Answer:
It is difficult to introduce products modification
A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questions. Note: Evaluate the properties of the wall as cross grain oak at 300 K.
What is the surface temperature, in °C, after 400 s?
T (0,400 sec) =
Will the surface of the wall reach the ignition temperature of oak (400°C) after 400 s?
What is the temperature, in °C, 1 mm from the surface after 400 s? T (1 mm, 400 sec) =
Answer:
a) What is the surface temperature, in °C, after 400 s?
T (0,400 sec) = 800°C
b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s
c) What is the temperature, in °C, 1 mm from the surface after 400 s?
T (1 mm, 400 sec) = 798.35°C
Explanation:
oak initial Temperature = 25°C = 298 K
oak exposed to gas of temp = 800°C = 1073 K
h = 20 W/m².K
From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k
Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.
From energy balance: [tex]\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t[/tex]
Initial temperature wall = [tex]T_i[/tex]
Surface temperature = T
Gas exposed temperature = [tex]T_{\infty}[/tex]
The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the boy throws three bricks. The boy’s weight is 80lbs, and the weight of the wago 20 lbs. The bricks weigh 10 lbs each, and he throws them with a horizontal velocity of 10 ft/s relative to the wagon. Neglect horizontal forces on the wagon's wheels.
A) What velocity does the boy attain if he throws the bricks one at a time?
B) What velocity does the boy attain if he throws all three bricks at once?
C) Compare the results of parts (a) and (b). What does this suggest about rocket propulsion? Why are these results different?
Q:What velocity does the boy attain if he throws the bricks one at a time?
Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.
On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. How do you explain this behavior?
Answer:
Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero
Explanation:
On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.
For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.
True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity
Answer:
An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration
Explanation:
Given the unity feedback system
G(s)= K(s+4)/s(s+1.2)(s+2)
Find:
a. The range of K that keeps the system stable
b. The value of K that makes the system oscillate
c. The frequency of oscillation when K is set to the value that makes the system oscillate
Answer:
A.) 0 > K > 9.6
B.) K = 9.6
C.) w = +/- 2 sqrt (3)
Explanation:
G(s)= K(s+4)/s(s+1.2)(s+2)
For a closed loop stability, we can analyse by using Routh - Horwitz analysis.
To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.
Please find the attached files for the solution
An inventor claims to have developed a device requiring no work or heat transfer input yet able to produce hot and cold air streams at steady state.
a. True
b. False
Answer:
a. True
Explanation:
Apply the principle of conservation of mass.
and also the expression for the steady flow energy equation.
kinetic and potential energy effects can be neglected.
The given statement by the inventor who is claiming the development of a device that requires no work or heat transfer input yet is able to produce hot and cold air streams at a steady state is definitely false.
What is heat transfer?Heat transfer may be characterized as a type of process which involves the migration of heat from one object or component to another by numerous mechanisms like conduction, convection, and/or radiation.
The process of heat transfer may occur where there is a temperature difference between two objects exist. It significantly utilizes the mechanism of exchanging thermal energy between two or more physical systems.
According to the concept of physics, no object or thing has the ability to perform its function without the utilization of any source of heat or energy. Then, how it is possible for that device to produce hot and cold air streams at a steady state.
Therefore, the given statement by the inventor is absolutely false.
To learn more about Heat transfer, refer to the link:
https://brainly.com/question/16055406
#SPJ2
Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is inflated to a gage pressure of 78 kPa.
Answer:
The normal stress is 0.7821 MPa
Explanation:
The external diameter D = 160 mm
The thickness t = 3.8 mm = 3.8 x 10^-3 m
gauge pressure P = 78 kPa = 78 x 10^3 Pa
The maximum shear stress τmax = ?
The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm
The internal radius of the shell r = R - t
==> 80 - 3.8 = 76.2 mm
Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm
==> d = 152.4 x 10^-3 m
The normal stress σ = [tex]\frac{Pd}{4t}[/tex] = [tex]\frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} }[/tex] = 782052.63 Pa
==> σ = 0.7821 MPa
The system is stimulated, via the voltage source, with a pulse of height 2 and width 4 s. Determine the voltage across the resistor.
Answer:
Voltage across resistor = 2 v
Explanation:
Given data
pulse height = 2 v
pulse width = 4s
calculate voltage across resistor ( the free hand sketch attached below explains more )
pulse height is also = amplitude of voltage ) = 2v
The voltage across the resistor = 2v Since the voltage from the source of the circuit is equal to the amplitude voltage in the circuit ( assuming no loss of voltage )
also the graphical representation of the problem is attached below
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD pivoted at C. A pin at end D of the rod fits into one of several holes drilled in the edge of the lid. For α 5 50°, determine (a) the magnitude of the force exerted by rod CD, (b) the reactions at the hinges. Assume that the hinge at B does not exert any axial thrust. Beer, Ferdinand. Vector Mechanics for Engineers: Statics (p. 219). McGraw-Hill Higher Education. Kindle Edition.
Answer:
(a) The magnitude of force is 116.6 lb, as exerted by the rod CD
(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.
Explanation:
Step by step working is shown in the images attached herewith.
For this given system, the coordinates are the following:
A(0, 0, 0)
B(26, 0, 0)
And the value of angle alpha is 20.95°
Hope that answers the question, have a great day!
If you are involved in a collision where there is injury, you must report the incident within .......
48 hours
3 days
24 hours
72 hours
Answer:
24 hours
Explanation:
you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.
A four-cylinder four-stroke engine is modelled using the cold air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), maximum cycle temperature (T3), and engine speed in RPM, determine the efficiency and other values listed below. The specific heats for air are given as Cp 1.0045 kJ/kg-K and Cv-0.7175 kJ/kg-K.
--Given Values--
T1 (K) 325
P1 (kPa)= 185
V1 (cm^3) = 410
r=8
T3 (K) 3420
Speed (RPM) 4800
Answer:
56.47%
Explanation:
Determine the efficiency of the Engine
Given data : T1 (k) = 325, P1 (kpa) = 185,
V1 (cm^3) = 410 , r = 8, T3(k) = 3420
speed ( RPM) = 4800
USING THIS FORMULA
efficiency ( n ) = [tex]1 - (\frac{1}{(rp)^{r-1} })[/tex]
= 1 - [tex](\frac{1}{(8)^{r-1} })[/tex] = 1 - (1/8^1.4-1 )
= 0.5647 = 56.47%
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the ______________ speed and varies with weight
Answer:
Maneuvering speed.
Explanation:
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the maneuvering speed and varies with weight.
In aeronautical engineering, the maneuvering speed (Va) of an aircraft such as an aeroplane, helicopter, or jet is an airspeed limitation which is mainly selected by an aircraft designer.
Generally, at speeds higher or greater than the manoeuvring speed, aircraft pilots are advised not to attempt a full deflection of any flight control surface because it's capable of resulting in a damage to the structure of an aircraft.
If you're a pilot, to find the maneuvering speed of an aircraft, you should look at the flight manual of the aircraft or on the cockpit placard in the aircraft. The maneuvering speed of an aircraft is a calibrated speed and should not be exceeded by any pilot.
While having a discussion, Technician A says that you should never install undersized tires on a vehicle. The vehicle will be lower, and the speedometer will no longer be accurate. Technician B says that the increase in engine rpm for a given speed will result in a decrease in fuel economy. Who is correct
Answer:
Both technician A and technician B are correct.
Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.
It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.
In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very different register transfers to take place. Why does this not happen
Answer:
no need for that
Explanation:
they are not the same at all
Match the words in the left column to the appropriate blanks in the sentences on the right. Note that some words may be used more than once and some may not be used.
1. breaking
2. forming
3. positive
4. negative
5. twice
6. half
A. The reaction involves___five blue-blue and twenty blue-red bonds and then____twenty blue-red bonds. Enthalpies for bond breaking are always_____.
B. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are___as many molecules of in the products, delta S is___for this reaction
Answer:
A. Breaking; forming; positive.
B. Twice; half.
Explanation:
A. The reaction involves breaking five blue-blue and twenty blue-red bonds and then forming twenty blue-red bonds. Enthalpies for bond breaking are always positive.
B. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are twice as many molecules of gas in the products, delta S is half for this reaction.
Bond enthalpy refers to the amount of energy that is required to break a mole of a particular bond and it's an endothermic reaction, therefore it is always positive.
However, bond forming is always an exothermic reaction because energy is released, therefore, it is always negative. This simply means that, the breaking of a bond is an endothermic reaction (positive) while the formation of a bond is an exothermic reaction (negative).
A state of stress that occurs at a point on the free surface of the of a solid body is = 50 MPa σ x , =10 MPa σ y , and = −15 MPa xy τ .
a. Evaluate the two principal normal stresses and the one principal shear stress that can be found by coordinate system rotations in the x-y plane and give the coordinate system rotations.
b. Determine the maximum normal stress and the maximum shear stress at this point.
c. Sketch the rotations and stress magnitudes
Answer:
A) 5 MPa , 55 MPa
B) maximum stress = 55 MPa, maximum shear stress = 25 MPa
Explanation:
using the given Data
free surface of a solid body
α[tex]_{x}[/tex] = 50 MPa, α[tex]_{y}[/tex] = 10 MPa , t[tex]_{xy}[/tex] = -15 MPa
attached below is the detailed solution to the question
Answer:
I think its A) 5 MPa , 55 MPa
B) ms = 55 MPa, mss= 25 MPa
α = 50 MPa, α = 10 MPa , t = -15 MPa
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where it is traveling is 60°F. Determine the drag on the fin when the submarine is traveling at 2.5 ft/s.
Answer:
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
Explanation:
Given that:
The height of a triangular stabilizing fin on its stern is 1 ft tall
and it length is 2 ft long.
Temperature = 60 °F
The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.
From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.
The diagram can be found in the attached file below.
If we recall ,we know that;
Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]
the density of water ρ = 62.36 lb /ft³
[tex]Re_{max} = \dfrac{Ux}{v}[/tex]
[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]
[tex]Re_{max} = 414078.6749[/tex]
[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵
Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:
[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]
where;
[tex](2-x) dy[/tex] = strip area
[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]
Therefore;
[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]
[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]
Let note that y = 0.5x from what we have in the diagram,
so , x = y/0.5
By applying the rule of integration on both sides, we have:
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]
Let U = (2-2y)
-2dy = du
dy = -du/2
[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]
[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]
[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]
[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]
[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]
[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]
[tex]F_D = 1.071031071 \ lbf[/tex]
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
what are the conditions for sheet generator to build up its voltage?
Answer:
There are six conditions
1. Poles should contain some residual flux.
2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.
3. Resistance of field winding must be less than critical resistance.
4. Speed of prime mover of generator must be above critical speed.
5. Generator must be on load.
6. Brushes must have proper contact with commutators.
Explanation:
Describe the similarities and differences between circuits with resistors combined in series and circuits with resistors combined in parallel
Answer:
from the below explanation... we can say that, in the series circuit, flowing current remains the same at each part of the circuit. While in parallel circuits, the voltage across two endpoints of the branches is the same as the supplied voltage.
Explanation:
1.
The components in a series circuit are arranged in a single path from one end of supply to another end. However, the multiple components in a parallel circuit are arranged in multiple paths wrt the two end terminals of the battery.
2.
In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit.
3.
In the series circuit, different voltage exists across each component in the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.
4.
A fault in one of the components of the series circuit causes hindrance in the operation of a complete circuit. As against fault in a single component in a parallel network do not hinder the functioning of another part of the circuit.
5.
The detection of a fault in case of a series circuit is difficult, but it is quite easy in parallel circuits.
6.
The equivalent resistance in case of a series circuit is always more than the highest value of resistance in the series connection. While the equivalent resistance in the parallel circuit is always less than any of the individual resistances in parallel combination.
An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 750-lb weight is attached to it and dropped from the tower.
Determine:
a. The required spring constant k of the cable.
b. How close to the ground a 185-lb man will come if he uses this cable to jump from the tower?
Answer:
a) The spring constant is 50 lb/ft
b) The man is 26.3 ft close to the ground.
Explanation:
Height of tower is 130 ft
Specification calls for a cable of length 85 ft
the maximum this length stretches is 100 ft when subjected to a load of 750 lb
The extension of the cable is calculated from the formula from Hooke's law
F = kx
where F is the load or force on the cable
k is the spring constant of the cable
x is the extension on the cable
a) The extension on the cable is
x = 100 ft - 85 ft = 15 ft
substituting into the formula above, we'll have
750 = k*15
k = 750/15 = 50 lb/ft
b) for a 185 lb man, jumping down will give an extension gotten as
F = kx
185 = 50*x
x = 185/50 = 3.7 ft
The total length of the cable will be extended to 100 ft + 3.7 ft = 103.7 ft
closeness to the ground = 130 ft - 103.7 ft = 26.3 ft
Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object. B. Forces that act perpendicular to the surface and squeeze an object exert a tensile stress on the object. C. Forces that act parallel to the surface exert a tensile stress on the object. D. Forces that decrease the length of the material exert a tensile stress on the object.
Answer:
A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.
Explanation:
Tensile stress is due to tension forces on a material. Tensile force acts perpendicularly away from the surface of the substance. The pull on the material due to the tensile force exerts tensile stress on the material, that tends to pull the material apart. The magnitude of the tensile stress is given as
σ = [tex]\frac{P}{A}[/tex]
where σ is the tensile stress
P is the tensile force pulling the material apart
A is the cross-sectional area through which the tensile force acts perpendicularly.
what is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?
Answer:
P=11 kW
Explanation:
Given that
Number of poles= 8
I.E.C. 180L motor frame
From data book , for 8 poles motor at 50 Hz
Speed = 730 rpm
Power factor = 0.75
Efficiency at 100 % load= 89.3 %
Efficiency at 50 % load= 89.1 %
Output power = 11 kW
Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.
P=11 kW
The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm and 6 mm pitch.
force required at the rim of a 300mm diameter hand wheel, if there is a collar
bearing of 50 mm mean diameter provided in the arrangement to exert axial
load. Assume the coefficient of friction for the collar as 0.2.
Answer:
See calculation below
Explanation:
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
1. Force required at the rim of handwheel
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = [tex]\frac{do + dc}{2}[/tex] = (75 + 69) / 2 = 72 mm
and
tan α = p / πd = 6 / (π x 72) = 0.0265
∴ Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T = [tex]W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}[/tex]
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
2. Maximum compressive stress in the screw
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
3. efficiency of the straightener
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
∴Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%
By assuming the coefficient of friction for the collar as 0.2. efficiency of straightner is 11.6%.
What is Wheel?Wheels are circular frames or disks that are mounted on machines or vehicles and are designed to rotate around an axis.
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
1. Force required at the rim of handwheel :
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = = (75 + 69) / 2 = 72 mm and
tan α = p / πd = 6 / (π x 72) = 0.0265
Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
2. Maximum compressive stress in the screw :
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
Therefore, efficiency of the straightener :
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%
Learn more about Wheel, here;
https://brainly.com/question/29297499
#SPJ2
You subjected a rod under the cyclic stress with the maximum stress of 200 MPa and minimum stress of 20 MPa. The fatigue limit was determined to be ~100 MPa. How many cycles can this materials sustain before failure?
Answer:
The material will not fail
Explanation:
A rod subjected under cyclic stress will fail if the cyclic stress it is subjected to is a constant maximum value that is above the fatigue limit of the rod. but in this problem the Rod is subjected to a cyclic stress that ranges from 200 MPa(maximum stress) and 20 MPa ( minimum stress). this simply means that at all times the Rod will not experience maximum stress of 200 MPa and its Fatigue limit is also set at ~100 MPa
attached is the diagram showing the cyclic stress the rod is subjected to
1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
Answer:
Following are the answer to this question:
Explanation:
1)
Following the four operations in the airlines:
Landside operations:
In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.
Airside Operations:
It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.
Billing and invoicing Operations:
This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.
Information management:
In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.
2)
It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.
Carbon dioxide (CO2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit state of 10 bar, 520 K. The CO2 is modeled as an ideal gas, and kinetic and potential energy effects are negligible. For the compressor, determine (a) the work input, in kJ per kg of CO2 flowing, (b) the rate of entropy production, in kJ/K per kg of CO2 flowing, and (c) the isentropic compressor efficiency.
Answer:
A.) 0.08 kJ/kg.K
B.) 207.8 KJ/Kg
C.) 0.808
Explanation:
From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.
Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.
Please find the attached files for the solution and the remaining explanation.
The following liquids are stored in a storage vessel at 1 atm and 25°C. The vessels are vented with air. Determine whether the equilibrium vapor above the liquid will be flammable. The liquids are:________.
a. Acetone
b. Benzene
c. Cyclohexane
d. Toluene Problem
Answer:
The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )
Explanation:
Liquids stored at : 1 atm , 25⁰c and they are vented with air
Determining whether the equilibrum vapor above the liquid will be flammable
We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable
A) For acetone
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 16.6513
B = 2940.46
C = -35.93
T = 298 k input values into Antoine equation
therefore ; [tex]p^{out}[/tex] = 228.4 mg
calculate volume percentage using Dalton's law
= V% = (saturation vapor pressure / pressure ) *100
= (228.4 mmHg / 760 mmHg) * 100 = 30.1%
The liquid is not flammable because its UFL = 12.8%
B) For Benzene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A= 15.9008
B = 2788.52
C = -52.36
T = 298 k input values into the above equation
[tex]p^{out}[/tex] = 94.5 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (94.5 / 760 ) * 100 = 12.4%
Benzene is not flammable under the given conditions because its UFL =7.1%
C) For cyclohexane
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 15.7527
B = 2766.63
c = -50.50
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 96.9 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= ( 96.9 mmHg /760 mmHg) * 100 = 12.7%
cyclohexane not flammable under the given conditions because its UFL= 8%
D) For Toluene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten from appendix E ( chemical process safety (3rd edition) )
A = 16.0137
B = 3096.52
C = -53.67
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 28.2 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (28.2 mmHg / 760 mmHg) * 100 = 3.7%
Toluene is flammable under the given conditions because its UFL= 7.1%
Searches related to Probability questions - A person frequents one of the two restaurants KARIM or NAZEER, choosing Chicken's item 70% of the time and fish's item 30% of the time. Regardless of where he goes , he orders Afghani Chicken 60% of his visits. (a) The next time he goes into a restaurants, what is the probability that he goes to KARIM and orders Afghani Chicken. (b) Are the two events in part a independent? Explain. (c) If he goes into a restaurants and orders Afghani Chicken, what is the probability that he is at NAZEER. (d) What is the probability that he goes to KARIM or orders Afghani Chicken or both?
Answer:
a) 0.42
b) Independent
c) 30%
d) 0.88
Explanation:
Person chooses Chicken's item : 70% = 0.7
Person chooses fish's item : 30% = 0.3
Visits in which he orders Afghani Chicken = 60% = 0.6
a) Probability that he goes to KARIM and orders Afghani Chicken:
P = 0.7 * 0.6 = 0.42
b) Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.
c) P = 0.30 because he orders Afghani Chicken regardless of where he visits.
d) Let A be the probability that he goes to KARIM:
P(A) = 0.7 * ( 1 - 0.6 ) = 0.28
Let A be the probability that he orders Afghani Chicken:
P(B) = 0.3 * 0.6 = 0.18
Let C be the probability that he goes to KARIM and orders Afghani chicken:
= 0.7 * 0.6 = 0.42
So probability that he goes to KARIM or orders Afghani Chicken or both:
P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88
The data listed below are claimed for power cycles operating between hot and cold reservoirs at 1000 K and 400 K, respectively. For each case determine whether such a cycle is in keeping with the first and second laws of thermodynamics.
a) Qh=300 kJ, W(cycle)=160kJ, Qc=140 kJ
b) Qh=300 kJ, W(cycle)=180kJ, Qc=120 kJ
c) Qh=300 kJ, W(cycle)=170kJ, Qc=140 kJ
d) Qh=300 kJ, W(cycle)=200kJ, Qc=100 kJ
Answer:
a) This cycle observes the First and Second Laws of Thermodynamics, b) This cycle observes the First and Second Laws of Thermodynamics, c) This cycle does not observe the First Law of Thermodynamics, d) This cycle does not observe the Second Law of Thermodynamics.
Explanation:
The Carnot cycle offers a reliable criterion to determine the maximum theoretical efficiency ([tex]\eta_{th,max}[/tex]) for a power cycle in term of the temperatures of hot and cold reservoirs and expressed in percentage:
[tex]\eta_{th, max} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]T_{L}[/tex], [tex]T_{H}[/tex] - Temperatures of cold and hot reservoirs, measured in kelvins.
If [tex]T_{L} = 400\,K[/tex] and [tex]T_{H} = 1000\,K[/tex], the maximum theoretical thermal efficiency is:
[tex]\eta_{th, max} = \left(1-\frac{400\,K}{1000\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{th,max} = 60\,\%[/tex]
In addition, the real efficiency of the heat engine is described by the following formula:
[tex]\eta_{th} = \left(1-\frac{Q_{L}}{Q_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]Q_{H}[/tex] - Heat absorbed by the heat engine from hot reservoir, measured in kilojoules.
[tex]Q_{L}[/tex] - Heat released by the heat engine to the cold reservoir, measured in kilojoules.
The following conditions must be observed by all heat engines:
First Law of Thermodynamics
[tex]W = Q_{H}-Q_{L}[/tex]
Second Law of Thermodynamics
[tex]\eta_{th} \leq \eta_{th,max}[/tex]
Now, each cycle is checked:
a) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 160\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-140\,kJ[/tex]
[tex]W = 160\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 53.333\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
b) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 120\,kJ[/tex] and [tex]W = 180\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-120\,kJ[/tex]
[tex]W = 180\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{120\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 60\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
c) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 170\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-140\,kJ[/tex]
[tex]W = 160\,kJ[/tex]
This cycle does not observe the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 53.333\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
d) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 200\,kJ[/tex] and [tex]W = 100\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-100\,kJ[/tex]
[tex]W = 200\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{100\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 66.667\,\%[/tex]
This cycle does not observe the Second Law of Thermodynamics.
Based on the first parameters, the power cycle obeys both the First and Second Law of Thermodynamics.
Given the following data:
Temperature of hot reservoir = 1000 K.Temperature of cold reservoir = 400 K.How to verify which law a power cycle obeys.In order to verify a power cycle obeys the first and second laws of thermodynamics, we would use the Carnot cycle.
Mathematically, the maximum theoretical efficiency for a power cycle in terms of the temperature is given by this formula:
[tex]\eta_{th, max}=(1-\frac{T_c}{T_h} ) \times 100[/tex]
Where:
[tex]T_c[/tex] is the temperature of cold reservoir.[tex]T_h[/tex] is the temperature of hot reservoir.Substituting the given parameters into the formula, we have;
[tex]\eta_{th, max}=(1-\frac{400}{1000} ) \times 100\\\\\eta_{th, max}=60\%[/tex]
Similarly, the real efficiency of a power cycle, we have:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100[/tex]
The conditions for a power cycle.According to the First Law of Thermodynamics, the following condition must be met:
[tex]W_{cycle}=Q_h-Q_c[/tex]
According to the Second Law of Thermodynamics, the following condition must be met:
[tex]\eta_{th }\leq \eta_{th, max }[/tex]
Next, we would determine whether or not each obeys the first and second laws of thermodynamics:
When [tex]Q_h=300\; kJ, \;W_{(cycle)}=160kJ, \;and\;Q_c=140\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\160=300-140\\\\160=160[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the second parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=180kJ, \;and\;Q_c=120\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\180=300-120\\\\180=180[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{120}{300} ) \times 100\\\\\eta_{th}=60\%\\\\\eta_{th }\leq \eta_{th, max } =60\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the third parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=170kJ, \;and\;Q_c=140\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\170=300-140\\\\170\neq 160[/tex]
Therefore, the power cycle doesn't obey the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the fourth parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=200kJ, \;and\;Q_c=100\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\200=300-100\\\\200=200[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{100}{300} ) \times 100\\\\\eta_{th}=66.67\%\\\\\eta_{th }\leq \eta_{th, max } \neq 63.33\%\geq 60\%[/tex]
Therefore, the power cycle doesn't obey the Second Law of Thermodynamics.
Read more on thermodynamics here: brainly.com/question/11628413
You are the curator of a museum. The museum is running short of funds, so you decide to increase revenue. Should you increase or decrease the price of admission? Explain
Answer:
Explanation:
If the museum is running short of funds, and you decide to increase revenue. An increase or decrease in the price of admission into the museum depends on the following:
1. If demand for admission into the museum is elastic there are two possible outcomes
a. An increase in the price of admission leads to a decrease in the quantity demand of admission into the museum
b. A decrease in price of admission into the museum leads to an increase in the quantity demand of admission into the museum.
This follows the law of demand which states that "the higher the price, the lower the quantity demanded and the lower the price, the higher the quantity demanded".
2. If the demand for admission into the museum is inelastic, then an increase in price will lead to an increase in revenue of the museum.
Therefore, before the curator increase the price of admission into the museum, he should first determine the price elasticity of demand of the museum.