Answer:
Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.
Explanation:
Step1
Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.
Step2
All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.
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]
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There is 120 lbm. of saturated liquid water in a steel tank at 177 oF. What is the pressure and volume of the tank?
Answer:
P=7.027psi
V=1.9788ft ^3
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
For the first part of this problem, we must find the water saturation pressure at a temperature of 177F
Psat @(177F)=7.027psi
For the second part of this problem, we calculate the specific volume of the water knowing the temperature and that the state is saturated liquid, then multiply by the mass to know the volume
v@177F=0.01649ft^3/lbm
V=mv
V=(0.01649ft^3/lbm)(120lbm)
V=1.9788ft ^3
Dfine factor of safety and its significance
Explanation:
Step1
Factor of safety is the constant factor which is taken for the safe design of any product. It is the ratio of maximum stress induced in the material or the failure stress from tensile test to the allowable stress.
Step2
It is an important parameter for design of any component. This factor of safety is taken according to the type of material, environment condition, strength needed, type of component etc.
The expression for factor of safety is given as follows:
[tex]FOS=\frac{\sigma_{f}}{\sigma_{a}}[/tex]
Here, [tex]\sigma_{f}[/tex] is fracture stress and [tex]\sigma_{a}[/tex] is allowable stress.
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 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
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.
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.
A person, 175 lbm, wants to fly (hoover) on a 4 lbm skateboard of size 2 ft by 0.8 ft. How large a gauge pressure under the board is needed?
Answer:
[tex]p = 15260.643 \ lbf/ft^2[/tex]
Explanation:
person weight is 175 lbm
weight of stake board 4lbm
size of stakeboard = 2ft by 0.8 ft
area of stakeboard is [tex]2*0.8 ft^2 = 1.6 ft^2[/tex]
gauge pressure is given as
[tex]p =\frac{ w_p+w_s}{A} g[/tex]
where is g is acceleration due to gravity = 32.17 ft/sec^2
puttng all value to get pressure value
[tex]= \frac{175 +4}{1.6}* 32.17[/tex]
[tex]p = 15260.643 \ lbf/ft^2[/tex]
Radioactivity of C-14 is used for dating of ancient artifacts. Archeologist determined that 20% of initial amount of C-14 has remained. Estimate the age of this artifact.
Answer:
13282.3 years
Explanation:
The C-14 decays exponentially:
[tex]\frac{dN}{dt} =λ*N[/tex]
The solution for this equation is
[tex]N= N_{o}*e^{- λt}[/tex]
Where:
No = atom number of C-14 in t=0
N = atom number of C-14 now
I= radioactive decay constant
clearing t this equation we get:
[tex]t=-\frac{1}{ λ}*ln\frac{N}{N_{o}}[/tex]
The term 1/I is called half-life and the value for C-14 is 8252 years.
N for this exercise is 0.2No
[tex] t= -8033 * ln \frac{0.2N_{o} }{N_{o}}[/tex]
t = 13282.3 years
What is the maximum % carbon for structural steel?
Answer:
The large percentage of steel includes less than 0.35% carbon
Explanation:
Carbon is perhaps the most important business alloy of steel. Raising carbon material boosts strength and hardness and enhances toughness. However, carbon often increases brittleness.
The large percentage of steel includes less than 0.35% carbon. Any steel with a carbon content range of 0.35% to 1.86% can be considered as hardened.
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%
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)
A rectangular plate casting has dimensions 200mm x 100mm x 20mm. The riser for this sand casting mold is in the shape of a sphere. The casting takes 3.5mins to solidify. Calculate the diameter of the riser so that it takes 20% longer for the riser to solidify
Answer:
Diameter of riser =6.02 mm
Explanation:
Given that
Dimensions of rectangular plate is 200mm x 100mm x 20mm.
Volume of rectangle V= 200 x 100 x 20 [tex]mm^3[/tex]
Surface area of rectangle A
A=2(200 x 100+100 x 20 +20 x 200)[tex]mm^2[/tex]
So V/A=7.69
We know that
Solidification times given as
[tex]t=K\left(\dfrac{V}{A}\right)^2[/tex] -----1
Lets take diameter of riser is d
Given that riser is in spherical shape so V/A=d/6
And
Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.
Lets take [tex]\dfrac{V}{A}=M[/tex]
[tex]\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}[/tex]
Now from equation 1
[tex]\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2[/tex]
So by solving this d=6.02 mm
So the diameter of riser is 6.02 mm.
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-position and 2.5m/s when 225mm from mid-position. Find : i) It's max amplitude ii) Max acceleration iii) The periodic time iv) The frequency of oscillation.
Answer:
1) A=282.6 mm
2)[tex]a_{max}=60.35\ m/s^2[/tex]
3)T=0.42 sec
4)f= 2.24 Hz
Explanation:
Given that
V=3.5 m/s at x=150 mm ------------1
V=2.5 m/s at x=225 mm ------------2
Where x measured from mid position.
We know that velocity in simple harmonic given as
[tex]V=\omega \sqrt{A^2-x^2}[/tex]
Where A is the amplitude and ω is the natural frequency of simple harmonic motion.
From equation 1 and 2
[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex] ------3
[tex]2.5=\omega \sqrt{A^2-0.225^2}[/tex] --------4
Now by dividing equation 3 by 4
[tex]\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}[/tex]
[tex]1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}[/tex]
So A=0.2826 m
A=282.6 mm
Now by putting the values of A in the equation 3
[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]
[tex]3.5=\omega \sqrt{0.2826^2-0.15^2}[/tex]
ω=14.609 rad/s
Frequency
ω= 2πf
14.609= 2 x π x f
f= 2.24 Hz
Maximum acceleration
[tex]a_{max}=\omega ^2A[/tex]
[tex]a_{max}=14.61 ^2\times 0.2826\ m/s^2[/tex]
[tex]a_{max}=60.35\ m/s^2[/tex]
Time period T
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{14.609}[/tex]
T=0.42 sec
In laminar now, fluid particles are constrained to motion in (parallel —perpendicular opposite) paths by the action of (temperature- viscosity —pressure).
Answer:
parallel ; Viscosity
Explanation:
Laminar is flow is the flow of fluid layer in which motion of the liquid particle is very slow and there is no intermixing of the layer of fluid takes place.
There no perpendicular movement of particles takes place, no eddies are formed or swirl of fluid.
so, the first option will be Parallel. Fluid particles flow parallel to each other in laminar flow.
This path of flow is by the action of Viscosity of the fluid.
A water filled vertical u-tube manometer is used to measure pressure changes in a reaction vessel. Assuming that the change in height of the manometer can be measured to a precision of 1/16" of an inch, how accurately can pressure changes (in psi) be measured?
Answer:
The pressure will be measured to a precision of 0.073 psi.
Explanation:
Since the relation between the measurement of pressure and height is given by
[tex]dP=\rho gh[/tex]
For water we have
[tex]\rho _{water}=62.4lb/ft^3[/tex]
[tex]g=32.17ft/s^2[/tex]
Applying the given values we get
[tex]dP=62.4\times 32.17\times \frac{1}{16\times 12}=5.433lb/ft^2\\\\=\frac{10.455}{144}lb/in^2=0.073psi[/tex]
Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by hot air. The hot air enters at 300°F and is to leave at 200°F. Calculate the total lb mol air/h needed. The mean heat capacity of the oil is 0.45 btu/lbm · °F.
Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT. Finally, divide the heat transfer by the heat capacity of air to find the total lb mol air/h needed.
Explanation:To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, we need to calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the temperature difference. We know that the heat exchanger transfers 2200 lbm/h of oil from 100°F to 150°F, so the total heat transfer is Q = 2200 lbm/h * (150°F - 100°F) * 0.45 btu/lbm • °F. Next, we can calculate the lb mol of air needed by dividing the heat transfer by the heat capacity of air, which is 0.24 btu/lbm • °F. Therefore, the total lb mol air/h needed is Q / (0.24 btu/lbm • °F).
Give a reason why fighter aircraft use mid-wing design.
Explanation:
Mid-wing configuration places wings exactly at midline of airplane which means at half of height of fuselage. These airplanes are well balanced and also they have a large control surface area.It is the best option aerodynamically as these planes are streamlined much more and also has low interference drag as compared to the high and the low wing configurations.
The mid-wing has almost neutral roll stability that is further good from prespective of the combat as well as the aerobatic aircraft as mid-wing allows for performance of the rapid roll maneuvers with the minimum yaw coupling.
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]
If the shearing stress is linearly related to the rate of shearing strain for a fluid, it is a_____ fluid. What are other types of fluids and how does their rate of shearing strain relate to shearing stress?
Answer:
If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.
Explanation:
The other types of fluids are:
1) Non-Newtonian fluids which are further classified as
a) Thixotropic Fluid: Viscosity decreases with shearing stress over time.
b) Rehopactic Fluid: Viscosity increases with shearing stress over time.
c) Dilatant Fluid: Apparent viscosity increases with increase in stress.
d) Pseudoplastic: Apparent viscosity decreases with increase in stress.
Answer:
If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.
Explanation:
A rigid, sealed cylinder initially contains 100 lbm of water at 70 °F and atmospheric pressure. Determine: a) the volume of the tank (ft3 ). Later, a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibriu
Answer:
Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...
What is 100 kPa in psia?
Answer:
The pressure in pounds per square inch (psi) equals 14.50 psi.
Explanation:
Since psi stands for 'pounds per square inch' and Pascals stands for 'Newtons per square meters'
We can write the given pressure as
[tex]Pressure=100kPa\\\\Pressure=100\times 10^{3}N/m^{2}[/tex]
Now since 1 Newton of force equals 0.22481 Pound force also
since 1 meter equals 39.37 inches
Using the above values we get
[tex]Pressure=100\times 10^{3}\times \frac{0.22481lb}{(39.37)^{2}inch^{2}}[/tex]
thus the pressure becomes
[tex]Pressure=100\times 10^{3}\times \frac{0.22481}{(39.31)^{2}}pounds/inch^{2}\\\\Pressure=14.50psi[/tex]
The temperature at the outlet of the turbocharger turbine is 260°C, with an exhaust flow rate of 2 kg/min. Estimate the drop in temperature across the turbine, given that the turbine output power is 1kW. Assume exhaust gas to have a specific heat of 1.05 kJ/kg.K, and an ambient temperature of 25°C
Answer:
[tex]\Delta T = 28.57°C[/tex]
Explanation:
given data:
temperature at outlet of turbine = 260°C
Flow rate = 2 kg/min = 0.033 kg/s
output power = 1 kW
specific heat = 1.05 kJ/kg.K
We know that power generated by turbine is equal to change in enthalpy
[tex]W = h_2 - h_1[/tex]
[tex]= mCp(\Delta T)[/tex]
[tex]1*10^3 = 0.0333*1.05*10^3 * \Delta T[/tex]
[tex]\Delta T = \frac{10^3}{0.033*1.05*10^3}[/tex]
[tex]\Delta T = 28.57°C[/tex]
What is the primary structural difference between cantilever wing and semi-cantilever wing?
Answer:
The main component of fixed wing aircraft
1.Wings
2.Fuselages
3.Landing gear
4.Stabilizers
5.Flight control surface
Cantilever wing:
A cantilever wing is directly attached to the fuselages and do not have any external support.
Semi cantilever wing:
A cantilever wing does not directly attached to the fuselages and have any external support.The external support may be one two.
An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust from the engine is along the flight direction. Given the weight of the aircraft as 50,000 lb and L/D of 10, determine the lift, drag and thrust required for the given equilibrium flight. Assume g =32.2 ft/s^2
Answer:
L= 50000 lb
D = 5000 lb
Explanation:
To maintain a level flight the lift must equal the weight in magnitude.
We know the weight is of 50000 lb, so the lift must be the same.
L = W = 50000 lb
The L/D ratio is 10 so
10 = L/D
D = L/10
D = 50000/10 = 5000 lb
To maintain steady speed the thrust must equal the drag, so
T = D = 5000 lb
Which of the two materials (brittle vs. ductile) usually obtains the largest modulus of toughness and why?
Answer:
The modulus of toughness is greater for ductile material.
Explanation:
Modulus of toughness is defined as the amount of strain energy that a material stores per unit volume. It is equal to the area under stress-strain curve of the material up to the point of fracture.
As we know that the area under the stress-strain curve of a ductile material is much more than the area under the stress strain curve of a brittle material as brittle materials fail at a lesser load hence we conclude that the modulus of toughness is greater for ductile material than a brittle material.
In our experience we can also relate that for same volume a substance such as glass sheet (brittle) fails at lower load as compared to a sheet of steel (ductile) with identical dimensions.
Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters at 400 kPa and 300 K. Stream 3 leaves the control volume at 150 kPa and 270 K. The control volume does 3 kW of work on the surroundings while losing 5 kW of heat. Find the mass flow rate of stream 2. Neglect changes in kinetic and potential energy.
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Why is the process for making flat glass called the float process?
Explanation:
Step1
Float glass is the process of glass manufacturing on the flat surface of metal like tin. In this method molten glass is allowed to float on the surface of metal.
Step2
Float glass gives the uniform and flat surface of glass product. The thickness of the glass produced is uniform throughout. This process of glass making is very cheap and has negligible distortion. Flat glass process is called the float process because of producing high quality flat surface.
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]
Fluid flowing through a constriction or partially opened valve in a pipe system may develop localized low pressures due to increased velocity. If a pipe contains water at 210 °F, what is the minimum absolute pressure that the water can be reduced to and not have cavitation occur?
Answer:
P=41.84psi
Explanation:
Cavitation is a phenomenon that arises when the pressure of a liquid fluid reaches the vapor pressure, this means that if at any point in a system of pipes and pumps the water pressure is equal to or less than the vapor pressure it begins to evaporate, producing explosive bubbles that can cause damage and noise in the system.
Therefore, for this problem, all we have to do is find the water vapor pressure at a temperature of 270F, using thermodynamics tables
Pv@270F=41.84psi