Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=[tex]\frac{s}{(s+3)^2}[/tex]
U(s)=[tex]\frac{1}{s}[/tex] for unit step function
output =H(s)×U(s)
=[tex]\frac{s}{(s+3)^2}[/tex]×[tex]\frac{1}{s}[/tex]
=[tex]\frac{1}{(s+3)^2}[/tex]
taking inverse laplace of output
output=t×[tex]e^{-3t}[/tex]
at t=0 putting the value of t=0 in output
output =0
In shaft design, this of the following has the least influence on the shaft diameter to be determined: (a) Shaft deflection (b) Bearing type (c) Factor of safety (d) Rotational speed of the shaft
Answer: (d)Rotational speed of the shaft
Explanation: Shaft design is the design of a shaft that is used for defining the stresses at certain critical part of shaft. The shaft design has shaft diameter as a major part and this is determined by several factors like type of bearing , deflection, torque,safety factor etc.
But the least important factor for determining of the diameter is the rotational speed because it defines the rotation of an object around a particular axis, where is it states about the number of turns divisible by time. Therefore option(d) is the correct option.
Water flows in a pipe of diameter 0.5 m. The dianeter of the to 1,0 m. A U-tube manometer is of the enlargement with joining ercury levels 5 mm. Determine the flow rate as well as the pressure 3 Water pipe suddenly enlarges connected to either side pipes which contain water. The difference in m head loss as a result of the enlargement.
Answer:
Q = 0.943[tex]m^{3}/s[/tex]
[tex]h_{L}[/tex] = 0.6605 m
Explanation:
Given :
Diameter, d₁ = 0.5 m
Area, A₁ = [tex]\frac{\pi }{4}\times 0.5^{2}[/tex]
= 0.19625 [tex]m^{2}[/tex]
Enlargement diameter, d₂ = 1 m
Enlargement Area, A₂ = [tex]\frac{\pi }{4}\times 1^{2}[/tex]
= 0.785 [tex]m^{2}[/tex]
Manometric difference, h = 35 mm
=35 X [tex]10^{-3}[/tex] m
From manometer , we get
[tex]P_{1}+\rho _{w}.g.z_{1}+\rho _{m}.g.h=P_{2}+\rho _{w}.g.(z_{1}+h)[/tex]
[tex]P_{1}-P_{2}=(\rho _{w}-\rho _{m}).g.h[/tex]
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=(1-\frac{\rho _{m}}{\rho _{w}})\times h[/tex]
= [tex](1-13.6)\times 35\times 10^{-3}[/tex]
= -0.441
Now from newtons first law,
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=\frac{V_{2}^{2}-V_{1}V_{2}}{g}[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{A_{2}^{2}}-\frac{1}{A_{1}A_{2}})[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{0.785^{2}}-\frac{1}{(0.19625\times 0.785)^{2}})[/tex]
Therefore. Q = 0.943 [tex]m^{3} /s[/tex]
Now V₁ = [tex]\frac{Q}{A_{1}}[/tex]
= [tex]\frac{0.943}{0.19625}[/tex]
= 4.80 m/s
V₂ = [tex]\frac{Q}{A_{2}}[/tex]
= [tex]\frac{0.943}{0.785}[/tex]
= 1.20 m/s
Therefore, heat loss due to sudden enlargement is given by
[tex]h_{L}=\frac{(V_{1}-V_{2})^{2}}{2g}[/tex]
[tex]h_{L}=\frac{(4.80-1.20)^{2}}{2\times 9.81}[/tex]
= 0.6605 m
If a host system is 80% efficient, the minimum horsepower rating of the motor should be ? if the hoist is to provide 20,000 ft-lb/s. a) 45.45hp b) 23,530hp c) 36.36hp d) 42.78hp
Answer:
answer is option A i.e.45.45 hp
Explanation:
Given data:
load =20000 ft lb/s
efficiency = 80%
we know that
1 hp = 550 ft lb/s
minimum horsepower rating can be obtained by using following formula
minimum horse power rating = [tex]\frac{load}{efficiency * 1 horse power} \\[/tex]
= [tex]\frac{20000}{0.8*550} = 45.45 hp[/tex]
The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.
If water is boiling at 130 degrees Celsius inside a pressure cooker, the pressure is approximately 2.7 atmospheres
To determine the pressure inside a pressure cooker where water is boiling at 130 degrees Celsius, we use the properties of water and its boiling point at various pressures.
For water, the pressure at which the boiling point is elevated to 130 degrees Celsius can be found using steam tables.
At 130 degrees Celsius, the saturation pressure of water is approximately 2.7 atmospheres (atm). This is equivalent to :
= 2.7 x 101.325 kPa
= 273.5775 kPa, or about 2.7 bar (since 1 bar = 100 kPa)
Give two causes that can result in surface cracking on extruded products.
Answer:
1. High friction
2. High extrusion temperature
Explanation:
Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.
Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.
Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.
What is pre-flush and post flush in petroleum engineering?
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Answer:
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Explanation:
A horizontal jet of water strikes a vertical surface on a stationary cart that has a mass of 2.8 kg. The jet has a mass flow rate of 0.13 kg/s. The force required to hold the cart in place is 8N. What is the diameter of the nozzle? a. 1.6 mm b. 5.1 mm c. 3.4 mm d. 7.3 mm
Answer:
option a is correct answer i.e. d = 1.6 mm
Explanation:
From conservation principle
force by jet = force required to hold the jet
force by the jet is written as
[tex]force = \dot{m}(v_{1}-v_{2})\[/tex]
force required to hold the jet = 8N
[tex]\dot{m}(v_{1})\ = 8[/tex]
[tex]\dot{m}=0.13 kg/s[/tex]
[tex]v_{1} = 61.58 m/s[/tex]
we know that mass flow rate is given as
[tex]\dot{m}=\rho Av_{1}[/tex]
substituting value to get required diameter of the nozzle
[tex]0.13=1000*\frac{\pi }{4}d^{2}*61.58[/tex]
d = 1.6 mm
Three point bending is better than tensile for evaluating the strength of ceramics. a)-True b)- False
Answer:
a)-True
Explanation:
Three point bending is better than tensile for evaluating the strength of ceramics. It is got a positive benefit to tensile for evaluating the strength of ceramics.
A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2
Answer:
1590 m^2
Explanation:
Given data in this question
Diameter = 45 m
power = 700 kW
wind speed = 12 m/s
turbine speed = 1500 rpm
To find out
swept area of the wind turbine
Solution
we know wind turbine is rotate circular form
and diameter is given so by the area of circular swept we will calculate it
we know area = [tex]\pi /4[/tex] × d²
put the value of d here
area = [tex]\pi /4[/tex] × 45²
swept area = 1590 m^2
Fluids at rest possess no flow energy. a)- True b)- False
Answer:
True.
Explanation:
According the engineering flow they don not possess flow energy when they are in rest.
When they are in motion they show a translation energy.
The features if fluids may be different according the variables of pressure and temperature.
What are beats? Determine the terms decibel and octave.
Answer:
Explained below
Explanation:
Beats are interference pattern between two sounds of slightly different frequencies perceived as periodic vibration in volume whose rate is difference of the two.
Both octave and decibel are the terms of measurement.
Octave(In electronics) is a logarithmic unit for ratio between frequencies,with one octave corresponding to doubling of frequency. For example frequency one octave is from 40 Hz to 80 Hz.
Whereas decibel is a unit of sound intensity. It is one-tenth of A bel. In electronics it is used measure power level of an electrical signal by comparing it with given level of logarithmic scale.
A pendulum has an oscillation frequency (T) which is assumed to depend upon its length (L), load mass (m) and the acceleration of gravity (g). Determine the relationship between oscillation frequency, length, load mass and acceleration of gravity. Differentiate as well which variable does not affect the oscillation frequency.
Answer:
Mass does not affect oscillation frequency.
Explanation:
Let the bob of the pendulum makes a small angular displacement θ. When the pendulum is displaced from the equilibrium position, a restoring force tries to act upon it and it tries to bring the pendulum back to its equilibrium position. Let this restoring force be F.
Therefore, F = -mgsinθ
Now for pendulum, for small angle of θ,
sinθ[tex]\simeq[/tex]θ
Therefore, F = -mgθ
Now from Newton's 2nd law of motion,
F = m.a = -mgθ
[tex]\Rightarrow m.\frac{d^{2}x}{dt^{2}} = - mg\Theta[/tex]
Now since, x = θ.L
[tex]\Rightarrow L.\frac{d^{2}\Theta }{dt^{2}}= -g\Theta[/tex]
[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}= -\frac{g}{L}.\Theta[/tex]
[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}+\frac{g}{L}.\Theta =0[/tex]
Therefore, angular frequency
[tex]\omega ^{2}[/tex] = [tex]\frac{g}{L}[/tex]
ω = [tex]\sqrt{\frac{g}{L}}[/tex]
Also we know angular frequency is , ω = 2.π.f
where f is frequency
Therefore
2πf = [tex]\sqrt{\frac{g}{L}}[/tex]
f = [tex]\frac{1}{2 \pi }\sqrt{\frac{g}{L}}[/tex]
So from here we can see that frequency,f is independent of mass, hence it does not affect frequency.
Define spring stiffness and damping constant.
Answer Explanation:
SPRING STIFFNESS :The stiffness of a body is a measure of resistance offered by an elastic body to deformation.it is denoted by K every object has some stiffness for spring the spring stiffness is the force required to cause unit deflection
DAMPING CONSTANT: the damping constant is a number decided by manufacturer that describes the material property. Damping is an influence within an oscillatory system that has the effect of restricting its oscillation
Heat in the amount of 100 KJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600K.Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.
Answer:
0.0833 k J/k
Explanation:
Given data in question
total amount of heat transfedded (Q) = 100 KJ
hot reservoir temperature R(h) = 1200 K
cold reservoir temperature R(c) = 600 k
Solution
we will apply here change of entropy (Δs) formula
Δs = [tex]\frac{Q}{R(h)}+\frac{Q}{R(c)}[/tex]
Δs = [tex]\frac{-100}{1200}+\frac{100}{600)}[/tex]
Δs = [tex]\frac{1}{12}[/tex]
Δs = 0.0833 K J/k
this change of entropy Δs is positive so we can say it is feasible and
increase of entropy principle is satisfied
Answer:
0.0837 kJ/K
Explanation:
Given:
Temperature of the cold reservoir T,cold = 600 K
Temperature of the hot reservoir T,hot= 1200 K
Heat transferred , Q=100 kJ
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,cold=-\frac{Q}{T,cold}[/tex]
[tex]\bigtriangleup S,cold=-\frac{-100}{600}[/tex]
[tex]\bigtriangleup S,cold=0.1667 kJ/K[/tex]
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,hot=-\frac{Q}{T,hot}[/tex]
[tex]\bigtriangleup S,hot=-\frac{100}{600}[/tex]
[tex]\bigtriangleup S,hot=-0.0833 kJ/K[/tex]
Therefore, the total entropy change for the two reservoir is
[tex]\bigtriangleup S=\bigtriangleup S,hot +\bigtriangleup S,cold[/tex]
thus,
ΔS=0.1667-0.0833
ΔS=0.0833 kJ/K
Since, the change of entropy is positive thus we can say it is possible and
increase of entropy principle is satisfied
A 5Kw solar system may produce enough energy to power your home. a)-True b)- False
Answer: True
Explanation:
Yes, it is true that 5 Kw solar system may produce enough energy to power your home as, on an average good quality of 5 KW solar system can produced 22 units per day enough to power all home appliances. As, a 5 KW solar system produced energy is basically depends on the three main factor that are:
Quality of the solar panel system.Location from where the solar system generated its energy.And also on the positioning of the solar system.What is a substitutional solid solution? And what factors favor the formation of a substitu- tional solid solution?
Answer:
A substitutional solid solution is a kind of alloying process used to improve or strengthen a purer metal by alloying it. this process works out by adding atoms of an alloy element to the atoms of the crystal lattice of the parent or base element, thus forming a substitutional solid solution. This process generates local non uniformity in the lattice due to the presence or mixing of an alloy element with the base element which impedes the plastic deformation.
Factors that favor the formation of substitutional solid solution are:
Alloying beyond the solubility limitChemical affinity of the elementsRelative atomic size of the particlesCrystal structure of the elementsValenceAnswer:
Substitutional solids solution are formed when there is change in position of atom named as A in lattice by some other atoms of different atoms named as B.
Explanation:
Substitutional solids solution are formed when there is change in position of atom named as A in lattice by some other atoms of different atoms named as B.
Strengthening of this solution occurs when solute atom is lager than solvent atom so that it can replace its position
factor affect the extent of solid solution is
1) Relative atomic size
2) crystal structure
3) chemical affinity
4) valency
A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the top, the pipe has a diameter of 30 mm and a pressure gauge indicates a pressure of 860 kPa. At the bottom the diameter is 85 mm and a pressure gauge reading is 1 MPa. Assume the losses are negligible and determine the flov rate. Does the flow direction matter?
Answer:
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
Explanation:
Given that
At top[tex]d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm[/tex]
[tex]\rho =900\dfrac{Kg}{m^3}[/tex]
We know that
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2[/tex]
[tex]A_1V_1=A_2V_2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2[/tex]
[tex]V_2=8.02V_1[/tex]
[tex]Z_2=12 sin60^{\circ}[/tex]
[tex]\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}[/tex]
So [tex]V_1=1.30[/tex]m/s
We know that flow rate Q=AV
[tex]Q=A_1V_1[/tex]
By putting the values
[tex]A_1=\dfrac{\pi}{4}d^2[/tex]
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.
What are the x and y coordinates of the centroid of the area?
Answer:
[tex]\\X_{C.G}=\frac{\int xdA}{A}\\Y_{C.G}=\frac{\int ydA}{A}[/tex]
Explanation:
The x co-ordinate of the centroid is given by:
[tex]x_{C.G}=\frac{\int xdA}{A}[/tex]
The y co-ordinate of the centroid is given by:
[tex]Y_{C.G}=\frac{\int ydA}{A}[/tex]
where
[tex]x^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
and [tex]y^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
See attached figure
Vertical axis wind turbine generates more electrical power compared to horizontal axis wind turbine due to differences in wind speed. a)-True b)- False
Answer: False
Explanation: Horizontal-axis wind turbine is the major part of wind industry. Horizontal axis have rotating axis of wind turbine in horizontal direction.Horizontal wind produces more electricity from a particular amount of wind that is provided and thus is preferred more.
vertical-axis wind turbines have the rotational axis of the turbine in vertical or perpendicular direction and have easy installation even where wind conditions are not predictable and is lighter in weight but are not able to produce the large amount of electricity as compared to horizontal-axis wind turbine.
Thus the statement given is false.
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service factor, k=2.75 specify the torque rating for the clutch in lb-in.
Answer:
[tex](M_t)_{rated}=61.11lb-in[/tex]
Explanation:
speed of motor (N)=1500 rpm
power=4 hp = [tex]4 \times 0.7457 [/tex] =2.9828 KW
service factor(k)= 2.75
now,
[tex]KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }[/tex]
[tex]M_t= 18,989.09 \ N-mm= 168.06 lb-in[/tex]
torque rating
[tex](M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in[/tex]
An unknown gas (k=1.4, c v= 0.716 kJ/kg.K, c_p=1 kJ/kg.K, R = 0287 kJ/kg K) is trapped in a 1 m^3 piston-cylinder device at 1000 KPa and 1000 K. It then undergoes an isothermal (constant-temp) process in which 696 kJ of boundary work is delivered (positive work transfer). Determine the heat transfer. (Use the PG model).
Answer:
So heat transfer is 696 kJ
Explanation:
Given:
K = 1.4
[tex]C_{v}[/tex] = 0.716 kJ/kg
[tex]C_{p}[/tex] = 1 kJ/kg
R = 0.287 kJ/kg
V = 1 [tex]m^{3}[/tex]
P = 1000 kPa
T = 1000 K
Work delivered, δW = 696 kJ
It is isothermal process, so the initial and final temperature are same, that is T₁ = T₂ and the internal energy is zero (dU =0)
Therefore from 1st law of thermodynamcis,
δQ = dU + δW
= 0 + 696
= 696 kJ
So heat transfer is 696 kJ
What are the air-standard assumptions?
Answer:
The air-standard assumptions are:
The working fluid is air assumed to be perfect and it behave as an ideal gas .All process are internally reversible.The cycle is modeled as closed cycle with air cooled in the chiller heat exchanger and then re-circulated to the compressor. To avoid the complications, the combustion container are replaced by combustion heat exchanger .
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Answer:
Repairable component
Explanation:
Repairable component is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures.
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Repairable component
In a nuclear reactor, more fuel is burned than is consumed. a)-True b)- False
Answer:
true
Explanation:
yes, it is true more fuel is burned than is consumed.
nuclear reactor generate electricity from nuclear fission and while nuclear fission nucleus splits into small parts to generate energy.
these energies can be used for my purposes mainly it is used for power generation.
we can also say that in nuclear reactor more fuel is consumed because of metal acting as a fissionable material.
Harvesting wind energy using kites is: a) possible but currently very expensive. b) possible and currently inexpensive compared to wind turbines c) possible using a single kite.
Answer: b) possible and currently inexpensive as compared to wind turbines.
Explanation: Wind harvesting through kites is a process that will require less expenses in making and maintenance of it .Kites can be termed as the wind generator that is unconventional. The set up for the kite wind generator is also easy to install and also less costly as compared with turbines. So the correct option is option(a) .
A stainlesss steel cylinder diameter 60 mm is cooled in a air with h = 10 W/m^2-K. If the thermal conductivity is 20 W/m-K, the Biot number for this sphere is a)-0.005 b)-0.03 c)-8 d)-30
Answer:
The Biot number for this sphere is 0.03
Explanation:
Given data in equation
diameter (d) = 60 mm = 60/1000 = 0.06 m
heat transfer coefficient (h) = 10 w/m²-K
thermal conductivity (K) = 20 W/m-K
To find out
Biot number for sphere
Solution
we will find boit number by this given formula
Biot Number = ( h × S ) / K ..................1
here h and K value is given in question
and S = diameter/2
S= 60/2 = 30 mm = 0.03 m
Now put these value S, h and K in equation 1
Biot Number = ( 10 × 0.03 ) / 10
Biot Number = 0.03
An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due required that the amplitude of motion should not exceed 0.1 mm per 1 N of unbalance force. The mass of the motor is 120 kg and the operating speed is 720 rpm Use the concept of transfer function to determine the required stiffhess coefficient of each mount.
GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
[tex]\frac{A}{F}[/tex] = [tex]\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}[/tex]
Formula Used:
[tex]A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}[/tex]
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
[tex]\omega = \frac{2 \pi\times N}{60}[/tex]
so,
[tex]\omega = \frac{2 \pi\times 720}{60}[/tex] = 75.39 rad/s
Using the given formula:
Damping is negligible, so, [tex]\zeta = 0[/tex]
[tex]\frac{A}{F}[/tex] will give the tranfer function
Therefore,
[tex]\frac{A}{F}[/tex] = [tex] \frac{1}{\sqrt{(4K - 120\ ^{2})}}[/tex]
[tex]0.1\times 10^{-3}[/tex] = [tex] \frac{1}{\sqrt{(4K - 120\ ^{2})}}[/tex]
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
It is not a practical proposition to take direct measurements in nanoscale, but we can estimate variations in position and momentum of particles by a)-Scanning Electron Microscopes b)-Transmission Electron Microscope c)-Heisenberg Uncertainly Principle d)- None of the above
Answer:
Answer is c Heisenberg's uncertainty principle
Explanation:
According to Heisenberg's uncertainty principle there is always an inherent uncertainty in measuring the position and momentum of a particle simultaneously.
Mathematically
[tex]\Delta x\times \Delta \overrightarrow{p}\geq \frac{h}{4\pi }[/tex]
here 'h' is planck's constant
A water pipe with a 5 cm inner diameter is designed for to have a flow rate of 75 Umin. What is the entrance length for this pipe (in cm)?
Answer:
121.20 cm
Explanation:
Given data in question
inner dia = 5 cm
flow rate = 75 umin = 1.25 ×10³ cm³/sec
Solution
First we calculate Re by this formula
Re= [tex]\frac{V × D}{v)}[/tex] = [tex]\frac{Q × D}{π/4 × D²× v)}[/tex]
Re= [tex]\frac{4Q }{π × D× v)}[/tex]
here we know Q is flow rate and D is dia of pipe and v is kinematic viscosity that is 1.14 × [tex]10^{-2}[/tex] cm ² / sec
so Re= [tex]\frac{4Q }{π × D× v)}[/tex]
Re = [tex]\frac{4×1.25× 10³ }{π × 5 × 1.14 × 10^{-2} )}[/tex]
So Re will be 27936 that is greater than 4000 thats why it is turbulent flow
and we know [tex]\frac{Length}{dia)}[/tex] ≡ 4.4 [tex]Re^{1/6}[/tex]
so [tex]\frac{Length}{dia)}[/tex] ≡ 24.24
length will be 121.20 cm
Convert 25 mm into in.
Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
1 mm = 1/25.4 inches
From the question, we have to convert 25 mm into inches
Thus,
25 mm = (1/25.4)*25 inches
So,
[tex]25 mm=\frac{25}{25.4} inches[/tex]
Thus, solving we get:
25 mm = 0.984252 inches