Assume that the flowrate. Q, of a gas from a smokestack is a function of the density of the ambient air,rhoarho a , the density of the gas,rhogrho g , within the stack, the acceleration of gravity, g, and the height and diameter of the stack, h and d, respectively. Use rhocrho c , d, and g as repeating variables to develop a set of pi terms that could be used to describe this problem.
Answers
Answer:
hello your question poorly written attached below is the well written question
answer : ∅[tex]( \frac{Pg}{Pa} , \frac{h}{d})[/tex][tex]( \frac{p_{g} }{p_{a} } , \frac{h}{d} )[/tex]
Explanation:
Develop a set of pi terms that could be used to describe the problem
attached below is the required solution
Related Questions
Compute the volume percent of graphite, VGr, in a 3.9 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answers
Answer:
Vgr = 0.122 = 12.2 vol %
Explanation:
Density of ferrite = 7.9 g/cm^3
Density of graphite = 2.3 g/cm^3
compute the volume percent of graphite
for a 3.9 wt% cast Iron
W∝ = (100 - 3.9) / ( 100 -0 ) = 0.961
Wgr = ( 3.9 - 0 ) / ( 100 - 0 ) = 0.039
Next convert the weight fraction to volume fraction using the equation attached below
Vgr = 0.122 = 12.2 vol %
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%
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath
Answers
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]
[tex]\rho = 7900 \ kg/m^3[/tex]
[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]
[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]
Here, we can't apply the lumped capacitance method, since Bi > 0.1
[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]
[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]
[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]
However, on a single rod, the energy extracted is:
[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]
Hence, for centerline temperature at 50 °C;
The surface temperature is:
[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]
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 )
Which two technologies were combined to create product life cycle management (PLM) software?
CAD and a database
spreadsheets and graphics
a database and spreadsheets
CAD and spreadsheets
Answers
Answer:
CAD and a database
Explanation:
The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.
In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B. the ratio of a machine's power to the force of its input Ос. the rate at which a machine performs work D. the rate at which a machine consumes energy E. the ratio of the work output of a machine to the work input
Answers
Answer:
E
Explanation:
I have a big brain and I just took the test and got it correct.
How should backing plates, struts, levers, and other metal brake parts be cleaned?
Answers
Answer: Cleaning of mechanical parts is necessary to remove contaminants, and to avoid clogging of wastes which could restrict the functioning of the machine.
Explanation:
There are different agents used for cleaning different machine instruments to prevent their corrosion and experience proper cleaning.
Backing plates must be dry cleaned using a cotton cloth to remove the dirt, dust or any other dry contaminant.
Struts can be wet cleaned by applying alcoholic solvent.
Levers can be cleaned using a mineral spirit.
Metallic plates can be cleaned using water based solution or water.
A rectangular block of 1m by 0.6m by 0.4m floats in water with 1/5th of its volume being out of water. Find the weight of the block.
Answers
Answer:
Weight of block is 191.424 Kg
Explanation:
The volume of rectangular block = [tex]1*0.6*0.4 = 0.24[/tex] cubic meter
1/5th of its volume being out of water which means water of volume nearly 4/5 th of the volume of rectangular block is replaced
Volume of replaced water = [tex]\frac{4}{5} * 0.24 = 0.192[/tex] cubic meter
Weight of replaced water = weight of rectangular block = [tex]0.192 * 997[/tex] Kg/M3
= 191.424 Kg
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
What Modern-Day Railroad Runs From Chicago to Seattle?
Answers
Answer:
The empire builder
Explanation:
Can some one help me with this plumbing question. Even just a guess.
Plz no shady links
Answers
Answer:
true
Explanation:
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
List six possible valve defects that should be included in the inspection of a used valve?
Answers
Answer:
Valvular stenosis , Valvular prolapse , Regurgitation,
Explanation:
MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False
Answers
The standard procedure for dimensioning the location of a house on a site is to dimension ____ of the house from adjacent lot lines. A one side b two sides c two corners d one corner
Answers
Answer:
One corner ( D )
Explanation:
when dimensioning the location of a house on site the standard and the acceptable procedure is to ; Dimension One corner of the house
Adjacent lots is a term used to describe parcels of the site that meet each other along their boundary lines. and they also include parcels that may be separated by streets
What regulates the car engines temperature
Answers
Answer:
car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.
Explanation:
Answer:
Coolant
Explanation:
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
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
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
Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.
Answers
Answer:
The issues related to the privacy are:
1. Informational privacy
2. Discrimination factors
3. Biased grouping on the basis of Data mining
4. Lack of consent
5. Morally wrong
6. Illegal distribution of information risks
7. Possibility of threat to life
Let's look at some major concerns:
1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.
2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.
3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:
A coal fired powerplant emits 0.5 kg/s of SO2 into the atmosphere from a stack that has a physical height of 100 meters. There is no temperature inversion and the atmosphere is characterized by class B stability for open country conditions. Exhaust emissions are 120 degrees C, ambient temperature is 25 degrees C, the stack diameter is 2.5 meters, and the volumetric flow of the exhaust is 35 cubic meters per second. The wind speed is 2 m/s.
Calculate the plume rise of the emissions using the Holland equation and determine the effective emission height.
Answers
pelo o que diz na database é que você n é ser humano normal por perguntar isso!!
Plumbing
The should indicate the materials, fixtures, and faucets to be
used.
A. specifications
B. plumbing code
C. mechanical code
D. plumbing instructor
Answers
Answer:
B....................
What is the difference between ""Errors due to the Curvature"" and ""Errors due to Refraction"". Support your answer with sketches
Answers
Answer:
In " errors due to the curvature " the points appears to be lower than they are in reality while
In " errors due to Refraction " The points appears to be higher than they are in reality .
Explanation:
The difference between both errors
In " errors due to the curvature " the points appears to be lower than they are in reality while
In " errors due to Refraction " The points appears to be higher than they are in reality .
when the effects of both Errors are combined the points will appear lower and this is because the effect of "error due to the curvature" is greater
attached below are the sketches
1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter of 0.620 inches and 0.750 inches, respectively. Cooling water enters the tubes at 60.0°F and leaves at 75.0°F at a velocity of 6.00ft/s. (HINT: You may assume laminar condensate flow.You many also assume that the mean bulk temperature of the cooling water is equal to the wall temperature on the outside of the pipe, T".You may also neglect the viscosity correction in your calculations.)a)What are the inside
Answers
Answer:
hi = 7026.8 W/m^2.k
Explanation:
Given data :
pressure of saturated steam = 1.2 bar
Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches
temperature of water at entry = 60°F
temperature of water at exit = 75°F
velocity of water = 6 ft/s
Calculate the Inside convective heat transfer coefficient ( hi )
mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C
next : find the properties of water at this temperature ( 19.727°C )
thermal conductivity = 0.598 w/m.k
density = 1000 kg/m^3
specific heat ( Cp ) = 4.18 KJ/kg.k
viscosity = 0.001 pa.s
velocity of water = 6 ft/s ≈ 1.8288 m/s
∴ Re ( Reynolds number ) = 28712.16
and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598 = 6.989
finally to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation
hi = 7026.8 w/m^2.k
attached below is the remaining solution
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
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:
A 5% upgrade on a six-lane freeway (three lanes in each direction) is 1.25 mi long. On this segment of freeway, the directional peak-hour volume is 3800 vehicles with 2% large trucks and 4% buses (no recreational vehicles), the peak-hour factor is 0.90, and all drivers are regular users. The lanes are 12 ft wide, there are no lateral obstructions within 10 ft of the roadway, and the total ramp density is 1.0 ramps per mile. A bus strike will eliminate all bus traffic, but it is estimated that for each bus removed from the roadway, seven additional passenger vehicles will be added as travelers seek other means of travel. What are the density, volume-to-capacity ratio, and level of service of the upgrade segment before and after the bus strike?
Answers
Answer:
a) Density of the road segment
i) Before bus strike = 23.02 pc/mi/In
ii) after bus strike = 25.76 pc/mi/In
b) Volume to capacity ratio
i) Before bus strike = 0.69
ii) After bus strike = 0.78
C) level of service of the upgrade segment
i) Before bus strike = LOS C
ii) after bus strike = LOS D
Explanation:
a) Density of the road segment
i) before bus strike ( D1 )
D1 = 1662 / 72.18 = 23.02 pc/mi/In
ii) After bus strike ( D2 )
D1 = 1859 / 72.18 = 25.76 pc/mi/In
b) Volume to capacity ratio
i) Before bus strike ( v 1 )
V1 = 1662 / 2400 = 0.69
ii) After bus strike ( V2 )
V2 = 1859 / 2400 = 0.78
C) level of service of the upgrade segment ( Gotten from " LOS Criteria for basic freeway segments " )
i) Before bus strike = ( LOS C )
ii) After bus strike = LOS D
Attached below is the detailed solution to the question above
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
Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.12 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.5 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.
Answers
Answer:
F[tex]_D[/tex] for A > F[tex]_D[/tex] for B
Hence, Bearing A can carry the larger load
Explanation:
Given the data in the question,
First lets consider an application which requires desired speed of n₀ and a desired life of L₀.
Lets start with Bearing A
so we write the relation between desired load and life catalog load and life;
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
where F[tex]_R[/tex] is the catalog rating( 2.12 kN)
L[tex]_R[/tex] is the rating life ( 3000 hours )
n[tex]_R[/tex] is the rating speed ( 500 rev/min )
F[tex]_D[/tex] is the desired load
L[tex]_D[/tex] is the desired life ( L₀ )
n[tex]_D[/tex] is the the desired speed ( n₀ )
Now as we know, a = 3 for ball bearings
so we substitute
[tex]2.12( 3000 * 500 * 60 )^{1/3[/tex] = [tex]F_D( L_0n_060)^{1/3[/tex]
950.0578 = [tex]F_D( L_0n_0)^{1/3} 3.914867[/tex]
950.0578 / 3.914867 = [tex]F_D( L_0n_0)^{1/3}[/tex]
242.6794 = [tex]F_D( L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for A = (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Therefore the load that bearing A can carry is (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Next is Bearing B
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
F[tex]_R[/tex] = 7.5 kN, [tex](L_Rn_R60) = 10^6[/tex]
Also, for ball bearings, a = 3
so we substitute
[tex]7.5(10^6)^{1/3[/tex] = [tex]F_D(L_0n_060)^{1/3}[/tex]
750 = [tex]F_D(L_0n_0)^{1/3} 3.914867[/tex]
750 / 3.914867 = [tex]F_D(L_0n_0)^{1/3}[/tex]
191.5773 = [tex]F_D(L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for B = ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Therefore, the load that bearing B can carry is ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Now, comparing the Two results above,
we can say;
F[tex]_D[/tex] for A > F[tex]_D[/tex] for B
Hence, Bearing A can carry the larger load