4.54 Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0.008 kg/s and exits as saturated vapor (see Fig. P4.54). It then flows into a superheater also at 600 kPa, where it exits at 600 kPa, 280 K. Find the rate of heat transfer in the boiler and the superheater.
Answers
Answer:
hello the figure attached to your question is missing attached below is the missing diagram
answer :
i) 1.347 kW
ii) 1.6192 kW
Explanation:
Attached below is the detailed solution to the problem above
First step : Calculate for Enthalpy
h1 - hf = -3909.9 kJ/kg ( For saturated liquid nitrogen at 600 kPa )
h2- hg = -222.5 kJ/kg ( For saturated vapor nitrogen at 600 kPa )
second step : Calculate the rate of heat transfer in boiler
Q1-2 = m( h2 - h1 ) = 0.008( -222.5 -(-390.9) = 1.347 kW
step 3 : find the enthalpy of superheated Nitrogen at 600 Kpa and 280 K
from the super heated Nitrogen table
h3 = -20.1 kJ/kg
step 4 : calculate the rate of heat transfer in the super heater
Q2-3 = m ( h3 - h2 )
= 0.008 ( -20.1 -(-222.5 ) = 1.6192 kW
Related Questions
Which of the following identifies three advantages of PLM software?
reduced cost, faster marketing, and process transparency
faster marketing, process transparency, and outsourcing
process transparency, outsourcing, and reduced cost
outsourcing, reduced cost, and faster marketing
Answers
Answer:
reduced cost, faster marketing, and process transparency
Explanation:
The correct answer is reduced cost, faster marketing, and process transparency. PLM software helps coordinate tasks during implementation.
Which option identifies the specialized field of engineering that would best suit Erik in the following scenario?
Erik is an accounting major, but he fears that working with numbers all day may be a bit constrictive for him. His roommate is an engineering
major and is planning to specialize in chemical engineering, Erik is not interested in chemistry, but he does find engineering fascinating.
O structural engineering
O materials engineering
industrial engineering
O aerospace engineering
Answers
Answer:
industrial engineering
Explanation:
Calculate the relative pipe roughness for a plastic pipe with absolute roughness 0.0025 mm and internal diameter of pipe is 0.157 inches.
Answers
Answer:
6.27 × 10⁻⁴
Explanation:
Relative roughness, k = ε/D where ε = absolute roughness = 0.0025 mm and D = internal pipe diameter = 0.157 in = 0.157 × 25.4 mm = 3.9878 mm
So, k = ε/D
= 0.0025 mm/3.9878 mm
= 6.27 × 10⁻⁴
The relative pipe roughness for a plastic pipe will be:
"6.27 × 10⁻⁴".
Relative roughness of pipeAccording to the question,
Absolute roughness, ε = 0.0025 mm
Internal pipe diameter, D = 0.157 in or,
= 0.157 × 25.4 mm
= 3.9878 mm
We know that,
The relative roughness be:
→ k = [tex]\frac{Absolute \ roughness}{Diameter}[/tex]
or,
k = [tex]\frac{\varepsilon }{D}[/tex]
By substituting the above values,
= [tex]\frac{0.0025}{3.9878}[/tex]
= 6.27 × 10⁻⁴
Thus the above approach is correct.
Find out more information about diameter will be:
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A thick steel slab ( 7800 kg/m3 , c 480 J/kg K, k 50 W/m K) is initially at 300 C and is cooled by water jets impinging on one of its surfaces. The temperature of the water is 25 C, and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach 50 C at a distance of 25 mm from the surface
Answers
Answer:
1791 secs ≈ 29.85 minutes
Explanation:
( Initial temperature of slab ) T1 = 300° C
temperature of water ( Ts ) = 25°C
T2 ( final temp of slab ) = 50°C
distance between slab and water jet = 25 mm
Determine how long it will take to reach T2
First calculate the thermal diffusivity
∝ = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s
next express Temp as a function of time
T( 25 mm , t ) = 50°C
next calculate the time required for the slab to reach 50°C at a distance of 25mm
attached below is the remaining part of the detailed solution
List six possible valve defects that should be included in the inspection of a used valve?
Answers
Answer:
Valvular stenosis , Valvular prolapse , Regurgitation,
Explanation:
When an arbitrary substance undergoes an ideal throttling process through a valve at steady state, (SELECT ALL THAT APPLY) inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet pressures will be equal. inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet temperatures will be equal.
Answers
Answer:
15x
Explanation:
Wattage is:
Select one:
a.
A measure of the total electrical work being performed per unit of time.
b.
Expressed as P = R × A.
c.
Both A and B.
d.
Neither A nor B.
Answers
Answer:
c.Both A and B.
Explanation:
the wattage is c and d
Technician A says when you push the horn button, electromagnetism moves an iron bar inside the horn, which opens and closes contacts in the horn circuit. Technician B says most vehicle horn circuits use a relay. Which technician is correct?
Answers
Answer: Both technicians A and B
Explanation:
I took the pf test
A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Answers
Answer:
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation )
Explanation:
Yield strength = 275 Mpa
Tensile strength = 380 Mpa
elastic modulus = 103 GPa
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation ) .
Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given
strain = yield strength / elastic modulus
= 0.0027
Help meeeeeeeee plzzzzz need explanation
Answers
the picture is blank for me what does it say i can comment the answer plz mark brainlyist
Discuss in detail the manners of interaction with opposite gender
Answers
Answer:
8 Tips on Better Communication with the Opposite Sex
Put emotions away. Ladies, this one is more aimed at us, for the most part. ...
Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .
Put yourself in their shoes. .
Listen. ...
Respond. ...
Actually communicate. ...
Be detailed. ...
Don't communicate too much.
Explanation:
For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answers
Answer:
a) the maximum load is 88,040 N
b)
the maximum length to which the specimen may be stretched is 0.12032148 mm
Explanation:
Given the data in the question;
the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa
modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa
a)
Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )
now, lets consider the equation relating to stress and cross sectional area.
σ = F / A₀
hence, maximum load F = σA₀
so we substitute
F = (2.84 × 10⁸) × (310 × 10⁻⁶)
F = 88,040 N
Therefore, the maximum load is 88,040 N
b)
Initial length specimen l₀ = 120 mm = 120 × 10⁻³ m
using engineering strain, ε = (l₁ - l₀)/l₀
Also from Hooke's law, σ = Eε
so from the equation above;
l₁ = l₀( ε + 1 )
l₁ = l₀( σ/E + 1 )
so we substitute
l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )
l₁ = (120 × 10⁻³) ( 1.002679 )
l₁ = 0.12032148 mm
Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm
consider a stead flow ideal carnot cycle using steam as the working fluid in which the high temperature constant pressure heat addition process starts with a saturated liquid and ends with a saturated vapor. plot this cycle in t-s coordinates showing the steam dome. calculate the thermal efficiency for this cycle if the pressure of the high temperature steam is 6 mpa and the low temperature heat rejection process occurs at 300 k.
Answers
Answer:
45.32%
Explanation:
Given data:
pressure of high temperature steam = 6 MPa
low temperature heat rejection process ( Tr ) = 300 k
A) plot of cycle in t-s coordinates showing steam dome
attached below
B) Calculate thermal efficiency
thermal efficiency = 1 - (Tr / Tsat )
Tsat = 275.59°C ≈ 548.59 K ( from steam table at Pa = 6 MPa )
back to equation 1
1 - (300 / 548.59 )
1 - 0.5468 = 0.4532 = 45.32%
a tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN
the minimum diameter at fracture is 10mm
determine the engineering stress at maximum load and the true fracture stress.
Answers
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
Calculate the engineering stress at Maximum load and the True fracture stress
i) Engineering stress at maximum load = P/ A
= P / [tex]\pi \frac{D^2}{4}[/tex] = 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
ii) True Fracture stress = P/A
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2
The volume of a right circular cone of radius r and height h is V = 1 3 πr 2h (a) (i) Find a formula for the instantaneous rate of change of V with respect to r if r changes and h remain constant. (ii) Suppose that h = 2 is fixed but r varies. Find the rate of change of V w. R. To r at the point where r = 4.
Answers
Answer:
(i) [tex]\frac{2}{3}[/tex][tex]\pi[/tex]rh
(ii) [tex]\frac{16}{3}[/tex][tex]\pi[/tex]
Explanation:
Given:
V = [tex]\frac{1}{3}[/tex][tex]\pi[/tex]r²h
Where;
V = volume of a right circular cone.
r = radius of the cone
h = height of the cone.
(i) The rate of change of V with respect to r if r changes and h remains constant is [tex]\frac{dV}{dr}[/tex], and is given by finding the differentiation of V with respect to r as follow:
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{d}{dr}[/tex][[tex]\frac{1}{3}[/tex][tex]\pi[/tex]r²h]
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{2}{3}[/tex][tex]\pi[/tex]rh --------------------(i)
(ii)
Given;
h = 2
r = 4
Substitute these values into equation (i) as follows;
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{2}{3}[/tex][tex]\pi[/tex](4 x 2)
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{2}{3}[/tex][tex]\pi[/tex](8)
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{16}{3}[/tex][tex]\pi[/tex]
[tex]\frac{dV}{dr}[/tex] = [tex]\frac{16}{3}[/tex][tex]\pi[/tex]
A right circular cone is one where the axis of cones is the line connecting the vertex to circular base's midway, the volume of right circular cone as follows:
Volume calculation:Formula:
[tex]V = \frac{1}{3} \pi r^2h[/tex]
Where;
V = right circular cone volume
r = Cone radius.
h = Cone height.
The calculation for part 1:
[tex]\frac{dV}{dr}[/tex] is indeed the rate of change of V with reference to r when r changes but h remains constant, and it is calculated via calculating the differentiation of V with respect to r as follows:
[tex]\to \frac{dV}{dr} =\frac{d}{d}r [ \frac{1}{3} \pi r^2h] =\frac{2}{3} \pi r h[/tex]
The calculation for part 2:
When h = 2 and r = 4 then substituting the value into the part 1 equation then:
[tex]\to \frac{dV}{dr} = \frac{2}{3} \pi (4 \times 2) = \frac{2}{3} \pi (8) = \frac{16}{3} \pi[/tex]
Find out more about the volume here:
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Can some one help me with this plumbing question. Even just a guess.
Plz no shady links
Answers
Answer:
true
Explanation:
An industrial boiler consists of tubes inside of which flow hot combustion gases. Water boils on the exterior of the tubes. When installed, the clean boiler has an over all heat transfer coefficient of 300 W/m^2 . K. Based on experience, i is anticipated that the fouling factors on the inner and outer surfaces will increase linearly with time as Ra,t and Ryo-at where a, 2.5 x 10^-11 m2 K/W s and a,-1.0 x 10^-11 m^2 - K/W s for the inner and outer tube surfaces, respectively. If the boiler is to be cleaned when the overall heat transfer coeffi- cient is reduced from its initial value by 25%, how long after installation should the first cleaning be scheduled?
Answers
Answer:
the first cleaning be scheduled 1.006 years after installation
Explanation:
Given the data in the question;
U[tex]_{clean[/tex] = 300 W/m².K
first we determine the heat coefficient of the dirt surface;
overall heat transfer coefficient is reduced from its initial value by 25%
U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]
U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300
U[tex]_{dirt[/tex] = 0.75 × 300
U[tex]_{dirt[/tex] = 225 W/m².K
next we find the inner fouling factor
[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]
[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t
for the outer fouling water;
[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]
[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t
now, we determine the total heat transfer coefficient
[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]
we substitute
[tex]\frac{1}{U}[/tex] = (3.5 × 10⁻¹¹)t
so the first cleaning duration after insulation will be;
[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]
we substitute
(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]
(3.5 × 10⁻¹¹)t = 0.001111
t = 0.001111 / (3.5 × 10⁻¹¹)
t = 31742857.142857 seconds
t = 31742857.142857 / 3.154 × 10⁷
t = 1.006 years
Therefore, the first cleaning be scheduled 1.006 years after installation
what is the answer to life the universe and everything
(worth 95 points!)
Answers
Answer:
In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...
Explanation:
I think it goes on forever.
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.
Answers
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v[tex]_f[/tex] = 0.001057 m³/kg
v[tex]_g[/tex] = 1.0037 m³/kg
u[tex]_f[/tex] = 486.82 kJ/kg
u[tex]_g[/tex] 2524.5 kJ/kg
h[tex]_g[/tex] = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v[tex]_g[/tex]
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m[tex]_{out[/tex] = m₁ - m₂
m[tex]_{out[/tex] = 1.89414 - 0.003985
m[tex]_{out[/tex] = 1.890155 kg
so, Initial internal energy will be;
U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]
U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex] + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E[tex]_{in[/tex] - E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]
QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁
QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Answers
Answer:
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Explanation:
#carryonml
Answer:
using communication tools
Explanation:
The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.
A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).
Answers
Answer
a) 62 percent
b) 40 percent
Explanation:
Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm
Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm
diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm
New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm
Calculate ductility in terms of
a) percent reduction in area
percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100
A[tex]_{i}[/tex] ( initial area ) = π /4 di^2
= π /4 * ( 10.33 )^2 = 83.81 mm^2
A[tex]_{f}[/tex] ( final area ) = π /4 df^2
= π /4 ( 6.38)^2 = 31.97 mm^2
hence : %reduction = ( 83.81 - 31.97 ) / 83.81
= 0.62 = 62 percent
b ) percent elongation
percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]
= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
immediate possession at any time when operating a motor vehicle.
True
False
Answers
Answer:
true
Explanation:
the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once
An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
the correct distance?
Answers
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Which of the following combinations of bends can be used in a conduit run?
A. One 90-degree, three 45-degree, and four 30-degree
B. TWO 90-degree, two 45-degree, and four 30-degree
C. Two 90-degree, four 45-degree, and one 30-degree
D. One 90-degree, six 45-degree, and one 30-degree
Answers
Answer:
Answer is A
Explanation:
Doesn't break 360 degrees of bend.
The bend that can be used in conduit run is one 90-degree, three 45-degree, and four 30-degree.
Bend that can be used in conduit run is equal or less than 360 degree.
One 90-degree, three 45-degree, and four 30-degree, the total sum is 345 degree.Two 90-degree, two 45-degree, and four 30-degree, the total sum is 390 degree.Two 90-degree, four 45-degree, and one 30-degree, the total sum is 390 degree.One 90-degree, six 45-degree, and one 30-degree, the total sum is 390 degree.So, One 90-degree, three 45-degree, and four 30-degree is the bend that can be used in conduit run.
Learn more: brainly.com/question/21199524
Using a small cube as a representative volume of material, illustrate graphically all the six independent components of strains that determine the state of strain in the volume of material that is enclosed by the cube. Indicate in your schematics, the change in shape produced by each of the normal strain and shear strain components.
Answers
Answer:
Attached below
Explanation:
Attached below is the schematic diagram of a small cube representing volume of material and all the six independent components of strains that will determine the state of strain
Note : Attached below is a free hand diagram of the cube indicating the six independent components of strains and a screen shot of the change in shape produced by each of the normal strain and shear strain components
( drawn with online tools )
pacing pieces of information into groups to remember them better is called
a.
Visualizing
c.
Rhyming
b.
Keywording
d.
Categorizing
Please select the best answer from the choices provided
Answers
Answer:
D. Categorizing
Explanation:
pls mark me as your brainlist
Answer:
D
Explanation:
A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.
Answers
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
[tex]E^z[/tex] = - ( εˣ / v )
[tex]E^z[/tex] = - ( -0.00088 / 0.34 )
[tex]E^z[/tex] = - ( - 0.002588 )
[tex]E^z[/tex] = 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife
Answers
Answer: i got you its d
Explanation:had the smae question as you
5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction (d) none of the above 6. The support at B does not allow _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction
Answers
Answer: Diagram associated with your question is attached below
5) B
6) C
Explanation:
5) The pin support at A allows ; Rotation about its central axis
This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis
6) The support at B does not allow displacement in y direction
This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction
You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip formation is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.
Answers
Answer:
A)
shear plane angle = 31.98°
shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) shear strength = 7339.78
Explanation:
a) Determine the shear plane angle and shear strain
Given data :
Chip thickness before chip formation = 0.5 inches
Chip thickness after separation = 1.125 inches
rake angle ( ∝ ) = 10°
shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex] ----- ( 1 )
r = chip thickness ratio = 0.5 / 1.125 = 0.4444
back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10
Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736 = 0.5296
hence ∅ = tan^-1 ( 0.5296 ) = 31.98°
shear strain : R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )
R = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) determine the shear strength of the material
cutting force = 1559 N
thrust force = 1271 N
width of cut ( diameter ) = 3.0 mm
shear strength = c + σ.tan ∅
c = cohesion force = 1271 * 3 = 3813
σ = normal stress = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94
hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78