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.
What is the ratio between the maximum elastic moment, MY, and the maximum plastic moment, MP, for a solid rectangular section made from a ductile, elastoplastic material? What is this ratio referred to?
Answer:
Shape factor
Explanation:
Shape factor is the ratio of maximum plastic moment to maximum elastic moment.Shape factor is denoted by K.
Shape factor can be given as
[tex]K=\dfrac{M_p}{M_y}[/tex]
[tex]K=\dfrac{\sigma _yZ_p}{\sigma _y Z}[/tex]
[tex]K=\dfrac{Z_p}{ Z}[/tex]
For a solid rectangular section made from ductile material shape factor is 1.5 .
A rocket developed by an amateur was traveling upwards at a velocity given by v = (11 + 0.2s) m/s, where s is in meters. Determine the time for the rocket to reach an altitude of s = 80 m. Initially, s = 0 when t = 0. [Hint: obtain initial velocity].
Answer:
Time taken to reach 80 meters equals 4.4897 seconds.
Explanation:
We know that velocity is related to position as
[tex]v=\frac{ds}{dt}[/tex]
Now it is given that [tex]v=(11+0.2v[/tex]
Using the given velocity function in the above relation we get
[tex]\frac{ds}{dt}=(11+0.2s)\\\\\frac{ds}{(11+0.2s)}=dt\\\\\int \frac{ds}{(11+0.2s)}=\int dt\\\\[/tex]
Now since the limits are given as
1) at t = 0 , s=0
Using the given limits we get
[tex]\int_{0}^{80} \frac{ds}{(11+0.2s)}=\int_{o}^{t} dt\\\\\frac{1}{0.2}[ln(11+0.2s)]_{0}^{80}=(t-0)\\\\5\times (ln(11+0.2\times 80)-ln(11))=t\\\\\therefore t=4.4897seconds[/tex]
What is the relation between Poisson's ration, young's modulus, shear modulus for an material?
Answer:
Explanation:
Poisson's ration= is the ratio between the deformation that occurs in the material in the direction perpendicular to the force applied with the deformation suffered by the material in the same direction of the force.
young's modulus= is the ratio between the stress applied to a material with respect to its unit deformation when this material complies with the hooke's law.
shear modulus=change in the way an elastic material experiences when subjected to shear stresses
How much would a 10.0 inch long, 0.25 inch diameter AISI 1020 Q&T bolt stretch when loaded with 2000 lbs?
Answer:
0.014 in
Explanation:
The Young's module of steel is of E = 210 GPa = 30*10^6 psi
The section of the bolt would be:
A = π/4 * D^2
The stiffness would be:
k = E * A / L
k = π * D^2 * E / (4 * L)
k = π * 0.25^2 * 30*10^6 / (4 * 10) = 147000 lb / in
If I apply a force of 2000 lb, I calculate the stretching with Hooke's law:
Δx = f / k
Δx = 2000 / 147000 = 0.014 in
why HF (hydrogen fluoride) has higher boiling temperature than HCl (hydrogen chloride), even thought HF has lower molecular weight?
Answer:
Boiling point of HF is higher as compared to HCl because of presence of hydrogen bonding in it.
Explanation:
In HF, intermolecular force of attraction is hydrogen bonding.
Hydrogen bonding is a type of electrostatic force of attraction existing between H atom and electronegative atom.
For a molecule to have hydrogen bonding, H atom must be bonded to electronegative atom, O, N and F.
Hydrogen bonding can be intermolecular and intramolecular.
So, in HF hydrogen bonding present.
In HCl, only van der Waals force exists. van der Waals forces are weak as compared to hydrogen bonding.
Because of presence of hydrogen bonding, HF molecules are held tightly and so requires more heat to boil.
Therefore, boiling point of HF is more as compared to HCl.
To 3 significant digits, what is the change of entropy of air in kJ/kgk if the pressure is decreased from 400 to 300 kPa and the temperature is increased from 300 to 900 K? DO NOT ASSUME constant specific heats.
Answer:
The change of entropy is 1.229 kJ/(kg K)
Explanation:
Data
[tex] T_1 = 300 K [/tex]
[tex] T_2 = 900 K[/tex]
[tex] p_1= 400 kPa[/tex]
[tex] p_2= 300 kPa[/tex]
[tex] R= 0.287 kJ/(kg K)[/tex] (Individual Gas Constant for air)
For variable specific heats
[tex]s(T_2, p_2) - s(T_1, p_1) = s^0(T_2) - s^0(T_1) - R \, ln \frac{p_2}{p_1}[/tex]
where [tex] s^0(T) [/tex] is evaluated from table attached
[tex] s^0(900 K) = 2.84856 kJ/(kg K)[/tex]
[tex] s^0(300 K) = 1.70203 kJ/(kg K)[/tex]
Replacing in equation
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 2.84856 kJ/(kg K) - 1.70203 kJ/(kg K) - 0.287 kJ/(kg K) \, ln \frac{300 kPa}{400 kPa}[/tex]
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 1.229 kJ/(kg K)[/tex]
A 4,000-km^2 watershed receives 102cm of precipitation in one
year.The avg. flow of the river draining the watershed is 34.2
m^3/s.Infiltration is est. to be 5.5 x 10^(-7) cm/s
andevapotranspiration is est. to be 40 cm/y. Determine the change
instorage in the watershed over one year. The ratio of runoff
toprecipitation (both in cm) is termed the runoff
coefficient.Compute the runoff coefficient for this
watershed.
Answer:
1) The change in storage of the catchment is 707676800 cubic meters.
2) The runoff coefficient of the catchment is 0.83.
Explanation:
The water budget equation of the catchment can be written as
[tex]P+Q_{in}=ET+\Delta Storage+Q_{out}+I[/tex]
where
'P' is volume of precipitation in the catchment =[tex]Area\times Precipitation[/tex]
[tex]Q_{in}[/tex] Is the water inflow
ET is loss of water due to evapo-transpiration
[tex]\Delta Storage[/tex] is the change in storage of the catchment
[tex]Q_{out}[/tex] is the outflow from the catchment
I is losses due to infiltration
Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get
[tex]4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365[/tex]
[tex]\therefore \Delta Storage=707676800m^3[/tex]
Part b)
The runoff coefficient C is determined as
[tex]C=\frac{P-I}{P}[/tex]
where symbols have the usual meaning as explained earlier
[tex]\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83[/tex]
Define a) Principal Plane b) Principal Stress c) anelasticity d) yield point e) ultimate tensile stress f) hardness g) toughness h) elastic limit
Answer:
Principal Plane: It is that plane in a stressed body over which no shearing stresses act. As we know that in a stressed body on different planes 2 different kind of stresses act normal stresses acting normal to the plane ans shearing stresses acting in the plane. The special planes over which no shearing stresses act and only normal stresses are present are termed as principal planes.
Principal Stress: The stresses in the principal planes are termed as normal stresses.
Anelasticity: It is the behavior of a material in which no definite relation can found to exist between stress and strain at any point in the stressed body.
Yield Point: It is the point in the stress-strain curve of a body at which the stress in the body reaches it's yield value or the object is just about to undergo plastic deformation if we just increase value of stress above this value. It is often not well defined in high strength materials or in some materials such as mild steel 2 yield points are observed.
Ultimate tensile strength: It is the maximum value of stress that a body can develop prior to fracture.
Hardness: it is defined as the ability of the body to resist scratches or indentation or abrasion.
Toughness: It is the ability of the body to absorb energy and deform without fracture when it is loaded. The area under the stress strain curve is taken as a measure of toughness of the body.
Elastic limit: The stress limit upto which the body regains it's original shape upon removal of the stresses is termed as elastic limit of the body.
Why should a toolpath be verified on the screen of a CAM system prior to creating the program code?
Answer:
The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons
1. analyze the machining strategy and identify which one is better for each piece.
2.Avoid the collision of the tool holder with the work piece.
3.Avoid the shock of the tool with the piece.
4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.
How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the same size carrying the same load?
Answer:
The volume of the extra water is [tex]2.195 ft^{3}[/tex]
Solution:
As per the question:
Mass of the canoe, [tex]m_{c} = 175 lb + w[/tex]
Height of the canoe, h = 21.5 ft
Mass of the kevlar canoe, [tex]m_{Kc} = 38 lb + w[/tex]
Now, we know that, bouyant force equals the weight of the fluid displaced:
Now,
[tex]V\rho g = mg[/tex]
[tex]V = \frac{m}{\rho}[/tex] (1)
where
V = volume
[tex]\rho = 62.41 lb/ft^{3}[/tex] = density
m = mass
Now, for the canoe,
Using eqn (1):
[tex]V_{c} = \frac{m_{c} + w}{\rho}[/tex]
[tex]V_{c} = \frac{175 + w}{62.41}[/tex]
Similarly, for Kevlar canoe:
[tex]V_{Kc} = \frac{38 + w}{62.41}[/tex]
Now, for the excess volume:
V = [tex]V_{c} - V_{Kc}[/tex]
V = [tex]\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}[/tex]
In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler and 4 psi in the condenser. What is the minimum temperature required at the turbine inlet, so that the quality of the steam at the turbine outlet is not less than 85%. What would be the thermal efficiency of the cycle?
Answer:
Explanation:
The pressures given are relative
p1 = 2000 psi
P1 = 2014 psi = 13.9 MPa
p2 = 4 psi
P2 = 18.6 psi = 128 kPa
Values are taken from the steam pressure-enthalpy diagram
h2 = 2500 kJ/kg
If the output of the turbine has a quality of 85%:
t2 = 106 C
I consider the expansion in the turbine to adiabatic and reversible, therefore, isentropic
s1 = s2 = 6.4 kJ/(kg K)
h1 = 3500 kJ/kg
t2 = 550 C
The work in the turbine is of
w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg
The thermal efficiency of the cycle depends on the input heat.
η = w/q1
q1 is not a given, so it cannot be calculated.
Can you carry 1 m3 of liquid water? Why or why not? (provide the weight to support your answer)
Answer:
No we cannot carry 1 cubic meter of liquid water.
Explanation:
As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.
No, I cannot carry 1 cubic meter (1 m³) of liquid water. To understand why, let's calculate the weight of 1 cubic meter of water.
1 cubic meter (m³) of water is equivalent to 1000 liters (L). The density of water is approximately 1 kilogram per liter (kg/L). Therefore, the weight of 1 cubic meter of water can be calculated as:
[tex]\[ 1 \, \text{m}^3 \times 1000 \, \text{L/m}^3 \times 1 \, \text{kg/L} = 1000 \, \text{kg} \][/tex]
So, 1 cubic meter of water weighs 1000 kilograms, or about 2204.62 pounds.
This weight is far beyond the carrying capacity of an average human. For comparison, most people can carry only a few tens of kilograms comfortably for a short period, so carrying 1000 kilograms is not feasible for any human.
A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling at 98 m/s. If the acceleration of gravity is 9.81 m/s^2, what is the radius of curvature of the cannon balls path at this instant?
Answer:
The radius of curvature is 979 meter
Explanation:
We have given velocity of the canon ball v = 98 m/sec
Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]
We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity
Acceleration due to gravity is given by [tex]a_c=\frac{v^2}{r}[/tex], here v is velocity and r is radius of curvature
So [tex]\frac{98^2}{r}=9.81[/tex]
r = 979 meter
So the radius of curvature is 979 meter
A freight train and a passenger train share the same rail track. The freight train leaves station A at 8:00 A.M. The train travels at a speed of 45 km/h for the first 10 minutes and then continues to travel at a speed of 60 km/h. At 8:35 A.M., the passenger train leaves station A. The pas¬senger train travels first at a speed of 75 km/h for 5 minutes and then continues to travel at a speed of 105 km/h. Determine the location of the siding where the freight train will have to be parked to allow the faster passenger train to pass through. As a safety precaution, it is determined that the time headway between the two trains should not be allowed to fall below 5 minutes.
Answer:
74.2 km from station A.
Explanation:
We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.
The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.
For this problem it is better to convert the speeds to km/min
45 km/h = 0.75 km/min
The equation for position under constant speed is:
X(t - t0) = X0 + v0 * (t - t0)
Since we know the time it will stop moving at this speed:
X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km
After it ran those 7.5 km it will keep running at 60 km/h.
60 km/h = 1 km/min
The position equation for it is now:
X(t - 10) = 7.5 + 1 * (t - 10)
The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.
75 km/h = 1.25 km/min
After those 5 minutes it will have traveled:
X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km
Then it travels at 105 km/h
105 km/h = 1.75 km/min
Its position equation is now:
X(t - 40) = 6.25 + 1.75 * (t - 40)
Equating both positions we find the time at which they would meet:
7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)
7.5 + t - 10 = 6.25 + 1.75*t - 70
t - 1.75*t = 6.25 - 70 +10 - 7.5
-0.75*t = -61.25
t = 61.25 / 0.75
t = 81.7 minutes
The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.
At that moment the freight train will be at:
X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km
Find the power production (in MW) of a 25 m radius wind turbine if the average wind speed is 12 m/s and the efficiency of this turbine in converting kinetic energy of air to mechanical work is 10%? The density of air is 1.20 kg/m^3
Answer:
shaft power 0.2034 MW
Explanation:
given details
radius of turbine = 25 m
average wind velocity = 12 m/s
density of air = 1.20 kg/m^2
Total power is calculated as
[tex]P = \frac{1}{2} \rho AV^3[/tex]
[tex]= \frac{1}{1} \rho \pir^2 V^3[/tex]
[tex]= \frac{1}{2} 1.20\times \pi \times 625\times 12^3 = 2034,720 watt[/tex]
P = 2.034 MW
shaft power [tex] = \eta \times P[/tex]
[tex]= 0.10 \times 2.034[/tex]
= 0.2034 MW
Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?
Answer:
Free convection:
When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.
Force convection:
When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.
Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.
We know that convective heat transfer given as
q = h A ΔT
h=Heat transfer coefficient
A= Surface area
ΔT = Temperature difference
Calculate the surface temperature of a black surface, 1.6 m^2 in area if the rate of heat transfer is 632 kW. The Stefan-Boltzmann constant is σ = 5.67 x 10^-8 W/m^2 K^4 a) 1734 °C b) 273 °C c) 1625 K d) 1640 K e) 1682 K
Explanation:
From Stefan's formula
P=A&T^4
T=(P/A&)^1/4
T=(632000W/1.6m^2 x 5.67E-8W/m^2K^4)^1/4
T=
Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three
significantfigures and express the answer in SI units using an
appropriateprefix.
Answer:
the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]
Explanation:
We have evaluate [tex]\frac{(204mm\times 0.00457kg)}{34.6N}[/tex]
We know that 1 mm [tex]=10^{-3}m[/tex]
So 240 mm [tex]=204\times 10^{-3}m[/tex]
Newton can be written as [tex]kgm/sec^2[/tex]
So [tex]\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}[/tex]
So the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]
Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ
Answer:
C)185,500 KJ
Explanation:
Given that
Latent heat fusion = 333.23 KJ/kg
Latent heat vaporisation = 333.23 KJ/kg
Mass of ice = 100 kg
Mass of water = 40 kg
Mass of vapor=60 kg
Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .
Sensible heat for water Q
[tex]Q=mC_p\Delta T[/tex]
For water
[tex]C_p=4.178\ KJ/Kg.K[/tex]
Q=4.178 x 40 x 100 KJ
Q=16,712 KJ
So total heat
Total heat =100 x 333.23+16,712 + 60 x 2257 KJ
Total heat =185,455 KJ
Approx Total heat = 185,500 KJ
So the answer C is correct.
A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow rope immediately as he leaves the end of the ramp. Determine the maximum height he attains. Determine his velocity and direction of his velocity at that maximum height. Why is one of the components equal to zero at that point? How far does he travel before landing on the water? How long does it take him to land? What is his velocity when he lands? And finally, at what angle does he land?
Answer:
At highest point:
y1 = 10.4 ft
v1 = (26.5*i + 0*j) ft/s
When he lands:
x2 = 31.5 ft (distance he travels)
t2 = 1.19 s
V2 = (26.5*i - 25.9*j) ft/s
a2 = -44.3°
Explanation:
Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.
20 mi / h = 29.3 ft/s
If the ramp has an angle of 25 degrees, the speed is
v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s
v0 = (26.5*i + 12.4*j) ft/s
I set up the coordinate system with the origin at the base of the ramp under its end, so:
R0 = (0*i + 8*j) ft
The equation for the horizontal position is:
X(t) = X0 + Vx0 * t
The equation for horizontal speed is:
Vx(t) = Vx0
The equation for vertical position is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
The equation for vertical speed is:
Vy(t) = Vy0 + a * t
In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.
In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.
0 = Vy0 + a * t1
a * t1 = -Vy0
t1 = -Vy0 / a
t1 = -12.4 / -32.2 = 0.38 s
y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft
The velocity at that moment will be:
v1 = (26.5*i + 0*j) ft/s
When he lands in the water his height is zero.
0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2
-16.1 * t2^2 + 12.4 * t2 + 8 = 0
Solving this equation electronically:
t2 = 1.19 s
Replacing this time on the position equation:
X(1.19) = 26.5 * 1.19 = 31.5 ft
The speed is:
Vx2 = 26.5 ft/s
Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s
V2 = (26.5*i - 25.9*j) ft/s
a2 = arctg(-25.9 / 26.5) = -44.3
What is the weight in pounds of a gallon of oil that has a specific gravity of .86
Answer:
Mass of oil will be 7.176 pound
Explanation:
We have given specific gravity of oil = 0.86
We know that specific gravity is given by [tex]specific\ gravity=\frac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]0.86=\frac{density\ of\ oil}{1000}[/tex]
Density of oil = [tex]860kg/m^3[/tex]
We have given volume of oil = 1 gallon
We know that 1 gallon = 0.003785 [tex]m^3[/tex]
So mass of oil = volume ×density
mass = 0.003785×860 = 3.2551 kg
We know that 1 kg = 2.2046 pound
So 3.2551 kg = 3.2551×2.2046 = 7.176 pound
Specific cutting energy increases with increasing the cutting speed. a) True b) False
Answer:
b)false
Explanation:
Specific cutting energy:
Energy required to remove unit volume of material is called specific energy.In other words we can say that the ratio of energy to the volume removal rate is called specific cutting energy.
When cutting speed is increases then specific energy goes to decrease.As well as when depth of cut and feed of tool is increase then specific cutting energy will decrease.
Yield and tensile strengths and modulus of elasticity with increasing temperature. (increase/decrease/independent)
Answer:
Decrease
Explanation:
Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.
So we can say that when with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.
Decrease
Answer:
Decreases
Explanation:
Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.
The modulus of elasticity is proportional to its tensile strength.
Thermal conductivity of AISI 316 Stainless Steel at 90ºC is 14.54 W/m K. Convert this value to IP system.
Answer:
the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]
Explanation:
Given that
Thermal conductivity K=14.54 W/m.K
This above given conductivity is in SI unit.
SI unit IP unit Conversion factor
m ft 0.3048
W Btu/hr 0.293
The unit of conductivity in IP is Btu./ft.hr.F.
Now convert into IP divided by 1.73 factor.
[tex]0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}[/tex]
So
[tex]0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]
[tex]8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]
So the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]
What are the two types of furnaces used in steel production?
Explanation:
The two types of furnaces used in steel production are:
Basic oxygen furnace
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
Electric arc furnace
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.
When is it appropriate to model a structural element as a beam?
It is convenient to model a structural element like a beam when a significant amount of forces produce the stress called flexion.
Flexion occurs when an element is supported on one or more supports and a force is presented between them, driving a bending moment in the element.
Define ""acidity"" of an aqueous solution. How do you compare the strength of acidity of solutions ?
Answer with Explanation:
The acidity of an aqueous solution is a term used to identify how acidic the solution is. An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. In the other case around if the concentration of hydrogen ions is lesser than the concentration of hydroxide ions the solution is termed to be basic or alkaline. For a solution with equal concentration of hydrogen and hydroxide ions the solution is termed to be neutral.
The acidity of solutions is compared on the basis of the concentration of the hydrogen ions reduced to log of base 10 to ease calculations. The comparison is made in terms of 'pH' value which is defined as
[tex]pH=-log[H^+][/tex]
where
[tex][H^+][/tex] is the hydrogen ion concentration of the solution in moles per liter of solution.
If the pH is < 7 the solution is acidic and the closer the pH value to 1 the higher is the acidity of the solution.
A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated power density of the wind that is hitting the turbine. Calculate the mechanical power developed at the shaft connecting rotor and generator. Assume rotor diameter 100 m and air density 1.225 kg/m^3.
Answer:
Rated power = 1345.66 W/m²
Mechanical power developed = 3169035.1875 W
Explanation:
Wind speed, V = 13 m/s
Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3
Rotor diameter, d = 100 m
or
Radius = 50 m
Air density, ρ = 1.225 kg/m³
Now,
Rated power = [tex]\frac{1}{2}\rho V^3[/tex]
or
Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]
or
Rated power = 1345.66 W/m²
b) Mechanical power developed = [tex]\frac{1}{2}\rho AV^3C_p[/tex]
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]
or
Mechanical power developed = 3169035.1875 W
A Carnot heat engine receives heat at 900 K and rejects the waste heat to the environment at 300 K. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at –15°C at a rate of 295 kJ/min and rejects it to the same environment at 300 K. Determine the rate of heat supplied to the heat engine. (Round the final answer to one decimal place. You must provide an answer before moving to the next part.).The rate of heat supplied to the engine is ___ kJ/min.
Answer:
The rate of heat supplied to the engine is 71.7 kJ/min
Explanation:
Data
Engine hot temperature, [tex] T_H [/tex] = 900 K
Engine cold temperature, [tex] T_C [/tex] = 300 K
Refrigerator cold temperature, [tex] T'_C [/tex] = -15 C + 273 = 258 K
Refrigerator hot temperature, [tex] T'_H [/tex] = 300 K
Heat removed by refrigerator, [tex] Q'_{in} [/tex] = 295 kJ/min
Rate of heat supplied to the heat engine, [tex] Q_{in} [/tex] = ? kJ/min
See figure
From Carnot refrigerator coefficient of performance definition
[tex] COP_{ref} = \frac{T'_C}{T'_H - T'_C} [/tex]
[tex] COP_{ref} = \frac{258}{300 - 258} [/tex]
[tex] COP_{ref} = 6.14 [/tex]
Refrigerator coefficient of performance is defined as
[tex] COP_{ref} = \frac{Q'_{in}}{W} [/tex]
[tex] W = \frac{Q'_{in}}{COP_{ref}} [/tex]
[tex] W = \frac{295 kJ/min}{6.14} [/tex]
[tex] W = 48.04 kJ/min [/tex]
Carnot engine efficiency is expressed as
[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]
[tex] \eta = 1 - \frac{300 K}{900 K}[/tex]
[tex] \eta = 0.67[/tex]
Engine efficiency is defined as
[tex] \eta = \frac{W}{Q_{in}} [/tex]
[tex] Q_{in} = \frac{W}{\eta} [/tex]
[tex] Q_{in} = \frac{48.04 kJ/min}{0.67} [/tex]
[tex] Q_{in} = 71.7 kJ/min [/tex]
Rounding to one decimal place, the rate of heat supplied to the engine is 147.5 kJ/min.
First, we need to calculate the coefficient of performance (COP) of the Carnot refrigerator using the formula:
[tex]\[ \text{COP} = \frac{T_C}{T_H - T_C} \][/tex]
where:
[tex]- \( T_C \)[/tex] is the absolute temperature of the cold sink (300 K)
[tex]- \( T_H \)[/tex] is the absolute temperature of the heat source (900 K)
Substituting the given values, we get:
[tex]\[ \text{COP} = \frac{300}{900 - 300} = \frac{300}{600} = 0.5 \][/tex]
Next, we use the COP of the refrigerator to find the rate of heat supplied to the engine:
[tex]\[ \text{Rate of heat supplied to the engine} = \text{COP} \times \text{Rate of heat removed by the refrigerator} \][/tex]
Given that the rate of heat removed by the refrigerator is 295 kJ/min, we can calculate the rate of heat supplied to the engine:
[tex]\[ \text{Rate of heat supplied to the engine} = 0.5 \times 295 = 147.5 \, \text{kW} \][/tex]
Rounding to one decimal place, the rate of heat supplied to the engine is 147.5 kJ/min.
The complete question is here.
A carnot heat engine receives heat at 900K and rejects the waste heat to the enviroment at 300K. The entire work output of the heat engine is used to drive a carnot refrigerator that removes heat from the cooled space at -150C at a rate of 250 kJ/min and rejects it to the same enviroment at 300 K. Determine (a) the rate of heat supplied to the heat engine and (b) the total rate of heat rejection to the enviroment.
Water has a density of 1.94 slug/ft^3. What is the density expressed in SI units? Express the answer to three significant figures
The density of water in SI units, converted from 1.94 slug/ft^3, is approximately 998.847 kg/m^3 when expressed to three significant figures.
Explanation:The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:
(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)
This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.
Water's density conversion to SI units is 1000 kg/m³.
The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.
Water's density conversion to SI units is 1000 kg/m³.
The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.
The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:
(1.94 slug/ft3) * (14.5939 kg/slug) * ((1 ft/0.3048 m)3)
This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.
The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:
(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)
This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.