"A three-phase, 60-Hz, 4-pole, 440-V (line-line, rms) induction-motor drive has a full-load (rated) speed of 1750 rpm. The rated torque is 40 Nm. Keeping the air gap flux-density peak constant at its rated value, (a) plot the torque-speed characteristics (the linear portion) for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz. (b) This motor is supplying a load whose torque demand increases linearly with speed, such that it equals the rated torque of the motor at the rated motor speed. Calculate the speeds of operation at the four values of frequency in part (a)."
Answers
Answer:
A) slip(s1) = (1800 - 1750) / 1800 = 0.0277
B) 1800 rpm, 1350 rpm, 900 rpm, 450 rpm
Explanation:
Given data
frequency = 60 Hz,
Line - line rms = 440 - V
3 phase induction-motor drive
number of poles = 4
Full-load rated speed = 1750
rated torque = 40 Nm
A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below
first we calculate the rated speed:
Ns = [tex]\frac{120f}{p}[/tex]
f = 60 Hz . p (number of poles) = 4
Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800
full loaded rated value = 1750.
slip(s1) = (1800 - 1750) / 1800 = 0.0277
considering a linearly condition the slip is low
[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]
S1 = 0.0277
f1 = 60 Hz
hence s2 = 0.018 therefore Ns2 = 1500
B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz
for 60 Hz :
Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm
for 45 Hz:
Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm
for 30 Hz:
Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm
for 15 Hz:
Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm
Related Questions
what's the maximum shear on a 3.0 m beam carrying 10 kN/m?
Answers
Answer:
max shear = R = V = 15 kN
Explanation:
given:
load = 10 kn/m
span = 3m
max shear = R = V = wL / 2
max shear = R = V = (10 * 3) / 2
max shear = R = V = 15 kN
An AISI/SAE 4340-A steel rod with the yield strength of 450 MPa, 2.0 m long will be subjected to a tensile force, must have the minimum weight possible, and must behave elastically for this load. The elastic modulus of steel is 207 GPa. What is the engineering strain of the rod
Answers
Answer: 0.002174
Explanation:
Given that the
Yield strength rho = 450 MPa
Length = 2 m
Elastic modulus E= 207 GPa
According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə
E = rho/ə
Make ə the subject of formula
ə = rho/ E
ə = (450 × 10^6) / (207 × 10^9)
ə = 2.174 × 10^-3
Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3
Elastic Strain has no S.I Units.
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
Answers
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.
how can we prevent chemical hazards in labotary
Answers
Answer:
We can prevent it by:
a) By wearing GOOGLES.
b) By wearing our Lab coat.
c) Fire extinguisher should always be present in the lab.
d) Hand Gloves must be worn.
e) No playing in the lab.
f) No touching of things/equipment's e.g bottles, in the lab.
g) No eating/snacking in the lab.
h) Always pay attention, no gisting.
i) Adult/qualified person must be present in the lab with pupils/students.
Explanation:
Hope it helps.
Under conditions for which the same roojm temperature is mainteined bt a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) y by considering a room whose air temperature is maintained at 20 Dagree C throughout the year, while the walls of the room are nominally at 27 Dagree C and 14 Dagree C in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of 32 Dagree C throughout the year and to have an emissivity of 0.90. The coefficient associated with heat transfer by natural convection between the person and the room air is approximately 2 W/m2 Middot K.
Answers
Answer:
Net heat transfers: during summer = 52.253 W/m^2 during winter = 119.375 w/m^2
Explanation:
Given data :
Room temperature throughout the year = 20⁰c = 293 k
Room temperature during summer = 27⁰c = 300 k
Room temperature during winter = 14⁰c = 287 k
surface temperature of a person throughout the year = 32⁰c = 305 k
coefficient of heat transfer by natural convection (h)= 2 w /m^2 k
emissivity = 0.9
An explanation to the condition of feeling chilled during the winter and comfortable during summer can be explained with the calculation below
The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as
q = hΔt = 2 * ( 305 - 293) = 2 * 12 = 24 w /m^2
also calculate heat transfer through radiation using this formula
[tex]q_{rad} =[/tex] εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]
ε = 0.90,(emissivity) σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)
during summer :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2
therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2
During winter :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3
therefore the net heat transfer during the winter
= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2
When you shift your focus, everything you
see is still in perfect focus.
True or false
Answers
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes provide slip resistance. Who is correct
Answers
Given:
We have given two statements.
Statement 1: Proper footwear may include both leather and steel-toed shoes.
Statement 2: Leather-soled shoes provide slip resistance.
Find:
Which statement is true.
Solution:
A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.
Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.
Leather-soled shoes don't provide slop resistance.
Therefore, both the Technicians are wrong.
From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.
We are given the statements made by both technicians;
Technician A: Proper footwear may include both leather and steel-toed shoes.
Technician B: Leather-soled shoes provide slip resistance.
Now, they are talking about safety shoes to be worn in workshops.
A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.
This is the type of shoe that should be worn by technicians in the workshop.
Thus, Technician A is wrong because proper footwear does not include leather shoes.
Similarly, technician B is also wrong because leather shoes are not safety shoes.
Read more about slip resistant shoes at; https://brainly.com/question/17411739
1000 lb boulder B is resting on a 500 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?
a. The tension in the cord connected to the truck is 200 lb.
b. The tension in the cord connected to the truck is 1200 lb.
c. The tension in the cord connected to the truck is greater than 1200 lb.
d. The normal force between A and B is 1000 lb.
e. The normal force between A and B is 1200 lb.
f. None of the above are true.
Answers
Answer:
c. The tension in the cord connected to the truck is greater than 1200 lb
e. The normal force between A and B is 1200 lb.
Explanation:
The correct question should be
A 1000 lb boulder B is resting on a 200 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?
A free body diagram is shown below.
The normal force between the the boulder and the platform will be the sum of the force, i.e 1000 lb + 200 lb = 1200 lb
For the combination of the bodies to accelerate upwards, then the tension must be greater than the normal force, i.e T > 1200 lb
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.
Answers
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]
True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity
Answers
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:
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
Answers
Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]
= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s
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
Answers
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
An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pressure ratio (rp) for constant volume heat addition, determine the efficiency and other values listed below. The gas constant for air is R = 0.287 kJ/ kg.K
T1= 310K
P1(kpa)= 100
r=11.5
rp =1.95
Required:
a. Determine the specific internal energy (kJ/kg) at state 1.
b. Determine the relative specific volume at state 1.
c. Determine the relative specific volume at state 2.
d. Determine the temperature (K) at state 2.
e. Determine the specific internal energy (kJ/kg) at state 2.
Answers
Answer:
A) 222.58 kJ / kg
B) 0.8897 M^3/ kg
c) 0.7737 m^3/kg
D) 746.542 k
E) 536.017 kj/kg
efficiency = 58% ( approximately )
Explanation:
Given Data :
Gas constant (R) = 0.287 kJ/ kg.K
T1 = 310 k
P1 ( Kpa ) = 100
r = 11.5 ( compression ratio )
rp = 1.95 ( pressure ratio )
A ) specific internal energy at state 1
= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg
B) Relative specific volume at state 1
= P1*V1 = R*T1 ( ideal gas equation )
V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3
V1 = 88.97 / 100 = 0.8897 M^3/ kg
C ) relative specific volume at state 2
Applying r ( compression ratio) = V1 / V2
11.5 = 0.8897 / V2
V2 = 0.8897 / 11.5 = 0.7737 m^3/kg
D) The temperature (k) at state 2
since the process is an Isentropic process we will apply the p-v-t relation
[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]
hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k
e) specific internal energy at state 2
= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg
efficiency = output /input = 390.3511 / 667.5448 ≈ 58%
attached is a free hand diagram of an Otto cycle is attached below
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
Answers
In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:
[tex]\sigma_b = \frac{MC}{I}[/tex]
Where,
[tex]\sigma_b[/tex] is the compressive stress or tensile stress[tex]M[/tex] is the B.M [tex]C[/tex] is the N.A distance[tex]I[/tex] is the moment of interiorSo making this formula for the max, we have:
[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]
With all this information we can put the formula as:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
See more about stress in the beam at brainly.com/question/23637191
One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 120 in., the height of the first bounce of the ball must be in the range 55 in. <= h <= 60 in. Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement. Any ideas on this?
Answers
Answer:
At temperature is and relative humidity is 86% therefore, the humidity ratio is 0.0223 and the specific volume is 14.289
At temperature is and Relative humidity is 40% therefore, the humidity ratio is 0.0066 and the specific volume is 13.535.
To calculate the mass of air can be calculated as follows:
Now , we going to calculate the volume,
The time which is required to fill the cistern can be calculated as follows:
Now, putting the value in above formula we get,
Therefore, the hours required to fill the cistern is 4.65 hours.
Explanation:
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answers
Answer:
the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Explanation:
From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
The Critical Stress for a maximum internal crack can be expressed by the formula:
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
where;
[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation
[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa
Y = dimensionless parameter
a = length of the internal crack
given that ;
the maximum internal crack length is 8.6 mm
half length of the internal crack will be 8.6 mm/2 = 4.3mm
half length of the internal crack a = 4.3 × 10⁻³ m
From :
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]
[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]
[tex]Y= \dfrac{26}{13.01748802}[/tex]
[tex]Y=1.99731315[/tex]
[tex]Y \approx 1.997[/tex]
For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
when the length of the internal crack a = 3mm
half length of the internal crack will be 3.0 mm / 2 = 1.5 mm
half length of the internal crack a =1.5 × 10⁻³ m
From;
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]
[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]
[tex]\sigma_c =189.6595506[/tex]
[tex]\sigma_c =[/tex] 189.66 MPa
Thus; the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Answers
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]
So, the potential energy of the mass is 2940 J.
Some characteristics of clay products such as (a) density, (b) firing distortion, (c) strength, (d) corrosion resistance, and (e) thermal conductivity are affected by the extent of vitrification. Will they increase or decrease with increasing degree of vitrification?
1. (a) increase (b) decrease (c) increase (d) decrease (e) increase
2. (a) decrease (b) increase (c) increase (d) increase (e) decrease
3. (a) decrease (b) decrease (c) increase (d) decrease (e) decrease
4. (a) increase (b) increase (c) increase (d) increase (e) increase
5. (a) increase (b) decrease (c) decrease (d) increase (e) decrease
Answers
Explanation:
1. increase This due to increase in the pore volume.
2.increase . This is due to the fact that more liquid phase will be present at the firing.
3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.
4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.
5. Increase , glass has higher thermal conductivity than the pores it fills.
Which of these properties generally applies to transfer lines? A. It is difficult to introduce product modifications. B. It is difficult to store products between individual machines. C. The process can run even if one of the machines fails. D. Products can take different paths through the system.
Answers
Answer:
It is difficult to introduce products modification
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
Answers
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%
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
Answers
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%
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?
Answers
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
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?
Answers
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
A car radiator is a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of 0.05 kg/s, enters the radiator at 400 K and is to leave at 330 K. The water is cooled by air that enters at 0.75 kg/s and 300 K. If the overall heat transfer coefficient is 200 W/m2-K, what is the required heat transfer surface area?
Answers
Answer:
Explanation:
Known: flow rate and inlet temperature for automobile radiator.
Overall heat transfer coefficient.
Find: Area required to achieve a prescribed outlet temperature.
Assumptions: (1) Negligible heat loss to surroundings and kinetic and
potential energy changes, (2) Constant properties.
Analysis: The required heat transfer rate is
q = (m c)h (T h,i - T h,o) = 0.05 kg/s (4209J / kg.K) 70K = 14,732 W
Using the ε-NTU method,
Cmin = Ch = 210.45 W / K
Cmax = Cc = 755.25W / K
Hence, Cmin/Cmx(Th,i - Th,o) = 210.45W / K(100K) = 21,045W
and
ε=q/qmax = 14,732W / 21,045W = 0.700
NTU≅1.5, hence
A=NTU(cmin / U) = 1.5 x 210.45W / K(200W) / m² .K) = 1.58m²
1. the air outlet is..
Tc,o = Tc,i + q / Cc = 300K + (14,732W / 755.25W / K) = 319.5K
2. using the LMTD approach ΔTlm = 51.2 K,, R=0.279 and P=0.7
hence F≅0.95 and
A = q/FUΔTlm = (14,732W) / [0.95(200W / m².K) 51.2K] = 1.51m²
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
Answers
An exothermic reaction releases 146 kJ of heat energy and 3 mol of gas at 298 K and 1 bar pressure. Which of the following statements is correct?
A) ΔU=-138.57 kJ and ΔH=-138.57 kJ
B) ΔU=-153.43 kJ and ΔH=-153.43 kJ
C) ΔU=-138.57 kJ and ΔH=-146.00 kJ
D) ΔU=-153.43 kJ and ΔH=-146.00 kJ
Answers
Answer:
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Explanation:
Given;
heat energy released by the exothermic reaction, ΔH = -146 kJ
number of gas mol, n = 3 mol
temperature of the gas, T = 298 K
Apply first law of thermodynamic
Change in the internal energy of the system, ΔU;
ΔU = ΔH- nRT
where;
R is gas constant = 8.314 J/mol.K
ΔU = -146kJ - (3 x 8.314 x 298)
ΔU = -146kJ - 7433 J
ΔU = -146kJ - 7.433 kJ
ΔU = -153.43 kJ
Therefore, the enthalpy change of the reaction ΔH is -146 kJ and change in the internal energy of the system is -153.43 kJ
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2. Deflection.
Answers
Answer:
1. Slope = 53.3 x 10⁻⁶
2. Deflection = -0.00016m
Explanation:
given:
let L = 4 m (span of cantilever beam)
let w = 300 N/m (distributed load)
let EI =60 MNm² (flexural stiffness)
dy w * L³ 300 x 4³
1. slope = ------- = --------- = ------------------- = 53.3 x 10⁻⁶
dx 6EI 6 x 60x10⁶
wL⁴ 300 x 4⁴
2. Deflection = y = - ----------- = - ------------------ = -0.00016m
8EI 8 x 60x10⁶
therefore the deflection is 0.16mm downwards.
the partner who can lose only what he or she has invested in business is the
Answers
Answer:
partner itself
Explanation:
Answer:
limited partner
hope this answer correct :)
Define the coefficient of determination and discuss the impact you would expect it to have on your engineering decision-making based on whether it has a high or low value. What do high and low values tell you
Answers
Answer and Explanation:
The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.
In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.
Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answers
Answer:
The volume percent of graphite is 91.906 per cent.
Explanation:
The volume percent of graphite ([tex]\% V_{Gr}[/tex]) is determined by the following expression:
[tex]\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%[/tex]
[tex]\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%[/tex]
Where:
[tex]V_{Gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{Fe}[/tex] - Volume occupied by the ferrite phase, measured in cubic centimeters.
The volume of each phase can be calculated in terms of its density and mass. That is:
[tex]V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}[/tex]
[tex]V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}[/tex]
Where:
[tex]m_{Gr}[/tex], [tex]m_{Fe}[/tex] - Masses of the graphite and ferrite phases, measured in grams.
[tex]\rho_{Gr}[/tex], [tex]\rho_{Fe}[/tex] - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.
Let substitute each volume in the definition of the volume percent of graphite:
[tex]\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%[/tex]
[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%[/tex]
Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:
[tex]m_{Gr} = \frac{2.5}{100}\times (100\,g)[/tex]
[tex]m_{Gr} = 2.5\,g[/tex]
[tex]m_{Fe} = 100\,g - 2.5\,g[/tex]
[tex]m_{Fe} = 97.5\,g[/tex]
If [tex]m_{Gr} = 2.5\,g[/tex], [tex]m_{Fe} = 97.5\,g[/tex], [tex]\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}[/tex] and [tex]\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}[/tex], the volume percent of graphite is:
[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%[/tex]
[tex]\% V_{Gr} = 91.906\,\%[/tex]
The volume percent of graphite is 91.906 per cent.