Draw the Pressure - Temperature Diagram showing the liquid and vapor phases, along with the saturation line and the critical point (no need to include the solid section).
Answers
Answer:
Pressure- temperature diagram of the fluid is the phase lines that separate all the phases.
Explanation:
Step1
Pressure temperature diagram is the diagram that represents the all the phases of the fluid by separating a line. There is no phase change region in the pressure temperature diagram out of 15 possible diagrams. There are three lines that separate the phase of the fluid. These three lines are fusion, vaporization and sublimation.
Step2
The intersecting point of these lines is triple point of fluid. Out of 15 possible phase diagram, only pressure temperature diagram has triple point as a point. In other diagrams phase change region is present and triple point is not a point. Critical point is the point in all possible property diagrams.
Pressure temperature diagram is shown below:
Related Questions
A fluid flows through a pipe of constant circular cross-section diameter 1.1m with a mean velocity of 50cm/s. The mass flow rate of the flow through the pipe, given by the formula mass flow rate = (density) (mean velocity) (cross - sectional area), is measured to be 110 kg/s. Find the specific gravity, specific weight and specific volume of the fluid.
Answers
Answer:
Explanation:
Given
diameter(d)=1.1 m
mean velocity(u)=50 cm/s
mass flow rate(m)=110 kg/s
[tex] A=0.950 m^2[/tex]
[tex]m=\rho \cdot A\cdot u[/tex]
where [tex]\rho [/tex] =density of fluid
[tex]110=\rho \times 0.95\times 0.5[/tex]
[tex]\rho =231.46 kg/m^3[/tex]
Therefore specific gravity[tex]=\frac{\rho }{\rho _{water}}[/tex]
[tex]=\frac{231.46}{1000}=0.231[/tex]
Specific weight[tex]=\rho \cdot g=231.46\times 9.81=2270.70 kg/m^2-s^2[/tex]
specific volume[tex]=\frac{1}{density}[/tex]
[tex]=\frac{1}{231.46}=0.00432 m^3/kg[/tex]
A thin film of oil (v=0.001 m^2/s) 2mmthick flows down to
a surface inclined at 30 degrees to thehorizontal. What is the
maximum and mean velocity offlow?
Answers
Answer:
Maximum and mean velocity will be equal i.e 0.5 m/s
Explanation:
given data:
viscosity = 0.001 m^2/s
Thickness of thin film = 2 mm = 0.002 m
Neglecting the body weight hence no shear stress
Mean velocity is given V
[tex]V = \frac{\nu}{D}[/tex]
[tex]V = \frac{0.001}{2*10^{-3}}[/tex]
V = 0.5 m/s
Maximum and mean velocity will be equal i.e 0.5 m/s
A motocycle starts from rest at t = 0.0 on a circular track
of400 meter radious. The tangential component of acceleration
isat=2+0.2t m/sec^2. At time t =10 sec determine
thedistance the motorcycle has moved along the track and the
magnitudeof its acceleration. (Knowing that a=at
+an).
Answers
Answer:
[tex]S_{10}=133.3m\\a_{10}=4.589m/s^{2}[/tex]
Explanation:
In order to know the value of the speed at any time t, we need to integrate the acceleration. Once we get the speed vs time, we need to integrate again to get the distance traveled by the motorcycle vs time. So let's start with the speed first:
[tex]V(t) = \int {a(t)} \, dt = \int {2+0.2*t} \, dt = 2*t + 0.2*\frac{t^{2}}{2}[/tex]
We will integrate once again to get distance:
[tex]S(t) = \int {V(t)} \, dt = \int {2*t+0.2*\frac{t^{2}}{2} } \, dt = t^{2}+0.2*\frac{t^{3}}{6}[/tex]
Now we just need t evaluate S(10s):
S(10) = 133.3m
For the acceleration, we know that:
[tex]a = \sqrt{a_{t}^{2}+a_{N}^{2}}[/tex]
where [tex]a_{t}(10)=2+0.2*(10)=4m/s^{2}[/tex]
and [tex]a_{N}(10)=\frac{V(10)^{2}}{R}=2.25m/s^{2}[/tex]
So, finally:
[tex]a = \sqrt{a_{t}^{2}+a_{N}^{2}} = 4.589m/s^{2}[/tex]
How do you define a renewable energy source? Why are coal, oil, and natural gas not renewable energy sources?
Answers
Answer:
Explanation:
A renewable energy source is one that can not depelted in a reasonable timeframe.
Sun energy will last until the Sun dies, which is expected to haen in about 5 billion years. Because it will last so long it is considered practiacally eternal and undepleteable.
Fossil fuels on the other hand are material deposits in the crust of Earth that might be depleted within centuries.
A renewable energy source can be replenished rapidly and is not permanently consumed by human use, examples include solar and wind energy. Fossil fuels like coal, oil, and natural gas are non-renewable because they take millions of years to form and are finite. Sustainable energy use does not exhaust resources and can continue over time.
Explanation:A renewable energy source is defined as one that can be replenished in a short amount of time and is not permanently depleted by human usage. Renewable resources include things like solar and wind energy, which are continually available through natural processes such as sunlight and wind currents.
Coal, oil, and natural gas are not considered renewable energy sources because they are formed over millions of years from the buried remains of plants and animals. These fossil fuels are finite and once consumed, they do not replenish in a timeframe that is practical for human use, thereby classifying them as non-renewable.
When we use renewable resources, they offer a sustainable way to generate energy. Sustainable means using methods that do not completely use up or destroy natural resources, therefore capable of continuing for a long time. Examples of methods to harness renewable resources include solar panels for sunlight, wind turbines for wind, and hydroelectric power stations for flowing water.
A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is 75 F. If the surface area of the hot tub is 7.5 m2 , what is the heat loss (heat rate) due to convection? The temp of Hot Tube is 102F.
Answers
Answer:4050 W
Explanation:
Given
Heat transfer Coefficient(h)=[tex]20 W/m^2-K[/tex]
Air temperature =75 F
surface area(A)=[tex]7.5 m^2[/tex]
Temperature of hot tube is 102 F
We know heat transfer due to convection is given by
[tex]Q=hA\left ( \Delta T\right )[/tex]
[tex]Q=20\times 7.5\left ( 102-75\right )=4050 W[/tex]
Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one end of the beam is free and one end is fixed. The cross section has a base of 100 mm and height of 300 mm, length of the beam is 5.9 m, E=20.5x10^10 N/m2 and density of 7830 kg/m3. Write your answer in rad/sec with 2 decimal points.
Answers
Answer:
The natural angular frequency of the rod is 53.56 rad/sec
Explanation:
Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure
We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by
[tex]\Delta x=\frac{PL^3}{3EI}[/tex]
Hence we can write
[tex]P=\frac{3EI\cdot \Delta x}{L^3}[/tex]
Comparing with the standard spring equation [tex]F=kx[/tex] we find the cantilever analogous to spring with [tex]k=\frac{3EI}{L^3}[/tex]
Now the angular frequency of a spring is given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
where
'm' is the mass of the load
Thus applying values we get
[tex]\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}[/tex]
[tex]\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec[/tex]
Answer:
its a lil confusing
Explanation:
Determine the activation energy for diffusing copper into gold if you know that the diffusion coefficient is 3.98 x 10-13 m2/s at 980 degreeC and the diffusion coefficient is 3.55 x 10-16 m2/s at 650degreeC.
Answers
Answer:
[tex]Q_d = 0.0166 J/mol[/tex]
Explanation:
given data:
diffusion coefficient [tex]= 3.98 \times 10^{-13} m^2/s[/tex]
[tex]T_1 = 980 Degree\ celcius = 1253 K[/tex]
[tex]T_2 = 650 Degree\ celcius = 923 K[/tex]
[tex]D_2 = 3.55\times 10^{-16} m^2/s[/tex]
Activation energy is given as[tex] = -2.3R \frac{ \Delta log D}{\Delta\frac{1}{T}}[/tex]
[tex] Q_d = -2.3 R [\frac{logD_1 - logD_2}{ \frac{1}{T_1} - \frac{1}{T_2}}][/tex]
[tex]Q_d = -2.3 \times 8.31 \frac{log(3.98*10^{-13} - log(3.55*10^{-16}}{ \frac{1}{1253} - \frac{1}{923}}[/tex]
[tex]Q_d = 0.0166 J/mol[/tex]
Consider a single pane window. The dimensions of the ¼-inch-thick glass window pane are about 0.9 m by 1.65 m. The surface temperatures of the glass will not be equal to the air temperature on the respective sides, as will be seen later in the course. For these conditions, the glass surface temperatures are approximately 1°F and -7°F. (At this time, we do not have the HT skills to calculate the surface temperatures.) Determine the heat transfer rate vector through the glass for these surface temperatures. Express your answer in vector notation. Your sketch should identify your coordinate system and your vector answer should agree with your sketch. For consistency, set your coordinate system such that the inside glass surface is at the origin and the outside surface is in the positive x direction (typically to the right!).
Answers
Answer:
q = (709*K*i + 0*j + 0*k)
Explanation:
1/4 inch is 6.35 mm = 0.00635 m
A = 0.9 * 1.65 = 1.48 m^2
t1 = 1 F = -17.2 C
t2 = -7 F = -21.7 C
The heat will be conducted through the glass. We use a frame of reference with the origin on the warmer side of the glass (t1) and the positive X axis pointing perpendicular to the glass towards the colder side.
The Fourier equation for plates (one dimensional form) is:
q = -K * dT / dx
We can simplify dT/dx to ΔT/Δx if we assume the temperature changes linearly through the glass.
q = -K * ΔT/Δx
K is the thermal conductivity of the glass
ΔT = t2 - t1
ΔT = -21.7 + 17.2 = -4.5 C
Δx is the thickness of the glass
q = -K * -4.5 / 0.00635
q = 709 * k
The vector is: q = (709*K*i + 0*j + 0*k)
It has no components in Y or Z because there if no variation of temperature in those directions (the temperature of the glass is given as each face having a constant temperature throughout its area).
Draw and write shear strain equation
Answers
Answer:
[tex]\phi =\dfrac{\Delta L}{h}[/tex]
Explanation:
Normal strain :
It is the ratio of deformation to the original dimension.Normal strain occurs due to longitudinal force.
Normal strain = ΔL/L
Shear strain:
It is the ratio of deformation to perpendicular dimension.Shear strain occurs due to tangential force.
From the second diagram:
[tex]tan\phi =\dfrac{\Delta L}{h}[/tex]
if angle Φ is small then we can write
[tex]tan\phi =\phi[/tex]
So
[tex]\phi =\dfrac{\Delta L}{h}[/tex]
This is the shear strain equation.
Write a matrix, that is a lower triangular matrix.
Answers
Answer:
[tex]\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right][/tex]
Explanation:
A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.
If there is any question, just let me know.
What is the range of a 32-bit signed integer?
Answers
Answer:
From -2147483647 to 2147483648.
Explanation:
A 32 bit integer has 2^32 = 4294967296 possible values. A signed integer has positive and negative values as well as the zero.
Of these values (2^32)/2 will be positive, (2^32)/2 - 1 will be negative and one will be the zero.
Therefore the range is from -(2^32)/2 + 1 to (2^32)/2.
This can also be expressed as from -2147483647 to 2147483648.
Momentum is a conserved quantity for any general control volume. a)- True b)- False
Answers
Answer:
The given statement is false.
Explanation:
The basic equation of motion for a control volume is as follows
[tex]\frac{d\overrightarrow{p}}{dt}=\int_{c.v}\frac{\partial }{\partial t}(\rho v)dV+\int_{cs}(\rho \overrightarrow{v})\cdot \overrightarrow{v_{r}}.\widehat{n}dS[/tex]
the symbols have the usual meaning as
[tex]\rho [/tex] is density of the fluid
[tex]v [/tex] is the velocity of the fluid
[tex]\widehat{n}[/tex] is the direction vector of area over which the integration is carried out
As we see that the terms in the right hand side of the equation is not zero if the flow is unsteady or the velocity is changing in the control volume the term in the left is non zero hence the momentum is not conserved.
It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?
Answers
Answer with Explanation:
The crown will be pure if it's specific gravity is 19.3
Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume
Since it is given that the weight of the crown is 11.8 N we need to find it's volume
Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown
Mathematically this is the difference between the weight of the crown in air and weight when immersed in water
Thus Buoyant force is [tex]F_{B}=11.8-10.9=0.9N[/tex]
Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown
Thus the specific gravity of the crown equals
[tex]S.G=\frac{11.8}{0.9}=13.11[/tex]
As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.
Given:
[tex]\to SG_{gold} = 19.3 \\\\[/tex]
weight of air [tex]W_{air} = 11.8\ N \\\\[/tex]
water weight [tex]W_{water} = 10.9 \ N \\\\[/tex]
To Find:
using the buoyancy B that calculates the difference of weight=?
Solution:
Using formula:
[tex]\to B = W_{air} - W_{water} \\\\[/tex]
[tex]= 11.8\ N - 10.9\ N\\\\ = 0.9 \ N\\\\[/tex]
Using a formula for calculating weight:
[tex]\to W_{air} = SG_{\gamma water}\times V_{crown}\\\\ \to W_{water} = B(SG - 1)[/tex]
calculating the [tex]\bold{SG_{crown}}[/tex]:
[tex]\to SG_{crown}=1+\frac{W_{water}}{B} \\\\[/tex]
[tex]= 1 + \frac{10.9\ N}{0.9\ N}\\\\= 1 + 12.11\\\\= 13.11[/tex]
Knowing that the [tex]SG_{gold}= 19.3[/tex] indicates that the crown is not made of pure gold.
[tex]\to \bold{SG_{crown}= 13.1}[/tex] The crown is not composed entirely of gold.
Find out more information about the gold here:
brainly.com/question/900003
A circular piston exerts a pressure of 80kPa on a fluid
whenthe force applied to the piston is 0.2kN . Find the diameter of
thepiston.
Answers
Answer:
Diameter of the piston will be [tex]3.1847\times 10^{-3}m[/tex]
Explanation:
We have given force F = 0.2 kN = 200 N
Pressure P = 80 kPa = 80000 pa
We have to find the diameter of the piston
We know that [tex]pressure = \frac{force}{area }=\frac{200}{80000}=2.5\times 10^{-3}m^2[/tex]
We know that area of circular piston is given by [tex]A=\pi r^2[/tex]
So [tex]2.5\times 10^{-3}=3.14\times r^2[/tex]
[tex]r=1.5923\times 10^{-3}m[/tex]
We know that diameter d = 2×radius =[tex]d=1.5923\times 10^{-3}\times 2=3.1847\times 10^{-3}m[/tex]
Write the chemical equation of the polymerization reaction to synthesize PS
Answers
Explanation:
Styrene is a vinyl monomer in which there is a carbon carbon double bond.
The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).
The equation is shown below as:
[tex]\begin{matrix}& C_6H_5 \\&|\\n H_2C & =CH\end{matrix}[/tex] ⇒ [tex]\begin{matrix}&C_6H_5 \\&|\\ -[-H_2C & -CH-]-_n\end{matrix}[/tex]
Vernier calipers are capable of taking readings to the nearest 0.001 in. a)- True b)- false
Answers
Answer:
False
Explanation:
Vernier caliper is used for measuring very small distances. It is very precise instrument for measuring very small distances.
With the help of vernier caliper we can measure distances from 6 inch to about 12 inch
But in question we have given 0.001 in so it is impossible to measure 0.001 inch from vernier caliper so it is a false statement
Under a construction management (CM) contract, should the CM firm's responsibilities normally begin at the construction phase, the design phase, the planning phase, or the conceptual phase?
Answers
The CM firm's responsibilities under a construction management contract should begin at the planning phase. This early engagement allows for a cohesive approach to project management, involving definition of scope, budgeting, and risk assessment. Construction managers work closely with designers and contractors throughout the project, ensuring it meets design specifications and safety standards.
Explanation:Under a construction management (CM) contract, the CM firm's responsibilities should begin at the planning phase, which precedes the design phase, construction phase, and conceptual phase. During the planning phase, the CM firm is involved in defining project objectives, scope, and feasibility, providing valuable input that shapes the overall project. The expertise of construction managers is utilized to help with site selection, project scheduling, budgeting, and risk management. Their early involvement ensures that the construction phase flows more smoothly, as potential issues may have been identified and addressed beforehand. Construction managers play a crucial role in maintaining the integrity of the design and adherence to specifications throughout the construction process.
Designers, on the other hand, continue their involvement even after generating detailed technical drawings and digital models by acting as on-site supervisors during construction. Their responsibility remains to ensure that contractors follow the design's specifications, indicating direct involvement in the contract construction phase. The construction site can present various risks, including the exposure of workers to dangerous materials which designers need to consider while specifying project materials. The collaborative effort between the CM firm, designers, and contractors is fundamental to achieving a successful and safe project completion, fulfilling the client's expectations, and managing any equity issues related to contract allocations.
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−1. The exhaust from the turbine is carried through a 15-cm-diameter pipe and is at 50 kPa and 100°C [state 2]. What is the power output of the turbine?
H1 = 3150.7 kJ/kg V1 = 0.2004 m3/kg
H2 = 2682.6 kJ/kg V2 = 3.4181 m3/kg
Answers
Answer:
Power output, [tex]P_{out} = 178.56 kW[/tex]
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, [tex]T = 350^{\circ}C[/tex]
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, [tex]\dot{m} = 0.1 kg.s^{- 1}[/tex]
Diameter of exhaust pipe, [tex]d_{h} = 15 cm = 0.15 m[/tex]
Pressure at exhaust, P' = 50 kPa
temperature, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:
[tex]\dot{m} = \frac{Av_{i}}{V_{1}}[/tex]
[tex]0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]v_{i} = 3.986 m/s[/tex]
Now, eqn for compressible fluid:
[tex]\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}[/tex]
Now,
[tex]\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}[/tex]
[tex]v_{e} = 19.33 m/s[/tex]
Now, the power output can be calculated from the energy balance eqn:
[tex]P_{out} = -\dot{m}W_{s}[/tex]
[tex]P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}[/tex]
[tex]P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW[/tex]
The power output of the Turbine is; 225.69 kW
What is the Power Output?We are given
Pressure of steam; P = 1400 kPa
Temperature of steam at state 1; T = 350°C
Diameter of pipe; d₁ = 8 cm = 0.08 m
Mass flow rate; m' = 0.1 kg/s
Diameter of exhaust pipe; d₂ = 15 cm = 0.15 m
Pressure at exhaust; P' = 50 kPa
Temperature at state 2; T' = 100°C
Area; A = πd²/4
A = π * 0.08²/4
A = 0.0016π m²
We can find the find initial velocity from the formula;
v₁ = m' * V₁/A
v₁ = (0.1 * 0.2004/(0.0016π))
v₁ = 3.986 m/s
From equation of compressible fluid, we know that;
(A₁ * v₁)/V₁ = (A₂ * v₂)/V₂
A₂ = πd₂²/4
A₂ = π * 0.15²/4
A₂ = 0.005625π m²
(0.0016π * 3.986)/0.2004 = (0.005625π * v₂)/3.4181
Solving for v₂ gives;
v₂ = 19.33 m/s
Finally power output is gotten from the expression;
P_out = -m'(H₂ - H₁) + (v₂² - v₁²)/2
P_out = -0.1(2682.6 - 3150.7) + (19.33² - 3.986²)/2
P_out = 225.69 kW
Read more about Power Output at; https://brainly.com/question/25573309
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated with an external surface finish. c)- both answers 1 and 2 d)- all of the above.
Answers
Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
[tex]Losses=K\dfrac{V^2}{2g}[/tex]
Air at 20' C in the cylinder of a diesel engine is compressed from an initial pressure of 1 atm and volume of 800 cm3 to 60 cm3. Assume that air behaves as an ideal gas with γ-1.40 and that the compression is adiabatic. Find the final pressure and temperature of the air.
Answers
Answer:
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
Explanation:
Given that
[tex]T_1=20C[/tex]
[tex]P_1=1\ atm[/tex]
[tex]V_1=800\ cm^3[/tex]
[tex]V_2=60\ cm^3[/tex]
So r=800/60=13.33
γ=1.4
For process 1-2
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{273+20}=13.33^{1.4-1}[/tex]
[tex]T_2=825.73\ K[/tex]
[tex]\dfrac{P_2}{P_1}=r^\gamma[/tex]
[tex]\dfrac{P_2}{1}=13.33^{1.4}[/tex]
[tex]P_2=37.56\ atm[/tex]
So
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
The hoist of a crane consists of a 10 kW electric motor running at 1440 rpm drivine 300 mm diameter drunn through a 60:1 gear reduction unit. If the efficiency is 90% . calculate the load, in tonnes that can be lifted at the rated motor capacity, and the lifting speed.
Answers
Answer:
Load = 2.42 tons
Lifting speed = 24 RPM
Explanation:
Given that
Power=10 KW
Efficiency = 90%
So the actual power,P=0.9 x 10 =9 KW
Speed of motor = 1440 RPM
Diameter of drum = 300 mm
radius =150 mm
G=60:1
Lets take speed of drum =N
We know that
[tex]G=\dfrac{Speed\ of\ motor}{Speed\ of\ drum}[/tex]
[tex]60=\dfrac{1440}{N}[/tex]
N=24 RPM
We know that
[tex]P=\dfrac{2\pi NT}{60}[/tex]
Where T is the torque
[tex]9000=\dfrac{2\pi \times 24\times T}{60}[/tex]
T=3580.98 N.m
Lets take load =F
So T= F x r
3580.98 = F x 0.15
F=23.87 KN
We know that
1 KN=0.109 tons
So 23.87 KN= 2.42 tons
So the load = 2.42 tons
Lifting speed = 24 RPM
Determine the maximum weight of the flowerpot that can besupported without exceeding a cable tension of 50 lb in eithercable AB or AC.
Answers
Answer:
Maximum weight = 76.64 lb
Explanation:
In the second figure attached it can be seen the free body diagram of the problem.
The equation of equilibrium respect x-axis is
[tex] \sum F_x = 0 [/tex]
[tex] F_{AC} \times sin 30 - F_{AB} \times \frac{3}{5} = 0 [/tex]
[tex] F_{AC} = 1.2 \times F_{AB} [/tex]
So, cable segmet AC supports bigger tension than segment AB. If [tex] F_{AC} = 50 lb [/tex] then
[tex] F_{AB} = \frac{F_{AC}}{1.2} [/tex]
[tex] F_{AB} = 41.67 lb [/tex]
The equation of equilibrium respect y-axis is
[tex] \sum F_y = 0 [/tex]
[tex] F_{AB} \times \frac{4}{5} + F_{AC} \times cos 30 - W = 0 [/tex]
[tex] 41.67 lb \times \frac{4}{5} + 50 lb \times cos 30 - W = 0 [/tex]
[tex] W = 76.64 lb [/tex]
Relation between Poisson's ration, young's modulus, shear modulus for an isotropic material
Answers
Answer:
E=2 G(1+2μ)
Explanation:
Isotropic material :
Those material have same property in all direction is known as isotropic material.
Homogeneous material :
Those material have same property through out the volume is known as homogeneous material.
Relationship between Poisson ratio ,young modulus and shear modulus:
E=2 G(1+2μ)
Relationship between Poisson ratio ,bulk modulus and shear modulus:
E=2 K(1-μ)
Where
E is young modulus.
G is shear modulus.
μ is Poisson ratio.
For an orthogonal cutting operation, tool material is HSS, rake angle is 15° , chip thickness is 0.5 mm, speed is 55 m/min and feed is 0.3 mm/rev. The shear plane angle (in degrees) is
Answers
Answer:
The shear plane angle is 34.45°.
Explanation:
Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the edge of the surface. For orthogonal cutting operation feed is the chip thickness.
Given:
Rake angle is 15°.
Uncut chip thickness is 0.5 mm.
Speed is 55 m/min.
Feed or the chip thickness is 0.3 mm.
Calculation:
Step1
Chip reduction ratio is calculated as follows:
[tex]r=\frac{t_{c}}{t}[/tex]
[tex]r=\frac{0.3}{0.5}[/tex]
r = 0.6
Step2
Shear angle is calculated as follows:
[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]
Here, [tex]\phi[/tex] is shear plane angle, r is chip reduction ratio and [tex]\alpha[/tex] is rake angle.
Substitute all the values in the above equation as follows:
[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]
[tex]tan\phi=\frac{0.6cos15^{\circ}}{1-0.6sin15^{\circ}}[/tex]
[tex]tan\phi=\frac{0.57955}{0.8447}[/tex]
[tex]\phi=34.45^{\circ}[/tex]
Thus, the shear plane angle is 34.45°.
A rigid tank holds 10 lbm of 160 °F water. If the quality of the water is 0.5 then what is the pressure in the tank and volume of the tank?
Answers
Answer:
The pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.
Explanation:
Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 160°F.
Given:
Mass of the water is 10 lb.
Dryness fraction is 0.5.
Temperature of water is 160°F.
From steam table at 160°F:
The pressure in the tank is 4.74703 psi.
Specific volume of saturated water is 0.0163918 ft³/lb.
Specific volume of saturated steam is 77.1773 ft³/lb.
Calculation:
Step1
From steam table at 160°F:
The pressure in the tank is 4.74703 psi.
Step2
Specific volume of tank is calculated as follows:
[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]
[tex]v=0.0163918 +0.5(77.1773 -0.0163918)[/tex]
[tex]v=0.0163918 +38.58045[/tex]
v=38.5968 ft³/lb.
Step4
Volume is calculated as follows:
[tex]V=v\times m_{t}[/tex]
[tex]V=38.5968\times10[/tex]
V=385.968 ft³.
Thus, the pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.
The temperature at which the polymer becomes viscous is called: glass transition temperature (Tg). a)-True b)- false?
Answers
Answer:
a)True
Explanation:
The temperature at which the polymer becomes viscous is called glass transition temperature.
When polymer is heated then it will become viscous and temperature is called glass transition temperature.
Glass temperature depends on the chemical structure of the polymer.
Polymer are made of long change structure and these change are connected together one by one and form a complex structure.
Example -Rubber or elastomer,Plastics.
By increasing the cross-sectional area of the restriction, one can significantly increase the flow velocity. a) True b) False
Answers
Answer:
b)false
Explanation:
As we know that
Volume flow rate Q
Q = A x V
For constant volume flow rate,if velocity will increase then automatically area will decrease and vice versa.
Generally nozzle are used to increase the velocity and diffuser are used to decrease the exit velocity of flow.
So by increasing the cross sectional area of the restriction ,the velocity of the flow will decrease.
Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).
Answers
Answer:
Total elongation will be 0.012 m
Explanation:
We have given diameter of the cylinder = 2.1 mm
Length of wire [tex]L=3.2\times 10^4mm[/tex]
So radius [tex]r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m[/tex]
Load F = 280 N
Elastic modulus = 207 Gpa
Area of cross section [tex]A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2[/tex]
We know that elongation in wire is given by [tex]\delta =\frac{FL}{AE}[/tex], here F is load, L is length, A is area and E is elastic modulus
So [tex]\delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m[/tex]
Which of carbon will have higher strength but less ductility, 0.49% or 0.19%
Answers
Answer:
0.49%
Explanation:
Assuming this is about steel. Higher carbon steel tend to be harder and stronger, but that hardness makes them more brittle and less ductile. This is because with higher carbon concentrations there is more formation of pearlite, which has higher hardness than ferrite. Pearlite contains layers of cementite, which is a fragile material, so the presence of these microstructures will make the whole thing more brittle.
The three major network category are_____,_____ and_____
Answers
Answer:
The three major network categories are LAN (Local Area Network), MAN (Metropolitan Area Network) and WAN (Wide Area Network)
Explanation:
A Local Area Network (LAN) is the simplest form of a network. It is usually limited to a geographical area and consists of a group of computers in the same organisation linked together. In most cases, the same technology is used for all computers in this network (eg. ethernet).
A Metropolitan Area Network (MAN) can connect multiple LANs as long as they are close to each other (geographically). MANs allow high speed remote connection as if the devices were on the same LAN.
A Wide Area Network (WAN) can connect multiple LANs across large geographical distances. Unlike MANs, the speed might differ based on factors such as distance. The internet is the most popular example of a WAN.