Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
[tex]\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }[/tex] ...........1
put here all these value and we get t2
[tex]\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }[/tex]
t2 = 424.8
final temperature is 424.8 K
so correct option is e
A material’s chemical property could be described as it’s reaction with other chemicals (gases, liquids and solid materials). a) True b) False
Answer:
The given statement is true.
Explanation:
A chemical property of any material be it solid liquid or gas is defined as how it interacts chemically with other substances after an interaction takes place between them. The interaction in language of chemistry is known as chemical reaction. Different materials in nature show different interactions with various other substances.The 2 substances can react with each other and form a different compound or may not react with each other and are termed as inert chemicals. Infact it is this interaction between the various chemicals that we can group different into classes based on their behavior with different chemicals. The interaction of different materials act as their signatures that help us in identifying them.
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is [tex]1.557\times10^{-7}[/tex]m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
Here, v is velocity, [tex]\nu[/tex] is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
[tex]Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}[/tex]
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
A heat pump with a 2 kW motor is used to heat a building at 30 deg C. The building loses heat at a rate of 0.5 kW per degree difference to the colder ambient at T amb. The heat pump has a coefficient of performance that is 50 % of a carnot heat pump. What is the maximum ambient temperature for which the heat pump is sufficient?
Answer:
T=5.3° C
Explanation:
Given that
Power input to the pump = 2 KW
Building loses heat rate = 0.5 KW/C
So rate of heat transfer = 0.5(273+30-T)
rate of heat transfer = 0.5(303-T)
T=Ambient temperature
Building temperature = 30° C
We know that ,heat pump is used to heat the building.
COP of pump = 0.5 COP of Carnot heat pump
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{273+30}{303-T}[/tex]
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{303}{303-T}[/tex]
[tex]COP\ of\ pump=\dfrac{303-T}{Power}[/tex]
[tex]COP\ of\ pump=0.5\times \dfrac{303-T}{2}[/tex] -----1
And also
[tex]COP\ of\ pump=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex] ----2
So from now equation 1 and 2
[tex]\dfrac{303-T}{4}=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]
So T= 278.38 K=5.3° C
T=5.3° C
Ambient temperature =5.3° C.
An inventor proposes an engine that operates between the 27 deg C warm surface layer of the ocean and a 10 deg C layer a few meters down. The claim is that this engine can produce 100 kW at a flow of 20 kg/s. Is this possible?
Answer:
Engine not possible
Explanation:
source temperature T1 = 300 K
sink temperature T2= 283 K
therefore, carnot efficiency of the heat engine
η= 1- T_2/T_1
[tex]\eta= 1-\frac{T_1}{T_2}[/tex]
[tex]\eta= 1-\frac{283}{300}[/tex]
= 0.0566
= 5.66%
claims of work produce W = 100 kW, mass flow rate = 20 kg/s
[tex]Q=mc_p(T_1-T_2)[/tex]
[tex]Q=20\times4.18(300-283)[/tex]
= 1421.2 kW
now [tex]\eta= \frac{W}Q}[/tex]
now [tex]\eta= \frac{100}{1421.2}[/tex]
=7%
clearly, efficiency is greater than carnot efficiency hence the engine is not possible.
a valueable preserved biological specimen is weighed by suspeding it from a spring scale. it weighs 0.45 N when it is suspendedin air and 0.081 N when it is suspended in a bottle of alchol what is its dencity?
Answer:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. Determine the minimum required wire diameter assuming a factor of safety 5 and that the yield strength of the steel is 900MPa.
Answer:
11.6 mm
Explanation:
With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:
σadm = strength / fos
σadm = 900 / 5 = 180 MPa
The stress is the load divided by the section:
σ = P / A
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
[tex]d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm[/tex]
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Answer:
1) g=31.87ft/s^2
2)m=120
W=119.64lbf
Explanation:
first part
the weight of a body with mass is calculated by the following equation
W=mg
we convert 120lb to slug
m=120lbx1slug/32.147lb=3.733slug
solving for g
g=W/m
g=119/3.733=31.87ft/s^2
second part
the mass is the same
m=120lb=3.733slug
Weight
W=3.733slug*32.05ft/s^2=119.64lbf
A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m^3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.
Answer:
The mass of the air is 4.645 kg.
The work done or heat transfer is 277.25 kj.
Explanation:
In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.
Given:
Initial pressure of the gas is 2 bar.
Initial volume of the gas is 2 m³.
Initial temperature is 300 K.
Final pressure is 1 bar.
Calculation:
Step1
Apply ideal gas equation for air as follows:
PV=mRT
[tex]2\times10^{5}\times2=m\times 287\times300[/tex]
m = 4.645 kg.
Thus, the mass of the air is 4.645 kg.
Step2
For isothermal process work done is same as heat transfer.
Work done or the heat transfer is calculated as follows:
[tex]W=P_{i}V_{i}ln\frac{P_{i}}{P_{f}}[/tex]
[tex]W=2\times10^{5}\times2\times ln\frac{2}{1}[/tex]
[tex]W=2.7725\times10^{5}[/tex] j
Or,
W=277.25 kj.
Thus, the work done or heat transfer is 277.25 kj.
What is 220 C in degrees Fahrenheit (F)?
Answer:
428°F
Explanation:
The equation to convert degrees Celsius to degrees fahrenheit is
°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32
°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)
Derive the following conversion factors: (a) Convert a viscosity of 1 m^2/s to ft^2/s. (b) Convert a power of 100 W to horsepower. (c) Convert a specific energy of 1 kJ/kg to Btu/lbm.
Answer:
(a) 10.76 [tex]ft^2/sec[/tex]
(b) 0.134 lbm
(c) 0.4308 Btu/lbm
Explanation:
We have to do conversion
(a) conversion of viscosity [tex]1m^2/sec\ to\ ft^2/sec[/tex]
We know that [tex]1m^2=10.76feet^2[/tex]
So [tex]1m^2/sec=10.76ft^2/sec[/tex]
(b) We have to convert power 100 W in hp
We know that 1 W = 0.00134 hp
So 100 W = 100×0.00134=0.134 hp
(c) We have to convert 1 KJ/kg to Btu/lbm
We know that 1 KJ = 0.94780 Btu
And 1 kg = 2.20 lbm
So [tex]1KJ/kg=\frac{0.9478Btu}{2.20lbm}=0.4308Btu/lbm[/tex]
The head difference between the inlet and outlet of a 1km long pipe discharging 0.1 m^3/s of water is 0.53 m. If the diameter is 0.6m, what is the friction factor? Is a) 0.01 b) 0.02 c) 0.03 d) 0.04 e) 0.05
Answer:
The correct option is 'e': f = 0.05.
Explanation:
The head loss as given by Darcy Weisbach Equation is as
[tex]h_{l}=\frac{flv^{2}}{2gD}[/tex]
where
[tex]h_{l}[/tex] is head loss in the pipe
'f' is the friction factor
'l' is the length of pile
'v' is the velocity of flow in pipe
'D' is diameter of pipe
From equation of contuinity we have [tex]v=\frac{Q}{A}[/tex]
Thus using this in darcy's equation we get
[tex]h_{l}=\frac{flQ^{2}}{2gDA}[/tex]
where
'Q' is discharge in the pipe
'A' is area of the pipe [tex]A=\frac{\piD^2}{4}[/tex]
Applying the given values we get
[tex]h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}[/tex]
Solving for 'f' we get
[tex]f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05[/tex]
Define the hydraulic diameter for a rectangular duct
Answer with Explanation:
Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.
It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.
The hydraulic diameter is 4 times the hydraulic radius of any section.
For a rectangular duct as shown in the attached figure
[tex]R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)} [/tex]
Where
[tex]D_{h}[/tex] is the hydraulic diameter of the duct with depth 'd' and width 'b'
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A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The system is heated until the volume is 0.2 m3 and the pressure is 180 kPa. What is the work done by the air? External pressure is 100 kPa.
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = [tex]\frac{60+180}{2}[/tex] = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
How much heat (Btu) is prod uced by a 150-W light bulb that is on for 20-hours?
Answer:
heat produced by 150 watt bulb is 510 Btu
Explanation:
Given data:
Power of bulb is 150 W
Duration for light is 20 hr
We know that 1 Watt = 3.4 Btu
hence using above conversion value we can calculate the power in Btu unit
so for [tex]150 W\ bulb = 150\times 3.4 = 510 Btu[/tex]
therefore, 150 W bulb consist of 510 Btu power for 20 hr
output heat by 150 watt bulb is 510 Btu
calculate the density of air(as a perfect gas) when the pressure
is1.01325*105 N/m2 and the temperature
is288.15 K.
Answer:
1.2253 kg/m3
Explanation:
from ideal gas equation we know that
[tex]PV = n\timesRT[/tex]
[tex]\frac{n}{V} = \frac{P}{(RT)}[/tex]
where,
pressure P = 1.01325 x 10^5 N/m2 = 101325 Pascals
gas constant R = 8.314 J/mol/K
T = 288.15 K
[tex]\frac{n}{V} = \frac{101325}{(8.314*288.15K)}[/tex]
[tex]\frac{n}{V} = 42.2948967 Pa*mol/J[/tex]
we know that 1 Pa = 1 J/m3
so[tex]\frac{n}{V} = 42.2948967\ mol/m3[/tex]
as we know, Air consist of 21% oxygen and 79% nitrogen
therefore
Molar mass of air:
[tex]0.21 *(32 g/mol) + 0.79 *(28 g/mol) = 28.97 g/mol[/tex]
So [tex](42.2948967\ mol/m3) \times 28.97\ g/mol[/tex]
= 1 225.27 g/m^3
= 1.22528316 kg/m^3 ~ 1.2253 kg/m3
Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance +/- 0.000005 m^3/kg)?
Answer:
[tex]specific\ volume=0.00097\ m^3/kg[/tex]
Explanation:
Given that
Specific gravity of sea water = 1.025
So density of sea water = 1.025 x 1000 [tex]kg/m^3[/tex]
Density of sea water = 1025 [tex]kg/m^3[/tex]
We know that
[tex]Density=\dfrac{mass}{Volume}[/tex] ---1
Specific volume
[tex]specific\ volume=\dfrac{Volume}{mass}[/tex] ---2
From equation 1 and 2
We can say that
[tex]specific\ volume=\dfrac{1}{density}\ m^3/kg[/tex]
[tex]specific\ volume=\dfrac{1}{1025}\ m^3/kg[/tex]
[tex]specific\ volume=0.00097\ m^3/kg[/tex]
What is the definition of diameter pitch?
Answer:
Diameter pitch is the parallel thread of the imaginary cylinder of the diameter that basically intersect in the surface of thread. It is also known as effective diameter. The diameter pitch is basically used to determine the threated parts.
The pitch diameter is the essential component for determining the basic compatibility in the externally and internally thread parts. The diameter pitch is the sensitive measuring tool for determine the specific measurements.
What is specific gravity? How is it related to density?
Answer:
Specific gravity is defined as the ratio of the Densities of the two substances.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
Explanation:
Specific gravity is defined as the ratio of the Densities of the two substances.
For the standardization, the density ration of densities is calculated with respect to the density of water i.e the denominator is the density of water.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water}}[/tex]
Also,
At STP density of water is 1000 Kg/m³
Therefore the relation between the specific gravity and density is,
The Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water at STP}}[/tex]
or
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
What are the parameters that affect life and drag forces on an aerofoil?
Answer:
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Explanation:
Drag force:
Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.
Drag force given as
[tex]F_D=\dfrac{1}{2}\rho\ A\ V^2[/tex]
So we can say that drag force depends on following properties
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.
Answer:
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
Explanation:
Given that
Diameter(d)=62 mm
Thickness(t)= 300 μm=0.3 mm
Internal pressure(P)=100 KPa
Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .
The hoop stress
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
Longitudinal stress
[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]
Now by putting the values
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
[tex]\sigma _h=\dfrac{100\times 62}{2\times 0.3}[/tex]
[tex]\sigma _h=10.33MPa[/tex]
[tex]\sigma _l=5.16MPa[/tex]
So the principle stress are
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
What impact does modulus elasticity have on the structural behavior of a mechanical design?
Answer with Explanation:
The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:
1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.
2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.
3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity the resistance it offer's to shear forces and the torques is more.
While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.
What is the function of air preheater? How does air preheating save fuel?
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.
A room with dimensions of 3 x 10 x 20 m is estimated to have outdoor air brought in at an infiltration rate of 1/4 volume changes per hour. Determine the infiltration rate in m^3/s.
Answer:
The infiltration rate is of 0.042 m^3/s.
Explanation:
The total volume of the room is:
V = 3 * 10 * 20 = 600 m^3
If the air infiltration rate is of 1/4 volume per hour:
v = V/4 * 1 hour /3600 seconds
Replacing:
v = 600/4 * 1/3600 = 0.042 m^3/s
The infiltration rate would then be of 0.042 m^3/s.
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to convert the ice to water at 200°F?
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice [tex]C_p=2.03\ KJ/kg.K[/tex]
Latent heat for ice H=336 KJ/kg
Specific heat for ice [tex]C_p=4.187\ KJ/kg.K[/tex]
We know that sensible heat given as
[tex]Q=mC_p\Delta T[/tex]
Heat for -15F to 32 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 2.03(32+15) KJ[/tex]
Q=858.69 KJ
Heat for 32 Fto 200 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 4.187(200-32) KJ[/tex]
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.
A convenient and cost-effective way to store biogas is to use light-weight, rigid gas containers. The pressure and temperature of biogas stored in a 10 m^3 container are 1.5 bar and 5°C, respectively, as measured at an early morning time (state 1). The temperature of the biogas is expected to increase to 40°C at noon on the same day, without any significant change in the volume of the container or amount of methane in the container (state 2). The biogas can be approximated as methane (CH4) modelled as an ideal gas with constant specific heat ratio, k = 1.4. The effects of gravity and motion are negligible. The reference environment pressure and temperature are, respectively, po = 100 kPa and To = 0 °C. The molecular weight of methane is M = 16.04 gmol-1. The universal gas constant is R= 8.31 J mol-1K-1. (a) Calculate the mass, in kg, and amount of substance, in mol, for the methane in the container.
Answer:
10.41 kg
Explanation:
The gas state equation is:
p * V = n * R * T
For this equation we need every value to be in consistent units
1.5 bar = 150 kPa
5 C = 278 K
n = p * V / (R * T)
n = 150000 * 10 / (8.31 * 278) = 649 mol
Multiplying the amount of moles by the molecular weight of the gas we obtain the mass:
m = M * mol
m = 16.04 * 649 = 10410 g = 10.41 kg
A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine the force R which the elevator floor exerts on her feet and the lifting force L which she exerts on the package during the acceleration in-travel. If the elevator supports cables suddenly and completely fail, what values would R and L acquire
Answer:
force R = 846.11 N
lifting force L = 110.36 N
if cable fail complete both R and L will be zero
Explanation:
given data
mass woman mw = 60 kg
mass package mp = 9 kg
accelerates rate a = g/4
to find out
force R and lifting force L and if cable fail than what values would R and L acquire
solution
we calculate here first reaction R force
we know elevator which accelerates upward
so now by direction of motion , balance the force that is express as
R - ( mw + mp ) × g = ( mw + mp ) × a
here put all these value and a = g/4 and use g = 9.81 m/s²
R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4
R = ( 69 ) × 9.81/4 + ( 69 ) 9.81
R = 69 ( 9.81 + 2.4525 )
force R = 846.11 N
and
lifting force is express as here
lifting force = mp ( g + a)
put here value
lifting force = 9 ( 9.81 + 9.81/4)
lifting force L = 110.36 N
and
we know if cable completely fail than body move free fall and experience no force
so both R and L will be zero
The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotational speed of 620 rpm. Determine the torque of the input shaft shaft, in N-m.
Answer:
Torque at input shaft will be 176.8695 N-m
Explanation:
We have given input power [tex]P_{IN}=42.6KW=42.6\times 10^3W[/tex]
Angular speed = 2300 rpm
For converting rpm to rad/sec we have multiply with [tex]\frac{2\pi }{60}[/tex]
So [tex]2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec[/tex]
We have to find torque
We know that power is given by [tex]P=\tau \omega[/tex], here [tex]\tau[/tex] is torque and [tex]\omega[/tex] is angular speed
So [tex]42.6\times 10^3=\tau \times 240.855[/tex]
[tex]\tau =176.8695N-m[/tex]
So torque at input shaft will be 176.8695 N-m
For two 0.2 m long rotating concentric cylinders, the velocity distribution is given by u(r) = 0.4/r - 1000r m/s. If the diameters are 2 cm and 4 cm, respectively, calculate the fluid viscosity if the torque on the inner cylinder is measured to be 0.0026 N*m.
Answer:
5.9*10^-3 Pa*s
Explanation:
The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.
τ = μ * du/dr
u(r) = 0.4/r - 1000*r
The derivative is:
du/dr = -1/r^2 - 1000
On the radius of the inner cylinder this would be
u'(0.02) = -1/0.02^2 - 1000 = -3500
So:
τ = -3500 * μ
We don't care about the sign
τ = 3500 * μ
That is a tangential force per unit of area.
The area of the inner cylinder is:
A = h * π * D
And the torque is
T = F * r
T = τ * A * D/2
T = τ * h * π/2 * D^2
T = 3500 * μ * h * π/2 * D^2
Then:
μ = T / (3500 * h * π/2 * D^2)
μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. What is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?
Answer:
0.2 m/s
Explanation:
The velocity of a point on the edge of a disk rotating disk can be calculated as:
[tex]v=\omega*r[/tex]
Where [tex]\omega[/tex] is the angular velocity and r the radius of the disk. This leads to:
[tex]v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s[/tex]
Answer:
0.2 m/s
Explanation:
Step 1: identify the given parameters
angular velocity, ω = 0.5 rad/s
radius of the disk, r = 0.4m
Note: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.
Step 2: calculate the velocity of the disk at the outer edge in m/s
Velocity = Angular velocity (rad/s) X radius (m)
Velocity = 0.5 rad/s X 0.4 m
Velocity = 0.2 m/s
An open tank contains ethylene glycol at 25°C. Compute the pressure at a depth of 3.0m.
Answer:
pressure at depth 3 m is 33.255 kPa
Explanation:
given data
temperature = 25°C
depth = 3 m
to find out
pressure
solution
we know pressure formula that is
pressure = ρ g h ................1
and we know specific gravity of Ethylene glycol is 1.13
so
1.13 = [tex]\frac{\rho (e)}{\rho (water)}[/tex]
ρ (Ethylene glyco) = 1130 Kg/m³
so
pressure will be by equation 1
pressure = 1130 × 9.81 × 3
pressure = 33255.9 Pa
so pressure at depth 3 m is 33.255 kPa