Answer:
Air preheater:
Air preaheater is a heat exchanger which take heat from flue gases and increase the temperature of the air before entering into the combustion process.
As we know that ,flue gases contains high temperature and that temperature can be used by air preheater and it will reduce the amount of heat addition in the combustion process and that leads to increase the efficiency of plant.When heat addition reduces for the same out put of power it means that less amount of fuel is required for the same out put of power as without using air preheater. The main purpose of air preheater is to increase the thermal efficiency.
By increasing the temperature of air it will reduce the amount of fuel reduce for the combustion.
For a rod of annealed AISI 1018 steel with a cross sectional area of 0.65 in^2?; what is the maximum tensile load Pmax that should be placed on it given a design factor of 3 to avoid yielding?
Answer:
maximum tensile load Pmax is 11.91 ksi
Explanation:
given data
area = 0.65 in²
design factor of safety = 3
to find out
what is the maximum tensile load Pmax
solution
we know here area is 0.65 in² and FOS = 3
so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi
so
we use here design factor formula that is
[tex]\frac{ \sigma y}{FOS} = \frac{Pmax}{area}[/tex] .............1
put here all these value we get Pmax in equation 1
[tex]\frac{55}{3} = \frac{Pmax}{0.65}[/tex]
Pmax = 11.91 ksi
so maximum tensile load Pmax is 11.91 ksi
What is the ideal cooling system for low horsepower motor? For example1hp motor
Answer:
Air cooling.
Explanation:
Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.
Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.
The primary heat transfer mechanism that warms me while I stand next to a campfire is: a)- Conduction b)- Impeadance c)- Convection d)- Radiation
Answer:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Explanation:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.
Is refrigerator with an ice-maker an open or a closed system? Explain your answer
Answer:
Open system
Explanation:
In refrigerator there is a interaction between refrigerator and environment.
In refrigerator heat moves from the system to the outer environment means there is transfer of heat from one system to environment.
We know that whenever there is transfer of energy or mass from system then it is known as pen system
So refrigerator is a open system
A cylindrical tank with a radius of 2-m is filled with oil and water. The water has a density of rho = 1000 kg/m3 while the oil has a density of rho = 800 kg/m^3 . If the depth of the water is 2.5 m, and the pressure difference between the top of the oil and the bottom of the water is 80 kPa, determine the depth of the oil, in m. Assume that gravity is 9.81 m/s^2 .
Answer:
Height of oil is 7.06 meters.
Explanation:
The situation is shown in the attached figure
The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by
[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}[/tex]
Applying the given values we get
[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m[/tex]
A disk with radius of 0.4 m is rotating about a centrally located axis with an angular acceleration of 0.3 times the angular position theta. The disk starts with an angular velocity of 1 rad/s when theta = 0. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf=[tex]\sqrt{24.687}[/tex] =4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²
A(n)_____is an interconnected collection of computers.
Answer:
Network is a collection of computers
Explanation:
A network is a gathering of at least two gadgets that can convey. Practically speaking, a system is included various distinctive PC frameworks associated by physical or potentially remote associations.
The scale can run from a solitary PC sharing out fundamental peripherals to huge server farms situated far and wide, to the Internet itself. Despite extension, all systems permit PCs or potentially people to share data and assets.
Draw and label a typical true stress-strain curve for a ductile material.
Answer:
this is a typical stress-strain curve for a ductile material
Explanation:
A: proportional limit
B:Elastic limit
C:upper yield point
D:lower yield point
E:ultimate strength
F:rupture strength
How does fouling affects the performance of a heat exchanger?
Answer:
Fouling :
When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.
Effect of heat exchanger are as follows"
1.It decreases the heat transfer.
2.It increases the thermal resistance.
3.It decreases the overall heat transfer coefficient.
4.It leads to increase in pressure drop.
5.It increases the possibility of corrosion.
Is CO, an air pollutant? How does it differ from other emissions resulting from the combustion of fossil fuels?
Answer:
Explanation:
CO, carbon monoxide is a toxic gas. It casues asphixiation on people and animals by interfering with hemoglobin, not allowing blood to transport oxygen to the cells in the body.
The normal emissions resulting from the combustion of fussil fuels are CO2 (carbon dioxide) and H2O (water). Carbon monoxide is formed by an incomplete combustion of fossil fuels or carbon containing fuels in general, this not only produces toxic gas, but also is an inefficient combustion that wastes energy.
An air-conditioned room at sea level has an indoor design temperature of 80°F and a relative humidity of 60%. Determine a) Humidity ratio, b) Enthalpy, c) Density, d) Dew point, and e) Thermodynamic wet bulb temperature (Use Psychrometric charts)
Answer:
a)Humidity ratio =0.013 kg/kg
b) Enthalpy=60.34 KJ/kg
c) Density = 1.16 [tex]Kg/m^3[/tex]
d) Dew point temperature = 64.9°F
e) Wet bulb temperature = 69.53°F
Explanation:
Given that
Dry bulb temperature = 80°F
relative humidity = 60%
Given air is at sea level it means that total pressure will 1 atm.
So P= 1 atm.
As we know that psychrometric charts are always drawn at constant pressure.
Now from charts
We know that dry bulb temperature line is vertical and relative humidity line is curve and at that point these two line will meet ,will us property all property.
a)Humidity ratio =0.013 kg/kg
b) Enthalpy=60.34 KJ/kg
c) Density = 1.16 [tex]Kg/m^3[/tex]
d) Dew point temperature = 64.9°F
e) Wet bulb temperature = 69.53°F
There is 200 kg. of saturated liquid water in a steel tank at 97 C. What is the pressure and volume of the tank?
Answer:
[tex]V = 0.208 m^3[/tex]
saturated pressure = 0.91 bar 0r 91 kPa
Explanation:
Given data:
Mass of liquid water 200 kg
Steel tank temperature is 97 degree celcius = 97°C
At T = 97 degree celcius, saturated pressure = 0.91 bar 0r 91 kPa
And also for saturated liquid
Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg
Volume of tank is given as
[tex]V = \nu * m[/tex]
[tex]V = 0.00104 * 200[/tex]
[tex]V = 0.208 m^3[/tex]
Determine the factor of safety for a 9 foot long hollow steel
column 3.5 inches on a side that has a wall thickness of 0.225
inches and is loaded with a 22 kip load. Use the steel E of 29 *10
^6 psi and assume the column is pin connected at each end.
Answer:
factor of safety for A36 structural steel is 0.82
Explanation:
given data:
side of column = 3.5 inches
wall thickness = 0.225 inches
load P = 22 kip
Length od column = 9 ft
we know that critical stress is given as
[tex]\sigma_{cr} = \frac{\pi^2 E}{(l/r)^2}[/tex]
where
r is radius of gyration[tex] = \sqrt{\fra{I}{A}}[/tex]
Here I is moment od inertia [tex]= \frac{b_1^2}{12} - \frac{b_2^2}{12}[/tex]
[tex] I == \frac{3.5^2}{12} - \frac{3.05^2}{12} = 5.294 in^4[/tex]
For hollow steel area is given as [tex]A = b_1^2 -b_2^2[/tex]
[tex]A = 3.5^2 -3.05^2 = 2.948 in^2[/tex]
critical stress [tex]\sigma_{cr} = = \frac{\pi^2\times 29\times 10^6}{((9\times12)/(1.34))^2}[/tex]
[tex]\sigma_{cr} = 44061.56 lbs/inc^2[/tex]
considering Structural steel A36
so A36[tex] \sigma_y = 36ksi[/tex]
factor of safety [tex]= \frac{yield\ stress}{critical\ stress}[/tex]
factor of safety =[tex]\frac{36\times10^3}{44061.56} = 0.82[/tex]
factor of safety for A36 structural steel is 0.82
The student's query involves calculating the factor of safety for a hollow steel column, but without complete material strength data, the calculation cannot be precisely done.
Explanation:The student's question pertains to determining the factor of safety for a hollow steel column with specified dimensions, material properties (steel E), and loading conditions. To calculate the factor of safety for the column, we would need to compare the column's actual stress under load to its maximum allowable stress. However, due to insufficient data about material yield strength or ultimate strength and considering the provided information does not match the examples given, this calculation cannot be accurately completed without further specifics on the steel's properties.
An element has two naturally occurring isotopes, isotope 1 with an atomic weight of 78.918 amu and isotope 2 with an atomic weight of 80.916 amu. If the average atomic weight for the element is 79.903 amu, calculate the fraction of occurrences of these two isotopes.
Answer:
The fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.
Explanation:
Given that
Weight of isotope 1 = 78.918 amu
Weight of isotope 2 = 80.916 amu
Average atomic weight= 79.903 amu
Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.
The total weight will be summation of these two isotopes
79.903 = 78.918 x + 80.916(1-x)
By solving above equation
80.916 - 79.903 = (80.916-78.918) x
x=0.507
So the fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.
Describe the physics associated with the concept of thermal resistance.
Answer:
Thermal resistance:
Thermal resistance is the property which oppose the the heat transfer.It is the property for measurement of heat flow.
As we know that heat transfer take place from the high temperature to low temperature.Heat transfer Q given as
[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]
Where Δt is the temperature difference
[tex]{R_{th}}[/tex] is the thermal resistance.
Connection:
1. Series connection
[tex]R_{total}={R_1}+{R_2}+{R_1}----{R_n}[/tex]
2. Parallel connection
[tex]\dfrac{1}{R_{total}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}-----\dfrac{1}{R_n}[/tex]
Discuss the difference between the observed and calculated values. Is this error? If yes, what is the source?
Measurement error is the difference between observed and calculated values, including both random and systematic errors. Adjusting parameters such as K can help fit experimental data to models, with potential errors addressed through improved methods and investigation of discrepancies.
Explanation:The difference between the observed and calculated values is known as measurement error or observational error. This error can be classified into two types: random error, which occurs naturally and varies in an unpredictable manner, and systematic error, which is consistent and typically results from a flaw or limitation in the equipment or the experimental design. To identify the sources of these errors, one can compare the experimental data with calculated models and adjust parameters, such as the constant K, to find the best fit.
If the experimental values do not match the accepted values, sources of error should be investigated and procedures modified to improve accuracy. For example, in a physics experiment, one could compare the experimental acceleration to the standard acceleration due to gravity (9.8 m/s²) and identify factors contributing to any discrepancy. Common sources of error might include environmental factors, instrument calibration issues, or procedural mistakes.
Water at room temperature of 20.0°C is poured into an aluminum cylinder which has graduation markings etched on the inside. The reading in the graduations is 300.0 cc. The cylinder with the water in it is then immersed in a constant temperature bath at a temperature of 100°C. What is the reading for the level of water on the graduations of the cylinder after the water and the cylinder reach thermal equilibrium with the bath? The volume coefficient of expansion of water is 2.07 × 10 -4 K -1, and the linear coefficient of expansion of aluminum is 23.0 × 10 -6 K -1. 305.0 cc 303.5 cc 304.5 cc 304.0 cc 303.3 cc
Answer:
304.42 cc
Explanation:
When the aluminum expands the markings will be further apart. If the 300 cc mark was at a distance l0 of the origin at 20 C, at 100 C it will be
l = l0 * (1 + a * (t - t0))
l = l0 * (1 + 23*10^-6 * (100 - 20))
l = l0 * 1.0018
The volume of water would have expanded by
V = V0 (1 + a * (t - t0))
V = 300 (1 + 2.07*10^-4 * (100 - 20))
V = 304.968 cc
Since the markings expanded they would measure
304.968/1.0018 = 304.42 cc
The velocity of a particle along the s-axis is given by v = 14s^7/6 where s is in millimeters and v is in millimeters per second. Determine the acceleration when s is 5.5 millimeters.
Answer:
The acceleration is [tex]2220.00m/s^{2}[/tex]
Explanation:
We know that the acceleration is given by
[tex]a=v\frac{dv}{ds}........................(i)[/tex]
The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]
Thus the acceleration as obtained from equation 'i' becomes
[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]
Hence acceleration at s = 5.5 equals
[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]
How much power would you need to cool down a closed, 1 Liter container of water from 100°C to 20°C in 5 minutes? (a) 1.1W (b)1.1kW (c)67kW (d)334 kJ
Answer:
The power required to cool the water is 1.11Kw.
Hence the correct option is (b).
Explanation:
Power needed to cool down is equal to heat extract from the water.
Given:
Volume of water is 1 liter.
Initial temperature is 100C.
Final temperature is 20C.
Time is 5 minutes.
Take density of water as 100 kg/m3.
Specific heat of water is 4.186 kj/kgK.
Calculation:
Step1
Mass of the water is calculated as follows:
[tex]\rho=\frac{m}{V}[/tex]
[tex]1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}[/tex]
m=1kg
Step2
Amount of heat extraction is calculated as follows:
[tex]Q=mc\bigtriangleup T[/tex]
[tex]Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)[/tex]
Q=334880 j.
Step3
Power to cool the water is calculated as follows:
[tex]P=\frac{Q}{t}[/tex]
[tex]P=\frac{334880}{(5min)(\frac{60s}{1min})}[/tex]
P=1116.26W
or
[tex]P=(1116.26W)(\frac{1Kw}{1000 W})[/tex]
P=1.11 Kw.
Thus, the power required to cool the water is 1.11Kw.
Hence the correct option is (b).
The boiler pressure is 38bar and the condenser pressure 0.032 bar.The saturated steam is superheated to 420 oC before entering the turbine. a) Calculate the cycle efficiency of the Rankine cycle. b) Calculate the work ratio of the Rankine cycle.
Answer:
a)38.65%
b)1221KJ/kg
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
first we calculate the enthalpies in all states
h1=3264kJ/Kg
h2=2043kJ/Kg
h2=h3=105.4kJ/Kg
a)we use the ecuation for efficiency
Efficiency = (h1-h2) / (h1-h4)
Efficiency = (3264-2043) / (3264-105.4)
=0.3865=38.65%
b)we use the ecuation for Wout
Wout = m (h1-h2)
for work ratio=
w = (h1-h2)
w=(3264-2043)=1221KJ/kg
Given two resistors R1=40 ohm and R2=30 ohm connected in series, what is the total resistance of this configuration? Enter the value in the box below without the unit. Round the result to two decimal places if necessary. For example if the answer is 10.333 ohm, put 10.33 in the box.
Answer:
Rt=70.00 ohm
Explanation:
ohm's theory tells us that the connected resistors in series add up directly
Rt=R1+R2+R3+R4.......
Therefore, for this case, the only thing we should do is add the resistance directly
Rt=R1+R2
Rt=40.00 ohm+30.00 Ohm.
Rt=70.00 ohm
the total resistance of this configuration is 70.00 ohm
The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2250 K. Assume constant specific heats evaluated at 300 K. Determine the cut-off ratio. There is a +/- 5% tolerance.
Answer:
Cut off ratio=2.38
Explanation:
Given that
[tex]T_1=300K[/tex]
[tex]P_1=100KPa[/tex]
[tex]P_2=P_3=7200KPa[/tex]
[tex]T_3=2250K[/tex]
Lets take [tex]T_1[/tex] is the temperature at the end of compression process
For air γ=1.4
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]
[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]
[tex]T_2=1070K[/tex]
At constant pressure
[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{V_3}{V_2}=2.83[/tex]
So cut off ratio
[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]
Cut off ratio=2.38
A water jet that leaves a nozzle at 60 m/s at a certain flow rate generating power of 250 kW by striking the buckets located on the perimeter of a wheel. Determine the mass flow rate of the jet.
Answer:
The mass flow rate of jet =69.44 kg/s
Explanation:
Given that
velocity of jet v= 60 m/s
Power P=250 KW
As we know that force offered by water F
[tex]F=\rho\ A \ v^2[/tex]
Power P= F.v
So now power given as
[tex]P=\rho\ A \ v^3[/tex]
We know that mass flow rate = ρAv
[tex]P=mass\ flow\ rate\ \times v^2[/tex]
250 x 1000 = mass flow rate x 3600
mass flow rate = 69.44 kg/s
So the mass flow rate of jet =69.44 kg/s
The absolute pressure of an automobile tire is measured to be 320 kPa before a trip and 349 kPa after the trip. Assuming the volume of the tire remains constant at 0.022 m^3, determine the percent increase in the absolute temperature of the air in the tire. The percent increase in the absolute temperature of the air in the tire is_____ %.
Answer:
9%
Explanation:
An ideal gas is one that has its molecules widely dispersed and does not interact with each other, studies have shown that air behaves like an ideal gas, so the state change equation for ideal gases can be applied.
P1V1T2 = P2V2T1
where 1 corresponds to state 1 = 320kPa
and 2 is state 2 = 349kPa.
Given that the volume remains constant the equation is:
P1T2=P2T1
SOLVING for T2/T1
[tex]\frac{T2}{T1} =\frac{P2}{P1} =\frac{349}{320} =1.09\\\\[/tex]
The equation to calculate the percentage increase is as follows
%ΔT=[tex]\frac{(T2-T1)100}{T1} =(\frac{T2}{T1} -1)100=(1.09-1)100=(0.09)100=9%[/tex]
In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain is 0.226. Determine strain hardening exponent, n, in the flow curve for the plastic region of this steel.
Answer:n=0.973
Explanation:
Given
When True strain[tex]\left ( \epsilon _T_1\right )=0.171[/tex]
at [tex]\sigma _1=263.8 MPa[/tex]
When True stress[tex]\left ( \sigma _2\right )[/tex]=346.2 MPa
true strain [tex]\left ( \epsilon _T_2\right )[/tex]=0.226
We know
[tex]\sigma =k\epsilon ^n [/tex]
where [tex]\sigma [/tex]=True stress
[tex]\epsilon [/tex]=true strain
n=strain hardening exponent
k=constant
Substituting value
[tex]263.8=k\left ( 0.171\right )^n------1[/tex]
[tex]346.2=k\left ( 0.226\right )^n-----2[/tex]
Divide 1 & 2 to get
[tex]\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n[/tex]
[tex]1.312=\left ( 1.3216\right )^n[/tex]
Taking Log both side
[tex]ln\left ( 1.312\right )=nln\left ( 1.3216\right )[/tex]
n=0.973
A car enters a circular off ramp (radius of 180 m) at 30 m/s. The car's on - bond accelerometers sense a total acceleration magnitude of 7.07 m/s2. What is the car's acceleration magnitude a2 (the component tangent to its path)?
Answer:
Horizontal acceleration will be [tex]2.07m/sec^2[/tex]
Explanation:
We have given radius of track = 180 m
velocity = 30 m/sec
Total acceleration [tex]a_t=7.07m/sec^2[/tex]
The centripetal;l acceleration which is always normal to the velocity also known as normal acceleration is given by [tex]a_n=\frac{v^2}{r}=\frac{30^2}{180}=5m/sec^2[/tex]
So the tangent acceleration will be [tex]a_t=7.07-5=2.07m/sec^2[/tex]
A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine the final pressure of the water.
Answer:
final pressure is 6847.41 kPa
Explanation:
given data:
[tex]P_1 = 10,000 kPa[/tex]
[tex]T_1 =520\ degree\ celcius = 793 K[/tex]
[tex]T_2 = 270 degree celcius = 543 K[/tex]
as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream
for ischoric process we know that
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
[tex]\frac{10*10^6}{793} = \frac{P_2}{543}[/tex]
[tex]P_2 = 6.84741*10^6 Pa[/tex]
final pressure is 6847.41 kPa
In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm. Roll diameter is 600mm and coefficient of friction between rolls and strip 0.15. The specification is that the draft is to be equal on each pass. Determine a) Minimum number of passes required? b) Draft for each pass?
Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip ([tex]\nu [/tex])=0.15
maximum draft[tex](d_{max})=\nu ^2R[/tex]
[tex]=0.15^2\times 300=6.75 mm[/tex]
Minimum no of passes[tex]=\frac{45-20}{6.75}=3.70\approx 4[/tex]
(b)draft per each pass
[tex]d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}[/tex]
[tex]d=\frac{45-20}{4}=6.25 mm[/tex]
Evaluate each of the following and express with an appropriate prefix: (a) (430 kg)^2 (b) (0.002 mg)^2, and (c) (230 m)^3
The expressions squared or cubed with appropriate metric prefixes are: (a) 184.9 Mg^2, (b) 4 x 10^-18 kg^2, and (c) 12.167 km^3. Metric prefixes are used for clarity in representing large quantities.
Explanation:The question deals with the operation of squaring and cubing given quantities and expressing them with the appropriate metric prefixes. Let's evaluate each expression step by step:
(a)It is important to use the proper metric prefixes when
expressing large numbers to ensure clarity and avoid confusion.
If the local atmospheric pressure is 14.6 psia, find the absolute pressure (in psia) in a column of glycerin (rho = 74.9 lbm/ft^3) at depth of 27in.
Answer:
52.2538 psia
Explanation:
The absolute pressure at depth of 27 inches can be calculated by:
Pressure = Local pressure + Gauge pressure
Also,
[tex]P_{gauge}=\rho\times g\times h[/tex]
Where,
[tex]\rho[/tex] is the density of glycerin ([tex]\rho=74.9\ lbm/ft^3[/tex])
g is the gravitational acceleration = 32.1741 ft/s²
h = 27 in
Also, 1 in = 1/12 ft
So,
h = 27 / 12 ft = 2.25 ft
So,
[tex]P_{gauge}=74.9\times 32.1741\times 2.25\ lbf/ft^2=5422.1402\ lbf/ft^2[/tex]
Also,
1 ft = 12 inch
1 ft² = 144 in²
So,
[tex]P_{gauge}=5422.1402\ lbf/ft^2=\frac {5422.1402\ lbf}{144\ in^2}=37.6538\ lbf/in^2=37.6538\ psia[/tex]
Local pressure = 14.6 psia
So,
Absolute pressure = 14.6 psia + 37.6538 psia=52.2538 psia