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
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.
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
At the beginning of the compression process of an air-standard Otto cycle, p1 = 1 bar and T1 = 300 K. The compression ratio is 8.5 and the heat addition per unit mass of air is 1400 kJ/kg. Determine the maximum temperature of the cycle in Kelvin (input a number ONLY). Do not assume specific heats are constant. There is a ±5% tolerance.
Answer:
Maximum temperature of the cycle is 2231.3 K
Explanation:
See table (values there do not assume constant specific heat) and figure attached.
Assuming ideal gas behaviour, p1*v1 = p2*v2, rearranging p2/p1 = v1/v2
Data
[tex]p_1 = 1 bar [/tex]
[tex]T_1 = 300 K [/tex]
[tex] \frac{v_1}{v_2} = 8.5 [/tex] (compression ratio)
[tex] \frac{Q_{23}}{m} = 1400 kJ/kg [/tex] (heat addition)
We can use the following relationship for air
[tex] \frac{v_1}{v_2} = \frac{v_{r1}}{v_{r2}} [/tex]
[tex] v_{r1} [/tex] is only function of temperature and can be taken from table. In this case:
[tex] v_{r1} = 621.2 [/tex]
Rearranging previous equation
[tex] v_{r2} = v_{r1} \times \frac{v_2}{v_1} [/tex]
[tex] v_{r2} = 621.2 \times \frac{1}{8}[/tex]
[tex] v_{r2} = 73.082 [/tex]
Interpolating from table
[tex] u_2 = 503.06 kJ/kg [/tex]
Energy balance in the process 2-3 gives
[tex] \frac{Q_{23}}{m} = u_3 - u_2 [/tex]
[tex] u_3 = \frac{Q_{23}}{m} + u_2 [/tex]
[tex] u_3 = 1400 kJ/kg + 503.06 kJ/kg [/tex]
[tex] u_3 = 1903.06 kJ/kg [/tex]
Interpolating from table
[tex] T_3 = 2231.3 K [/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.
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.
An experiment was set-up to measure an unknown fluid's viscosity. Two flat plates are separated by a gap of 4.5 mm and move relative to each other at a velocity of 5 m/s. The space between them is occupied by the unknown viscosity. The motion of the plates is resisted by a shear stress of 10 Pa due to the viscosity of the fluid. Assuming that the velocity gradient of the fluid is constant, determine the coefficient of viscosity of the fluid.
Answer:[tex]\mu =9\times 10^{-3}Pa-s[/tex]
Explanation:
Distance between Plates(dy)=4.5 mm
Relative Velocity(du)=5 m/s
We know shear stress is given by [tex]\tau =10 Pa[/tex]
[tex]\tau =\frac{\mu du}{dy}[/tex]
where du=relative Velocity
dy=Distance between Plates
[tex]10=\frac{\mu \times 5}{4.5\times 10^{-3}}[/tex]
[tex]\mu =9\times 10^{-3}Pa-s[/tex]
What is the total kinetic energy of a 2500 lbm car when it is moving at 80 mph (in BTU)?
Answer:
The kinetic energy will be 687.186 BTU
Explanation:
We have given mass of car = 2500 lbm
We know that 1 lbm = 0.4535 kg
So 2500 lbm = [tex]2500\times 0.4535=1133.75kg[/tex]
Speed = 80 mph
We know that 1 mile = 1609.34 meter
1 hour = 3600 sec
So [tex]80mph=\frac{80\times 1609.34}{3600}=35.763m/sec[/tex]
We know that kinetic energy [tex]E=\frac{1}{2}mv^2=\frac{1}{2}\times 1133.75\times 35.76^2=725.033KJ[/tex]
We know that 1 KJ = 0.9478 BTU
So 725.033 KJ = 725.033×0.9478 = 687.186 BTU
If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential energy is unchanged, calculate its change in specificinternal energy if its mass is 5.4 kg. Report your answer in kJ/kg to one decimal place.
Answer:
The change in specific internal energy is 3.5 kj.
Explanation:
Step1
Given:
Total change in energy is 15.5 kj.
Change in kinetic energy is –3.5 kj.
Change in potential energy is 0 kj.
Mass is 5.4 kg.
Step2
Calculation:
Change in internal energy is calculated as follows:
[tex]\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U[/tex][tex]15.5=-3.5+0+\bigtriangleup U[/tex]
[tex]\bigtriangleup U=19[/tex] kj.
Step3
Specific internal energy is calculated as follows:
[tex]\bigtriangleup u=\frac{\bigtriangleup U}{m}[/tex]
[tex]\bigtriangleup u=\frac{19}{5.4}[/tex]
[tex]\bigtriangleup u=3.5[/tex] kj/kg.
Thus, the change in specific internal energy is 3.5 kj/kg.
The position of a particle moving along a straight line is defined by the relation. s = t^3 – 6t^2 – 15t + 40, where s is expressed in feet and t in seconds. Determine:(a) s when t = 3 seconds.(b) v when t = 5 seconds. (c) a when t = 4 seconds.(d) the time when the velocity is equal to zero. What is important about this information?
Answer:
1) s(3) = -32 feet.
2)v(5) = 3 feet/sec
3)a(4) = 12[tex]feet/s^{2}[/tex]
4) Velocity becomes zero at t = 5 seconds
Explanation:
Given that position as a function of time is
[tex]s(t)=t^{3}-6t^{2}-15t+40[/tex]
Now by definition of velocity we have
[tex]v=\frac{ds}{dt}\\\\v=\frac{d}{dt}(t^{3}-6t^{2}-15t+40)\\\\\therefore v(t)=3t^{2}-12t-15[/tex]
Now by definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\a=\frac{d}{dt}(3t^{2}-12t-15)\\\\\therefore a(t)=6t-12[/tex]
Applying values of time in corresponding equations we get
1) s(3)=[tex]3^{3}-6\times (3)^{2}-15\times 3+40=-32feet[/tex]
2)v(5)=[tex]3\times {5}^{2}-12\times 5-15=3feet/sec[/tex]
3)a(4)=[tex]6\times 4-12=12ft/s^{2}[/tex]
4)To obatin the time at which velocity is zero equate the velocity function with zero we get
[tex]3t^{2}-12t-15=0\\\\t^{2}-4t-5=0\\\\t^{2}-5t+t-5=0\\\\t(t-5)+1(t-5)=0\\\\(t-5)(t+1)=0\\\\\therefore t=5\\\\or\\\therefore t=-1[/tex]
Thus the correct time is 5 seconds at which velocity becomes zero.
Micrometers with a vernier graduation are capable of taking readings to the nearest 0.0001 in. a)- True b)- false
Answer:
The micrometer with vernier graduation can measure reading to the nearest 0.0001 inches. So, the statement is true.
Explanation:
Micrometer is the measuring device that used to measure length with more accuracy. Micrometer can measure the length in metric as well as in English unit. Micrometer is generally used to measure diameter and length of the mechanical component.
Working:
Micrometer is a screwed device. It contains spindle, anvil and thimble. Object is placed between spindle and anvil. Thimble is rotated that rotates the spindle till it touches the component completely. Two types of scales are used to measure the reading of micrometer, one is sleeve scale and other is thimble scale. Spindle moves toward component by 0.5 mm in or 0.025 in on every one rotation of spindle. There are three types of micrometer
Least count of micrometer:
Minimum measurement of any measuring device is the least count of that device. So, the least count for normal micrometer is 0.01 mm or 0.001 inches.
The micrometer is called vernier micrometer if the micrometer is provided with the vernier scale. The least count of vernier micrometer scale is 0.0001 inches.
Hence the micrometer with vernier graduation can measure reading to the nearest 0.0001 inches.
Thu, the statement is true.
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 73.5 mm, find the cutting ratio. If the rake angle is 15 deg, what is the shear angle?
The cutting ratio is 0.027, and the shear angle is 88.46 degrees.
The Breakdown
- Initial diameter of the bar: 75 mm
- Final diameter of the bar after cutting: 73 mm
- Mean length of the cut chip: 73.5 mm
- Rake angle: 15 degrees
Calculate the cutting ratio.
Cutting ratio = (Initial diameter - Final diameter) / Mean length of the cut chip
Cutting ratio = (75 mm - 73 mm) / 73.5 mm
Cutting ratio = 0.027
Calculate the shear angle.
The shear angle (φ) can be calculated using the following formula:
tan(φ) = (1 - cutting ratio) / (cutting ratio × cos(α))
Where:
α = Rake angle (in radians)
Substituting the given values:
α = 15 degrees = 15 × π/180 = 0.2618 radians
Cutting ratio = 0.027
tan(φ) = (1 - 0.027) / (0.027 × cos(0.2618))
φ = tan-¹(0.9730 / 0.0265)
φ = 88.46 degrees
Therefore, the cutting ratio is 0.027, and the shear angle is 88.46 degrees.
What are the main renewable energy sources? Why are ocean, wave, and tidal energies not considered as main renewable sources?
Explanation:
Renewable energy -
The energy source that does not get exhaust after using it , and can naturally replenish themselves .
These source of energy is naturally available and can be used with out limitation of being getting over .
The major types of renewable energy sources are as follows -
Geothermal Solar Wind Hydropower BiomassOcean , tide and wave are not the main renewable source , because , these are available only for certain time period , as tidal energy can be used only during high tides , similarly with the ocean and wave .
Answer:
Tidal energy and wave energy are considered renewable resources because tides are controlled by the moon, and the moon will constantly raise and lower the water. This is why tidal energy and wave energy are considered renewable resources.
Explanation:
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.
Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth are on each gear?
Answer:50 , 20
Explanation:
Given
Diametrical Pitch[tex]\left ( P_D\right )=\frac{T}{D}[/tex]
where T= no of teeths
D=diameter
module(m) of gears must be same
[tex]m=\frac{D}{T}=\frac{1}{P_D}=0.1[/tex]
Let [tex]T_1 & T_2[/tex]be the gears on two gears
Therefore Center distance is given by
[tex]m\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]
thus
[tex]0.1\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]
[tex]T_1+T_2=70----1[/tex]
and Velocity ratio is given by
[tex]VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear} [/tex]
[tex]2.5=\frac{T_1}{T_2}----2[/tex]
From 1 & 2 we get
[tex]T_1=50, T_2=20[/tex]
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8 seconds? If the net tangential force applied to the racer is constant, what is its value?
Answer:
Impulse =14937.9 N
tangential force =14937.9 N
Explanation:
Given that
Mass of car m= 800 kg
initial velocity u=0
Final velocity v=390 km/hr
Final velocity v=108.3 m/s
So change in linear momentum P= m x v
P= 800 x 108.3
P=86640 kg.m/s
We know that impulse force F= P/t
So F= 86640/5.8 N
F=14937.9 N
Impulse force F= 14937.9 N
We know that
v=u + at
108.3 = 0 + a x 5.8
[tex]a=18.66\ m/s^2[/tex]
So tangential force F= m x a
F=18.66 x 800
F=14937.9 N
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=
In electric heaters, electrical energy is converted to potential energy. a)-True b)-false?
Answer:
False
Explanation:
In electric heater electric energy is converted into heat energy. In heater wires are present which have resistance and current is flow in heater when we connect the heater to supply.
And we know that whenever current is flow in any resistance then heat is produced so in electric heaters electric energy is converted into heat energy
So this is a false statement
Lets assume, a represents the edge length (lattice constant) of a BCC unit cell and R represents the radius of the atom in the unit cell. Draw a BCC unit cell and show the atoms in the unit cell. Derive the relationship between the a and R.
Answer:
[tex]4\ R=\sqrt 3\ a[/tex]
Explanation:
Given that
Lattice constant = a
Radius of unit cell cell =R
Atom is in BCC structure.
In BCC unit cell (Body centered cube)
1.Eight atoms at eight corner of cube which have 1/8 part in each cube.
2.One complete atom at the body center of the cube
So the total number of atoms in the BCC
Z= 1/8 x 8 + 1 x 1
Z=2
In triangle ABD
[tex]AB^2=AD^2+BD^2[/tex]
[tex]AB^2=a^2+a^2[/tex]
[tex]AB=\sqrt 2\ a[/tex]
In triangle ABC
[tex]AC^2=AB^2+BC^2[/tex]
AC=4R
BC=a
[tex]AB=\sqrt 2\ a[/tex]
So
[tex]16R^2=2a^2+a^2[/tex]
[tex]4\ R=\sqrt 3\ a[/tex]
So the relationship between lattice constant and radius of unit cell
[tex]4\ R=\sqrt 3\ a[/tex]
A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displacement if the material is structural steel.
Answer:
stress = 50MPa
Explanation:
given data:
Length of strain guage is 5mm
displacement[tex] \delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm[/tex]
stress due to displacement in structural steel can be determined by using following relation
[tex]E =\frac{stress}{strain}[/tex]
[tex]stress = E \times strain[/tex]
where E is young's modulus of elasticity
E for steel is 200 GPa
[tex]stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}[/tex]
stress = 50MPa
A container ship is 240 m long and 22 m wide. Assume that the shape is like a rectangular box. How much mass does the ship carry as load if it is 10 m down in the water and the mass of the ship itself is 30 000 tonnes?
Answer:
22800 tonne
Explanation:
Given:
Length of the container, L = 240 m
Width of the container, B = 22 m
Depth inside the water, H = 10 m
Mass of the ship, m = 30000 tonnes
Now,
Total immersed volume of the ship = LBH = 240 × 22 × 10 = 52800 m³
From the Archimedes principle, we have
Total mass of the ship (i.e mass of the ship along with the load carried)
= Mass of the volume of water displaced by ship
= 52800 × Density of water
also,
Density of water = 1000 kg/m³
thus,
Total mass of the ship (i.e mass of the ship along with the load carried)
= 52800 × 1000 kg
also,
1 tonne = 1000 kg
thus,
Total mass of the ship (i.e mass of the ship along with the load carried)
= 52800 tonne
Therefore,
the load carried by the ship = Total mass of the ship - mass of ship
or
the load carried by the ship = 52800 - 30000 = 22800 tonne
There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.12/kWh where 1 kWh=3600 kJ c. Oil Heating: $2.30/gallon where 1 gal of oil=138,500 kJ. Which option is the cheapest for this house?
Answer:
Option ‘a’ is the cheapest for this house.
Explanation:
Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.
Given:
Three methods are given to heat a particular house are as follows:
Method (a)
Through Gas, this gives energy of amount $1.33/therm.
Method (b)
Through electric resistance, this gives energy of amount $0.12/KWh.
Method (c)
Through oil, this gives energy of amount $2.30/gallon.
Calculation:
Step1
Change therm to kj in method ‘a’ as follows:
[tex]C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})[/tex]
[tex]C_{1}=1.2606\times10^{-5}[/tex] $/kj.
Step2
Change kWh to kj in method ‘b’ as follows:
[tex]C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})[/tex]
[tex]C_{2}=3.334\times10^{-5}[/tex] $/kj.
Step3
Change kWh to kj in method ‘c’ as follows:
[tex]C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})[/tex]
[tex]C_{3}=1.66\times10^{-5}[/tex] $/kj.
Thus, the method ‘a’ has least cost as compare to method b and c.
So, option ‘a’ is the cheapest for this house.
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.
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.
2–25 Consider a medium in which the heat conduction equation is given in its simplest form as
d^2T/dx^2 + d^2T/dy^2 = 1/a dT/dt
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or variable?
Answer:
d) Is the thermal conductivity of the medium constant or variable.
Explanation:
As we know that
Heat equation with heat generation at unsteady state and with constant thermal conductivity given as
[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]
With out heat generation
[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]
In 2 -D with out heat generation with constant thermal conductivity
[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]
Given equation
[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}[/tex]
So we can say that this is the case of with out heat generation ,unsteady state and with constant thermal conductivity.
So the option d is correct.
d) Is the thermal conductivity of the medium constant or variable.
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.
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
A piston-cylinder assembly contains a two-phase liquid-vapor mixture of H20 at 220 lbf/in^2 with a quality of 75%. The mixture is heated and expands at constant pressure until a final temperature of 475°F is reached. Determine the work for the process, in Btu per lb of H2O present.
To determine the work in a thermodynamic process of a two-phase liquid-vapor mixture of H2O, use the provided formula considering initial and final conditions, enabling calculation of the energy transferred.
Explanation:In this thermodynamic process, the work done can be calculated using the area under the constant pressure line on a P-v diagram. Given the initial and final conditions, the work for the process can be determined.
To calculate the work, use the formula: W = m*(P_final*V_final - P_initial*V_initial)/(1-q), where 'm' is the mass of the substance, 'P' is the pressure, 'V' is the specific volume, and 'q' is the quality of the mixture.
Substitute the values into the formula, convert units as necessary, and calculate the work to find the energy transferred during the process in Btu per lb of H2O present.
Convert the velocity of a mower v = 7,943 cm/min to inches/s.
Answer:
Velocity in inch per second will be 52.11 inch/sec
Explanation:
We have given velocity = 7943 cm/min
We have to convert this velocity into inches/sec
We know that 1 cm = 0.3937 inch
So 7943 cm = 7943×0.3937=3127.1193inch
And 1 minute = 60 sec
So [tex]7943cm/min =\frac{7943\times 0.3937inch}{60sec}=52.11inch/sec[/tex]
So velocity in inch per second will be 52.11 inch/sec
Can anyone answer this question
entirely,i.e. work it out and explain it?
A flywheel has a radius of 600 mm, a mass of 144 kg,and a
radius of gyration of 450 mm.. An 18 kg block A is attachedto a
wire that is wrapped around the flywheel, and the system isreleased
from rest. Neglecting the effect of friction,determine (a) the
acceleration of block A, (b) the speed ofblock A after it has moved
1.8 m.
Answer:
a) 2.18 m/s^2
b) 9.83 m/s
Explanation:
The flywheel has a moment of inertia
J = m * k^2
Where
J: moment of inertia
k: radius of gyration
In this case:
J = 144 * 0.45^2 = 29.2 kg*m^2
The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.
T = p * r
r is the radius of the wheel.
T = m1 * g * r
T = 18 * 9.81 * 0.6 = 106 N*m
The torque will cause an acceleration on the flywheel:
T = J * γ
γ = T/J
γ = 106/29.2 = 3.63 rad/s^2
SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:
at = γ * r
at = 3.63 * 0.6 = 2.81 m/s^2
Now that we know the acceleration of the block we can forget about the flywheel.
The equation for uniformly accelerated movement is:
X(t) = X0 + V0*t + 1/2*a*t^2
We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:
X(t) = 1/2*a*t^2
Rearranging
t^2 = 2*X(t)/a
[tex]t = \sqrt{\frac{2*X(t)}{a}}[/tex]
[tex]t = \sqrt{\frac{2*18}{2.81}} = 3.6 s[/tex]
It will reach the 1.8 m in 3.6 s.
Now we use the equation for speed under constant acceleration:
V(t) = V0 + a*t
V(3.6) = 2.81 * 3.6 = 9.83 m/s
Vibration analysis is a technique adopted under: Select one: 1. General Maintenance 2. Predictive Maintenance 3. Proactive Maintenance 4. Preventive Maintenance 5. Breakdown Maintenance
Answer:
2. Predictive Maintenance
Explanation:
Although definitions differ among authors, it is generally accepted that predictive maintenance uses different kinds of techniques to monitor critical machines to prevent them from failing unexpectedly and causing losses in production (or a service), and many more unpleasant events.
Among, thermography, tribology, ultrasonics, and others, vibration analysis is one of the techniques into predictive maintenance, and since most plant types of equipment are mechanical, this is the primary maintenance technique in predictive maintenance.
In general, vibration analysis first needs to acquire data (making use of vibration monitoring using transductors, like accelerometers). Then, the time-domain data is converted into frequency-domain data using a mathematical technique called Fast Fourier Transform (FFT).
Consequently, for each machine's anomaly, there will be a unique 'signature' in the frequency-domain data that corresponds to it.
For example, if the machine presents some imbalance, then there will be a typical frequency (primary frequency) and multiples of it (harmonics), in that frequency-domain data, unique for this imbalance, and so for other machine elements' anomalies, like misalignment, rolling-element bearings high vibrations, bent shafts, and many more.