Answer:
a) 333,33 b)291,67
Explanation:
you have both equations :
C=50+0.5V and V=2500-10C
and you want to know the variable V so, you can calculate V in function of C, and you have already clear the variable "C", then you replace for (a):
V=2500-10CV=2500-10(50+0.5V)V=2500-500-5VV+5V=2500-5006V=2000V=2000/6V=333,33for b) they tell you that you increase your constant of the equiation C=50+0.5V (remember that the constant is the one alone, wihout any variable, in this case "50") increase in 25 so, your equiation for this point of C is C=75+0.5V, then you do the same:
V=2500-10CV=2500-10(75+0.5V)V=2500-750-5VV+5V=2500-7506V=1750V=1750/6V=291,67It's a system of two equations and two variables, wich gives you a compatible define system, it gives you only one solution.
I hope it helps you.
To determine the volume of traffic across the bridge, we can solve an equation relating cost and traffic volume. The volume of traffic across the bridge is approximately 333 vehicles per day. If a toll of 25 cents is added, the volume of traffic across the bridge is approximately 292 vehicles per day.
Explanation:To determine the volume of traffic across the bridge, we need to solve the equation for V. The equation relating cost and volume of traffic is given by V = 2500 - 10C. Substituting the cost function C = 50 + 0.5V into the equation, we get V = 2500 - 10(50 + 0.5V). Simplifying the equation, we find V = 2500 - 500 - 5V. Combining like terms, we get 6V = 2000, which yields V = 333.33. Therefore, the volume of traffic across the bridge is approximately 333 vehicles per day.
To find the volume across the bridge if a toll of 25 cents is added, we can modify the cost function to be C = 50 + 0.5V + 25 cents. We again substitute this into the equation V = 2500 - 10C and solve for V. Substituting in the new cost function, we get V = 2500 - 10(50 + 0.5V + 25). Simplifying the equation, we find V = 2500 - 10(75 + 0.5V). Continuing to simplify, we have V = 2500 - 750 - 5V. Combining like terms, we get 6V = 1750, which yields V = 291.67. Therefore, the volume across the bridge with the toll of 25 cents is approximately 292 vehicles per day.
An aluminum rod if 20 mm diameter iselongated 3.5 mm along its
longitudinal direction by a load of 25KN. If the modulus of
elasticity of aluminum is E = 70 GPa,determine the original length
of the bar.
Answer:
3.0772 m
Explanation:
Given:
Diameter of the aluminium rod, d = 20 mm = 0.02 m
Length of elongation, δL = 3.5 mm = 0.0035 m
Applied load, P = 25 KN = 25000 N
Modulus of elasticity, E = 70 GPa = 70 × 10⁹ N/m²
Now,
we have the relation
[tex]\delta L=\frac{\textup{PL}}{\textup{AE}}[/tex]
Now,
Where, A is the area of cross-section
A = [tex]\frac{\pi}{4}d^2[/tex]
or
A = [tex]\frac{\pi}{4}\times0.02^2[/tex]
or
A = 0.000314 m²
L is the length of the member
on substituting the respective values, we get
[tex]0.0035=\frac{25000\times L}{0.000314\times70\times10^9}[/tex]
or
L = 3.0772 m
A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for a 1µm-diameter methanol droplet for complete evaporation at same conditions based on the scaling analysis?
Answer:
Time taken by the [tex]1\mu m[/tex] diameter droplet is 60 ns
Solution:
As per the question:
Diameter of the droplet, d = 1 mm = 0.001 m
Radius of the droplet, R = 0.0005 m
Time taken for complete evaporation, t = 1 min = 60 s
Diameter of the smaller droplet, d' = [tex]1\times 10^{- 6} m[/tex]
Diameter of the smaller droplet, R' = [tex]0.5\times 10^{- 6} m[/tex]
Now,
Volume of the droplet, V = [tex]\frac{4}{3}\pi R^{3}[/tex]
Volume of the smaller droplet, V' = [tex]\frac{4}{3}\pi R'^{3}[/tex]
Volume of the droplet ∝ Time taken for complete evaporation
Thus
[tex]\frac{V}{V'} = \frac{t}{t'}[/tex]
where
t' = taken taken by smaller droplet
[tex]\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}[/tex]
[tex]\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}[/tex]
t' = [tex]60\times 10^{- 9} s = 60 ns[/tex]
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.
A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h in a distance of 100m? What time is required to travel this distance?
First, write down the information given and the change units if necessary (we must have similar units to operate on).
Initial speed, u = 36 km/h = 10 m/s
Final speed, v = 72 km/h = 20 m/s
Distance, s = 100 m
We know that
[tex] {v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2} [/tex]
Now, we substitute v, u, and a in the formula
[tex]v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds[/tex]
Please mark Brainliest if this helps!
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
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.
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]
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 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:
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
True False. First angle projection type used in United states.
Answer:
FALSE.
Explanation:
the correct answer is FALSE.
Projection is the process of representing the 3 D object on the flat surface.
there are four ways of representing the projection
1) First angle projection
2) second angle projection
3) third angle projection
4) fourth angle projection.
Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.
In USA people uses the third angle of projection.
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
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.
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.
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 steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual work delivered by the turbine? A heat pump has a COP of 2.2. It takes 5 kW electric power. What is the heat delivery rate to the room being heated in kW? Heat pump is used for winter heating of a room. A refrigerator takes 5 kW electric power. It extracts 3 kW of heat from the space being cooled a. What is the heat delivery rate to the surroundings in kW? b. What is the COP of the refrigerator?
Answer:
80 kW; 11 kW; 8 kW; 0.6
Explanation:
Part 1
Isentropic turbine efficiency:
[tex]\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s} [/tex]
[tex]W_{real} = \eta_t*W_s [/tex]
[tex]W_{real} = 0.8*100 kW [/tex]
[tex]W_{real} = 80 kW [/tex]
Part 2
Coefficient of performance COP is defined by:
[tex]COP = \frac{Q_{out}}{W} [/tex]
[tex]Q_{out} = W*COP[/tex]
[tex]Q_{out} = 5 kW*2.2[/tex]
[tex]Q_{out} = 11 kW[/tex]
Part 3
(a)
Energy balance for a refrigeration cycle gives:
[tex]Q_{in} + W = Q_{out} [/tex]
[tex]3 kW + 5 kW = Q_{out} [/tex]
[tex]8 kW = Q_{out} [/tex]
(b)
[tex]COP = \frac{Q_{in}}{W} [/tex]
[tex]COP = \frac{3 kW}{5 kW} [/tex]
[tex]COP = 0.6 [/tex]
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.
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
The Viscosity of Fluid: a)- resistance to flow b)- a measure of internal shear c)- depends upon shear forces and velocity profile d)- answers 1 and 3
Answer:
a)Resistance to flow
Explanation:
Viscosity of fluid:
Resistance to flow is called as viscosity of fluid.It is also know as fluid friction.It try to stop the flow of fluid.Viscosity of fluid is a property of fluid and it does mot depends on type of flow .
If fluid viscosity is high it means that it have very low flow ability.And opposite to viscosity is called fluidity.If fluidity is high then it means that it have low viscosity.
Viscosity are of two type
1.Dynamic viscosity
2.Kinematic viscosity
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.
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.
To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are fJK = 0.40 and fik = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack of plywood to reach the end of the bed in 0.4 s.
Answer:
a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
Explanation:
The stack of plywood has a certain mass. The weight will depend on that mass.
w = m * g
There will be a normal reaction between the stack and the bed of the truck, this will be:
nr = -m * g * cos(θ)
Being θ the tilt angle of the bed.
The static friction force will be:
ffs = - m * g * cos(θ) * fJK
The dynamic friction force will be:
ffd = - m * g * cos(θ) * fik
These forces would produce accelerations
affs = -g * cos(θ) * fJK
affd = -g * cos(θ) * fik
affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)
affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)
These accelerations oppose to movement and must be overcome by another acceleration to move the stack.
The acceleration of the truck is horizontal, the horizontal component of these friction forces is:
affs = -3.92 * cos(θ)^2
affd = -2.94 * cos(θ)^2
The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
Assuming the bed has a lenght L.
The horizontal movement will be over a distance cos(θ) * L because L is tilted.
Movement under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.
If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.
L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2
2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)
a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
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]
Find the difference between the first and third angle projection type.
Answer:
First angle projection
1.Object is between observer and plane of projection.
2.Projection of object take on first quadrant.
3.Plane of projection is assume non transparent.
Third angle projection:
1.Plane of projection is between observer and object.
2.Projection of object take on third quadrant.
3.Plane of projection is assume transparent.
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
A 15,000lb freight car is pulled along a horizontal track.
Ifthe car starts from rest and attains a velocity of 40ft/s
aftertraveling a distance of 300ft.
determine the total work done on the car by the towing forcein
this distance if the rolling frictional force between the carand
the track is 80lb.
Answer:
12024000 lb*ft
Explanation:
The total work will be the sum of the energy consumed by friction plus the kinetic energy the car attained.
L = Ek + Lf
Lf = Ff * d
Ek = 1/2 * m * v^2
Therefore:
L = Ff * d + 1/2 * m * v^2
L = 80 * 300 + 1/2 * 15000 * 40^2 = 12024000 lb*ft
THe total work done on the car is of 12024000 lb*ft
Two wafer sizes are to be compared: a 156-mm wafer with a processable area = 150mm diameter circle and a 312-mm wafer with a processable area = 300mm diameter circle. The IC chips in both cases are square with 10 mm on a side? Assume the cut lines (streets) between chips are of negligible width. What is the percent increase in (a) Wafer diameter, (b) processable wafer area, and (c) number of chips for the larger wafer size?
Answer:
a) 100%
b) 300%
c) 301 %
Explanation:
The first wafer has a diameter of 150 mm.
The second wafer has a diameter of 300 mm.
The second wafer has an increase in diameter respect of the first of:
((300 / 150) - 1) * 100 = 100%
The first wafers has a processable area of:
A1 = π/4 * D1^2
The scond wafer has a processable area of:
A2 = π/4 * D2^2
The seconf wafer has a increase in area respect of the first of:
(A2/A1 * - 1) * 100
((π/4 * D2^2) / (π/4 * D1^2) - 1) * 100
((D2^2) / (D1^2) - 1) * 100
((300^2) / (150^2) - 1) * 100 = 300%
The area of a chip is
Ac = Lc^2
So the chips that can be made from the first wafer are:
C1 = A1 / Ac
C1 = (π/4 * D1^2) / Lc^2
C1 = (π/4 * 150^2) / 10^2 = 176.7
Rounded down to 176
The chips that can be made from the second wafer are:
C2 = A2 / Ac
C2 = (π/4 * D2^2) / Lc^2
C2 = (π/4 * 300^2) / 10^2 = 706.8
Rounded down to 706
The second wafer has an increase of chips that can be made from it respect of the first wafer of:
(C2 / C1 - 1) * 100
(706 / 176 - 1) *100 = 301%
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
Air enters a 34 kW electrical heater at a rate of 0.8 kg/s with negligible velocity and a temperature of 60 °C. The air is discharged at a height 50 m above the inlet at a temper-ature of 200 °C and a velocity of 50 m / s. What is the work done in the heater?
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
The roof of a car in a parking lot absorbs a solar radiant flux of 800 W/m2, and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is 12 W/m2·K. (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steady state conditions if the ambient air temperature is 20°C. (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is 0.8.
The roof temperature is calculated to be 93°C without radiation or 86.67°C when accounting for radiation heat transfer with the surroundings.
Given: q = 800 W/m2 h = 12 W/m2∙K T∞ = 293 K
Convection heat transfer equation: q = hA(Ts - T∞)
Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K)
Distribute the 12 W/m2∙K: 800 W/m2 = 12 W/m2∙K * Ts - 12 W/m2∙K * 293 K
Group the Ts terms: 800 W/m2 = 12 W/m2∙K * Ts - 3,516 W/m2
Add 3,516 W/m2 to both sides: 4,316 W/m2 = 12 W/m2∙K * Ts
Divide both sides by 12 W/m2∙K: Ts = 4,316 W/m2 / 12 W/m2∙K
Calculate:
Ts = 366 K = 93°C
(a) Without radiation: Heat transfer equation: q = hA(Ts - T∞)
Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K) 800 = 12(Ts - 293) Ts = 800/12 + 293 Ts = 366 K = 93°C
(b) With radiation: Heat transfer equation:
q = hA(Ts - T∞) + εσA(Ts4 - Tsur4)
Given: T∞ = 293 K ε = 0.8
σ = 5.67x10-8 W/m2∙K4
Radiation heat transfer equation: q = εσA(Ts4 - Tsur4)
Assume: Tsur = T∞ = 293 K
Plug in values: q = 0.8 * 5.67x10-8 * A * (Ts4 - (293)4)
(293)4 evaluation: (293)4 = 293 x 293 x 293 x 293 = 2.97 x 108
Plug this into equation:
q = 0.8 * 5.67x10-8 * A * (Ts4 - 2.97x108)
800 = 12(Ts - 293) + 0.8*5.67x10-8(Ts4 - 2.97x108)
Solve for Ts: Ts = 359.8 K = 86.67°C
Therefore, with radiation the roof temperature is 86.67°C.
Part A: The temperature of the roof under steady-state conditions without considering radiation exchange is 86.67°C.
Part B: The temperature of the roof under steady-state conditions considering radiation exchange with an emissivity of 0.8 is 81.17°C.
Part A: Neglecting Radiation Exchange
Step 1
Under steady-state conditions, the heat absorbed by the roof [tex](\( Q_{\text{absorbed}} \))[/tex] is equal to the heat lost through convection [tex](\( Q_{\text{convection}} \))[/tex].
Given:
- Solar radiant flux, [tex]\( q_{\text{solar}} = 800 \, \text{W/m}^2 \)[/tex]
- Convection coefficient, [tex]\( h = 12 \, \text{W/m}^2 \cdot \text{K} \)[/tex]
- Ambient air temperature, [tex]\( T_{\infty} = 20^\circ \text{C} \)[/tex]
The absorbed heat:
[tex]\[ Q_{\text{absorbed}} = q_{\text{solar}} \][/tex]
The heat loss by convection:
[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]
At steady state:
[tex]\[ q_{\text{solar}} = h (T_s - T_{\infty}) \][/tex]
Step 2
Solving for the surface temperature [tex]\( T_s \)[/tex]:
[tex]\[ 800 = 12 (T_s - 20) \][/tex]
[tex]\[ T_s - 20 = \frac{800}{12} \][/tex]
[tex]\[ T_s - 20 = 66.67 \][/tex]
[tex]\[ T_s = 86.67^\circ \text{C} \][/tex]
So, the temperature of the roof under steady-state conditions without radiation exchange is 86.67°C.
Part B: Considering Radiation Exchange
Step 1
When considering radiation exchange, the roof loses heat through both convection and radiation. The net radiative heat loss is given by the Stefan-Boltzmann law.
Given:
- Emissivity, [tex]\( \varepsilon = 0.8 \)[/tex]
- Stefan-Boltzmann constant, [tex]\( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \)[/tex]
The total heat loss (convection + radiation):
[tex]\[ Q_{\text{total}} = Q_{\text{convection}} + Q_{\text{radiation}} \][/tex]
For convection:
[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]
For radiation:
[tex]\[ Q_{\text{radiation}} = \varepsilon \sigma (T_s^4 - T_{\infty}^4) \][/tex]
At steady state:
[tex]\[ q_{\text{solar}} = h (T_s - 20) + \varepsilon \sigma (T_s^4 - 293.15^4) \][/tex]
[tex]\[ 800 = 12 (T_s - 20) + 0.8 \times 5.67 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]
Step 2
To simplify:
[tex]\[ 800 = 12 (T_s - 20) + 4.536 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]
This equation is nonlinear and needs to be solved iteratively. Let's outline the steps to solve it without detailed iteration steps:
1. Initial Guess: Start with an initial guess for [tex]\( T_s \)[/tex].
2. Iteration: Adjust [tex]\( T_s \)[/tex] until both sides of the equation are equal.
Through iteration or numerical methods, we find:
[tex]\[ T_s \approx 81.17^\circ \text{C} \][/tex]
In summary, the temperature of the roof is 86.67°C without considering radiation, and it drops to approximately 81.17°C when radiation exchange is taken into account.