The answer is : True
Which of these is not applicable to tuning for some Overshoot on Start-up? Select one: 1)-No overshoot during normal modulating control only 2)-KC-0.2 KU 3)- KC 0.33 KU 4)- KC 0.59 KU 5)- KC-0.78 KU
Answer: 5) KC-0.78 KU
Explanation:
KC-0.78 KU is not defined for tuning of overshoot on the Start-up as, this method is only applicable for there is some turning constants and also there is no overshoot during the normally modulating control. But some overshoot at start up are applicable. For the continuous cycling method, some overshoot is same as for closed loop tuning.
What is the output of a system having the transfer function G = 2/[(s + 3) x(s + 4)] and subject to a unit impulse?
Answer:
output=[tex]\frac{2}{(s+3)(s+4)}[/tex]
Explanation:
output =transfer function ×input
here transfer function G=[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
input = unit impulse
in S domain unit impulse =1
so output =[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
=[tex]\frac{2}{(s+3)(s+4)}[/tex]
An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?
Answer:
Explanation:
we have given E(t)=120 sin(12t)
R=5 ohm
L=0.2 H
ω=12 ( from expression of E)
[tex]X_L=0.2\times 12=2.4[/tex] ohm
[tex]X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm[/tex]
[tex]Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}[/tex]
[tex]Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}[/tex]
=5.021 ohm
so amplitude of current = [tex]\frac{v}{z}=\frac{120}{5.021}=23.89[/tex]
An automobile having a mass of 884 kg initially moves along a level highway at 68 km/h relative to the highway. It then climbs a hill whose crest is 69 m above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic energy, in kJ
Answer:
ΔK.E. = - 142.72 kJ
Explanation:
mass = 884 kg
initial velocity = 68 km/h = 68 \times \frac {5}{18} = 18.89 m/s
final velocity = 0 m/s
height = 69 m
change in kinetic energy :
ΔK.E. = [tex]\dfrac{1}{2}m(v_f^2-v_i^2)[/tex]
ΔK.E. =[tex]\dfrac{1}{2}\times 884 \times (0^2-18.89^2)[/tex]
ΔK.E. =-142,716.05 J
ΔK.E. =-142.72 kJ
hence change in kinetic energy of the automobile is -142.72 kJ
A gear pump has a 4.25in outside diameter, a 3.25in inside diameter, and a 2in width. If the actual pump flow rate is 1800rpm and rated pressure is 29gpm, what is the volumetric efficiency?
Answer:
volumetric efficeincy = 0.315%
Explanation:
Given data:
outside diameter = 4.25 inch
inside diameter = 3.25 inch
flow rate V = 1800 rpm
actual flow rate Qa = 29gpm = 6699 inch3/min
volume of pump can be determine by using below formula
[tex]VOLUME = \frac{\pi}{4}*(D_{0}^{2}-D_{1}^{2})L[/tex]
[tex]=\frac{\pi}{4}*(4.25^{2}-3.25^{2})2[/tex]
[tex]VOLUME = 11.78 inc^{3}[/tex]
theoretical flow rate is given as Qt
[tex]Q_{T} = V.N[/tex]
= 11.78*1800
=[tex]21204 inch ^{3}/ min[/tex]
[tex]volumetric efficeincy = \frac{Q_{A}}{Q_{T}}[/tex]
[tex]=\frac{6699}{21204}[/tex]
volumetric efficeincy = 0.315%
Why the inviscid, incompressible, and irrotational fields are governed by Laplace's equation?
Answer: Laplace equation provides a linear solution and helps in obtaining other solutions by being added to various solution of a particular equation as well.
Inviscid , incompressible and irrotational field have and basic solution ans so they can be governed by the Laplace equation to obtain a interesting and non-common solution .The analysis of such solution in a flow of Laplace equation is termed as potential flow.
The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
Answer:
Explained
Explanation:
Cold working: It is plastic deformation of material at temperature below recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.
Given melting point temp of lead is 327° C and lead recrystallizes at about
0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.
Primary Creep: slope (creep rate) decreases with time
Answer:
true
Explanation:
Creep is known as the time dependent deformation of structure due to constant load acting on the body.
Creep is generally seen at high temperature.
Due to creep the length of the structure increases which is not fit for serviceability purpose.
When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.
When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.
The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False
Answer:
a). TRUE
Explanation:
Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.
Mathematically, thermal efficiency is
η = net work done / heat input
While heat rate is the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.
Mathematically, heat rate is
Heat rate = heat input / net work done
Thus from above we can see that heat rate is the reciprocal of thermal efficiency.
Thus, Heat rate is reciprocal of thermal efficiency.
The contact angle between the mercury surface and capillary tube wall is______ A) Less than 90 B) Equal to 90 C) Greater than 90 D) Varying with mercury level
Answer:
The Answer to the question is :
Explanation:
The contact angle between the mercury surface and capillary tube wall is Greater than 90.
If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.
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.
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.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 Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical density (g/cm^3).
Solution:
Given :
atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m
atomic wieght, M = 95.96
Cell structure is BCC (Body Centred Cubic)
For BCC, we know that no. of atoms per unit cell, z = 2
and atomic radius, r =[tex]\frac{a\sqrt{3} }{4}[/tex]
so, a = [tex]\frac{4r}{\sqrt{3}}[/tex]
m = mass of each atom in a unit cell
mass of an atom = [tex]\frac{M}{N_{A} }[/tex],
where, [tex]N_{A}[/tex] is Avagadro Number = 6.02×10^{23}
volume of unit cell = a^{3}
density, ρ = [tex]\frac{mass of unit cell}{volume of unit cell}[/tex]
density, ρ = [tex]\frac{z\times M}{a^{3}\times N_{A}}[/tex]
ρ = [tex]\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}[/tex]
ρ = 10.215gm/[tex]cm^{3}[/tex]
What is the Thermodynamic (Absolute) temperature scale?
Answer:
0 K
Explanation:
The thermodynamic absolute temperature is that temperature at which there is an infinite cooling and so there is no movement of any molecules or particles it is given by kelvin. 0 K is that temperature at which there is no movement of molecule so 0 K or -273°C in degree Celsius is the absolute temperature in thermodynamic temperature scale.
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
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
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure c)- Atomic number d)- Microstructure
The answer is c) atomic number
Why degree of crystallinity affects polymer properties?
Answer:
Degree of crystallinity affects polymer properties as increasing in crystallinity means higher the thermal stability and harder the material. Basically, crystallinity strongly affects polymer properties as it defines the degree of long range order in a material. Crystallinity is also determined by the size as well as molecular chain orientation.
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
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.
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)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
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 wooden block (SG = 0.6) floats in oil (GS = 0.8). What fraction of the volume of the block is submerged in oil?
Answer:
3/4 th fraction of the volume of block is submerged in oil.
Explanation:
We know that
density of the block, ρ[tex]_{b}[/tex] =SG[tex]\times[/tex]density of water
= 0.6[tex]\times[/tex] 1000
= 600 kg/[tex]m^{3}[/tex]
density of the oil, ρ[tex]_{o}[/tex] =SG [tex]\times[/tex] denity of water
= 0.8[tex]\times[/tex] 1000
= 800 kg/[tex]m^{3}[/tex]
Let acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]
Volume of block submerged in oil is [tex]V_{1}[/tex]
Volume of block above the oil surface is [tex]V_{2}[/tex]
The total volume of the block is V = [tex]V_{1}[/tex]+[tex]V_{2}[/tex]
Therefore for a partially submerged body, we know that
Buoyant force = total weight of the block
and
total weight of the block = Weight of the fluid displaced by the block
ρ[tex]_{b}[/tex]\timesV\timesg = ρ[tex]_{o}[/tex]\times[tex]V_{1}[/tex]\times g
600([tex]V_{1}[/tex]+[tex]V_{2}[/tex]) = 800 [tex]V_{1}[/tex]
600[tex]V_{1}[/tex] + 600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex] - 600[tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 200[tex]V_{1}[/tex]
[tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{600}{200}[/tex]
Thus [tex]V_{1}[/tex] = 600
[tex]V_{2}[/tex] = 200
V = 600+200
= 800
Therefore fraction of volume of the block submerged in oil is,
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{600}{800}[/tex]
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{3}{4}[/tex]
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
In a horizontal pipeline a 150 mm diameter pipe is connected to a 250 mm diameter pipe. The flow rate in the pipeline is 0.15 m^3/s. Take the connection as a sudden enlargement and determine: (a) The pressure head loss vhen the vater flovs from the large pipe to the smaller pipe (take Cc=0.64) (b) The pressure head loss when vater flovs from the small pipe to the larger pipe (c) The loss of power in both cases
Answer:
a) [tex]h_L=1.17m[/tex]
b)[tex]h_L=1.52m[/tex]
c)[tex]P_1=1.721 kN[/tex]
[tex]P_2=2.236 kN[/tex]
Explanation:
velocities of the pipe;
velocity of small dia pipe
[tex]v_{small}=\frac{Q}{A_{small}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.15^2 }=8.52m/s[/tex]
velocity of larger dia pipe
[tex]v_{large}=\frac{Q}{A_{large}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.25^2 }=3.05m/s[/tex]
a) pressure head loss when water is flowing from large to smaller pipe
[tex]h_L=(\frac{1}{C_c} -1)^2 \times \frac{v^2}{2g}[/tex]
[tex]h_L=(\frac{1}{0.64} -1)^2 \times \frac{8.52^2}{2\times 9.81} = 1.17m[/tex]
b) pressure head loss when water flow from small pipe to large pipe
[tex]h_L= \frac{v^2}{2g}(1-\frac{A_{small}}{A_{large}} )^2[/tex]
[tex]h_L=\frac{8.52^2}{2\times9.81}(1-\frac{\frac{\pi}{4}\times 0.15^2}{\frac{\pi}{4}\times 0.25^2}} )^2= 1.52 m[/tex]
c) power loss in both cases are
[tex]P_1= \rho g Q h_{L1}= 1000 \times 9.81\times 0.15\times 1.17= 1.721 kN[/tex]
[tex]P_2= \rho g Q h_{L2}= 1000 \times 9.81\times 0.15\times 1.52= 2.236 kN[/tex]
To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; Series Ammeter; Parallel Voltmeter; Series Voltmeter; Parallel
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.
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]
Describe the design experiments method for product and process optimization
Answer:
Design of Experiment also known as DoE is a systematic methodology to understand how any process or parameters of any product affects the response variables like physical properties or performance of any product, etc. It serves the purpose of making the job easier.
It is technique to generate valuable information required with minimum experimentation with the use of these:
Statistical methodologyMathematical analysis to predict the output within the limits of experiments at any point.Experimental extremities
Process Optimization:
It includes the following:
Product prototype should be used in designingOptimization of significant value adding activitiesDetection and minimization of errorStrong built concept for the designTesting and validation of the process for further improvement in efficiencyTime, quality, environmental, manufacting and overall cost, safety and operational constraints should be duly noted.