Answer
given,
P₁ = 8 bar T₁ = 500 K V₁ = 150 m/s
P₂ = 1 bar T₂ = 320 K V₂ = 10 m/s
writing energy equation
h₁ + (KE)₁ + (PE)₁ + Q m = h₂ + (KE)₂ + (PE)₂ + W
[tex]W = (h_1 - h_2 ) + \dfrac{V_1^2-V_2^2}{2000}[/tex]
ideal gas property of air
T₁ = 500 K h₁ = 503.02 KJ/kg S₁ = 2.21952 kJ/kgK
T₂ = 320 K h₂ = 320.29 KJ/kg S₂ = 1.7679 kJ/kgK
[tex]W = (503.02-320.29) + \dfrac{150^2-10^2}{2000}[/tex]
W = 193.93 KJ/Kg
calculation of energy destruction
= [tex]T_0(S_2-S_1-Rln(\dfrac{P_2}{P_1}))[/tex]
= [tex]T_0(S_2-S_1+Rln(\dfrac{P_1}{P_2}))[/tex]
= [tex]300(1.7679-2.21952-\dfrac{8.314}{28.97}ln(\dfrac{8}{1}))[/tex]
=[tex]300 \times 0.145152[/tex]
=43.54 KJ/Kg
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
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.
What is the mass in both slugs and kilograms of a 1000-lb beam?
Answer:
31.1 slug, 453.4 Kg
Explanation:
given,
mass of the beam is 1000 lb
to convert mass of beam into slugs and kilograms.
1 lb is equal to 0.0311 slug
1000 lb = 1000 × 0.0311
= 31.1 slug
now, conversion of lb into kg
1 lb is equal to 0.4534 kg
so,
1000 lb = 1000 × 0.4534
= 453.4 Kg
hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.
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
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
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.
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.
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per hour c. 2000 km per hour d. 200 km per hour
Answer:d
Explanation:
Given
Temperature[tex]=200^{\circ}\approc 473 K[/tex]
Also [tex]\gamma for air=1.4[/tex]
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.[tex]=\frac{V}{\sqrt{\gamma RT}}[/tex]
(a)[tex]1000 km/h\approx 277.78 m/s[/tex]
Mach no.[tex]=\frac{277.78}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.63
(b)[tex]500 km/h\approx 138.89 m/s[/tex]
Mach no.[tex]=\frac{138.89}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.31
(c)[tex]2000 km/h\approx 555.55 m/s[/tex]
Mach no.[tex]=\frac{555.55}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=1.27
(d)[tex]200 km/h\approx 55.55 m/s[/tex]
Mach no.[tex]=\frac{55.55}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.
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]
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.
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]
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
A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?
Answer:43.70 MPa
Explanation:
Given
mass of engine [tex] 700 lb \approx 317.515 kg[/tex]
diameter of cable [tex]0.375 in.\approx 9.525 mm[/tex]
[tex]A=\frac{\pi d^2}{4}=71.26 mm^2[/tex]
we know stress([tex]\sigma [/tex])[tex]=\frac{load\ applied}{area\ of\ cross-section}[/tex]
[tex]\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa[/tex]
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
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.
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
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 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.
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
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.
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.
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]
How many joules are required to raise the temperature of a cubic meter of water by 10K?
Answer:
4.186 × 10⁷ J
Explanation:
Heat gain by water = Q
Thus,
[tex]m_{water}\times C_{water}\times \Delta T=Q[/tex]
For water:
Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)
Density of water= 1 kg/L
So, mass of the water:
[tex]Mass\ of\ water=Density \times {Volume\ of\ water}[/tex]
[tex]Mass\ of\ water=1 kg/L \times {1000\ L}[/tex]
Mass of water = 1000 kg
Specific heat of water = 4.186 kJ/kg K
ΔT = 10 K
So,
[tex]1000\times 4.186\times 10=Q[/tex]
Q = 41860 kJ
Also, 1 kJ = 1000 J
So, Q = 4.186 × 10⁷ J
What is the difference between point-to-point and continuous path control in a motion control system?
Answer:
Point to point control motion system:
In point to point control motion system tool perform specific task at a particular location.Point to point control motion system is also called positioning system.It perform intermittent operation.
Ex: Drilling operation is a point to point motion control system.
Continuous path control system:
Continuous path control system is continuous operation that perform by tool.The program used in continuous path control system is more complex than point to point motion control system.
Ex :Milling operation is a continuous path control system.
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
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]
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.
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.
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
What is a quasi-equilibrium process? What is its importance in engineering?
Answer:
Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.
Explanation:
Step1
Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.
Step2
All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.