Answer:
Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.
Is the beam design acceptable for a SF of 2? YES
Explanation:
Your factor of safety is 2, this means your stress allowed is:
σall = YS/FS = 100kpsi/2 = 50kpsiWhere:
σall => Stress allowedYS => Yield StrengthFS => Factor of safetyNow we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.
First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:
ΣFy = 0-100 + Ry = 0 thus,Ry = 100 lbfKnowing this reaction force you can already calculate the shear stress on the cantilever beam:
τ = F/Aτ = 100lbf/(2in*5in) τ = 10 psiNow, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.
ΣM = 0-100lbf*10in + M = 0M = 1000 lbf-inKnowing the maximum bending moment you can now calculate your bending stress as follows:
σ = M*c/IxWhere:
σ => Bending StressM => Bending Momentc => Distance from the centroid of your beam geometry to the outermost fiber.Ix => Second moment area of inertiaOut of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in
To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:
Ix = (b*h^3)/12 = (2in*5in^3)/12 = 20.83 in4Now you can use this numbers in your bending stress formula:
σ = M*c/Ixσ = 1000 lbf-in * 2.5in / 20.83 in4σ = 120 psiThe shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.
For a motor purchased for a tank project, it claims that at 12V it produces 6000 RPM. This motor is attached to a 50:1 gear reduction gearbox. The output of this gearbox is attached to a 2.5" wheel. Determine the speed of the tank under these conditions in miles/hour (mph).
Answer:
0.889 mph
Explanation:
Given:
Voltage = 12 V
RPM produced, N₁ = 6000
Gear reduction gearbox, G = 50:1
Diameter of the wheel = 2.5" = 2.5 × 0.0254 = 0.0635 m
Now,
The output speed = [tex]\frac{N_1}{\textup{G}}=\frac{6000}{50}[/tex] = 120 RPM
Thus,
The speed, v = [tex]\frac{\pi DN_2}{60}[/tex]
on substituting the respective values, we get
v = [tex]\frac{\pi\times0.0635\times120}{60}[/tex]
or
v = 0.39878 m/s
Also,
1 m/s = 2.23 mph
thus,
v = 0.39878 × 2.23 = 0.889 mph
The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k= 0.029 W/ m K. The measured temperature difference across a 25-mm-thick sheet of the material is T1 - T2 = 12°C. a. What is the heat flux through a 3 mx 3 m sheet of the insulation? b. What is the rate of heat transfer through the sheet of insulation? c. What is the thermal resistance of the sheet due to conduction?
Answer:
a. The heat flux through the sheet of insulation is 19.92 W/m^2
b. The rate of heat transfer through the sheet of insulation is 125.28 W
c. The thermal resistance of the sheet due to the conduction is 0.86 Km^2/W.
Explanation:
From the heat conduction Fourier's law it can be state for a wall of width e and area A :
q = Q/ΔT = k*A* (T2-T1)/e
Where q is the rate of heat transfer, k the conductivity constant, and T2 and T1 the temperatures on the sides of the wall. Replacing the values in the correct units, we obtained the rate of heat transfer:
q = 0.029 W/*mK * (3m*3m) * (12°K) / (0.025m)
(The difference in temperatures in Kelvin is the same than in Celcius degres).
q = 0.029 W/*mK * (9 m^2) * (12°K) / (0.025m) = 125.28 W
The heat flux is calculated by dividing q by the area of the wall:
q/A = k* (T2-T1)/e = 0.029 W/*mK * (12°K) / (0.025m) = 19.92 W/m^2
The thermal resistance of the sheet is defined as:
R = e / k
Replacing the values in the proper units:
R = 0.025 m / 0.029 W/*mK = 0.86 Km^2/W
The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.
Answer:
The air heats up when being compressed and transefers heat to the barrel.
Explanation:
When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.
In an adiabatic transformation:
[tex]P^{1-k} * T^k = constant[/tex]
For air k = 1.4
SO
[tex]P0^{-0.4} * T0^{1.4} = P1^{-0.4} * T1^{1.4}[/tex]
[tex]T1^{1.4} = \frac{P1^{0.4} * T0^{1.4}}{P0^{0.4}}[/tex]
[tex]T1^{1.4} = \frac{P1}{P0}^{0.4} * T0^{1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.4/1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.28}[/tex]
SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.
After it was compressed the hot air will exchange heat with the barrel heating it up.
An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inch and a standard deviation of 0.07 inch. Assume that the shaft diameter follows a Gaussian distribution. What percentage of the diameter of the total shipment of shafts will fall between 2.39inch and 2.60 inch?
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : [tex]\overline{x}=2.45 [/tex]
Standard deviation : [tex]\sigma=0.07[/tex]
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to 2.39 will be :-
[tex]z=\dfrac{2.39-2.45}{0.07}\approx-0.86[/tex]
z-value corresponds to 2.60 will be :-
[tex]z=\dfrac{2.60-2.45}{0.07}\approx2.14[/tex]
Using the standard normal table for z, we have
P-value = [tex]P(-0.86<z<2.14)=P(z<2.14)-P(z<-0.86)[/tex]
[tex]=P(z<2.14)-(1-P(z<0.86))=P(z<2.14)-1+P(z<0.86)\\\\=0.9838226-1+0.8051054\\\\=0.788928\approx0.7889=78.89\%[/tex]
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Phase difference is: Select one: 1. The angle difference of a point on waveform from the origin 2. The angle between points of three time wave forms 3. The offset between the crests of two time waveforms 4. The offset between the troughs of the two time waveforms 5. The offset between similar points of the two time waveforms
Answer:
#5
Explanation:
#3, #4 and #5 are correct answers, however #5 is more general, and therefore it is a better answer.
#1 would refer to argument of a wave function, equal to the product of the time by the pulsation plus the phase.
#2 seems to refer three phase signals, and they do have phase differences, but not all phase differences are three phase.
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter = 0.8 in. Yielding occurs at a load of 30,000 lb. The corresponding gage length = 1.8075 in, which is the 0.2 percent yield point. The maximum load of 56,050 lb is reached at a gage length = 2.35 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 2.5 in, determine the percent elongation. (e) If the specimen necked to an area=0.35 in^2, determine the percent reduction in area.
Answer:
a) 60000 psi
b) 1.11*10^6 psi
c) 112000 psi
d) 30.5%
e) 30%
Explanation:
The yield strength is the load applied when yielding behind divided by the section.
yield strength = Fyield / A
A = π/4 * D^2
A = 0.5 in^2
ys = Fy * A
y2 = 30000 * 0.5 = 60000 psi
The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).
k = E * L0 / A
And the stiffness is related to change of length:
Δx = F / k
Then:
Δx = F * A / (E * L0)
E = F * A / (Δx * L0)
When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was
Δx = L - L0 = 1.8075 - 1.8 = 0.0075
Then:
E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi
Tensile strength is the strees at which the material breaks.
The maximum load was 56050 lb, so:
ts = 56050 / 0.5 = 112000 psi
The percent elongation is calculated as:
e = 100 * (L / L0)
e = 100 * (2.35 / 1.8 - 1) = 30.5 %
If it necked with and area of 0.35 in^2 the precent reduction in area was:
100 * (1 - A / A0)
100 * (1 - 0.35 / 0.5) = 30%
An object at a vertical elevation of 20 m and a speed of 5 m/s decreases in elevation to an elevation of 1 m. At this location, the object has a velocity of 15 m/s. The mass of the object is 68 kg. Assuming the object is the system, determine if there is any work transfer associated with the object (there is no heat transfer). The object is solid, incompressible and its temperature does not change during the process. If there is work transfer, is work done on or by the object? Assume the acceleration of gravity g = 9.81 m/s2.
Answer with Explanation:
We know that from the principle of work and energy we have
Work done on/by a body =ΔEnergy of the body.
Now as we know that energy of a body is the sum of it's kinetic and potential energy hence we can find out the magnitude of the final and initial energies as explained under
[tex]E_{initial}=P.E+K.E\\\\E_{initail}=mgh_1+\frac{1}{2}mv_1^{2}\\\\68\times 9.81\times 20+\frac{1}{2}\times 68\times (5)^{2}=14191.6Joules[/tex]
Similarly the final energy is calculated to be
[tex]E_{final}=P.E+K.E\\\\E_{final}=mgh_2+\frac{1}{2}mv_2^{2}\\\\68\times 9.81\times 1+\frac{1}{2}\times 68\times (15)^{2}=8317.08Joules[/tex]
As we can see that the energy of the object has changed thus by work energy theorem we conclude that work transfer is associated with the object.
Part 2)
The change in the energy of the body equals [tex]8317.08-14191.6=-5874.52Joules[/tex]
Since the energy is lost by the system hence we conclude that work is done by the object.
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over the city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13600 kg/m3, respectively. Show this using the systematic solution method.
Answer:
1) The absolute pressure equals = 96.98 kPa
2) Absolute pressure in terms of column of mercury equals 727 mmHg.
Explanation:
using the equation of pressure statics we have
[tex]P(h)=P_{surface}-\rho gh...............(i)[/tex]
Now since it is given that at 6400 meters pressure equals 45 kPa absolute hence from equation 'i' we obtain
[tex]P_{surface}=P(h)+\rho gh[/tex]
Applying values we get
[tex]P_{surface}=45\times 10^{3}+0.828\times 9.81\times 6400[/tex]
[tex]P_{surface}=96.98\times 10^{3}Pa=96.98kPa[/tex]
Now the pressure in terms of head of mercury is given by
[tex]h_{Hg}=\frac{P}{\rho _{Hg}\times g}[/tex]
Applying values we get
[tex]h=\frac{96.98\times 10^{3}}{13600\times 9.81}=0.727m=727mmHg[/tex]
Area under the strain-stress curve up to fracture:______
Answer:
Area under the strain-stress curve up to fracture gives the toughness of the material.
Explanation:
When a material is loaded by external forces stresses are developed in the material which produce strains in the material.
The amount of strain that a given stress produces depends upon the Modulus of Elasticity of the material.
Toughness of a material is defined as the energy absorbed by the material when it is loaded until fracture. Hence a more tough material absorbs more energy until fracture and thus is excellent choice in machine parts that are loaded by large loads such as springs of trains, suspension of cars.
The toughness of a material is quantitatively obtained by finding the area under it's stress-strain curve until fracture.
Determine the specific volume of superheated water vapor at 15 Mpa and 1000C, using (a) the ideal-gas equation, and (b) the steam tables.
Answer:
a)[tex]v=25.56\ kg/m^3[/tex]
b) [tex]v=38.8\ kg/m^3[/tex]
Explanation:
Given that
Pressure P = 15 MPa
Temperature = 1000 C = 1273 K
a)If assume as ideal gas
The gas constant for super heated steam R is 0.461 KJ/kg.K.
We know that ideal gas equation
P v =RT
15 x 1000 x v =0.461 x 1273
[tex]v=25.56\ kg/m^3[/tex]
b) By using steam table
From steam table we can see that volume of super heated vapot at 15 MPa and 1273 K .
[tex]v=38.8\ kg/m^3[/tex]
What is the difference Plastic vs elastic deformation.
Answer:
What is the difference Plastic vs elastic deformation
Explanation:
The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.
The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.
A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving through the middle of a wind farm when an idea snapped into her head. She thought to herself, what if she mounted a small windmill generator on the roof of her Prius and wired it into the battery? She thought, with a little head wind to get her started she could just drive off under wind power! The faster she drove, the more power she would make, and the faster she could go… What’s wrong with her idea? a. As a side question, Priuses make use of regenerative braking. Why is it called regenerative braking? Why don’t cars advertise regenerative accelerators?
Answer:
Explanation:
It wouldn't work because the wind energy she would be collecting would actually come from the car engine.
The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.
u' = u + v
While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.
Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.
A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.
the relative velocity of two infinitely wide parallel plates
thatare 1cm apart is 10cm/s. If the dynamic viscosity of theliquid
between the plates is 0.001 N s/m2 what is theshear
stress between the plates?
Answer:[tex]0.01 N/m^2[/tex]
Explanation:
Given
Distance between Plates(dy)=1 cm
Relative Velocity(du)=10 cm/s
Dynamic viscosity[tex]\left ( \mu \right )=0.001 Ns/m^2[/tex]
We know shear stress is given by [tex]\tau [/tex]
[tex]\tau =\frac{\mu du}{dy}[/tex]
where du=relative Velocity
dy=Distance between Plates
[tex]\tau =\frac{0.001\times 10}{1}=0.01 N/m^2[/tex]
What are primary and secondary super-heaters?
Explanation:
Superheater has two types of parts which are:
The primary super-heater The secondary super-heaterPrimary super-heater is first heater which is passed by the steam after steam comes out of steam drum.
After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.
Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.
A 600 MW power plant has an efficiency of 36 percent with 15
percent of the waste heat being released to the atmosphere as stack
heat and the other 85 percent taken away in the cooling water.
Instead of drawing water from a river, heating it, and returning it
to the river, this plant uses an evaporative cooling tower wherein
heat is released to the atmoshphere as cooling water is
vaporized.
At what rate must 15C makeup water be provided from the river to
offset the water lost in the cooling tower?
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
Answer:
m = 367.753 kg.s
Explanation:
GIVEN DATA:
Power plant capacity W=600 MW
Plant efficiency is [tex]\eta = 36\%[/tex]
[tex]efficiency = \frac{W}{QH}[/tex]
[tex]0.36 = \frac{600}{QH}[/tex]
QH = 1666.6 MW
from first law of thermodynamics we hvae
QH -QR = W
Amount of heat rejection is QR = 1066.66 MW
As per given information we have 15% heat released to atmosphere
[tex]QR = 0.15 \tiimes 1066.66 = 159.99 MW[/tex]
AND 85% to cooling water
[tex]Q = 0.85 \times 1066.66 = 906.66 MW[/tex]
from saturated water table
at temp 150 degree c we have Hfg = 2465.4 kJ/kg
rate of cooiling water is given as = mhfg
[tex]906.66 \times 1000 KW = m \times 2465.4[/tex]
m = 367.753 kg.s
where m is rate of makeup water that is added to offset
A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the atmosphere is at 20 deg C, what is the maximum power output for a mass flow of 0.2 kg/s? Assume the water is cooled to atmospheric temperature.
Answer:
W = 12.8 KW
Explanation:
given data:
mass flow rate = 0.2 kg/s
Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)
we know that maximum possible efficiency is given as
[tex]\eta = 1- \frac{T_L}{T_H}[/tex]
[tex]\eta = 1 - \frac{ 293}{368}[/tex]
[tex]\eta = 0.2038[/tex]
rate of heat transfer is given as
[tex]Q_H = \dot m C_p \Delta T[/tex]
[tex]Q_H = 0.2 * 4.18 8(95 - 20)[/tex]
[tex]Q_H = 62.7 kW[/tex]
Maximuim power is given as
[tex]W = \eta Q_H[/tex]
W = 0.2038 * 62.7
W = 12.8 KW
In an oscillating LC circuit, L ! 25.0 mH and C ! 7.80 mF. At time t 0 the current is 9.20 mA, the charge on the capacitor is 3.80 mC, and the capacitor is charging.What are (a) the total energy in the circuit, (b) the maximum charge on the capacitor, and (c) the maximum current? (d) If the charge on the capacitor is given by q ! Q cos(vt $ f), what is the phase angle f? (e) Suppose the data are the same, except that the capacitor is discharging at t ! 0.What then is f?
Answer:
a) 926 μJ
b) 3.802 mC
c) 8.61 A
d) 0.0721
e) 3.2137
Explanation:
The energy in the inductor is
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J[/tex]
The energy store in a capacitor is
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
The voltage in a capacitor is
V = Q/C
[tex]V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V[/tex]
Therefore:
[tex]Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J[/tex]
The total energy is Et = 925.6 + 1.1 = 926.7 μJ
At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
V = Q/C
[tex]Ec = \frac{1}{2}*C*(\frac{Q}{C})^2[/tex]
[tex]Ec = \frac{1}{2}*\frac{Q^2}{C}[/tex]
[tex]Q^2 = 2*Ec*C[/tex]
[tex]Q = \sqrt{2*Ec*C}[/tex]
[tex]Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC[/tex]
When the capacitor is completely empty all the energy will be in the inductor and current will be maximum
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]I^2 = 2*\frac{El}{L}[/tex]
[tex]I = \sqrt{2*\frac{El}{L}}[/tex]
[tex]I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A[/tex]
At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC
q = Q * cos(vt + f)
q(0) = Q * cos(v*0 + f)
3.8 = 3.81 * cos(f)
cos(f) = 3.8/3.81
f = arccos(3.8/3.81) = 0.0721
If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmospheric pressure at the surface is the same as in the submarine. (S.G. of sea water =1.025).
Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:
[tex]S.G=\frac{\rho_{sw}}{\rho_{w}}[/tex]
Here, density of sea water is[tex]\rho_{sw}[/tex], surface gravity is S.G and density of water is [tex]\rho_{w}[/tex].
Substitute all the values in the above equation as follows:
[tex]S.G=\frac{\rho_{sw}}{\rho_{w}}[/tex]
[tex]1.025=\frac{\rho_{sw}}{1000}[/tex]
[tex]\rho_{sw}=1025[/tex] kg/m³.
Step2
Difference in pressure is calculated as follows:
[tex]\bigtriangleup p=rho_{sw}gh[/tex]
[tex]\bigtriangleup p=1025\times9.81\times320[/tex]
[tex]\bigtriangleup p=3217680[/tex] pa.
Or
[tex]\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})[/tex]
[tex]\bigtriangleup p=3217.68[/tex] kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.
A settling tank with 50 foot diameter and a SWD of 9 feet treats
aflow of 15, 000 gpd. What is the detention time?
Answer:
Detention time will be 9 days
Explanation:
We have given diameter of setting tank d = 50 feet
Radius of setting tank [tex]r=\frac{d}{2}=\frac{9}{2}=25 feet[/tex]
SWD = 9 FEET
Area [tex]A=\pi r^2=3.14\times 25^2=1962.5ft^2[/tex]
So volume [tex]V = area\times SWD=1962.5\times 9=17671.5ft^3[/tex]
We have given flow Q = 15000 gpd
So Q= 15000×0.13368 =2005.22 [tex]ft^3[/tex] per day
Detention time is given by [tex]=\frac{volume}{Q}=\frac{17671.5}{2005.22}=9days[/tex]
So detention time will be 9 days
When a 2-kg block is suspended from a spring, the spring is stretched a distance of 20 mm. Determine the natural frequency and the period of vibration for a 0.5-kg block attached to the same spring
Answer:
7.05 Hz
Explanation:
The natural frequency of a mass-spring system is:
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]
To determine the constant k of the spring we use Hooke's law:
Δl = F / k
k = F / Δl
In the first case the force was the weight of the 20 kg mass and the Δl was 20 mm.
F = m * a
F = 2 * 9.81 = 19.6 N
Then:
k = 19.6 / 0.02 = 980 N/m
Therefore:[tex]f = \frac{1}{2 \pi}\sqrt{\frac{980}{0.5}} = 7.05 Hz[/tex]
The atmospheric pressure at the top and the bottom of a building is read by a barometer to be 98.5 kPa and 100 kPa, respectively. If the density of air is 1.2 kg/m^3, calculate the height of the building (in m).
Answer:
127.42m
Explanation:
The air pressure can be understood as the weight exerted by the air column on a body, for this case we must remember that the pressure is calculated by the formula P=αgh, Where P=pressure, h=gravity, h= height,α=density
So what we must do to solve this problem is to find the length of the air column above and below the building and then subtract them to find the height of the building, taking into account the above the following equation is inferred
h2-h1= building height=H
[tex]H=\frac{P1-P2}{g\alpha }[/tex]
P1=100kPa=100.000Pa
P2=98.5kPa=98.500Pa
α=1.2 kg/m^3
g=9.81m/s^2
[tex]H=\frac{100000-98500}{(9.81)(1.2) }=127.42m[/tex]
A pump is put into service at the coast where the barometric pressure is 760 mm Hg. The conditions of service are : Flow rate 0,08 m3/s, suction lift 3,5 metres, suction pipe friction loss 0,9 metres, water temperature 65°C, water velocity 4 m/s. Under these conditions of service, the pump requires an NPSH of 2,1 metres. Assuming the density of water as 980,6 kg/m3, establish whether it will operate satisfactorily.
Answer:
The pump operates satisfactorily.
Explanation:
According to the NPSH available definition:
[tex]NPSHa = \frac{P_{a} }{density*g} + \frac{V^{2} }{2g} - \frac{P_{v}}{density*g}[/tex]
Where:
[tex]P_{a} absolute pressure at the inlet of the pump [/tex]
[tex]V velocity at the inlet of te pump = 4m/s[/tex]
[tex]g gravity acceleration = 9,8m/s^{2}[/tex]
[tex]P_{v} vapor pressure of the liquid, for water at 65°C = 25042 Pa[/tex]
The absolute pressure is the barometric pressure Pb minus the losses: Suction Lift PLift and pipe friction loss Ploss:
To convert the losses in head to pressure:
[tex]P = density*g*H [/tex]
So:
[tex]P_{b} = 760 mmHg = 101325 Pa[/tex]
[tex]P_{lift} = 33634,58 Pa[/tex]
[tex]P_{loss} = 8648,89 Pa[/tex]
The absolute pressure:
[tex]P_{a} = P_{b} - P_{lift} - P_{loss} = 59044,53 Pa[/tex]
replacing on the NPSH available equiation:
[tex]NPSHa = 6,14 m + 0,816 m - 2,6 m = 4,356 m [/tex]
As the NPSH availiable is higher than de required the pump should operate satisfactorily.
What is the specific volume of H2O at 1000F and 2000 psia?
Answer:
The specific volume of H2O at 1000F and 2000 psia is 0.3945 ft^3/lb
Explanation:
In figure you can see that for 2000 psia saturation temperature of water is 636 F, so at 1000 F water is at vapor phase. Then, we have to use superheated steam table. From the table the specific volume of H2O at 1000F and 2000 psia is 0.3945 ft^3/lb
Answer:
[tex]0.3945\frac{ft^{3}}{lb}[/tex]
Explanation:
The specific volume is the inverse of the density, so
[tex]v=\frac{1}{d}[/tex]
[tex]d=\frac{m}{V}[/tex]
[tex]v=\frac{V}{m}[/tex]
For superheated steam you can use the table and locate the temperature an pressure given by te problem, in its correspondent value that is [tex]0.3945\frac{ft^{3}}{lb}[/tex]
Two factors that can be modified to optimize the cutting processes are feed and depth, and cutting velocity. a) True b) False
Answer:
True
Explanation:
The process of cutting is used to cut an object with the help of physical forces.
This process includes cutting like shearing, drilling, etc.
The cutting process makes use of the mechanical tools to maintain the contact of cutter with the object.
This process can be optimized by modification of the cutting depth, cutting velocity and feed.
The process optimization also depends on the cutting fluid as these are the deterministic factors for the cutting condition.
Write the equation to estimate heat flow q due to convection.
Answer:
q= h ΔT
Explanation:
Heat flow due to convection given as follows
Q= h A ΔT
Heat flux per unit area given as
q= h ΔT
Where h is the heat transfer coefficient,A is the surface area and ΔT is the temperature difference.
Lets take one plate having temperature [tex]T_1[/tex] and air is flowing on the plate with temperature [tex]T_2[/tex] and air having heat transfer coefficient h.Dimensions of plate is l x b.
Area of plate
[tex]A=lbm^2[/tex]
So heat transfer
[tex]Q=h\times l\times b(T_1-T_2)[/tex]
If an elevator is to raise a 5000 lb a height of 30 ft in 320 seconds, what horsepower motor would be required?
Answer:
Force will be equal to 0.08522 horsepower
Explanation:
We given weight of elevator = 5000 lb
Height of the elevator = 30 ft
Time t = 320 sec
We have to calculate the power in horsepower
For let first find power in lb-ft/sec
So power in lb-ft/sec [tex]=\frac{5000lb\times 30ft}{320sec}=46.875lb-ft/sec[/tex]
Now we know that 1 horsepower = 550 lb-ft/sec
So 46.785 lb-ft/sec = [tex]\frac{46.875}{550}=0.08522horsepower[/tex]
A Carnot engineoperates between a heat source at
1200 F and a heat sink at 70 F.The engine delivers 200 hp. Compute
the heat supplied (Btu/s), theheat rejected (Btu/s), and the
thermal efficiency of the heatengine.
Answer:
Heat supplied = 208.82 BTU/s
Heat rejected = 66.82 BTU/s
Carnot thermal efficiency = 0.68
Explanation:
Data
Hot temperature,[tex] T_H [/tex] = 1200 F + 459.67 = 1659.67 R
Cold temperature,[tex] T_C [/tex] = 70 F + 459.67 = 529.67 R
Engine power, [tex] \dot{W} = 200 hp \times 0.71\frac{BTU/s}{hp} = 142 \frac{BTU}{s} [/tex]
Carnot thermal efficiency is computed by
[tex] \eta = 1 - \frac{T_C}{T_H} [/tex]
[tex] \eta = 1 - \frac{529.67 R}{1659.67 R} [/tex]
[tex] \eta = 0.68 [/tex]
Efficiency is by definition
[tex] \eta = \frac{\dot{W}}{\dot{Q_{in}}} [/tex]
[tex] \dot{Q_{in}} = \frac{\dot{W}}{\eta} [/tex]
[tex] \dot{Q_{in}} = \frac{142 \frac{BTU}{s}}{0.68} [/tex]
[tex] \dot{Q_{in}} = 208.82 \frac{BTU}{s} [/tex]
where [tex] \dot{Q_{in}} [/tex] is the heat supplied
Energy balance in the engine
[tex] \dot{Q_{in}} = \dot{W} + \dot{Q_{out}} [/tex]
[tex] \dot{Q_{out}} = \dot{Q_{in}} - \dot{W} [/tex]
[tex] \dot{Q_{out}} = 208.82 \frac{BTU}{s} - 142 \frac{BTU}{s} [/tex]
[tex] \dot{Q_{out}} = 66.82 \frac{BTU}{s} [/tex]
where [tex] \dot{Q_{out}} [/tex] is the heat rejected
The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as Re Durho/μ whereD is pipe diameter, u is fluid velocity, rho is fluid density, and μ is fluid viscosity.When the value of the Reynolds number is less than about 2100, the flow is laminar—that is, the fluid flows in smooth streamlines. For Reynolds numbers above 2100, the flow is turbulent, characterized by a great deal of agitation. Liquid methyl ethyl ketone (MEK) flows through a pipe with an inner diameter of 2.067 inches at an average velocity of 0.48 ft/s. At the fluid temperature of 20°C the density of liquid MEK is 0.805 g/cm3 and the viscosity is 0.43 centipoise [1 cP 1:00 103 kg/ m s]. Without using a calculator, determine whether the flow is laminar or turbulent. Show your calculations.
The Reynolds number is approximately 14,067, indicating turbulent flow of MEK through the pipe, as Re > 2100 denotes turbulence.
let's calculate the Reynolds number for the flow of liquid methyl ethyl ketone (MEK) through the pipe.
Given:
- Pipe diameter (D) = 2.067 inches = 0.0512 meters (converted from inches to meters)
- Fluid velocity (u) = 0.48 ft/s = 0.1463 meters/s (converted from feet per second to meters per second)
- Fluid density (ρ) = 0.805 g/cm³ = 805 kg/m³ (converted from grams per cubic centimeter to kilograms per cubic meter)
- Fluid viscosity (μ) = 0.43 centipoise = 0.43 x 10² kg/(m*s)(converted from centipoise to kg/(m*s))
The Reynolds number (Re) is calculated as:
[tex]\[ Re = \frac{(D \cdot u \cdot \rho)}{\mu} \][/tex]
Plugging in the values:
[tex]\[ Re = \frac{(0.0512 \, \text{m} \cdot 0.1463 \, \text{m/s} \cdot 805 \, \text{kg/m³})}{0.43 \times 10^{-3} \, \text{kg/(m*s)}} \][/tex]
[tex]\[\text{Re} = \frac{(0.0074991 \, \text{m}^2/\text{s}^2 \cdot 805 \, \text{kg/m}^3)}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} = \frac{6.0498755 \, \text{kg/(m*s)}}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} \approx \frac{6.0498755}{0.00043}\]\[\text{Re} \approx 14,067.35\][/tex]
Since the calculated Reynolds number is greater than 2100 (Re > 2100), the flow of MEK through the pipe is turbulent.
In an orthogonal cutting operation, the 0.250 in wide tool has a rake angle of 5º. The lathe is set so the chip thickness before the cut is 0.010 in. After the cut, the deformed chip thickness is measured to be 0.027 in. Calculate (a) the shear plane angle and (b) the shear strain for the operation.
Answer:
(a) Shear plane angle will be 21.9°
(b) Shear train will be 2.7913 radian
Explanation:
We have given rake angle [tex]\alpha =5^{\circ}[/tex]
Thickness before cut [tex]t_1=0.01inch[/tex]
Thickness after cutting [tex]t_2=0.027inch[/tex]
Now ratio of thickness before and after cutting [tex]r=\frac{t_1}{t_2}=\frac{0.01}{0.027}=0.3703[/tex]
(a) Shear plane angle is given by [tex]tan\Phi =\frac{rcos\alpha }{1-rsin\alpha }[/tex], here [tex]\alpha[/tex] is rake angle.
So [tex]tan\Phi =\frac{0.3703\times cos5^{\circ}}{1-sin5^{\circ}}=0.4040[/tex]
[tex]\Phi =tan^{-1}0.4040[/tex]
[tex]\Phi =21.9^{\circ}[/tex]
(b) Shear strain is given by [tex]\gamma =tan(\Phi -\alpha )+cot\Phi[/tex]
So [tex]\gamma =tan(21.9 -5 )+cot21.9=2.7913radian[/tex]
(a) The shear plane angle is approximately [tex]\( 20.9^\circ \)[/tex].
(b) The shear strain for the operation is approximately 2.90.
In an orthogonal cutting operation, we need to calculate the shear plane angle and the shear strain using the given parameters.
Given:
- Tool width [tex]\( w = 0.250 \)[/tex] in
- Rake angle [tex]\( \alpha = 5^\circ \)[/tex]
- Uncut chip thickness [tex]\( t_1 = 0.010 \)[/tex] in
- Deformed chip thickness [tex]\( t_2 = 0.027 \)[/tex] in
(a) Shear Plane Angle
The shear plane angle [tex]\( \phi \)[/tex] can be calculated using the chip thickness ratio r and the rake angle [tex]\( \alpha \)[/tex].
1. Calculate the chip thickness ratio r :
[tex]\[ r = \frac{t_1}{t_2} = \frac{0.010 \text{ in}}{0.027 \text{ in}} = 0.3704 \][/tex]
2. Use the relationship for shear plane angle [tex]\( \phi \)[/tex] :
[tex]\[ \tan(\phi) = \frac{r \cos(\alpha)}{1 - r \sin(\alpha)} \][/tex]
Substituting r and [tex]\( \alpha = 5^\circ \)[/tex] :
[tex]\[ \tan(\phi) = \frac{0.3704 \cos(5^\circ)}{1 - 0.3704 \sin(5^\circ)} \][/tex]
3. Calculate [tex]\( \cos(5^\circ) \)[/tex] and [tex]\( \sin(5^\circ) \)[/tex] :
[tex]\[ \cos(5^\circ) \approx 0.9962, \quad \sin(5^\circ) \approx 0.0872 \][/tex]
4. Substitute and calculate [tex]\( \tan(\phi) \)[/tex] :
[tex]\[ \tan(\phi) = \frac{0.3704 \times 0.9962}{1 - 0.3704 \times 0.0872} \approx \frac{0.3690}{0.9677} \approx 0.3816 \][/tex]
5. Find [tex]\( \phi \)[/tex] :
[tex]\[ \phi = \tan^{-1}(0.3816) \approx 20.9^\circ \][/tex]
(b) Shear Strain
The shear strain [tex]\( \gamma \)[/tex] in the shear plane can be calculated using:
[tex]\[ \gamma = \cot(\phi) + \tan(\phi - \alpha) \][/tex]
1. Calculate [tex]\( \cot(\phi) \)[/tex] :
[tex]\[ \cot(\phi) = \frac{1}{\tan(\phi)} = \frac{1}{0.3816} \approx 2.62 \][/tex]
2. Calculate [tex]\( \phi - \alpha \)[/tex] :
[tex]\[ \phi - \alpha = 20.9^\circ - 5^\circ = 15.9^\circ \][/tex]
3. Calculate [tex]\( \tan(15.9^\circ) \)[/tex] :
[tex]\[ \tan(15.9^\circ) \approx 0.2844 \][/tex]
4. Calculate [tex]\( \gamma \)[/tex] :
[tex]\[ \gamma = 2.62 + 0.2844 \approx 2.9044 \][/tex]
The shear plane angle and shear strain are crucial in determining the efficiency of the cutting operation. The shear plane angle is derived from the relationship between the undeformed and deformed chip thicknesses and the rake angle. The shear strain reflects the material deformation occurring during the cutting process, combining the geometric factors and material properties.
Briefly discuss the cooling system for motor.
Answer and Explanation:
COOLING SYSTEM :
Cooling systems are used for cooling the motor engines.There are mainly two type of cooling system air cooling system and water cooling system.In air cooling system air is used as coolant and in water cooling system water is sued as coolant.In this system water or air is passed in the engine block and heads and and it brings heat out of the engine.Due to cooling system efficiency of the system increases because it reduces the heat loss.