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
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
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select from a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Answer:
59°C
Explanation:
Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)
and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)
Therefore the minimum and maximum heat capacities are:
Cmin = Cc = 5803.2(W/K)
Cmax = Ch = 11634.3(W/K)
The capacity ratio is:
Cr = Cmin / Cmax = 0.499 = 0.5
The maximum possible heat transfer rate is:
Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)
And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99
Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)
Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C
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
Write torsion equation and explain the importance of each components.
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
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:
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
A quartz window of thickness L serves as a viewing port in a furnace used for annealing steel. The inner surface (x=0) of the window is irradiated with a uniform heat flux q''o due to emission from hot gases in the furnace. A fraction, β, of this radiation may be assumed to be absorbed at the inner surface, while the remaining radiation is partially absorbed as it passes through the quartz. The volumetric heat generation due to this absorption may be described by an expression of the form q˙(x)=(1−β)q''oα^e−αx where α is the absorption coefficient of the quartz. Convection heat transfer occurs from the outer surface (x=L) of the window to ambient air at T[infinity] and is characterized by the convection coefficient h. Convection and radiation emission from the inner surface may be neglected, along with radiation emission from the outer surface.
Required:
Determine the temperature distribution in the quartz, expressing your result in terms of the foregoing parameters.
Answer:
Assuming steady state condition the temperature distribution is calculated as expressed in the attached solution below
Explanation:
Given data :
thickness : L , inner surface (x) : 0, uniform flux : q"o
fraction : β
volumetric heat generation : q˙(x)=(1−β)q''oα^e−αx
determine the temperature distribution in the quartz
attached below is the detailed solution
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wire required to create a secondary impedance of 25 ohms
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.
A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one- dimensional. Two thermocouples are embedded in each sample some distance (L) apart, and a differential thermometer reads the temperature drop (Delta T) across this distance along each sample. When steady-state operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, rectangular samples (5 cm Times 5 cm on the side exposed to the heater and 10 cm long) are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15 degree C. Determine the thermal conductivity of the sample.
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
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%
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 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?
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.
Answer:
[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,
[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]
For the maximum value of internal torque
[tex]\dfrac{dt(x)}{dx}=0[/tex]
Therefore
[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]
Thus the maximum internal torque will be at x= 0.25 L
[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
. 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
A drilling operation is to be performed with a 12.7 mm diameter drill on cast iron. The hole depth is 60 mm and the drill point angle is 118∘. The cutting speed is 25 m/min and the feed is 0.30 mm/rev. Calculate:___________.
a) The cutting time (min) to complete the drilling operation
b) Material removal rate (mm3/min) during the operation, after the drill bit reaches full diameter.
Answer:
a. Tm = 0.3192min.
b. MRR = 396.91mm^{3}/s.
Explanation:
Given the following data;
Drill diameter, D = 12.7mm
Depth, L = 60mm
Cutting speed, V = 25m/min = 25,000m
Feed, F = 0.30mm/rev
To find the cutting time;
Cutting time, Tm =?
[tex]Tm = \frac{L}{Fr}[/tex] .......eqn 1
We would first solve for the feed rate (F);
[tex]Fr = NF[/tex] .......eqn 2
But we need to find the rotational speed (N);
[tex]N= \frac{V}{\pi *D}[/tex]
[tex]N= \frac{25000}{3.142*12.7}[/tex]
[tex]N= \frac{25000}{39.90}[/tex]
N = 626.57rev/min.
Substiting N into eqn 2;
[tex]Fr = NF[/tex]
Fr = 626.57 * 0.30
Fr = 187.97mm/min.
Substiting F into eqn 1;
[tex]Tm = \frac{L}{Fr}[/tex]
[tex]Tm = \frac{60}{187.97}[/tex]
Tm = 0.3192min.
Therefore, the cutting time is 0.3192 minutes.
For the material removal rate (MRR);
[tex]MRR = \frac{\pi *D^{2}Fr}{4}[/tex]
[tex]MRR = \frac{3.142*12.7^{2}*187.97}{4}[/tex]
[tex]MRR = \frac{3.142*161.29*187.97}{4}[/tex]
[tex]MRR = \frac{95258.16}{4}[/tex]
[tex]MRR = 23814.54mm^{3}/min[/tex]
Time in seconds, we divide by 60;
MRR = 23814.54/60 =396.91mm^{3}/s.
Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.
A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the maximum shear stress should not exceed 100 Mpa. b) Using the diameter, determine the maximum normal stress.
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
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.
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.
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 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.
When you shift your focus, everything you
see is still in perfect focus.
True or false
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 :)
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
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.
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:
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
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.
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?
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
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.
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.
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
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