du}{dy}[/tex]
[tex]\tau=\mu\times \dfrac{du}{dy}[/tex]..........(2)
Where
[tex]\tau[/tex] = Shear stress, [tex]\tau=\dfrac{F}{A}[/tex]............(3)
[tex]\mu[/tex] = viscosity
[tex]\dfrac{du}{dy}[/tex] = rate of shear deformation
On rearranging equation (1), (2) and (3) we get :
[tex]\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}[/tex]...............(4)
A is the area of the cylinder, [tex]A=2\pi rl[/tex]
Equation (4) becomes :
[tex]F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl[/tex]..............(5)
[tex]A=\pi d\times l[/tex]
[tex]A=\pi \times 0.08\ m\times 0.5\ m[/tex]
[tex]A=0.125\ m^2[/tex]
Now, equation (5) becomes :
[tex]F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl[/tex]
[tex]F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2[/tex]
F = 25.6 N
Learn moreKinematic viscosity : https://brainly.com/question/12947932
Keyword :Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.
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.
A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneumatic-to-current (P/I) B. voltage-to-pneumatic (V/P) C. current-to-pneumatic (I/P) D. pneumatic-to-voltage (P/V)
Answer:
The correct answer is
option C. current to pneumatic (V/P)
Explanation:
A current to pneumatic controller is basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.
Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.
If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250 K, what is the amount of heat added to the system? a)-5J b)-125J c)- 600 J d)-5000 J e)-8000 J
Solution:
Given:
Change in entropy of the system, ΔS = 12J/K
Temperature of the system, [tex]T_{o}[/tex] = 250K
Now, we know that the change in entropy of a system is given by the formula:
ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex]
Amount of heat added, ΔQ = [tex]\Delta S\times T_{o}[/tex]
ΔQ = 3000J
A piston-cylinder device contains 2.8 kg of water initially at 400 °C and 1.2 MPa. The water is allowed to cool at constant pressure until 28% ofits mass condenses into liquid. a) Evaluate the final temperature. b) Calculate the initial and final volumes (m3) c) Calculate the enthalpy at the initial and final states (kJ)
Answer:
a).Final temperature, [tex]T_{2}[/tex] = 180°C
b).Initial Volume, [tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
Final Volume, [tex]V_{2}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). Initial enthalpy,[tex]H_{1}[/tex] =9129.68 kJ
Final enthalpy, [tex]H_{1}[/tex] =6234.76 kJ
Explanation:
Given :
Total mass, m= 2.8 kg
Initial temperature, [tex]t_{i}[/tex] = 400°C
Initial pressure, [tex]p_{i}[/tex] = 1.2 MPa
Therefore from steam table at 400°C, we can find--
[tex]h_{1}[/tex] = 3260.6 kJ/kg
[tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
Now it is mentioned that 28% of the mass is condensed into liquid.
So, mass of liquid, [tex]m_{l}[/tex] = 0.28 of m
= 0.28 m
mass of vapour, [tex]m_{v}[/tex] = 0.72 m
∴ Dryness fraction, x = [tex]\frac{m_{v}}{m_{l}+m_{v}}[/tex]
= [tex]\frac{0.72 m}{0.28 m+0.72 m}[/tex]
= 0.72
a). The final temperature can be evaluated from the steam table at 1.2 MPa,
[tex]h_{2}[/tex] = 2226.7 kJ/kg
[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
Final temperature, [tex]T_{2}[/tex] = 180°C
b). We know [tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
∴ Initial Volume, [tex]V_{1}[/tex] = [tex]v_{1}[/tex] x m
[tex]V_{1}[/tex] = 0.25479 x 2.8
[tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
We know,[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
∴ Final Volume, [tex]V_{2}[/tex] = [tex]v_{2}[/tex] x m
[tex]V_{2}[/tex] = 0.1179 x 2.8
[tex]V_{1}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). We know,
[tex]h_{1}[/tex] = 3260.6 kJ/kg
∴ Initial enthalpy,[tex]H_{1}[/tex] = [tex]h_{1}[/tex] x m
= 3260.6 x 2.8
= 9129.68 kJ
[tex]h_{2}[/tex] = 2226.7 kJ/kg
∴ Final enthalpy, [tex]H_{1}[/tex] = [tex]h_{2}[/tex] x m
= 2226.7 x 2.8
= 6234.76 kJ
A rigid tank initially contains 2 kg of steam at 500 kPa and 350°C. The steam is then cooled until it is at 100°C. Determine the final pressure and the heat transferred during this process.
Answer:
299.36 kPa
Explanation:
given mass of steam =2 kg
initial pressure that is [tex]P_1=500kPa[/tex]
initial temperature that is [tex]T_1=350^{\circ} C=350+273=623 K[/tex]
final temperature that is [tex]T_2=100^{\circ} C=100+273=373 K[/tex]
it is a rigid tank so volume is constant
for constant volume process [tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
[tex]P_2=\frac{T_2}{T_1}\times P_1=\frac{373}{623}\times 500=299.36 kPa[/tex]
so the final pressure will be 299.36 kPa
A M14 x 2 hexagonal head bolt is used to clamp together two 15 mm steel plates. Young's modulus of the bolt and the plates can be taken as 207 GPa. (i)-Determine a suitable length for the bolt. (ii)-Determine the bolt stiffness (iii)-Determine the stiffness of the members
Answer:
(i) 50 mm
(ii) 874.62 m N/m
(iii) 3116.45 m N/m
Explanation:
Given data
hexagonal head bolt = M14 x 2
steel plate = 15 mm
Young's modulus = 207 Gpa
Solution
1st part
dia of bolt (D) = 14 mm and height (H) = 12.8 mm from table
we know grip length = thickness of this plate i.e 30 mm , i.e (15mm+15mm)
so hexagonal bolt length = grip length + H
hexagonal bolt length = 30 mm + 12.8 mm = 42.80 mm i.e = 45 mm (round off)
2nd part
bolt stiffness = [tex]\frac{Ad*At*young modulus}{Ad*Lt*At*Ld}[/tex]
here Lt is length of thread = 2d +6mm
d is 14 so length of thread = 2*14 +6 = 34 mm
At from table 8-L i.e. = 115 mm2
Ad area of without thread part = [tex]\pi /4[/tex]×[tex]d^{2}[/tex]
Ad= [tex]\pi /4[/tex]×[tex]14^{2}[/tex] = 153.94 mm2
Ld is length of bolt without thread = length of bolt - Lt = 45 -34 = 11 mm
last Lt thread part length = length of bolt - Ld = 30-11 = 19 mm
put all these value in bolt stiffness i.e.
bolt stiffness = [tex]\frac{153.94*115*207}{153.94*19+115*11}[/tex] = 874.62
3rd part
stiffness of member = [tex]\frac{0.5774 \pi Ed}{2 ln (5\frac{0.5774L +0.5d}{0.5774L +2.5d})}[/tex]
here l is 30 and d is 14
so
stiffness of member = [tex]\frac{0.5774 \pi (207) 14}{2 ln (5\frac{0.5774(30) +0.5(14)}{0.5774(30) +2.5(14)})}[/tex]
stiffness of member = 3116.45 m N/m
Carbon nanotubes can be a considered toxic agent due to: (a)- Its ability to produce toxic gases (b)Its ability to penetrate skin quickly due its very small diameter (c)Its ability to mix with water (d)Difficulties in handling it properly
Answer: b)Its ability to penetrate skin quickly due to its very small diameter
Explanation: Carbon nano tubes(CNT) are the material widely used in the medical field due to the atomic structure of it ans also have small size. Toxicity in the carbon nano tubes is because their small sized atomic particles which can enter the skin by penetration or inhalation. But are still preferred in the medicine because having unique properties like mechanical property, chemical property,surface property etc.
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box and then it is placed in a large tank containing 12 ft of water so that it will sink just to the bottom of the tank. Determine the mass of stones placed in the box, in tons.
Answer:
44.95 tonnes
Explanation:
According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces
It is given that empty weight of box = 40 tons
Let the mass of the stones to be placed be = M tonnes
Thus the combined mass of box and stones = (40+M) tonnes..........(i)
Since the box will displace water equal to it's volume V we have [tex]volume of box = 25ft*10ft*12ft= 3000ft^{3}[/tex]
[tex]Volume= 84.95m^{3}[/tex]
[tex]Since 1ft^{3} =0.028m^{3}[/tex]
Now the weight of water displaced = [tex]Weight =\rho \times Volumewhererho[/tex] is density of water = 1000kg/[tex]m^{3}[/tex]
Thus weight of liquid displaced = [tex]\frac{84.95X1000}{1000}tonnes=84.95 tonnes[/tex]..................(ii)
Equating i and ii we get
40 + M = 84.95
thus Mass of stones = 44.95 tonnes
Explain the nature of defect in casting process and identify the defects through various NDT techniques.
Answer Explanation:
DEFECT IN CASTING : Defects are undisireable its not matter which type of defect we want no defect in any process casting defect are observed in the material manufacturing in metal casting process.
THESE DEFECTS ARE CLASSIFIED INTO DIFFERENT CLASS :
MOLD MATERIAL DEFECT: This defect occur when the property of mold material used is not up to the markPOURING METAL DEFECT: This defect occur due to adoption of wrong techniques for pouringSHRINKAGE DEFECT: This defect occur when the metal used is not enough to compensate the shrinkage of liquid metal METALLURGICAL DEFECT : Hot tears and hot spots are two such defectsNDT (NON DISTRUCTIVE TESTING): It is a inspection of material without distroy the material
there are different types of NDT
VISUAL INSPECTIONHYDROSTATIC PRESSURE TESTMAGNETIC PARTICAL INSPECTIONULTRASONIC INSPECTIONAnswer:
Cold dies, weak metal temperature, dusty metal, lack of ventilation, and too much lubricant can cause casting defects.
Explanation:
Cold dies, weak metal temperature, dusty metal, lack of ventilation, and too much lubricant can cause casting defects. Gas permeability, shrinkage porosity, warm tears, and stream stains are other potential defects. Caused by poor gating, sharp edges or heavy lubricant, flow traces are mark left on the casting surface.
fiver techniques used to identify defects in material are shown below
1) Liquid Penetrant Testing – Liquid penetrant screening is among the simplest approaches used to identify component defects.
2) Electromagnetic Testing – Electromagnetic testing comprises Eddy Current Testing, Rotating Current Field Quantification and Wireless Field Testing. Above mentioned techniques can identify both surface &underground flaws.
3) Magnetic Particle Testing – Magnetic particle screening is commonly used for the ferromagnetic materials to detect ground and near-surface defects.
4) Ultrasonic testing – Ultrasonic testing makes it possible to identify large and extremely small surface defects
5) Thermal Infrared Testing – Infrared thermography testing is used to measures and maps thermal variations on a substance's surface through thermal imaging equipment.
Which of the following quenching materials is LEAST severe in its quenching action (slowest cooling)? a. Air b. Brine c. Oil d. Water
Answer:
(a) air
Explanation:
in the following given quenching materials air is the least serve quenching action quenching is a process of heating the material and then rapidly cooling it. Quenching freezes the structure of the material including stresses. Oil has the most sever in its quenching action the commonly used used oil for quenching process is peanut and canola oil
A steel rod 6mm in diameter is stretched with a tensile force of 400N. Calculate the tensile stress experienced by the steel rod
Answer:
Tensile stress is 14.15 N/[tex]mm^{2}[/tex]
Explanation:
When any object is subjected to an external force, the body offers a resisting force which is equal and opposite to the external load. This resisting force is called stress. Thus stress is defined as the force acting perpendicular to the given cross sectional area of the object.
Mathematically, stress , σ = [tex]\frac{force }{area}[/tex]
Given : Tensile force, F = 400 N
Diameter of the rod, d = 6 mm
Area of the rod is given by, A = [tex]\frac{\pi }{4}\times d^{2}[/tex]
= [tex]\frac{\pi }{4}\times 6^{2}[/tex]
=28.27 [tex]mm^{2}[/tex]
Therefore, the tensile tress is, σ = [tex]\frac{force }{area}[/tex]
= [tex]\frac{400 }{28.27}[/tex]
= 14.149 N/[tex]mm^{2}[/tex]
[tex]\simeq[/tex] 14.15 N/[tex]mm^{2}[/tex]
Thus, tensile stress experieced by the rod is 14.15 N/[tex]mm^{2}[/tex]
What is an atom? a. it is a property of material b. it is specification of any material c. it the part of a substance d. it is a grain boundary
Answer:
Out of the multiple options provided,
option (c) it is the part of a substance
is correct
Explanation:
An atom is basically the smallest particle of matter or we can call it as building block of matter. It is as a brick to the wall or a cell to the body.
Matter is build up of a large number of these atoms. Atoms carbon(C), hydrogen(H), oxygen(O) combines to form molecules such as carbon dioxide([tex]CO_{2}[/tex]), water([tex]H_{2}O[/tex]), methane([tex]CH_{4}[/tex]), etc and then these molecules combines to form a substance.
It cannot be property, neither can it be specification of any material as these two characterizes the material and it can't be grain boundary for sure as grain boundary is the interface between two grain or crystal particles.
This shows that the smallest unit of matter is atom and it is the part of a substance
In solid motors, HTPB and PBAN are two common types of plasticizers. a) True b) False
Answer:
The given statement is True
Explanation:
Solid motor uses plasticizers out of which Poly Butadiene Acrylic Acid acrylo Nitrile (PBAN) and Hydroxy Terminator Poly Butadiene (HTPB) are the most commonly used. These are used in space shuttles.
PBAN is comparatively stick, thick and it has a stinky smell as compared to HTPB. PBAN can be mixed and cured at high temperastures few days later wheras HTPB can be mixed and cured at room termperatures within a day.
The propellant can be ready to use for flight a day later.
Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.
Answer:
a)[tex]T_2=868.24 K[/tex] ,[tex]P_2=2.32 bar[/tex]
b) [tex]s_2-s_1=0.0206[/tex]KW/K
Explanation:
P=4200 KW ,mass flow rate=20 kg/s.
Inlet of turbine
[tex]T_1[/tex]=807°C,[tex]P_1=5 bar,V_1=100 m/s[/tex]
Exits of turbine
[tex]V_2=125 m/s[/tex]
Inlet of diffuser
[tex]P_3=1 bar,V_3=15 m/s[/tex]
Given that ,use air as ideal gas
R=0.287 KJ/kg-k,[tex]C_p[/tex]=1.005 KJ/kg-k
Now from first law of thermodynamics for open system at steady state
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here given that turbine is adiabatic so Q=0
Air treat ideal gas PV=mRT, Δh=[tex]C_p(T_2-T_1)[/tex]
[tex]w=\dfrac{P}{mass \ flow\ rate}[/tex]
[tex]w=\dfrac{4200}{20}[/tex]
w=210 KJ/kg
Now putting the values
[tex]1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210[/tex]
[tex]T_2=868.24 K[/tex]
Now to find pressure
We know that for adiabatic [tex]PV^\gamma =C[/tex] and for ideal gas Pv=mRT
⇒[tex]\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}[/tex]
[tex]\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}[/tex]
[tex]P_2=2.32 bar[/tex]
For entropy generation
[tex]s_2-s_1=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}[/tex]
[tex]s_2-s_1=0.00103[/tex]KJ/kg_k
[tex]s_2-s_1=0.00103\times 20[/tex]KW/K
[tex]s_2-s_1=0.0206[/tex] KW/K
We can recycle glass as many time as we want. a)-True b)-False
Answer: True
Explanation: Yes , we can recycle glass endlessly, as many times as we want without the degradation of quality and purity factor because they are made from the widely available domestic components for example:- sand , limestone etc. These material are responsible for the successful recycling process of glass which does not produce any loss.
What is an example of a linear actuator?
Answer and Explanation :
LINEAR ACTUATOR-In simple terms a linear actuator is a mechanical device that creates motion in a straight line. Linear actuators are used in machine tools and industrial machinery, in computers peripherals such as drives and printers in valve and dampers and it is used where linear motion is required.
EXAMPLE OF LINEAR ACTUATOR: There are many types of motors that are used in a linear system these include DC MOTOR (with brush or brushless) STEPPER MOTOR a linear actuator can be used to operate a large valve in refinery. Hydraulic pump is an also a good example of linear actuator.
A gas in a piston-cylinder asscmbly undcrgocs a process for which the rclationship bctwcen pressurc and volumc is pV^2=constant The initial pressurc is 1 bar, the initial volume is 0.4 m^3, and the final pressure is 9 bar. Determine the work for the process, in kJ constant.
Given:
[tex]pV^{2}[/tex] = constant (1)
⇒ [tex]p_{1}V_{1}^{2} = p_{2}V_{2}^{2}[/tex] (2)
[tex]p_{1} = 1 bar = 1\times 10^{5}[/tex]
[tex]p_{2} = 9 bar = 9\times 10^{5}[/tex]
[tex]V_{1} = 0.4 m^{3}[/tex]
[tex]V_{2} = ? m^{3}[/tex]
Solution:
Here, from eqn (1), the polytropic constant is '2' ( Since, here [tex]pV^{n}[/tex] = [tex]pV^{2}[/tex] )
Now, using eqn (2), we get
[tex]V_{2}^{2} =\frac{p_{2}}{p_{1}}\times V_{1}^{2}[/tex]
putting the values in above eqn, we get-
[tex]V_{2}^{2} =\frac{9}{1}\times 0.4^{2}[/tex]
[tex]V_{2} = 1.2 m^{3}[/tex]
Now, work for the process is given by:
[tex]W = \frac{p_{2}V_{2} - p_{1}V_{1}}{1 - n}[/tex] (3)
where,
n = potropic constant = 2
Using Eqn (3), we get:
[tex]W = \frac{9\times 10^{5}\times 1.2 - 1\times 10^{5}\times 0.4}{1 - 2}[/tex]
W = - 240 kJ
Which of the following will decrease the value of the time constant for an RL circuit? (a) increasing resistance (b) decreasing capacitance (c) decreasing resistance (d) increasing inductance
Answer:
(a) increasing resistance
Explanation:
TIME CONSTANT FOR RL CIRCUIT : In RL circuit resistor and inductor are present only time constant for RL circuit is given by
time constant =[tex]\frac{L}{R}[/tex] From this expression it is clear that is we increase the value of resistance then denominator value increases and so the overall value decreases because we know that in a fraction when denominator value is increase then overall value is decreases so option (a) will be the correct option
Citations must be contested within_____working days of the notice of proposed penalty. a)-15 b)-10 c)-30 d)-7
Answer:
30
Explanation:
Legally that's when you have to respond
Describe the operational principle of a unitary type air conditioning equipment with a suitable sketch.
Answer:
Operational Principle of a Unitary Type Air Conditioning Equipment:
A unitary air conditioning system is basically a room type air conditioning system which comprises of an outdoor unit, a compressor for compressing
coolant, a heat exchanger (outdoor) for heat exchange, an expander attached to the heat exchanger for expansion of coolant and a duct.
It continuously removes heat and moisture from inside an occupied space and cools it with the help of heat exchanger and condensor in the condensing unit and discharges back into the same occupied indoor space that is supposed to be cooled.
The cyclic process to draw hot air, cool it down and recalculation of ther cooled air keeps the indoor occupied space at a lower temperature needed for cooling at home, for industrial processes and many other purposes.
refer to fig 1
What are the two types of pumps in compressors?
Answer: The two different types of pumps in compressors are:
1) Centrifugal Pumps
2) Reciprocating Pumps
Explanation:
Compressor is defined as a mechanical device which increases the gas pressure by reducing its volume.The main action of a pump is to transport liquids and pressurize and it coverts rotational energy. As, centrifugal pumps are the most common pumps used for transfer of fluids and it works on simple mechanism.
Reciprocating Pumps is used for low volumes of flow at a high pressure. It is the positive displacement pump. As, it works on the principle of pushing of liquid which executes a reciprocating motion in a closed cylinder.
Answer:
Centrifugal Pumps and Positive Displacement Pumps
Explanation:
The two of pumps listed above is based on the mode of operation. The centrifugal pumps works by increasing the velocity of the liquid through the machine while the displacement pumps works by alternating, filling a cavity and then displace some amount of liquid.
The larger the Bi number, the more accurate the lumped system analysis. a)-True b)- False
Answer:
b). False
Explanation:
Lumped body analysis :
Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.
Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.
The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.
In general it is assume that for a lumped body analysis, Biot number [tex]\leq[/tex] 0.1
Therefore, the smaller the Biot number, the more exact is the lumped system analysis.
For a 4-bar linkage with ri =7-in, r2 =3-in, r3= 9-in, and r =8-in, determi the minimum and maximum transmission angles.
Answer:
[tex]\mu_{min}[/tex]=[tex]26.38^{\circ}[/tex]
[tex]\mu_{max}[/tex]=[tex]71.79^{\circ}[/tex]
Explanation:
[tex]r_{1}[/tex]=7 in, [tex]r_{2}[/tex]=3 in, [tex]r_{3}[/tex]=9in
,[tex]r_{4}[/tex]=8 in
Transmission angle (μ ):
It is the acute angle between coupler and the output (follower) link.
Here we consider link [tex]r_{1}[/tex] as fixed link ,[tex]r_{2}[/tex] as input link ,link [tex]r_{3}[/tex] as coupler and link [tex]r_{4}[/tex] as output link.
As we know that
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
When link [tex]r_{2}[/tex] will be horizontal in left side direction then transmission angle will be minimum and when link [tex]r_{2}[/tex] will be horizontal in right side direction then transmission angle will be maximum.
Now by putting the values we will find
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{max}=0.3125[/tex]
[tex]\mu_{max}=71.79^\circ[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=0.8958[/tex]
[tex]\mu_{min}=26.38^\circ[/tex]
Hence, The minimum and maximum angle of transmission angle is 26.38° and 71.79° respectively.
What is Differential Analysis in fluid mechanics?
Answer:
Differential analysis is used when it is needed to determine the detail information of the flow i.e. pressure or stress variation along any point.
Explanation:
In fluid mechanics, sometime situation arise in which we need to determine in detail about flow characteristics like stress and pressure variation.
To find these flow characteristics some relationship need to imply either at a point or at very small volume and analysis of flow at very small point is known as differential analysis.
Example: pressure and shear stress variation in a line of the wing of a plane.
Quantity of erystallization centres in crystallization may increase a) insert imitation of crystallization b) use mechanical mixing: c) change cooling rate; d) all of the above are correct.
Answer: d) All of the above
Explanation: Crystallization is the process in which a solid crystalline structured material is obtained from the liquid substance. the quantity of crystallization center in crystal may increase due to several reasons like changing the cooling rate or mechanical mixing of the substances or imitation of crystals etc.
These processes end up adding mineral atoms to get attached to the center of crystal and hence increasing the size. Thus, the correct option is option (d).