Answer:
a) 42 mm
b) 144.4 MPa
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]
[tex]d^{3}[/tex] = 7.638 x 10^-5
d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm
b) Normal stress = P/A
where A is the area
A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa
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
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
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:
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
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
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
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 concentric tube heat exchanger is used to cool a solution of ethyl alcohol flowing at 6.93 kg/s (Cp = 3810 J/kg-K) from 65.6 degrees C to 39.4 degrees C using water flowing at 6.30 kg/s at a temperature of 10 degrees C. Assume that the overall heat transfer coefficient is 568 W/m2-K. Use Cp = 4187 J/kg-K for water.
a. What is the exit temperature of the water?
b. Can you use a parallel flow or counterflow heat exchanger here? Explain.
c. Calculate the rate of heat flow from the alcohol solution to the water.
d. Calculate the required heat exchanger area for a parallel flow configuration
e. Calculate the required heat exchanger area for a counter flow configuration. What happens when you try to do this? What is the solution?
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
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
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.
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
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.
A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the core is 1.8 m and the joints are equivalent to an airgap of 0.1 mm. The value of the magnetic field strength for 1.1 T in the core is 400 A/m, the corresponding core loss is 1.7 W/kg at 50 Hz and the density of the core is 7800 kg/m3. If the maximum value of the flux density is to be 1.1 T when a p.D. Of 2200 V at 50 Hz is applied to the primary, calculate: a. The cross-sectional area of the core; b. The secondary voltage on no load; c. The primary current and power factor on no load.
Answer:
a) cross sectional area of the core = 0.0187 m²
b) The secondary voltage on no-load = 413 V
c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.
Explanation:
See attached solution.
4. ""ABC constriction Inc."" company becomes the lowest in the bed process to get a $21 million construction project for ""Northern Inc."". Now ""ABC construction Inc."" planning to make a formal contract agreement
Answer:
hello your question is incomplete here is the complete question
. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?
Answer : elements of the agreement
offeracceptancecapacity certaintyconsiderationintention to create legal relationExplanation:
Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid
Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.
capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .
certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions
Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement
Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement
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
what's the maximum shear on a 3.0 m beam carrying 10 kN/m?
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
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.
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
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
A 60-Hz 220-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real power is 1500 W, and the apparent power is 4600 VA.
a. Determine the value of the resistance.
b. Determine the value of the inductance.
Answer:
(a) The value of the resistance is 3.431 Ω
(b) The value of the inductance is 0.0264 H
Explanation:
Given;
frequency of the source, f = 60 Hz
rms voltage, V-rms = 220 V
real power, Pr = 1500 W
apparent power, Pa = 4600 VA
(a). Determine the value of the resistance
[tex]P_r = I_{rms}^2R[/tex]
where;
R is resistance
[tex]I_{rms} = \frac{Apparent \ Power}{V_{rms}} \\\\I_{rms} = \frac{P_a}{V_{rms}}\\\\I_{rms}= \frac{4600}{220} \\\\I_{rms}= 20.91 \ A[/tex]
Resistance is calculated as;
[tex]R = \frac{P_r}{I_{rms}^2} \\\\R = \frac{1500}{(20.91)^2} \\\\R = 3.431 \ ohms[/tex]
(b). Determine the value of the inductance.
[tex]Q_L = I_{rms}^2 X_L[/tex]
where;
[tex]Q_L[/tex] is reactive power
[tex]X_L[/tex] is inductive reactance
[tex]Apparent \ power = \sqrt{Q_L^2 + P_r^2} \\\\P_a^2 = Q_L^2 + P_r^2\\\\Q_L^2 = P_a^2 - P_r^2\\\\Q_L^2 = 4600^2 - 1500^2\\\\Q_L^2 = 18910000\\\\Q_L = \sqrt{18910000}\\\\Q_L = 4348.56 \ VA[/tex]
inductive reactance is calculated as;
[tex]X_L = \frac{Q_L}{I_{rms}^2} \\\\X_L = \frac{4348.56}{(20.91)^2} \\\\X_L = 9.95 \ ohms[/tex]
inductance is calculated as;
[tex]X_L = \omega L\\\\X_L = 2\pi f L\\\\L = \frac{X_L}{2\pi f} \\\\L = \frac{9.95}{2\pi *60} \\\\L = 0.0264 \ H\\\\L = 26.4 \ mH[/tex]
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.
Answer: 164.2253 MPa
Explanation:
First we find the half internal crack which is = length of surface crack /2
so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )
Now we find the dimensionless parameter using the critical stress crack propagation equation
∝ = K / Y√πα
stress level ∝ = 112Mpa
fracture toughness K = 26Mpa
dimensionless parameter Y = ?
SO working the formula
Y = K / ∝√πα
Y = 26 / 112 (√π × 4.3× 10⁻³)
Y = 1.9973
We are asked to find stress level for internal crack length of 4m
so half internal crack is = length of surface crack /2
4/2 = 2mm ( 2 × 10⁻³)
from the previous formula critical stress crack propagation equation
∝ = K / Y√πα
∝ = 26 / 1.9973 √(π × 2 × 10⁻³)
∝ = 164.2253 Mpa
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%
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!
steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a 1000 ft ramps at a high speed vo and travels 600ft in 7 s at constant deceleration before its speed is reduced to uo/2. Assuming the same constant deceleration.
Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.
Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.
A car travels from A, due north to a town B 4 km away. It then travels due east until it arrives town C 5 km from B. determine the distance of town C from A
Answer:
A to C = 6.4 km
Explanation:
A to B = 4 km
B to C = 5 km
A to C = using pythagorean theorem
a² + b² = c²
a = A to B = 4
b = B to C = 5
c = A to C
c² = 4² + 5²
c = 6.4 km (A to C)
According to the scenario, the distance between town C from town A is found to be 6.40 Km.
Which background does this question depend on?The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.
According to the question,
The distance between town A to town B = 4 km.
The distance between town B to town C = 5 km.
Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:
[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].
[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]
[tex]AC^2[/tex] = 16 + 25 = 41.
AC = √41 = 6.40 km.
Therefore, the distance between town C from town A is found to be 6.40 Km.
To learn more about Pythagoras' theorem, refer to the link:
https://brainly.com/question/343682
#SPJ5
What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
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.
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.
If you see a red, a green, and a white light on another boat, what does this tell you?
A boat is approaching you head on.
The red and green lights are sidelights that are positioned on the port side (red) (left as facing the bow) and starboard (green) (right as facing the bow) side of the boat. Various white lights are required depending on the size of the boat, but generally, a white masthead light and stern light are required. See the US Coast Guard site in the link below for more specific information.
Hope this helps
Who plays a role in the financial activities of a company?
O A. Just employees
O B. Just managers
O C. Only members of the finance and accounting department
O D. Everyone at the company, including managers and employees
Hey,
Who plays a role in the financial activities of a company?
O D. Everyone at the company, including managers and employees
Answer:
Everyone at the company, including managers and employees
Explanation:
1. How many PWM generator blocks are there in LM3S1968? What are they? 2. How many independent PWM outputs can be generated on an LM3S1968? 3. List at least two applications for PWM. 4. What does NVIC in a timer stand for? Explain its significance. 5. Where does the counter/timer derive its time period from? 6. Draw the waveforms (square wave) with duty cycles (on) 25%, 50%, 75%.One of the purpose of the lab is to generate a PWM signal in one of the ports using systick timer. a. Given a signal with 1 KHz, find out the time period of each cycle. Find out the time span of the high signal and the low signal given 10%, 20%, 30% and 90% duty cycles. b. We would like to generate a signal with a certain frequency (ex. 100 Hz, 1 KHz, etc.) and certain duty cycle (10%, 20%, etc.), find out the values we need to load into the timer register? Given that the XTAL = 8 MHz.
Answer:
1) There are three (3) PWM generator blocks in LM3S1968 and they are
PWM signal generatorADC trigger selectorPWM dead-band generator2) Two (2) independent PWM outputs can be generated on an LM3S1968
3) Applications for PWM
Control Brightness of LED using Duty Cycle controlSpeed Control of DC Motor4) NVIC in a timer stand for ; Nested Vectored Interrupt Controller
its significance is that it is used to handle and give priorities to exception and Interrupts
5) The counter/timer derive its time period from counting the output pulses for one cycle which is the duration over which gate is open
Explanation:
1) There are three (3) PWM generator blocks in LM3S1968 and they are
PWM signal generatorADC trigger selectorPWM dead-band generator2) Two (2) independent PWM outputs can be generated on an LM3S1968
3) Applications for PWM
Control Brightness of LED using Duty Cycle controlSpeed Control of DC Motor4) NVIC in a timer stand for ; Nested Vectored Interrupt Controller
its significance is that it is used to handle and give priorities to exception and Interrupts
5) The counter/timer derive its time period from counting the output pulses for one cycle which is the duration over which gate is open
6) THE WAVEFORM DIAGRAMS IS ATTACHED BELOW
it can be seen that 50 % rises and goes down at half interval. 75 % goes down at half more of 50% and 25% goes down at half less of 50%
Describe the meaning of the different symbols and abbreviations found on the drawings/documents that they use (such as BS8888, surface finish to be achieved, linear and geometric tolerances, electronic components, weld symbols and profiles, pressure and flow characteristics, torque values, imperial and metric systems of measurement, tolerancing and fixed reference points)
Answer:
Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.
There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.
Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.
Explanation:
. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 rpm and the mixer must operate as close to 250 rpm as possible. Select an appropriate belt size, commercially available sheaves, and a belt for this application. Also calculate the actual belt speed and the center distance.
Answer:
Hello the table which is part of the question is missing and below are the table values
For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1
Answers:
belt size = 140 in with diameter of 20.1n
actual speed of belt = 288.49 in/s
actual center distance = 49.345 in
Explanation:
Given data :
Electric motor (driver sheave) speed (w1) = 950 rpm
Driven sheave speed (w2) = 250 rpm
pick D1 ( diameter of driver sheave) = 5.8 in ( from table )
To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first
VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250
also since the speed of belt would be constant then ;
Vb = w1r1 = w2r2 ------- equation 1
r = d/2
substituting the value of r into equation 1
equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex] = VR
Appropriate belt size ( d2) can be calculated as
d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04
From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value
next we have to determine the belt length /size
[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]
inputting all the values into the above equation including the value of C as calculated below
L ≈ 140 in
Calculating the center distance
we use this equation to get the ideal center distance
[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]
22.04 < c < 3 ( 5.8 + 20.1 )
22.04 < c < 77.7
the center distance is between 22.04 and 77.7 but taking an average value
ideal center distance would be ≈ 48 in
To calculate the actual center distance we use
[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3
B = [tex]4L -2\pi (d2 + d1 )[/tex]
inputting all the values into (B)
B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )
B ≈ 399.15 in
inputting all the values gotten Back to equation 3 to get the actual center distance
C = 49.345 in ( actual center distance )
Calculating the actual belt speed
w1 = 950 rpm = 99.48 rad/s
belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]
= 99.48 * 5.8 / 2 = 288.49 in/s
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.
An inventor claims to have developed a heat pump that produces a 200-kW heating effect for a 293 K heatedzone while only using 75 kW of power and a heat source at 273 K. Justify the validity of this claim.
Answer:
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
Explanation:
Heat generated Q = 200 kW
power input W = 75 kW
Temperature of heated region [tex]T_{h}[/tex] = 293 K
Temperature of heat source [tex]T_{c}[/tex] = 273 K
For this engine,
coefficient of performance COP = Q/W = 200/75 = 2.67
The maximum theoretical COP obtainable for a heat pump is given as
COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] = [tex]\frac{293 }{293 - 273 }[/tex] = 14.65
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.