Answer:
Q = 0.118 [tex]m^{3}[/tex]/s
Explanation:
Given :
diameter of the pipe, d = 150 mm
= 0.15 m
Pitot tube co efficient, [tex]C_{v}[/tex] = 1.05
manometer reading is given, x = 167 mm
= 0.167 m
From manometer reading,we can find the difference between the manometer height, h
[tex]h =x\times\left [ \frac{S_{m}}{S_{w}}-1 \right ][/tex]
[tex]h =0.167\times\left [ \frac{13.6}{1}-1 \right ][/tex]
h = 2.1042 m
Now, average velocity is v = [tex]C_{v}[/tex][tex]\sqrt{2.g.h}[/tex]
= [tex]1.05\times \sqrt{2\times 9.81\times 2.1042}[/tex]
= 6.74 m/s
Area of the pipe, A = [tex]\frac{\pi }{4}\times d^{2}[/tex]
= [tex]\frac{\pi }{4}\times 0.15^{2}[/tex]
= 0.0176 [tex]m^{2}[/tex]
Therefore, flow rate is given by, Q = A.v
= 0.0176 X 6.74
= 0.118[tex]m^{3}[/tex]/s
An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the gravitational potential energy of the object, in kJ, relative to the surface of the earth.
Answer:
Gravitational Potential =58.914 KJ
Explanation:
We know that
[tex]Gravitational Potential Energy = mass\times g\times Height[/tex]
Given mass = 251 kg
Height= 24 m
g is acceleration due to gravity = [tex]9.78m/s^{2}[/tex]
Applying values in the equation we get
[tex]Gravitational Potential Energy=251X9.78X24 Joules[/tex]
[tex]Gravitational Potential Energy=58914.72 Joules[/tex]
[tex]Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ[/tex]
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
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.
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.
What are the three elementary parts of a vibrating system?
Answer:
the three part are mass, spring, damping
Explanation:
vibrating system consist of three elementary system namely
1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.
2) Spring - the part that has elasticity and help to hold mass
3) Damping - this part considered to have zero mass and zero elasticity.
Name three major heat transfer mechanisms giving one example of each from day-to-day life. Also explain the physical mechanism behind these modes of heat transfer.
Answer:
The major heat transfer mechanisms are:
Conduction: When a body at higher temperature comes in direct contact with a body at lower temperature flow of heat takes place from higher temperature to lower temperature due to the Kinetic energy of particles in motion and this motion continues till equilibrium is reached. Heat transfer by this method is called conduction. For example: When a hot metal comes in direct contact with a cold metal, heat is transferred by conduction.Convection: This method of heat transfer applies to fluid motion of particles. Here, the heat transfer is due to thermal energy of the fluid particles. Due to the differences in their density the liquid over hot surface expands and rises up and heat flows from high to low temperatures. For example: When earth's surface is heated by the sun, the warm air comes up and cool air comes in Radiation: Thermal radiations are generated by electromagnetic waves. These are the result of motion of random molecules in the matter which carry the energy from the emitting body and provides motion to the charge particles of EM waves. At high temperature, shorter wavelength is produced and vice-versa. For example: visible light, UV, IR, etcThe 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]
A 350 gal air storage tank is initially at 100 psig. For how long can the tank supply 30 cfm of air to a machine that requires at least 80 psig to operate?
Answer:
93.8 sec
Explanation:
it is given that tab has 350 gallon
we know that 1 gallon = 0.134 cubic foot
350 gallon = 350×0.134=46.9 cubic foot
the delivery pressure is 100 psi which is greater than 80 psi to operate machine
it is given that supply volume is 30 cubic foot per minute
= [tex]\frac{30}{60}=0.5[/tex] [tex]ft^{3}/sec[/tex]
[tex]time\ required\ =\frac{tab\ air }{supply\ volume}[/tex]
[tex]time\ required\= [tex]\frac{46.9}{0.5}[/tex]
=93.8 sec
A finished, tapered workpiece has an included angle of 70 degrees. If the taper is nade vith the compound, how many degrees should tha setting be on the swivel base? A. 70 B. 140 C. 20 D. 35
Answer:
Tapering is basically the process of thinning or reducing a work piece according to the set standards. and the final product after tapering is known as tapered workpiece.
Solution:
Included angle = 70 degrees
setting on the swivel base is given by:
2α = 70°
α = 35°
Therefore, the setting on the swivel base should be 35°
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 work done by the system decreases the energy of the system. a)- True b)- False
Answer:
a. true
Explanation:
Define 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.
Different types of steels contain different elements that alter the characteristics of the steel. For each of the following elements, explain what the element does when alloyed with steel.
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.
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
Convert the temperature of 451 degree Fahrenheit to the units requested: a. Rankine b. Kelvin c. Celsius
Answer:
(a) 910.67°R (b) 505.9277 (c) 232.777
Explanation:
FAHRENHIET TO RANKINE: T(R°)=T(F°)+459.67
we have to change 451°F
T(R°)=451+459.67
=910.67°R
FAHRENHET TO KELVIN: T(K)=(T(F°)+459.67) ×[tex]\frac{5}{9}[/tex]
we have to convert 451°F
T(K)=(451+459.67)×[tex]\frac{5}{9}[/tex]
=505.9277
FAHRENHET TO CELSIUS: T(C°)=[tex]\frac{F-32}{9}[/tex]×5
we have to convert 451°F
T(C°)=[tex]\frac{451-32}{9}[/tex]×5
=232.777
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.
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.
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow rate is 5.5 kg/s, and the power developed is 1120 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine
(a) The temperature of the air at the turbine exit, in K.
(b) The isentropic turbine efficiency.
Answer:
a) [tex]T_{2}=837.2K[/tex]
b) [tex]e=91.3[/tex] %
Explanation:
A) First, let's write the energy balance:
[tex]W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})[/tex] (The enthalpy of an ideal gas is just function of the temperature, not the pressure).
The Cp of air is: 1.004 [tex]\frac{kJ}{kgK}[/tex] And its specific R constant is 0.287 [tex]\frac{kJ}{kgK}[/tex].
The only unknown from the energy balance is [tex]T_{2}[/tex], so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.
[tex]T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K[/tex]
B) The isentropic efficiency (e) is defined as:
[tex]e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}[/tex]
Where [tex]{h_{2s}[/tex] is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because [tex]h_{2}-h_{1}[/tex] can be obtained from the energy balance [tex]\frac{W}{m}=h_{2}-h_{1}[/tex]
[tex]h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}[/tex]
An entropy change for an ideal gas with constant Cp is given by:
[tex]s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})[/tex]
You can review its deduction on van Wylen 6 Edition, section 8.10.
For the isentropic process the equation is:
[tex]0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})[/tex]
Applying logarithm properties:
[tex]ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}[/tex]
Then,
[tex]T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K[/tex]
So, now it is possible to calculate [tex]h_{2s}-h_{1}[/tex]:
[tex]h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}[/tex]
Finally, the efficiency can be calculated:
[tex]e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3[/tex] %
A centrifugal pump provides a flow rate of 0.03 m/s when operating at 1750 rpm against 60 m head. Determine the pump's flow rate and developed head if the pump speed is increased to 3500 rpm.
Answer:240m
[tex]Q=0.06m^3/s[/tex]
Explanation:
Given rpm increases from 1750 rpm to 3500 rpm
initial head 60 m and flow rate=[tex]0.03 m^{3}/s[/tex]
Since unit speed remains same
therefore
[tex]N_u=\frac{N}{\sqrt{H}}[/tex]
[tex]\frac{1750}{\sqrt{60}}[/tex]=[tex]\frac{3500}{\sqrt{H}}[/tex]
H=240m
Also unit Flow remains same
[tex]\frac{Q}{\sqrt{H}}[/tex]=[tex]\frac{Q}{\sqrt{H}}[/tex]
[tex]\frac{0.03}{\sqrt{60}}[/tex]=[tex]\frac{Q}{\sqrt{240}}[/tex]
[tex]Q=0.06m^3/s[/tex]
It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca criterion; b) Distortion energy or von Mises criterion; (c) Mohr-Coulomb criterion; (d) Any of the above
Answer:
(b)Distortion energy theory.
Explanation:
The best suitable theory for ductile material:
(1)Maximum shear stress theory (Guest and Tresca theory)
It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.
(2)Maximum distortion energy theory(Von Mises henkey's theory)
It states that maximum shear train energy per unit volume at any point is equal to strain energy per unit volume under the state of uni axial stress condition.
But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.
Name 3 types of hydraulic cylinder mountings.
Answer:
Flanges MountingTrunnions MountingClevises MountingAnswer:
1.Flange mounting:
2.Foot mounting:
3.Mounting on end joint:
extra.4. Trunnion mounting
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.
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
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
Describe the process that you would use to hot forge an automotive connecting rod, indicating why each of the steps is used.
Answer:
Hot forging is a process which is carried at a temperature that is higher than the recrystalization temperature.
Explanation:
A connecting rod is used in a reciprocating engine which links the piston to the crankshaft. Connecting rods are made of steel which are hot forged.
The various steps that are used to hot forged a connecting rod are :
1. Rods are made to cut in the required size from the billet by billet shearing machine or saw band.
2. Heating of the billets in the furnace upto its recrystalization temperature.
3. Placing the billets in both upper and lower dies and doing the forging operation.
4. Rolling forging : it is important for the quality of the forged component.
5. Finishing and trimming : finishing is done to improve the surface quality and provide a smooth finish.
6. Inspection : Visual inspection is done for any defects.
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
'
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 cavitationThe 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 systemTool life testing on a lathe under dry cutting conditions gauge 'n' and 'C' of Taylor tool life equation as 0.12 and 130 m/min. respectively. When a coolant was used, 'C' increased by 10%. The increased tool life with the use of coolant at a cutting speed of 90 m/min is
Answer:
So % increment in tool life is equal to 4640 %.
Explanation:
Initially n=0.12 ,V=130 m/min
Finally C increased by 10% , V=90 m/min
Let's take the tool life initial condition is [tex]T_1[/tex] and when C is increased it become [tex]T_2[/tex].
As we know that tool life equation for tool
[tex]VT^n=C[/tex]
At initial condition [tex]130\times (T_1)^{0.12}=C[/tex]------(1)
At final condition [tex]90\times (T_2)^{0.12}=1.1C[/tex]-----(2)
From above equation
[tex]\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}[/tex]
[tex]T_2=47.4T_1[/tex]
So increment in tool life =[tex]\dfrac{T_2-T_1}{T_1}[/tex]
=[tex]\dfrac{47.4T_1-T_1}{T_1}[/tex]
So % increment in tool life is equal to 4640 %.
What is the difference between pump and turbine? Write the first law of thermodynamics for both (pump & turbine)?
Answer:
Pumps converts mechanical energy into hydraulic energy while turbines convert hydraulic energy into mechanical energy.
Explanation:
The machines which converts and transfers mechanical energy in the form of torque on the shaft into hydraulic energy in the form of water under pressure are called pumps whereas those machines which converts water pressure or hydraulic energy into mechanical energy that is further converted into electrical energy are called turbines.
The pump impeller rotates in the opposite direction to the turbine runner.
A turbine delivers work as output whereas a pump consumes work.
First law of thermodynamics for a pump :
W = ( H₁-H₂) +Q , where H₁ > H₂
First law of thermodynamics for a turbines :
W = ( H₂-H₁) +Q , where H₁ < H₂