Answer:
a)[tex]v=25.56\ kg/m^3[/tex]
b) [tex]v=38.8\ kg/m^3[/tex]
Explanation:
Given that
Pressure P = 15 MPa
Temperature = 1000 C = 1273 K
a)If assume as ideal gas
The gas constant for super heated steam R is 0.461 KJ/kg.K.
We know that ideal gas equation
P v =RT
15 x 1000 x v =0.461 x 1273
[tex]v=25.56\ kg/m^3[/tex]
b) By using steam table
From steam table we can see that volume of super heated vapot at 15 MPa and 1273 K .
[tex]v=38.8\ kg/m^3[/tex]
Define Plastic vs elastic deformation.
Answer:
Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.
Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.
Name the point of intersection, where the axis meet.
Answer:
origin
Explanation:
The point of intersection of axis is called origin.
In 2D origin is the intersection point of x-axis and y-axis if we go right to the origin then it is positive x axis, if we go left side of origin then it is negative x- axis
Similarly when we go above the origin then it positive y axis , and if we go bellow the origin then it is negative x axis
In 3D origin is the intersection of x-axis, y-axis and z-axis
NOTE- For defining i take here x axis as horizontal axis and y-axis as vertical axis
Why do overhung rotors need to be balanced on or near resonance?
Explanation:
Balancing:
Generally balancing are of two types
1.Static balancing:In this only force balancing is done.
2.Dynamic balancing:in this force as well as moment balancing is done.
Balancing become compulsory for over hanging rotor because unbalance force produce lots of vibration and lots of sound due to this rotor or the whole system in which rotor is attached can be damage.
A(n)______topology is a mixture of more than one type of topology.
Answer:
Hybrid topology is the connection of one or more than one topology.
Explanation:
Topology:
Topology is the arrangement of network.These network connects by line and nodes.
Type of topology:
1.Bus topology
2.Star topology
3.Ring topology
4.Mesh topology
Along with given above topology one topology is also used is known as hybrid topology.Hybrid topology is the connection of one or more than two one above given topology.
The horizontal component of acceleration, ay during a projectile motion is usually assumed to be_________ a)-9.81 m/s^2 b)-Zero c)- Constant d)- 32.2 m/s^2
Answer:
b) zero
Explanation:
The horizontal component of acceleration during projectile motion is usually assumed to be zero.Because in projectile motion horizontal component of velocity will remain the constant and we know that rate of change of velocity with time is called acceleration.So when velocity is constant then acceleration will be zero.
In projectile motion ,gravitational acceleration will be in only vertical direction.
A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft is 0.7R. a) If both the shafts are subjected to the same torque, compare their shear stresses, angle of twist and mass. b) Find the strength to weight ratio for both the shafts.
Answer with Explanation:
By the equation or Torque we have
[tex]\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}[/tex]
where
T is the torque applied on the shaft
[tex]I_{p}[/tex] is the polar moment of inertia of the shaft
[tex]\tau [/tex] is the shear stress developed at a distance 'r' from the center of the shaft
[tex]\theta [/tex] is the angle of twist of the shaft
'G' is the modulus of rigidity of the shaft
We know that for solid shaft [tex]I_{p}=\frac{\pi R^4}{2}[/tex]
For a hollow shaft [tex]I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}[/tex]
Since the two shafts are subjected to same torque from the relation of Torque we have
1) For solid shaft
[tex]\frac{2T}{\pi R^4}\times r=\tau _{solid} [/tex]
2) For hollow shaft we have
[tex]\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4} [/tex]
Comparing the above 2 relations we see
[tex]\frac{\tau _{solid}}{\tau _{hollow}}=0.76[/tex]
Similarly for angle of twist we can see
[tex]\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316[/tex]
Part b)
Strength of solid shaft = [tex]\tau _{max}=\frac{T\times R}{I_{solid}}[/tex]
Weight of solid shaft =[tex]\rho \times \pi R^2\times L[/tex]
Strength per unit weight of solid shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}[/tex]
Strength of hollow shaft = [tex]\tau '_{max}=\frac{T\times R}{I_{hollow}}[/tex]
Weight of hollow shaft =[tex]\rho \times \pi (R^2-0.7R^2)\times L[/tex]
Strength per unit weight of hollow shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}[/tex]
Thus [tex]\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16[/tex]
A window air conditioner unit is placed on a laboratory bench and tested in cooling mode using 750 W of electric power with a COP of 1.75. What is the cooling power capacity and what is the net effect on the laboratory?
Answer:
Q=1312.5 W
Explanation:
Given that
Cooling load or power input = 750 W
COP=1.75
We know that COP can be given as
COP is the ratio of cooling effect to the input power .
Lets take cooling effect is Q.So now by using COP formula
COP=Q/750
1.75=Q/750
Q=1312.5 W
So the cooling effect produce by air conditioning will be 1312.5 W.
The net effect on laboratory,the laboratory temperature will reduce.
What % of Nickel is needed to increase toughness?
Answer:
2% to 20% Ni
Explanation:
If we will talk about steel then ,for increasing the toughness property of steel generally 2% to 20% Ni added .Ni also increase resistance to corrosion and oxidation, impact strength and strength.
We know that steel is an alloy of iron and carbon.But to improve the property of steel different alloying elements added by this steel become desirable to use at different situations.
A cylindrical specimen of some metal alloy having an elastic modulus of 126 GPa and an original cross-sectional diameter of 4.0 mm will experience only elastic deformation when a tensile load of 2380 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.44 mm.
Answer:
The maximum length of the specimen is 0.2927 m or 292.7 mm
Solution:
Modulus of elasticity, E = 126 GPa = [tex]126\times 10^{9}[/tex]
Diameter of the cross-section, D = 4.0 mm = [tex]4.0\times 10^{- 3} m[/tex]
Force due to tension, F = 2380 N
Maximum elongation, [tex]\Delta L = 0.44 mm = 0.44\times 10^{- 3} m[/tex]
Now,
The maximum length of the specimen, [tex]L_{m}[/tex] can be calculated as follows:
The cross-sectional area, [tex]A_{c} = \frac{\pi D^{2}}{4} = \frac{\pi\times (4.0\times 10^{- 3})^{2}}{4} = 1.256\times 10^{- 5} m^{2}[/tex]
Now, the stress on the specimen, [tex]\sigma_{s} = \frac{F}{A_{c}} = \frac{2380}{1.256+\times 10^{- 5}}[/tex]
[tex]\sigma_{s} = 1.89\times 10^{8} N/m^{2}[/tex]
Now,
The strain on the specimen, [tex]\epsilon_{s}[/tex]:
[tex]\epsilon_{s} = \frac{\Delta L}{L_{m}}[/tex]
Also, from Hooke's law:
[tex]E = \frac{\sigma_{s}}{epsilon_{s}}[/tex]
⇒ [tex]E = \frac{1.89\times 10^{8}}{\frac{\Delta L}{L_{m}}}[/tex]
⇒ [tex]L_{m} = \frac{\Delta Ltimes E}{1.89\times 10^{8}}[/tex]
⇒ [tex]L_{m} = \frac{0.44\times 10^{- 3}\times 126\times 10^{9}}{1.89\times 10^{8}} = 0.2927 m[/tex]
The maximum length of the specimen before deformation is: 292.72 mm (0.2927 m).
Tensile PropertiesFor solving this question, it's necessary to know some concepts about the material's properties.
The tensile stress (σ) is determined from the ratio between load and original area before the load applied (σ=[tex]\frac{F}{Ao}[/tex]). Depending on the load applied, the material can have an elastic deformation (temporary deformation) and plastic deformation (permanent deformation). Both deformations can be calculated by the equation: ε=ΔL/Lo, where ΔL= deformation elongation and Lo= the original length before the load applied.
Elastic DeformationWhen the material is in the elastic portion, there is a linear relationship between stress and strain given by: σ=Eε. Due to this relationship, it is possible to find the elastic deformation (ε) when we know the stress (σ) and elastic modulus (E).
Now you have the necessary information to solve your question.
The question gives:
E (elastic modulus) =126 GPA
d (original cross-sectional diameter)=4 mm
F (tensile load)=2380 N
ΔL (maximum allowable elongation) =0.44 mm
1. Find the area of the cylindrical specimen.
[tex]Ao=\frac{\pi *d^2}{4} =\frac{\pi *4^2}{4}=\pi *4=12.57 mm^2[/tex]2. Find the tensile stress.
σ= [tex]\frac{F}{Ao} =\frac{2380 N }{12.57 mm^2} =189.39 MPa[/tex]
3. Calculate the maximum length of the specimen before deformation.
Knowing that ε=ΔL/Lo and σ=Eε, you can rewrite these equations as:
σ= E * (ΔL/Lo)
σ= (E * ΔL)/Lo
The question asks the maximum length of the specimen before deformation, therefore you should find Lo. Thus,
Lo= (E * ΔL)/σ
[tex]Lo=\frac{126*10^3 MPa*0.44 mm}{189.39 MPa} =292.72 mm= 0.2927 m[/tex]
Read more about the tensile stress here:
https://brainly.com/question/19756298
Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is 0.9 cm. The water density is 1.0 kg/L. The empty weight of the metal pipe is 2500 N/m. In kN, what is the total weight (pipe plus water)?
Answer:
1113kN
Explanation:
The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:
Inside Diameter = Outside Diameter - Thickness
Inside Diameter = 61cm - 0.9cm = 60.1cm
Converting this diameter to meters, we have:
[tex]60.1cm*\frac{1m}{100cm}=0.601m[/tex]
This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:
[tex]V_{water}=\pi r^{2}h[/tex]
[tex]V_{water}=\pi (\frac{0.601m}{2})^{2}*120m[/tex]
[tex]V_{water}=113.28m^{3}[/tex]
The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:
[tex]d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}[/tex]
[tex]d_{water}=1000\frac{Kg}{m^{3}}[/tex]
Now, water density is given by the equation [tex]d=\frac{m}{V}[/tex], where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:
[tex]m_{water}=d_{water}.V_{water}[/tex]
[tex]m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}[/tex]
[tex]m_{water}=113280Kg[/tex]
With the water mass we can find the weight of water:
[tex]w_{water}=m_{water} *g[/tex]
[tex]w_{water}=113280kg*9.8\frac{m}{s^{2}}[/tex]
[tex]w_{water}=1110144N[/tex]
Finally we find the total weight add up the weight of the water and the weight of the pipe,so:
[tex]w_{total}=w_{water}+w_{pipe}[/tex]
[tex]w_{total}=1110144N+2500N[/tex]
[tex]w_{total}=1112644N[/tex]
Converting this total weight to kN, we have:
[tex]1112644N*\frac{0.001kN}{1N}=1113kN[/tex]
The "Crawler" developed to transport the Saturn V launch vehicle from the assembly building to the launch pad is the largest land 6 vehicle ever built, weighing 4.9 x 10 -Ibs at sea level. a- What is its mass in slugs ? b- What is its mass in kilograms ?
Answer:
a) 152000 slugs
b) 2220000 kg or 2220 metric tons
Explanation:
A body with a weight of 4.9*10^6 lbf has a mass of
4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm
This mass value can then be converted to other mass values.
1 slug is 32.17 lbm
Therefore:
4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs
1 lb is 0.453 kg
Therefore:
4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg
Polymers can be natural or synthetic. a)-True b)- false?
Answer:
TRUE
Explanation:
Polymers can be natural as well as synthetic
The polymer which are found in nature are called natural polymer tease polymer are not synthesized, they are found in nature
Example of natural polymers is cellulose, proteins etc
On the other hand synthetic polymers are not found in nature they are synthesized in market
There are many example of synthetic polymer
Example : nylon, Teflon etc
So it is a true statement
Solar energy is the most widely available renewable energy source and it is sufficient to meet entire needs of the world. However, it is not used extensively. Why?
Answer:
Explanation:
Solar energy is one of the Renewable sources of energy and it is sufficient to meet the entire needs of the world.
But it is not used extensively because of the inconsistency. Inconsistency means that sun rays are not consistent throughout the year. If the weather is not good or in winters earth does not receive sun rays continuously.
And another reason to not to set up solar energy is that it is very expensive not everyone can afford it.
The searchlight on the boat anchored 2000 ft from shore
isturned on the automobile, which is traveling along the
straightroad at a constant speed of 80 ft/s. Determine the
angularrate of rotation of the light when the automobile is r =
3000 ftfrom the boat.
The angular speed of a flywheel rotating at 300 revolutions per minute is calculated by converting revolutions to radians and minutes to seconds, resulting in an angular speed of 10π rad/s.
Explanation:The question asks to determine the angular rate of rotation of a searchlight on a boat when targeting an automobile that is 3000 ft away and moving at 80 ft/s. This involves understanding the relationship between linear velocity, radius, and angular velocity, which is a crucial concept in trigonometry and physics.
To find the angular speed of the flywheel rotating at 300 revolutions per minute, we first convert revolutions per minute (rpm) to radians per second, knowing that one revolution is 2π radians and there are 60 seconds in a minute. Therefore:
Angular speed = 300 rpm × (2π radians/revolution) × (1 minute/60 seconds) = 300 × 2π / 60 rad/s = 10π rad/s.
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is 0.95 rad/s and the natural time period of the axial vibration is found to be 0.35 sec. What is the mass of the object.
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T= [tex]2\pi\sqrt{\frac{L}{g}}[/tex]
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35= [tex]2\pi\sqrt{\frac{L}{9.81}}[/tex]
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K = [tex]\frac{\textup{AE}}{\textup{L}}[/tex]
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K = [tex]\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}[/tex]
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ = [tex]\sqrt{\frac{K}{m}}[/tex]
or
mass, m = [tex]\sqrt{\frac{K}{\omega_n^2}}[/tex]
or
mass, m = [tex]\sqrt{\frac{20.66\times10^8}{0.95^2}}[/tex]
mass, m = 22.90 × 10⁸ kg
A car starts out from rest (zero velocity) at an elevation of 500 m and drives up a hill to reach a final elevation of 2000m and a final velocity of 20 m/s. At the same time the entire car heats up so the Internal Energy of the car increases by 100 kJ. What is the total energy change of the car if its mass is 2000 kg?
Answer:29,930 kJ
Explanation:
Given
Car starts with an initial elevation of 500 m and drives up a hill to reach a final elevation of 2000 m
Final velocity (V)=20 m/s
Energy of car increases by 100 kJ
mass of car(m)=2000 kg
[tex]Total Energy =\Delta PE+\Delta KE+\Delta U[/tex]
[tex]\Delta PE=mg(\Delta h)=2000\times 9.81\times (2000-500)[/tex]
[tex]\Delta PE=29,430 kJ[/tex]
[tex]\Delta KE=m\frac{v_2^2-v_1^2}{2}[/tex]
[tex]\Delta KE=2000\times \frac{20^2-0^2}{2}[/tex]
[tex]\Delta KE=400 kJ[/tex]
[tex]\Delta U=100 kJ[/tex]
Total Energy=29,430+400+100=29,930 kJ
2.4 kg of nitrogen at an initial state of 285K and 150 kPa is compressed slowly in an isothermal process to a final pressure of 600 kPa. Determine the work.
Answer:
W=-280.67 KJ
Explanation:
Given that
Initial pressure = 150 KPa
Final pressure = 600 KPa
Temperature T= 285 K
Mass m=2.4 Kg
We know that ,work in isothermal process given as
[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]
Gas constant for nitrogen gas R=0.296 KJ/kgK
Now by putting the values
[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]
[tex]W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}[/tex]
W=-280.67 KJ
Negative sign indicates that it is compression process and work is done on the gas.
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a conical section at the bottom, right circular cylindrical mid-section and a hemispherical dome to cover the top. The radius of the tank is 1.5 m, the cylindrical side-walls will be 4.0 m in height, and the apex of the conic section at the bottom has an included angle of 60°. If the tank is filled to the top of the cylindrical side-walls, what is the tank capacity in liters?
Answer:
The volume up to cylindrical portion is approx 32355 liters.
Explanation:
The tank is shown in the attached figure below
The volume of the whole tank is is sum of the following volumes
1) Hemisphere top
Volume of hemispherical top of radius 'r' is
[tex]V_{hem}=\frac{2}{3}\pi r^3[/tex]
2) Cylindrical Middle section
Volume of cylindrical middle portion of radius 'r' and height 'h'
[tex]V_{cyl}=\pi r^2\cdot h[/tex]
3) Conical bottom
Volume of conical bottom of radius'r' and angle [tex]\theta [/tex] is
[tex]V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}[/tex]
Applying the given values we obtain the volume of the container up to cylinder is
[tex]V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}[/tex]
Hence the capacity in liters is [tex]V=32.355\times 1000=32355Liters[/tex]
What is the pressure inside a tire in (N/mm^2) if a pressure gauge indicates 29.35 psi?
Answer:
The pressure inside the tire is [tex]0.304\frac{N}{mm^{2}}[/tex]
Explanation:
The pressure gauge indicates the difference between the atmospheric pressure and the pressure inside the tire, so we have the following equation:
Pressure inside the tire = Gauge pressure + Atmospheric pressure
Where the gauge pressure is given in the problem and is 29.35psi and the atmospheric pressure is 14.7psi.
Replacing the values, we have:
Pressure inside the tire = 29.35psi + 14.7psi
Pressure inside the tire = 44.05psi
Now we have to convert from psi to [tex]\frac{N}{mm^{2}}[/tex], so:
44.05psi = [tex]44.05\frac{lbf}{in^{2}}[/tex]
[tex]44.05\frac{lbf}{in^{2}}*(\frac{1in}{25.4mm})^{2}*\frac{4.4482N}{lbf}=0.304\frac{N}{mm^{2}}[/tex]
Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot-water tank from 60°F to 110°F. The specific heat of water is approximated as a constant, whose value is 0.999 Btu/·lbmR at the average temperature of (60 + 110)/2 = 85ºF. In fact, c remains constant at 0.999 Btu/lbm·R (to three digits) from 60ºF to 110ºF. For this same temperature range, the density varies from 62.36 lbm/ft3 at 60ºF to 61.86 lbm/ft3 at 110ºF. We approximate the density as remaining constant, whose value is 62.17 lbm/ft3 at the average temperature of 85ºF.
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
[tex]1\ gallon = 0.13ft^3[/tex]
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state where P2= 6.8 bar. During the process, the relation between pressure and volume follows pv= constant. For the air as the closed system, determine the work, in kJ
Answer:
W=-940.36 KJ
Explanation:
Given that
[tex]P_1=1\ bar,V_1=4.25 {m^3}[/tex]
[tex]P_2=6.8\ bar[/tex]
Process follows pv=constant
So this is the isothermal process and work in isothermal process given as
[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]
Now by putting the values (1.4 bar =140 KPa)
[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]
[tex]W=140\times 4.25 \ln \dfrac{1.4}{6.8}[/tex]
W=-940.36 KJ
Negative sign indicates that this is a compression process and work will given to the system.
Describe the cartesain coordinate system.
Answer:
Cartesian coordinate system is used to specify any point on a plane.
In two dimensional plane,the two types of axes or coordinates are [tex]x[/tex] and [tex]y[/tex] axis.
Explanation:
Cartesian coordinate system specifies each point and every point on axes.
A Cartesian Coordinate system in two dimension also named as rectangular coordinate system.
The two types of axes or coordinates are [tex]x[/tex] and [tex]y[/tex] axis.
The horizontal axis is [tex]x[/tex]-axis and vertical axis is named as [tex]y[/tex]-axis.
The coordinate system specifies each point as a set of numerical coordinates in a plane which are signed from negative to positive. that is from ([tex]-\infty[/tex],[tex]\infty[/tex]).
In three dimensional plane, [tex]x[/tex], [tex]y[/tex] and [tex]z[/tex] coordinates are used and in two dimensional plane, [tex]x[/tex] and [tex]y[/tex]coordinates are used to address any point in the interval ([tex]-\infty[/tex],[tex]\infty[/tex]).
For defining both the coordinates, the two perpendicular directed lines [tex]x[/tex]- axis and [tex]x[/tex]-axis are drawn.
Now, for example [tex](3,4)[/tex] is a point in which indicates that the value of [tex]x[/tex] is [tex]3[/tex] and [tex]y[/tex] is [tex]4[/tex].
It is drawn by moving [tex]3[/tex] units right from the origin to positive [tex]x[/tex] axis and [tex]4[/tex] units upwards from the origin [tex](0,0)[/tex] to positive [tex]y[/tex]-axis.
When a fluid flows through a sharp bend, low pressures may
developin localized regions of the bend. Estimate the
minimumabsolute pressure (in psi) that can develop without
causingcavitation if the fluid is water at 160 oF.
To avoid cavitation for water at 160 °F flowing through a sharp bend, the minimum absolute pressure should be slightly above the vapor pressure of water at this temperature, which is approximately 0.363 psi.
Explanation:When water flows through a sharp bend, cavitation can occur if the local pressure falls to or below the fluid's vapor pressure. To estimate the minimum absolute pressure without causing cavitation for water at 160 °F, we must consider water's vapor pressure at this temperature. At 160 °F (about 71 °C), the vapor pressure of water is approximately 0.363 psi. Since fluids cannot have a negative absolute pressure, and to avoid cavitation, the absolute pressure must stay above this vapor pressure. Therefore, considering atmospheric pressure to be approximately 14.7 psi, to avoid cavitation, the minimum absolute pressure in the system should be slightly above 0.363 psi to ensure no cavitation occurs.
You want a pot of water to boil at 105celcius. How heavy a
lid should you put on the 15 cm diameterpot when Patm =
101 kPa?
Answer:
36 kg
Explanation:
For water to boil at 105 C it needs a pressure of 121 kPa (this is the vapor pressure of water at 105 C).
In the lid there will be a difference of pressure from one side to the other, this will be compensated by the weight of the lid.
Δp = pwater - patm
Δp = 121 - 101 = 20 kPa
The pressure caused by the weigh of the lid is:
Δp = w / A
Δp = m * g / A
Rearranging
m = Δp * A / g
m = Δp * π/4 * d^2 / g
m = 20000 * π/4 * 0.15^2 / 9.81 = 36 kg
Find the diameter of the test cylinder in which 6660 N force is acting on it with a modulus of elasticity 110 x 103 Pa. The initial length of the rod is 380 mm and elongation is 0.50 mm.
Answer:
The diameter of the test cylinder should be 7.65 meters.
Explanation:
The Hooke's law relation between stress and strain is mathematically represented as
[tex]Stress=E\times strain\\\\\sigma =e\times \epsilon[/tex]
Where 'E' is modulus of elasticity of the material
Now by definition of strain we have
[tex]\epsilon =\frac{\Delta L}{L_{o}}[/tex]
Applying values to obtain strain we get
[tex]\epsilon =\frac{0.5}{380}=0.001316[/tex]
Thus the stress developed in the material equals
[tex]\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}[/tex]
Now by definition of stress we have
[tex]\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}[/tex]
Solving for 'D' we get
[tex]D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters[/tex]
What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R
Answer:
-273.16 °C
-459.677 °F
0 °K
0 °R
Explanation:
The lowest temperature is the absolute zero.
Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.
Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.
Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.
The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.
The two basic network administration models are ____ and ____
The two basic network administration models are centralized and decentralized, each with distinct advantages and challenges that affect network performance, security, and scalability.
Explanation:The two basic network administration models are centralized and decentralized. In a centralized model, network control and decision-making are located at a single point, typically within a dedicated device or group of servers. Conversely, a decentralized model distributes control across multiple nodes, allowing for individual nodes to operate independently while still being part of the overall network.
Understanding these models is crucial for designing efficient networks that cater to specific organizational needs and for implementing dynamics on network models, such as discrete state/time models or continuous state/time models. Each model presents different advantages and challenges that can influence network performance, security, and scalability.
I have a 500 L tank that can hold a fluid. How much heavier or lighter will the tank be if a. The tank holds water at 80C compared to if the water tank holds 5C water? b. The tank holds air at 80C compared to if the water tank holds 5C air? Assume the tank is vented in a way that the pressure remains at atmospheric pressure, given as 101.3 KPa.
Answer:
water=14.1 kg
air=0.1348kg
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
To solve this exercise we must find the specific volume of water and air in the two states and then subtract them and multiply them by the volume of the tank to find the change in mass, the following equation is inferred
Δm=V(ρ2-ρ1)
Where V= tank volume=500L=0.5m^3
ρ2= density of fluid in state 2
ρ1=density of fluid in state 1
Δm=change of mass
for water
ρ1=971.8kg/m^3(80C)
ρ2=1000kg/m^3(5C)
Δm=0.5(1000-971.8)
Δm=14.1 kg
for air
ρ1=0.9994kg/m^3(80C)
ρ2=1.269kg/m^3(5C)
Δm=0.5(1.269-0.9994)
Δm=0.1348kg
A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 30°, what must be the length of the bar?
Answer:
The length of bar will be 2.82 m
Explanation:
Given that
d= 15 mm
r= 7.5 mm
Shear stress = 110 MPa
θ = 30° (30° = 30° x π/180° =0.523 rad)
θ = 0.523 rad
G for steel
G= 79.3 GPa
We know that
[tex]\dfrac{\tau}{r}=\dfrac{G\theta }{L}[/tex]
[tex]\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}[/tex]
L= 2. 82 m
The length of bar will be 2.82 m
an aluminum bar 125mm (5in) long and having a square cross section 16.5mm (.65in) on an edge is pulled in tension with a load of 66,700 N (15000lb) and experiences an elongation of .43mm (1.7*10^-2 in ). assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.
The modulus of elasticity for the aluminum bar is calculated using the relationship between stress and strain, given by Young's modulus formula. The given values are substituted in the formula to find that the modulus of elasticity for the aluminum is approximately 60 GPa.
Explanation:Calculating the Modulus of Elasticity for Aluminum
To calculate the modulus of elasticity (also known as Young's modulus) for the aluminum bar, we can use the relationship described by Hooke's Law for elastic deformation, where stress is directly proportional to strain. The formula to find Young's modulus (Y) is:
Y = (F \/ A) \/ (\u0394L \/ L0)
Here, F is the force applied, A is the cross-sectional area, \u0394L is the change in length, and L0 is the original length of the bar.
Using the given values:
Force (F) = 66,700 NCross-sectional area (A) = 16.5 mm x 16.5 mm = 272.25 mm2 (or 272.25 x 10-6 m2)Change in length (\u0394L) = 0.43 mm (or 0.43 x 10-3 m)Original length (L0) = 125 mm (or 0.125 m)The calculation is as follows:
Y = (66,700 N / 272.25 x 10-6 m2) / (0.43 x 10-3 m / 0.125 m)
When the calculations are done, we find that the modulus of elasticity for the aluminum bar is approximately 60 GPa (Gigapascals), which is within the typical range for aluminum.
Calculate the modulus of elasticity of aluminum given specific values for force, length, cross-sectional area, and elongation during tension.
Young's modulus is a measure of a material's stiffness and is calculated using the formula: Y = (F * Lo) / (A * AL). Substituting the given values, we can find the modulus of elasticity of aluminum to be approximately 68 GPa.