Answer:
The presence of element Carbon.
Explanation:
The diagrams Steel- Carbon usually show the percent the carbon vs the phases of the steel.
In the middle you increase the carbon percent the steels are not commercial because they are no malleables ( Hardennes).
By the other hand according the application of the steel you need to look the diagram Fe-Cr.
Calculate the change in the enthalpy of argon, in kJ/kg, when it is cooled from 75 to 35°C. If neon had under-gone this same change of temperature, would its enthalpy change have been any different?
Answer:
Enthalpy almost doubles.
Explanation:
Argon
Cp = Specific heat at constant volume = 0.520 kJ/kgK
T₁ = Initial temperature = 75°C
T₂ = Final temperature = 35°C
Enthalpy
Δh = CpΔT
⇒Δh = Cp(T₂-T₁)
⇒Δh = 0.520×(35-75)
⇒Δh = -20.8 kJ/kg
Neon
Cp = Specific heat at constant volume = 1.03 kJ/kgK
T₁ = Initial temperature = 75°C
T₂ = Final temperature = 35°C
Δh = Cp(T₂-T₁)
⇒Δh = 1.03×(35-75)
⇒Δh = -41.2 kJ/kg
Enthalpy will change because Cp value is differrent.
Enthalpy almost doubles.
What is meant by the critical Reynolds number?
Answer:
Critical Reynolds number is a number or the threshold or the limit at which the laminar flow changes to turbulent flow.
Explanation:
It is basically the ratio of inertial forces to viscous forces and helps to predict if the flow is laminar or turbulent.
from the experiments, for flow in a pipe of diameter D, critical Reynolds number=2300
For Laminar flow: Reynolds number is less than 2000
For transitional flow: Reynolds number is in between 2000-4000, the flow is unstable
For Turbulent flow: Reynolds number is greater than 3500
Reynolds number is different for different geometries
The universe is sometimes described as an isolated system. Why?
Answer and Explanation :The universe means it includes everything, even the things which we can not see is an isolated system because universe has no surroundings. an isolated system does not exchange energy or matter with its surroundings.Sometime universe is treated as isolated system because it obtains lots of energy from the sun but the exchange of matter or energy with outside is almost zero.
the total energy of an isolated system is always constant means total energy of universe is also constant there is no exchange of matter or energy in an isolated systemDefine full and partial well abandonment.
Answer:
Partial and Fully abandonment of well is a term used for oil well
Explanation:
Full Abandonment - This condition is depend on economics of the oil well. when production from oil well is come to that level when operating cost become higher than operating income, then it is need to shut down the well fully to prevent further loss.
Partial Abandonment - in this condition only the base part of the well is neglect permanently and from the upper portion of the well sidetracked well is build to acquire new target.
The current that flows through series elements of a circuit is the same in each element. (a) True (b) False
Answer:
The given statement is true.
Explanation:
In a circuit, in which all circuit elements are connected in series, carries equal or same amount of current through each circuit element as in series circuit current have only one path through which it can flow and the voltage drop across each element depends on the values of resistances or reactances connected in the series circuit and vary depending on their respective respective resistances and reactances whereas the current flowing remain same throughout. Series circuit works on three rules:Current remains same throughout Total resistance equals the sum of individual resistances.Supply voltage equals the sum of voltage drops at each element.In a gas turbine, air (kinematic viscosity of 1x104-5 m 2/s) flows over a 2 cm long turbine blade at 100 m/s. How long should the blade be in my lab's wind tunnel (air, kinematic viscosity of 1.5x10A-5 mA2/s, velocity of 10 m/s), to match the Reynolds number of the gas turbine? a)-2cm b)-30cm c)-0.3cm
Answer:
30 cm
Explanation:
For Reynold's number similarity between model and prototype we should have
[tex]R_{e} _{model} =R_{_{e prototype}} \\\\\frac{V_{model} L_{model} }{kinematic viscosity in model} =\frac{V_{proto}L_{proto} }{kinematic viscosity in prototype}[/tex]
Given L(prototype)= 2cm
V(prototype) = 100m/s
V(model) = 10m/s
Thus applying values in the above equation we get
[tex]\frac{100m/s^{} X2cm^{} }{1X10^{-5}m^{2}/s } =\frac{L_{M}X10m/s }{1.5X10^{-5}m^{2}/s }[/tex]
Solving for Lmodel we get Lm = 30cm
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop per meter of pipe (in Pa/m)
Answer:0.1898 Pa/m
Explanation:
Given data
Diameter of Pipe[tex]\left ( D\right )=0.15m[/tex]
Velocity of water in pipe[tex]\left ( V\right )=15cm/s[/tex]
We know viscosity of water is[tex]\left (\mu\right )=8.90\times10^{-4}pa-s[/tex]
Pressure drop is given by hagen poiseuille equation
[tex]\Delta P=\frac{128\mu \L Q}{\pi D^4}[/tex]
We have asked pressure Drop per unit length i.e.
[tex]\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}[/tex]
Substituting Values
[tex]\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}[/tex]
[tex]\frac{\Delta P}{L}[/tex]=0.1898 Pa/m
Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide particles in a metal matrix. Assume thermal conductivities of 20 and 66 W/m-K for the carbide and metal, respectively.
Answer:
Explanation:
k_max = 26.9 w/mk
k_min = 22.33 w/mk
Explanation:
a) the maximum thermal conductivity is given as
K_MAX = k_m v_m + k_p v_p
where k_m is thermal conductvitiy of metal
k_p is thermal conductvitiy of carbide
v_m = proportion of metal in the cement = 0.15
v_p = proportion of carbide in the cement = 0.85
[tex]K_MAX = k_m v_m + k_p v_p[/tex]
= 66*0.15 + 20*0.85
k_max = 26.9 w/mk
b) the minimum thermal conductivity is given as
[tex]k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}[/tex]
= \frac{20*66}{20*0.15 +66*0.85}
k_min = 22.33 w/mk
A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False )
The answer is : True
A cylindrical specimen of some metal alloy having an elastic modulus of 117 GPa and an original cross-sectional diameter of 3.6 mm will experience only elastic deformation when a tensile load of 2460 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.47 mm.
Answer:227.56 mm
Explanation:
Given data
Elastic modulus[tex]\left ( E\right )[/tex]= 117 GPa
Diameter[tex]\left ( d\right )[/tex]=3.6mm
force applied[tex]\left ( F\right )[/tex]=2460N
Area of cross-section[tex]\left ( A\right )[/tex]=[tex]\frac{\pi}{4}\times d^{2}[/tex]=10.18[tex]mm^{2}[/tex]
and change in length is given by
[tex]\Delta L[/tex]=[tex]\frac{FL}{AE}[/tex]
[tex]\Delta {0.47\times 10^{-3}}[/tex]=[tex]\frac{2460\times L}{10.180\times 117\times 10^{3}}[/tex]
L=[tex]0.47\times 10^{-3}\times 10.18\times 117\times 10^{3][/tex]
L=227.56 mm
The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 10^4 (Ω-m)1. The electron concentration is known to be 2.9x 10^22 m^-3. Given that the electron and hole mobilities are 0.14 and 0.023 m^2/N-s, respectively, calculate the hole concentration (in m^-3)
Given:
[tex]\sigma _{s} = 2.8\times 10^{4} \Omega-m[/tex]
electron concentration, n = [tex]2.9\times 10^{22} m^{-3}[/tex]
[tex]\mu _{h} = 0.14[/tex]
[tex]\mu _{e} = 0.023[/tex]
Solution:
Let holes concentration be 'p'
[tex]\sigma _{s}[/tex] = ne[tex]\mu _{e}[/tex] +pe[tex]\mu _{h}[/tex] (1)
substituting all given values in eqn (1):
[tex]2.8\times 10^{4} = 2.9\times 10^{22}\times 1.6 \times 10^{-19}\times 0.14 + p\times1.6 \times 10^{-19}\times 0.023[/tex]
The cocentration of holes is:
[tex]p = 7.432\times 10^{24} m^{-3}[/tex]
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical example of the appropriate area to use?
Answer:
Explanation:
We know that Drag force[tex]F_D[/tex]
[tex]F_D=\dfrac{1}{2}C_D\rho AV^2[/tex]
Where
[tex]C_D[/tex] is the drag force constant.
A is the projected area.
V is the velocity.
ρ is the density of fluid.
Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.
Example:
When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.
Name four rules that will control or eliminate cavitation of a pump
Answer:
Four rules that will control or eliminate pump cavitation are:
By reducing the speed of the motor(as in vaporization cavitation of pump) Ensure pump performance within safe limits(as in turbulence cavitation)Free space between impeller and its housing must be 4% or more of the impeller's diameter.(as in vane syndrome cabvitation)Regularly checking the system valves, joints, etc or the cavitation prone regions to ensure its prevention.(as in internal re-circulation cavitation and air aspiration cavitation).Explanation:
With reference to the above respective rules:
By reducing the motor speed the pump will slow down this will further slow down the flow rate and reduce head pressure.Do not exceed manufacturer performance guidellines as it will lead to pump system failure.Any less than 4% of diameter will lead to the initiation of cavitation Viscous and abrasive materials can lead to erosion and may pave way to air suction and proper check on discharge wall for head pressure should be checked as well to eliminate cavitationA fluid whose viscosity is unknown flow past a flare plate at a free stream velocity such that the the boundary layer thickness is 10mm at a distance of 2.2m from the leading edge. Assuming laminar flow throughout. Determine the boundary layer thickness at a distance 5m from the leading edge.
Answer:
δ₂ = 15.07 mm
Explanation:
Given :
When the leading edge, [tex]x_{1}[/tex] is 2.2 m, then boundary layer thickness,δ₁ = 10 mm = 0.01 m
[tex]x_{2}[/tex] = 5 m
Now we know that for a laminar flow, the boundary layer thickness is
δ = [tex]\frac{5.x}{\sqrt{Re_{x}}}[/tex] -------(1)
and Reyonlds number, Re is
[tex]Re = \frac{\rho .v.x}{\mu }[/tex]------(2)
where ρ is density
v is velocity
x is distance from the leading edge
μ is dynamic viscosity
from (1) and (2), we get
δ∝[tex]x^{1/2}[/tex]
Therefore,
[tex]\frac{\delta _{1}}{x_{1}^{1/2}}= \frac{\delta _{2}}{x_{2}^{1/2}}[/tex]
[tex]\frac{10}{2.2^{1/2}}= \frac{\delta _{2}}{5^{1/2}}[/tex]
δ₂ = 15.07 mm
Therefore, boundary layer thickness is 15.07 mm when the leading edge is 5 m.
What is a rotary actuator and give an example of how it is used?
Answer Explanation:
ROTARY ACTUATOR: A rotary actuator is an actuator that produces a rotary motion. An actuator requires a control signal and a source of energy.the linear motion in one direction gives rise to rotation.
EXAMPLE OF ROTARY ACTUATOR: the most used rotary actuators are rack and pinion, vane and helical
HOW IT IS USED: an actuator requires a control signal and its energy sources are current, fluid pressure when it receives a control signal it responds by converting signal energy into mechanical motion
A structural steel shaft with an outer diameter of 1.9 inches and an applied torque of 82.7 ft*lbs. Find: The maximum torsional shear stress in the shaft. Select one: a)- 736.88 ksi b)- 61.41 psi c)- 1473.76 ksi d)- 736.88 psi e)- 368.44 psi
Answer:
Answer is part d -736.88 psi
Explanation:
We know that for a bar subjected to pure torsion the shear stresses that are generated can be calculated using the following equation
[tex]\frac{T}{I_{P} } =\frac{t}{r}[/tex]....................(i)
Where
T is applied Torque
[tex]I_{P}[/tex] is the polar moment of inertia of the shaft
t is the shear stress at a distance 'r' from the center
r is the radial distance
Now in our case it is given in the question T =82.7 ft*lbs
converting T into inch*lbs we have T = 82.7 x 12 inch*lbs =992.4 inch*lbs
We also know that for a circular shaft polar moment of inertia is given by
[tex]I_{P}=\frac{\pi D^{4} }{32}[/tex]
[tex]I_{P}= \frac{\pi\ 1.9^{4} }{32} =1.2794 inch^{4}[/tex]
Since we are asked the maximum value of shearing stresses they occur at the surface thus r = D/2
Applying all these values in equation i we get
[tex]\frac{992.4 inch*lbs}{1.2794 inch^{4} } \frac{1.9 inches}{2}[/tex] = t
Thus t = 736.88 psi
What are the three basic types of positive displacement pumps.
Answer:
a) reciprocating pump
b) rotary pump
c) linear pump
Explanation:
Positive displacement pump has enlarged cavity on the suction side and decreasing cavity on the discharge side and in positive displacement pump
the amount of fluid captured and inside and then discharged is same.
There are three type of positive displacement pump
a) reciprocating pump is the pump which two valve during suction outlet valve is closed and for delivery the inlet valve is closed and pump is used for discharge
b) rotary pump fluid move with the help of rotary the rotation of rotary displaces water.
c) linear pump is the pump in which displacement is linear rope and chain pump is example of this type of pump.
Provide main reasons for the short shot during the injection molding.
Answer:
some cause of short shot is
1) due to the restriction in the flow
2) air pockets
3) high viscosity.
Explanation:
short shot is a word defined for major defect, it is actually occur when molten material does not fully occupy the cavities in a mold. Due to which mold remained incomplete after cooling. short shot may be because of restriction in the flow of molten material through the cavities and other main cause is present of large percentage of entrapped air.
Both equilibrium equations and constitutive models are needed to solve statically indeterminate problems. a)- True b)-False
Answer:
The answer is a) True
Explanation:
"Statically Indeterminate" means the number of unknowns you have exceed the number of available equations you can get from the static analysis, then the Static (equilibrium) analysis isn't enough to solve the problem.
Tracing constitutive models help to know where a load or reaction is located, also what kind of support settlements are used. This way you will know how the deflection behaves and also the momentum and torques location depending on the method you decide to use to solve the Statically Indeterminate system.
The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 rad/sec, what is the magnitudes of the normal component of acceleration of a point on the outer surface of the disk (0.4 m from the center) at the instant t=1 sec. a. 32. 4 m/sec^2 b. 81.8m/sec^2 c. 60.33 m/sec2 d. 4.33 m/sec2
Answer:32.4m/[tex]s^2[/tex]
Explanation:
Given data
[tex]radius\left ( r\right )[/tex]=0.4m
Intial angular velocity[tex]\left ( \omega_0\right )[/tex]=4rad/s
angular acceleration[tex]\left ( \alpha\right )[/tex]=5rad/[tex]s^2[/tex]
angular velocity after 1 sec
[tex]\omega[/tex]=[tex]\omega_0 [/tex]+[tex]\alpha\times\t[/tex]
[tex]\omega[/tex]=4+5[tex]\left ( 1\right )[/tex]
[tex]\omega[/tex]=9rad/s
Velocity of point on the outer surface of disc[tex]\left ( v\right )[/tex]=[tex]\omega_0\timesr[/tex]
v=[tex]9\times0.4[/tex] m/s=3.6m/s
Normal component of acceleration[tex]\left ( a_c\right )[/tex]=[tex]\frac{v^2}{r}[/tex]
[tex]a_c[/tex]=[tex]\frac{3.6\times3.6}{0.4}[/tex]=32.4m/[tex]s^2[/tex]
The work done by the system decreases the energy of the system. a)- True b)- False
Answer:
a. true
Explanation:
Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5 m above B. The pressure in the pipe at A is 70 kPa above atmospheric. Assuming no loss of energy between A and B, determine, a)- the velocity of A and B b)- Find pressure at B
Answer:
The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.
Explanation:
Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:
[tex]Aa.Va=Ab.Vb=Q[/tex]
Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:
[tex]Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3[/tex]
[tex]Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3[/tex]
Using the volume rate:
[tex]Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s[/tex]
[tex]Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s[/tex]
Assuming no losses, the energy equation for fluids can be written as:
[tex]Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb[/tex]
Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:
[tex]Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za[/tex]
Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:
[tex]Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m[/tex]
[tex]Pb = 70000\ Pa-27303\ Pa - 14715\ Pa[/tex]
[tex]Pb = 27,996\ Pa = 28\ kPa[/tex]
The total energy of a system remains unchanged during a cycle. a)- True b)- False
Answer: True
Explanation: It is correct that during a cycle the energy of the system remains unchanged. Total energy consist of the initial energy and the final energy in a system and both remain equal in a system whether they are in cycle or not ans also according to the law of conservation of energy ,total energy cannot be destroyed or created. Thus the while a cycle goes on energy will remain the same.
Why do we need the metering devices? List the basic types.
Answer:para comprar terreno por ejemplo
to buy land for example
Explanation: meter, centimeter
The relatonship between Kalven and celsius______
Answer:
[tex]K=C+273.15[/tex]
Explanation:
Kelvin's climbing represents the absolute temperature. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.
The Celsius scale has an artificial zero, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.
1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature is 0°C. Determine the work done during this cooling.
Answer:
the work done during this cooling is −28.7 kJ
Explanation:
Given data
mass (m) = 1 kg
r = 287 J/kg-K
pressure ( p) = 50 kPa
temperature (T) = 100°C = ( 100 +273 ) = 373 K
to find out
the work done during this cooling
Solution
we know the first law of thermodynamics
pv = mRT ....................1
here put value of p, m R and T and get volume v(a) when it initial stage in equation 1
50 v(a) = 1 × 0.287 × 373
v(a) = 107.051 / 50
v(a) = 2.1410 m³ .......................2
now we find out volume when temperature is 0°C
so put put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1
50 v(b) = 1 × 0.287 × 273
v(a) = 78.351 / 50
v(a) = 1.5670 m³ .......................3
by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.
work done = [tex]\int\limits^a_b {p} \, dv[/tex]
integrate it and we get
work done = p ( v(b) - v(a) ) ................4
put the value p and v(a) and v(b) in equation 4 and we get
work done = p ( v(b) - v(a) )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 (−0.574)
work done = −28.7 kJ
here we can see work done is negative so its mean work done opposite in direction of inside air
'
What are the general principles of DFA? What are the steps to minimize the number of parts for an assembly?
Answer Explanation : The general principles for design for assembly (DFA) are,
MINIMIZE NUMBER OF COMPONENTUSE STANDARD COMMERCIALLY AVAILABLE COMPONENTS USE COMMON PARTS ACROSS PRODUCT LINESDESIGN FOR EASE OF PART FABRICATION DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITYMINIMIZE USE OF FLEXIBLE COMPONENTDESIGN FOR EASE OF ASSEMBLYUSE MODULAR DESIGNREDUCE ADJUSTMENT REQUIREDSTEPS TO MINIMIZE THE NUMBER OF PARTS
USE OF INCORPORATE HINGSUSE OF INTEGRAL SPRINGSUSE OF SNAP FITSUSE OF GUIDES BEARINGS USE OF COVERSDescribe harmful effect associated with extraction of Aluminum, Gold and Copper. Discuss each individually.
Answer:
Harmful effect associated with extraction of Aluminum, Gold and Copper are:
During the melting of aluminium there is a released of per fluorocarbon are more harmful than carbon dioxide in the environment as they increased the level of green house gases and cause global warming. The process of transforming raw material into the aluminium are much energy intensive.
Gold mining industries destroyed land scopes and increased the amount of toxic level in the environment and they also dump there toxic waste in the natural water bodies, which increased the level of water pollution in the environment.
Copper mining causes the health problems like asthma and problem in respiratory system because of the inhalation of silica dust. It also increased the level of sulfur diode in the environment which cause acid rain and destroyed various trees and buildings in the nature.
What is entropy? how is it related to the environment? Also, what is the increase of entropy principle? Brief answer please (Not too lengthy and not too short)
Answer:
1). Entropy can be defined as a measure of sytem's thermal energy per unit temperature unavailable for ding useful work.
In other words, we can say that its a measure of the degree of randomness of a defined system.
Entropy-Environment relation:
Entropy is the fundamental concept that applies to the environment, humans or the entire universe.
The Second law of thermodynamics explains the environmental impact of entropy. Reversing a process requires work, so reversing environmental process takes energy. Environmental impacts are higher at higher entropies and harder to reverse. Entropic flows and entropy measures can be used to prioritize the impacts which need action.
Increase of Entropy Principle:
This principle states that the total change in entropy of a system with its enclosed adiabatic surrounding is always positive i.e., greater than or equal to zero.
Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high density b. good lubricity, high viscosity, low density c. chemically stable, compatible with system materials, good heat insulative capability d, readily available, high density, large bulk modulus e. fire resistant, inexpensive, non-toxic.
Answer:
e.Fire resistance,Inexpensive,Non-toxic.
Explanation:
Desirable hydraulic property of fluid as follows
1. Good chemical and environment stability
2. Low density
3. Ideal viscosity
4. Fire resistance
5. Better heat dissipation
6. Low flammability
7. Good lubrication capability
8. Low volatility
9. Foam resistance
10. Non-toxic
11. Inexpensive
12. Demulsibility
13. Incompressibility
So our option e is right.