Explanation:
The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.
Answer:
the liquid woulriekwvhrnsshsnekwb ndrhwmoadi
hmmmmmmmm i already put the photo as attachment its
Answer:
letse see
Explanation:
The boundary work is positive during an expansion process.
a.
False
b.
True
Answer:
True
Explanation:
During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system
Hence, the given statement is true
Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answer: hello your question lacks some data below is the missing data
Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol
H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.
H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min
In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be
Answer:
the surface temperature of the plates when they come out of the oven is approximately 445 °C
Explanation:
Given the data in the question;
thickness t = 3 cm = 0.03 m
so half of the thickness L = 0.015 m
thermal conductivity of brass k = 110 W/m°C
Density p = 8530 kg/m³
specific heat [tex]C_p[/tex] = 380 J/kg°C
thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s
Temperature of oven T₀₀ = 700°C
initial temperature T[tex]_i[/tex] = 25°C
time t = 10 min = 600 s
convection heat transfer coefficient h = 80 W/m².K
Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.
So, using analytical one-term approximation method, the Fourier number > 0.2.
now, we determine the Biot number for the process
we know that; Biot number Bi = hL / k
so we substitute
Bi = hL / k
Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109
Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )
The interpolation method used to find the
λ₁ = 0.1039 and A₁ = 1.0018
so
The Fourier number т = ∝t/L²
we substitute
Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²
т = 0.02034 / 0.000225
т = 90.4
As we can see; 90.4 > 0.2
So, analytical one-term approximation can be used.
∴ Temperature at the surface will be;
θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation
θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )
so we substitute
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )
θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998
θ(L,t)[tex]_{wall[/tex] = 0.3775
so we substitute into equation 1
θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)
0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )
0.3775 = ( T(L,t) - 700 ) / ( - 675 )
0.3775 × ( - 675 ) = ( T(L,t) - 700 )
- 254.8125 = T(L,t) - 700
T(L,t) = 700 - 254.8125
T(L,t) = 445.1875 °C ≈ 445 °C
Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C
what is the most common type of suspensions system used on body over frame vehicles?
Answer:
Engine
Explanation:
Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.
What is a Semi-independent suspension?Semi-independent suspension give the front wheels some individual movement.
This suspension only used in rear wheels.
Thus, the correct option is Semi-independent suspension
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The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
Values are gotten from the table named: blackbody radiation functions
a) Calculate the band emission fractions for the visible region
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
b)calculate wavelength corresponding to the maximum spectral intensity
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)
Please help me answer this engineering question
) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
3.Which of the following drawings are matched with the project specifications to form the bulk of the contract document?
If it took a 30m capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) water from the
Complete question:
If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.
Answer:
the flow rate of the water from the tank is 0.05 m³/min
Explanation:
Given;
volume of water in the tank, v = 30 m³
length of the waterproof faucet, L = 2cm = 0.02 m
duration of water flow through the tank, t = 10 hours
The flow rate of the water from the tank is calculated as;
[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]
Therefore, the flow rate of the water from the tank is 0.05 m³/min
Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.
Required:
Determine the dew point temperature of the products, in K.
_____ can be defined as the rate at which work is done or the amount of work done based on a period of time. (2 Points) voltage power resistant current
Answer: Power
Explanation:
The rate at which work is done or the amount of work done based on a period of time is referred to as power.
Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.
5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Batteries D) All of the above
I say it's D) All of the above
Answer:
D
Explanation:
All of the above
1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address
Can someone put each letter by the correct word for my automotive class !
Answer:
L = spindle
M = lower ball joint
part without the letter showing = steering knuckle
Explanation:
Which of the following are the same as 1545.5347
Select one:
a. 1.545534e: 3
b. 1545534.0e 4
c0.15455340-5
d. 154553.4e 3
Explanation:
D the answer is D and anyone can answer this Question
Based on the provided options, the scientific expression that is the same as 1545.5347 is 154553.4e3. Therefore, the correct option is option D.
A scientific expression, often known as scientific notation, is a method of representing extremely big or extremely small integers. It is often used to simplify the representation of such numbers in scientific and mathematical operations. A number is expressed in scientific notation as a product of two parts: a coefficient and a power of ten.
The coefficient is a value between 1 and 10, and the power of 10 determines how far the decimal point should be moved. The "e3" signifies multiplying the integer by 10 raised to the power of 3, which is similar to moving the decimal point three places to the right in this formula. As a result, 154553.4e3 equals 154553.4103, which simplifies to 154553400.
Therefore, the correct option is option D.
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A spherical balloon is filled with a gas. The outer diameter of the balloon is 20 in. and the thick-ness is 0.012 in. Calculate the maximum permissible pressure in the balloon if the allowable tensile stress and the allowable shear stress in the balloon are 1 ksi and 0.3 ksi, respectively.
Answer:
1.4 psi
Explanation:
Before diving into the solution to the question above, let's pick out the parameters needed in solving this problem from the question.
=> The measurement for the outer diameter of the balloon = 20 inches, the measurement for the thickness = 0.012 in, the allowable tensile stress = 1ksi and the allowable shear stress in the balloon = 0.3 ksi.
The first thing to determine is the inner diameter = 20 - 2 × 0.012 in = 19.976 in.
Therefore, the tensile stress:
1000 = k × [19.976/2]÷ 2 × 0.012 = 2.4 psi.
Also, the sheer stress which is also the maximum permissible pressure in the balloon can be calculated below as:
0.3 × 1000 = k × [19.976/2]/ 4 × 0.012 = 1.4 psi.
. Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Martensite to spheroidite (b) Spheroidite to martensite (c) Bainite to pearlite (d) Pearlite to bainite (e) Spheroidite to pearlite (f) Pearlite to spheroidite (g) Tempered martensite to martensite (h) Bainite to spheroidite
Answer:
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Explanation:
The heat treatment procedure is simply the heating of a metal to a high temperature and cooling the metal back. during this process the metal will undergo certain mechanical changes
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
