Answer:
The pressure inside the tire is [tex]0.304\frac{N}{mm^{2}}[/tex]
Explanation:
The pressure gauge indicates the difference between the atmospheric pressure and the pressure inside the tire, so we have the following equation:
Pressure inside the tire = Gauge pressure + Atmospheric pressure
Where the gauge pressure is given in the problem and is 29.35psi and the atmospheric pressure is 14.7psi.
Replacing the values, we have:
Pressure inside the tire = 29.35psi + 14.7psi
Pressure inside the tire = 44.05psi
Now we have to convert from psi to [tex]\frac{N}{mm^{2}}[/tex], so:
44.05psi = [tex]44.05\frac{lbf}{in^{2}}[/tex]
[tex]44.05\frac{lbf}{in^{2}}*(\frac{1in}{25.4mm})^{2}*\frac{4.4482N}{lbf}=0.304\frac{N}{mm^{2}}[/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
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
Answer:
[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]
Explanation:
Reynolds number:
Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .
Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.
[tex]Re=\dfrac{F_i}{F_v}[/tex]
For plate can be given as
[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]
Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.
Flow on plate is a external flow .The values of Reynolds number for different flow given as
[tex]Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.[/tex]
[tex]Reynolds\ number\is \ <\ 5 \times 10 ^5\ then\ flow\ will\ be\ laminar.[/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.
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]
What are the processes by which polymer foams are produced.
Answer:
1.Molding
a.Cold molding
b.Hot molding
2.Slab stock
Explanation:
Polymer foam:
When solid and gas phases are mixed then they produce polymer foam.The gas used during forming of polymer foam is know as blowing agent and can be physical or chemical.Physical agent will not react and act as inert while chemical agent take part in reaction.
Blowing agent-Carbon diaoxide and Hydrochlorofluorocarbons.
Example of polymer foams-Polyurethane ,Starch etc.
Polymer foam can be produce by following process
1.Molding
a.Cold molding
b.Hot molding
2.Slab stock
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.
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.
Please find the power (in kW) needed in accelerating a 1000 kg car from 0 to 100 km/hour in 10 seconds on a 5% gradient up-hill road.
Answer:
P= 45.384 kW
Explanation:
given data:
m = 1000 kg
u = 0
v = 100 km/hr = 250/9 m/s
t = 10 sec
g =9.81 m/s2
5% gradient
from figure we can have
[tex]tan\theta = \frac{0.05x}{x}[/tex]
[tex]\theta = tan^{-1}0.05 [/tex]
[tex]\theta = 2.86[/tex] from equation of motion we have
v = u + at
[tex]\frac{250}{9} = 0 + 9*10[/tex]
[tex]a = \frac{25}{9} m/s^2[/tex]
distance covered in 10 sec
from equation of motion
[tex]s = ut + \frac{1}{2}at^2[/tex]
[tex]s = 0*10+ \frac{1}{2}*\frac{25}{9}*10^2[/tex]
s = 138.8 m
from newton's 2nd law of motion along inclined position
[tex]F -mgsin\theta = ma[/tex]
solving for f
[tex]f = mgsin\theta +ma[/tex]
[tex]F = 1000*9.81*SIN2.8624 +1000*\frac{25}{9}[/tex]
F = 3267.67 N
work done is given as W
[tex]W = F* s[/tex]
and power [tex]P = \frac{W}{t}[/tex]
[tex]P = \frac{F*s}{t}[/tex]
[tex]P = \frac{3267.67*\frac{1250}{9}}{10}[/tex]
P = 45384.30 W
P= 45.384 kW
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]
What is referred to as "Pyroelectric" materials?
Answer and Explanation:
Pyroelectric material
Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.
The potential difference generated is for very less time
The generation of potential difference is due to change in position of atoms after heating or cooling.
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]
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]
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.
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.
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.
The engine of a 2000kg car has a power rating of 75kW. How long would it take (seconds) to accelerate from rest to 100 km/hr at full power on level road. Neglect drag and friction.
Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
If P= ΔK/Δt, Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.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%
Explain the two advantages and the two disadvantages of fission as an energy source.
Answer with Explanation:
1) The advantages of fission energy are:
a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.
b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c
2) The disadvantages of fission energy are:
a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.
b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.
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.
Draw the Pressure - Temperature Diagram showing the liquid and vapor phases, along with the saturation line and the critical point (no need to include the solid section).
Answer:
Pressure- temperature diagram of the fluid is the phase lines that separate all the phases.
Explanation:
Step1
Pressure temperature diagram is the diagram that represents the all the phases of the fluid by separating a line. There is no phase change region in the pressure temperature diagram out of 15 possible diagrams. There are three lines that separate the phase of the fluid. These three lines are fusion, vaporization and sublimation.
Step2
The intersecting point of these lines is triple point of fluid. Out of 15 possible phase diagram, only pressure temperature diagram has triple point as a point. In other diagrams phase change region is present and triple point is not a point. Critical point is the point in all possible property diagrams.
Pressure temperature diagram is shown below:
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.
What are factor of safety for brittle and ductile materials.
Answer:
For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.
Explanation:
Factor of safety:
When materials are subjected to stress then we have to prevent it from a failure so we multiple stress by a factor and that factor is called factor of safety.
Factor of safety can be given as
[tex]FOS=\dfrac{Maximum\ strength}{Applied\ stress}[/tex]
Factor of safety is not a fixed quantity is varies according to the situation.
For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.
We know that brittle material did not shows any yield point and gets break without showing a indication but ductile materials shows a yield point and gives indication before fracture.
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
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.
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.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]
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]
What is the difference between Rage pressure and absolute pressure?
Answer:
Rage pressure:
The rage pressure in terms of engineering is the protected shield hard plastic shell and made up of the hard material that is basically used in the protection.
It is the effective material and used in the sewing machine so the pressure developed due to the hard material is known as rage pressure and it has strong elasticity.
Absolute pressure:
The pressure which is relative to the perfect vacuum is known as absolute pressure. The absolute pressure is basically measured against the atmospheric pressure.
The absolute pressure is defined as the total pressure in the fluid at the point is equal to the sum of the atmospheric pressure and gauge.
A random sample of 5 hinges is selected from a steady stream of product from a punch press, and the a. b. proportion nonconforming is 0.10. Sampling is with replacement. What is the probability of zero nonconforming unit in the sample? What is the probability of one nonconforming unit in the sample? hat is the probability of 2 or more nonconforming units in the sample?
Given : Sample size : n= 5
The proportion nonconforming : p= 0.10
Binomial probability formula :-
[tex]P(x)=^nC_x p^{x}(1-p)^{n-x}[/tex]
The probability of zero nonconforming unit in the sample :-
[tex]P(0)=^5C_0 (0.10)^{0}(1-0.1)^{5}\\\\=(1)(0.9)^5\ \ [ \because\ ^nC_0=1]=0.59049[/tex]
∴ The probability of zero nonconforming unit in the sample= 0.59049
The probability of one nonconforming unit in the sample :-
[tex]P(1)=^5C_1 (0.10)^{1}(0.9)^{4}\\\\=(5)(0.1)(0.9)^5\ \ [ \because\ ^nC_1=n]=0.295245[/tex]
∴ The probability of one nonconforming unit in the sample=0.295245
The probability of 2 or more nonconforming units in the sample :-
[tex]P(X\geq2)=1-(P(0)+P(1))=1-(0.59049+0.295245)\\\\=1-0.885735=0.114265[/tex]
∴ The probability of 2 or more nonconforming units in the sample=0.114265
What is the specific volume of superheated steam at 300 °C and a pressure of 1.2 MPa?
Answer:
The specific volume of the super heated steam is 0.2139 m³/kg.
Explanation:
Super heated steam is the condition where only compressed vapor of water present. This is not containing any liquid form of water. Super heated steam has very high pressure depending upon temperature. If heat supplied increases after saturation vapor condition of the water, the water fuses into steam completely. Properties of super heated steam are taken from the super heated steam table.
Given:
Temperature of superheated steam is 300°C.
Pressure of the super heated steam is 1.2 Mpa.
Calculation:
Step1
Value of specific volume of the super heated steam is taken from superheated steam table at 300°C and 1.2 Mpa as follows:
Specific volume is 0.2139 m³/kg.
Step2
All other properties are also taken from the table as:
Internal energy is 2789.7 kj/kg.
Enthalpy is 3046.3 kj/kg.
Entropy is 7.034 kj/kgK.
The part of table at the temperature 300°C and pressure 1.2 Mpa is shown below:
Thus, the specific volume of the super heated steam is 0.2139 m³/kg.