Answer:
a) [tex]a=-28m/s^{2}[/tex]
b) [tex]\frac{dT}{dx}=-5 ^{o}C/m[/tex]
Explanation:
a)
In order to solve this problem, we need to start by remembering how the acceleration is related to the velocity of a particle. We have the following relation:
[tex]a=\frac{dv}{dt}[/tex]
in other words, the acceleration is defined to be the derivative of the velocity function with respect to time. So let's take our speed function:
u=20-2x
if we take its derivative we get:
du=-2dx
this is the same as writting:
[tex]\frac{du}{dt}=-2\frac{dx}{dt}[/tex]
we also know that velocity is defined to be:
[tex]u=\frac{dx}{dt}[/tex]
so we get that:
a=-2u
when substituting we get that:
a=-2(20-2x)
when expanding we get:
a=-40+4x
and now we can use this equation to find our acceleration at x=3, so:
a=-40+4(3)
a=-40+12
[tex]a=-28 m/s^{2}[/tex]
b)
the same applies to this problem with the difference that this will be the rate of change of the temperature per m. So we proceed and take the derivative of the temperature function:
T=200-5x
[tex]\frac{dT}{dx}=-5[/tex]
so the rate of change is [tex] -5^{o}C/m [/tex]
for an electromotive force to be induced across a vertical loop from the field of an infinite length line of fixed current in the z axis the loop must be moving to?
Answer:
The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".
Explanation:
Since along that same z-axis none electromagnetic field would be triggered as being in the same orientation loop movement of them across different line portions would allow some caused emf/voltage to be canceled. And the only logical choice seems to be either x or y-axes.The magnetic field of fluctuation should indeed be changed and changed across both X as well as Y directions.So that the above is the appropriate choice.
Light acoustical panels in fire rated assemblies generally:
a. Compromise the fire rating by one hour.
b. Require hold down clips
c. Require pressure cleaning.
d. Are not allowed.
Answer:
a. Compromise the fire rating by one hour.
Explanation:
One hour fire rating is given to materials that can resist the fire exposure. The Acoustical panels controls reverberation and they are used for echo controls. The Fire rating is the passive fire protection which can resist standard fire. The test for fire rating also consider normal functioning of the material.
A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.
Answer:
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Explanation:
A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:
[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]
Output power is cleared:
[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]
If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:
[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\dot W_{out} = 1752.825\,kW[/tex]
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Consider a refrigerator that consumes 400 W of electric power when it is running. If the refrigerator runs only one-quarter of the time and the unit cost of electricity is $0.13/kWh, what is the electricity cost of this refrigerator per month (30 days)
Answer:
Electricity cost = $9.36
Explanation:
Given:
Electric power = 400 W = 0.4 KW
Unit cost of electricity = $0.13/kWh
Overall time = 1/4 (30 days) (24 hours) = 180 hours
Find:
Electricity cost
Computation:
Electricity cost = Electric power x Unit cost of electricity x Overall time
Electricity cost = 0.4 x $0.13 x 180
Electricity cost = $9.36
Given:
Electric power = 400 W = 0.4 KW
Over all Time = 30(1/4) = 7.5 days
Unit cost of electricity = $0.13/kWh
Find:
Electricity cost.
Computation:
Electricity cost = Electric power x Unit cost of electricity x Over all Time
Electricity cost = 0.4 x 0.13 x 7.5
Electricity cost = $
Q: Draw shear and bending moment diagram for the beam shown in
the figure. EI= constant
Answer:
Explanation:
Please
A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the inlet temperature if the gas is argon, helium, or nitrogen.
Given Information:
Inlet velocity = Vin = 25 m/s
Exit velocity = Vout = 250 m/s
Exit Temperature = Tout = 300K
Exit Pressure = Pout = 100 kPa
Required Information:
Inlet Temperature of argon = ?
Inlet Temperature of helium = ?
Inlet Temperature of nitrogen = ?
Answer:
Inlet Temperature of argon = 360K
Inlet Temperature of helium = 306K
Inlet Temperature of nitrogen = 330K
Explanation:
Recall that the energy equation is given by
[tex]$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $[/tex]
Where Cp is the specific heat constant of the gas.
Re-arranging the equation for inlet temperature
[tex]$ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$[/tex]
For Argon Gas:
The specific heat constant of argon is given by (from ideal gas properties table)
[tex]C_p = 520 \:\: J/kg.K[/tex]
So, the inlet temperature of argon is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 119 + 300$[/tex]
[tex]$ T_{in} = 360K $[/tex]
For Helium Gas:
The specific heat constant of helium is given by (from ideal gas properties table)
[tex]C_p = 5193 \:\: J/kg.K[/tex]
So, the inlet temperature of helium is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 12 + 300$[/tex]
[tex]$ T_{in} = 306K $[/tex]
For Nitrogen Gas:
The specific heat constant of nitrogen is given by (from ideal gas properties table)
[tex]C_p = 1039 \:\: J/kg.K[/tex]
So, the inlet temperature of nitrogen is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 60 + 300$[/tex]
[tex]$ T_{in} = 330K $[/tex]
Note: Answers are rounded to the nearest whole numbers.
A piston-cylinder device initially at 0.45-m3 contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.2 m3. The work done on the gas during this compression process is _____ kJ.
Answer:
219kJ
Explanation:
The work done (W) on a gas in an isothermal process is given by;
W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex] -----------------(i)
Where;
P₁ = initial pressure of the gas
V₁ = initial volume of the gas
V₂ = final volume of the gas
From the question;
P₁ = 600kPa = 6 x 10⁵Pa
V₁ = 0.45m³
V₂ = 0.2m³
Substitute these values into equation (i) as follows;
W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]
W = -6 x 10⁵ x 0.45 x ln (0.444)
W = -6 x 10⁵ x 0.45 x -0.811
W = 2.19 x 10⁵
W = 219 x 10³
W = 219kJ
Therefore, the work done on the gas during the compression process is 219kJ
Using models helps scientists conduct research. How else can research using models save lives?
Answer: find the answer in the explanation.
Explanation:
Some of the methods of science are hypothesis, observations and experiments.
Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c
Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.
Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.
Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.
In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.
Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.
So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.
For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger shell (just like a shell and tube heat exchanger). Consider one tube inside such a reactor that is 2.5 m long with an inside diameter of 0.025 m. The catalyst is alumina spheres with a diameter of 0.003 m. The particle density is 1300 kg/m3 and the bed void fraction is 0.38. Compute the pressure drop seen for a superficial mass flux of 4684 kg/m2hr. The feed is methane at a pressure of 5 bar and 400 K. At these conditions the density of the gas is 0.15 mol/dm-3 and the viscosity is 1.429 x 10-5 Pa s.
Answer:
the pressure drop is 0.21159 atm
Explanation:
Given that:
length of the reactor L = 2.5 m
inside diameter of the reactor d= 0.025 m
diameter of alumina sphere [tex]dp[/tex]= 0.003 m
particle density = 1300 kg/m³
the bed void fraction [tex]\in =[/tex] 0.38
superficial mass flux m = 4684 kg/m²hr
The Feed is methane with pressure P = 5 bar and temperature T = 400 K
Density of the methane gas [tex]\rho[/tex] = 0.15 mol/dm ⁻³
viscosity of methane gas [tex]\mu[/tex] = 1.429 x 10⁻⁵ Pas
The objective is to determine the pressure drop.
Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³
SO; we have :
Density = 0.15 mol/dm ⁻³
Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol
Density = [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]
Density = 2400
Density [tex]\rho_f[/tex] = 2.4 kg/m³
Density = mass /volume
Thus;
Volume = mass/density
Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³
Volume of the methane gas = 1951.666 m/hr
To m/sec; we have :
Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec
[tex]Re = \dfrac{dV \rho}{\mu}[/tex]
[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]
[tex]Re=2276.317705[/tex]
For Re > 1000
[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]
[tex]\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}[/tex]
[tex]\Delta P=8575.755212*2.5[/tex]
[tex]\Delta = 21439.38803 \ Pa[/tex]
To atm ; we have
[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]
[tex]\Delta P =0.2115903087 \ atm[/tex]
ΔP ≅ 0.21159 atm
Thus; the pressure drop is 0.21159 atm
If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the resultant moment produced by all the weights about point A.
Answer:
hello the required diagram is missing attached to the answer is the required diagram
7.9954 kip.ft
Explanation:
AB = 1550-Ib ( weight acting on AB )
BCD = 190 - Ib ( weight of cage )
169-Ib = weight of man inside cage
Attached is the free hand diagram of the question
calculate distance [tex]x![/tex]
= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]
[tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft
calculate distance x
= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]
x = 30 * cos 75⁰ = 7.765 ft
The resultant moment produced by all the weights about point A
∑ Ma = 0
Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )
Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )
= 4014.5 + 1950.35 + 2030.535
= 7995.385 ft. Ib ≈ 7.9954 kip.ft
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You close the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
[tex]T_{min} =[/tex] 26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, [tex]d_c = 0.5 mm[/tex]
Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]
minimum time required to reduce the depth of the plate by 20 mm:
[tex]T_{min} =[/tex] number of passes * Time/pass
[tex]T_{min} =[/tex] n * Time/pass
[tex]T_{min} =[/tex] 40 * 40
[tex]T_{min} =[/tex] 1600 = 26 mins 40 secs
Answer:
the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds
Explanation:
From the given information;
Assuming the tool moves 100 mm/sec
The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:
Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]
Number of passes = 20
We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times
However; the time per passes is;
Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]
where;
length L = 400mm
velocity of the feed is assumed as 100
Time/pass [tex]=\dfrac{20*400}{100}[/tex]
Time/pass = 80 sec
Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:
[tex]T_{min} = Time/pass *number of passes[/tex]
[tex]T_{min} = 20*80[/tex]
[tex]T_{min} = 1600 \ sec[/tex]
[tex]T_{min}[/tex] = 26 minutes 40 seconds
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa (293400 psi). Assume that the parameter Y has a value of 1.14. (a) If the largest surface crack is 0.2 mm (0.007874 in.) long, determine the critical stress .
Answer:
Explanation:
The formula for critical stress is
[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]
[tex]\sigma_c =\texttt{critical stress}[/tex]
K is the plane strain fracture toughness
Y is dimensionless parameters
We are to Determine the Critical stress
Now replacing the critical stress with 54.8
a with 0.2mm = 0.2 x 10⁻³
Y with 1
[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]
The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa
Therefore, the specimen does not failure for surface crack of 0.2mm
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects calculate the viscosities of the oils.
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
Cathy works in a welding shop. While working one day, a pipe falls from scaffolding above and lands on her head, injuring her. Cathy complains to OSHA, but the company argues that because it has a "watch out for falling pipe" sign in the workplace that it gave fair warning. It also says that if Cathy wasn’t wearing a hardhat that she is responsible for her own injury. Which of the following is true?1. Common law rules could hold Cathy responsible for her own injury.2. Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.3. OSHA rules can hold Cathy’s employer responsible for not maintaining a hazard-free workplace.4. More than one answer is correct.
Answer:1 common law
Explanation:
It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.
What are OSHA rules?In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.
According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.
You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.
Therefore, more than one answer is correct.
Learn more about OSHA, here:
https://brainly.com/question/13127795
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Q#1: Provide an example of a software project that would be amenable to the following models. Be specific. a. Waterfall b. Prototype c. Extreme Programming
Answer:
Waterfall model
Explanation:
The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.
The waterfall model is very easy to use. This is the earliest approach of the SDLC.
There are different phase of the waterfall:
Requirement analysisSystem DesignImplementationTestingDeploymentMaintenanceYou want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = [tex]\frac{A*L}{Vo}[/tex] Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s
since the reaction is in first order
X = 1 - [tex]e^{-kt}[/tex]
[tex]e^{-kt}[/tex] = 1 - X
kt = In [tex]\frac{1}{1-X}[/tex]
k = In [tex]\frac{1}{1-X}[/tex] / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156 [tex]s^{-1}[/tex]
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]
therefore
[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
Waste cooking oil is to be stored for processing by pouring it into tank A, which is connected by a manometer to tank B. The manometer is completely filled with water. Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa. To what height h can waste oil be poured into tank A? If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height?
KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.
==================================
Answer:
(1). 1.2 metres.
(2). There is going to be the same pressure.
Explanation:
From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;
" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."
=> Also, the density of oil = 930
That is if Pressure, P in B > 18kpa there will surely be a burst.
The height, h the can waste oil be poured into tank A is;
The maximum pressure = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).
18 × 10^3 = (height, h × 10 × 930) + 10 × (2 - 1.25) × 1000.
When we make height, h the Subject of the formula then;
Approximately, Height, h = 1.2 metres.
(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:
a. The oil mean temperature.
b. The centerline temperature.
c. The axial gradient of the mean temperature.
d. The heat transfer coefficient.
Answer:
(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K
Explanation:
Solution
Given that:
The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s
Diameter of the tube, D = 1cm =0.01 m
Electrical heat rate, q =76 W/m
Wall Temperature, Ts = 370 K
Now,
From the properties table of engine oil we can deduce as follows:
thermal conductivity, k =0.139 W/m .K
Density, ρ = 854 kg/m³
Specific heat, cp = 2120 J/kg.K
(a) Thus
The wall heat flux is given as follows:
qs = q/πD
=76/π *0.01
= 2419.16 W/m²
Now
The oil mean temperature is given as follows:
Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)
Tb =370 - 11/24 * (2419.16 * 0.005/0.139)
Tb = 330.12 K
(b) The center line temperature is given below:
Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)
Tc =304.73 K
(c) The flow velocity is given as follows:
V = m/ρ (πR²)
Now,
The The axial gradient of the mean temperature is given below:
dTb/dx = 2 *qs/ρ *V*cp * R
=2 *qs/ρ*[m/ρ (πR²) *cp * R
=2 *qs/[m/(πR)*cp
dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120
dTb/dx = 19.81 K/m
(d) The heat transfer coefficient is given below:
h =48/11 (k/D)
=48/11 (0.139/0.01)
h =60.65 W/m². K
Eight switches are connected to PORTB and eight LEDs are connected to PORTA. We would like to monitor the first two least significant bits of PORTB (use masking technique). Whenever both of these bits are set, switch all LEDs of Port A on for one second. Assume that the name of the delay subroutine is DELAY. You do not need to write the code for the delay procedure.
Answer:
In this example, the delay procedure is given below in the explanation section
Explanation:
Solution
The delay procedure is given below:
LDS # $4000 // load initial memory
LDAA #$FF
STAA DDRA
LDAA #$00 //load address
STAA DDRB
THERE LDAA PORT B
ANDA #%00000011// port A and port B
CMPA #%0000011
BNE THERE
LDAA #$FF
STAA PORT A
JSR DELAY
LDAA #$00
STAA PORT A
BACK BRA BACK
Time, budget, and safety are almost always considered to be
1. Efficiency
2. Constraints
3. Trade-offs
4. Criteria
Answer:
The answer is option # 2. (Constraints).
Steam is contained in a closed rigid container which has a volume of 2 initially the the pressure and the temperature is the remeraturedrops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m3, occupied by saturated liquid at the final state?
The given question is incomplete. The complete question is as follows.
Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],
(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?
Explanation:
Using the property tables
[tex]T_{1} = 500^{o}C[/tex], [tex]P_{1}[/tex] = 10 bar
[tex]v_{1} = 0.354 m^{3}/kg[/tex]
(a) During the process, specific volume remains constant.
[tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]
T = [tex](150 - 160)^{o}C[/tex]
Using inter-polation we get,
T = [tex]154.71^{o}C[/tex]
The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].
(b) When the system will reach at state 3 according to the table at 0.5 bar then
[tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]
[tex]v_{g} = 3.24 m^{3} kg[/tex]
Let us assume "x" be the gravity if stream
[tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]
[tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]
= [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]
= 0.109
At state 3, the fraction of total mass condensed is as follows.
[tex](1 - x_{5})[/tex] = 1 - 0.109
= 0.891
The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.
(c) Hence, total mass of the system is calculated as follows.
m = [tex]\frac{v}{v_{1}}[/tex]
= [tex]\frac{1}{0.354}[/tex]
= 2.825 kg
Therefore, at final state the total volume occupied by saturated liquid is as follows.
[tex]v_{ws} = m \times v_{f}[/tex]
= [tex]2.825 \times 0.00103[/tex]
= [tex]2.9 \times 10^{-3} m^{3}[/tex]
The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, ν = 0.30). Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon, (c) the thickness of portion AB, (d) the cross- sectional area of portion AB.
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
PART (a)For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
PART (b)
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
PART(c)
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
PART (d)
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²
What is the stress and strain in the plate?Let us first find the cross sectional area of AB from the image attached;
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
We are given;
Tensile Load; F = 2.75 kN = 2.75 × 10³ N
Horizontal Stress is calculated from the formula;
σₓ = F/A
σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m
σₓ = 143.23 × 10⁶ Pa
Horizontal Strain is calculated from;
εₓ = σₓ/E
We are given E = 200 GPa = 200 × 10⁹ Pa
Thus;
εₓ = (143.23 × 10⁶)/(200 × 10⁹)
εₓ = 716.15 × 10⁻⁶
Formula for Vertical Strain is;
ε_y = -ν * εₓ
We are given ν = 0.30. Thus;
ε_y = -(0.3) * (716.15 × 10⁻⁶)
ε_y = -214.84 × 10⁻⁶
A) We are given;
Gauge Length; L = 0.05m
Change in gauge length is gotten from;
Δx = L * εₓ
Δx = 0.05 × 716.15 × 10⁻⁶
Δx = 35.808 × 10⁻⁶ m
B) From the attached diagram, the width is;
W = 0.012m
Change in width is;
Δy = W * ε_y
Δy = 0.012 * -214.84 × 10⁻⁶
Δy = -2.5781 × 10⁻⁶m
C) We are given;
Thickness of plate; t = 1.6 mm = 0.0016m
Change in thickness;
Δ_z = t * ε_z
where;
ε_z = ε_y
Thus;
Δ_z = t * ε_y
Δ_z = 0.0016 * -214.84 × 10⁻⁶
Δ_z = -343.74 × 10⁻⁹m
D) The change in cross sectional area is gotten from;
ΔA = A_final - A_initial
From calculating the areas, we have;
A = -8.25 × 10⁻⁹ m²
Read more about stress and strain in steel plates at; https://brainly.com/question/1591712
The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T is that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False
a. Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R.
b. Compute and compare linear density values for these same two directions for iron (Fe).
A) The linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R are;
i) LD_110 = √3/(4R√2)
ii) LD_111 = 1/(2R)
B) The linear density values for these same two directions for iron (Fe) are;
i) LD_110 = 2.4 × 10^(9) m^(-1)
ii) LD_111 = 4 × 10^(9) m^(-1)
Calculating Linear Density of Crystalline StructuresA) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;
√((4R)² - (4R/√3)²)
This reduces to; 4R√(1 - 1/3) = 4R√(2/3)
Now, the expression for the linear density of this direction is;
LD_110 =
Number of atoms centered on (110) direction/vector length of 110 direction
In this case, there is only one atom centered on the 110 direction. Thus;
LD_110 = 1/(4R√(2/3))
LD_110 = √3/(4R√2)
ii) The length of the vector for the direction 111 is equal to 4R, since
all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;
LD_111 = 2/(4R) = 1/(2R)
B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear density for the [110] direction is;
LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))
LD_110 = 2.4 × 10^(9) m^(-1)
ii) for the 111 direction, we have;
LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))
LD_111 = 4 × 10^(9) m^(-1)
Read more about Linear Density of Crystalline Structures at; https://brainly.com/question/14831455
An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.
Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?
Answer:
A.) 7.13 degree north east
B.) 806.23 km/h
C.) 7.44 hours
D.) 0.06 hours
Explanation:
Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction
Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.
A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula
Tan Ø = y/x
Substitute y = 100 km/h and x = 800km/h
Tan Ø = 100/800
Tan Ø = 0.125
Ø = Tan^-1(0. 125)
Ø = 7.13 degrees north east.
Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east
B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem
Speed = sqrt ( 800^2 + 100^2)
Speed = sqrt (650000)
Speed = 806.23 km/h
C.) Let us use the speed formula
Speed = distance / time
Substitute the speed and distance into the formula
806.23 = 6000/ time
Make Time the subject of formula
Time = 6000/806.23
Time = 7.44 hours
D.) If there is no cross wind,
Time = 6000/800
Time = 7.5 hour
Time difference = 7.5 - 7.44
Time difference = 0.06 hours
To determine the viscosity of a liquid of specific gravity 0.95, you fill to a depth of 12 cm a large container that drains through a 30 cm long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm3 /s. What is the fluid viscosity (assume laminar flow)
Answer:
Fluid viscosity, [tex]\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]
Explanation:
Container depth, D = 12 cm = 0.12 m
Tube length, l = 30 cm = 0.3 m
Specific Gravity, [tex]\rho[/tex] = 0.95
Tube diameter, d = 2 mm = 0.002 m
Rate of flow, Q = 1.9 cm³/s = 1.9 * 10⁻⁶ m³/s
Calculate the velocity at point 2 ( check the diagram attached)
Rate of flow at section 2, [tex]Q = A_2 v_2[/tex]
[tex]Area, A_2 = \pi d^{2} /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^{-6} m^2[/tex]
[tex]v_{2} = Q/A_{2} \\v_{2} =\frac{1.9 * 10^{-6}}{3.14 * 10^{-6}} \\v_{2} = 0.605 m/s[/tex]
Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:
[tex]\frac{p_{1} }{\rho g} + \frac{v_{1}^2 }{2 g} + z_1 = \frac{p_{2} }{\rho g} + \frac{v_{2}^2 }{2 g} + z_2 + h_f\\ z_1 = L + l = 0.12 + 0.3\\z_1 = 0.42\\p_1 = p_{atm}\\v_1 = 0\\z_2 = 0\\\frac{p_{atm} }{\rho g} + \frac{0^2 }{2 g} + 0.42= \frac{p_{atm} }{\rho g} + \frac{0.605^2 }{2 *9.8} +0 + h_f\\h_f = 0.401 m[/tex]
For laminar flow, the head loss is given by the formula:
[tex]h_f = \frac{128 Q \mu l}{\pi \rho g d^4} \\\\0.401 = \frac{128 * 1.9 * 10^{-6} * 0.3 \mu}{\pi *0.95* 9.8* 0.002^4}\\\\\\\mu = \frac{0.401 * \pi *0.95* 9.8* 0.002^4}{128 * 1.9 * 10^{-6} * 0.3} \\\\\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]
Answer:
0.00257 kg / m.s
Explanation:
Given:-
- The specific gravity of a liquid, S.G = 0.95
- The depth of fluid in free container, h = 12 cm
- The length of the vertical tube , L = 30 cm
- The diameter of the tube, D = 2 mm
- The flow-rate of the fluid out of the tube into atmosphere, Q = 1.9 cm^3 / s
Find:-
To determine the viscosity of a liquid
Solution:-
- We will consider the exit point of the fluid through the vertical tube of length ( L ), where the flow rate is measured to be Q = 1.9 cm^3 / s
- The exit velocity ( V2 ) is determined from the relation between flow rate ( Q ) and the velocity at that point.
[tex]Q = A*V_2[/tex]
Where,
A: The cross sectional area of the tube
- The cross sectional area of the tube ( A ) is expressed as:
[tex]A = \pi \frac{D^2}{4} \\\\A = \pi \frac{0.002^2}{4} \\\\A = 3.14159 * 10^-^6 m^2[/tex]
- The velocity at the exit can be determined from the flow rate equation:
[tex]V_2 = \frac{Q}{A} \\\\V_2 = \frac{1.9*10^-^6}{3.14159*10^-^6} \\\\V_2 = 0.605 \frac{m}{s}[/tex]
- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.
[tex]\frac{P_1}{p*g} + \frac{V^2_1}{2*g} + z_t_o_p = \frac{P_2}{p*g} + \frac{V^2_2}{2*g} + z_d_a_t_u_m + h_L[/tex]
- The free surface conditions apply at atmospheric pressure and still ( V1 = 0 ). Similarly, the exit of the fluid is also to atmospheric pressure. Where, z_top is the total change in elevation from free surface to exit of vertical tube.
- The major head losses in a circular pipe are accounted using Poiessel Law:
[tex]h_L = \frac{32*u*L*V}{S.G*p*g*D^2}[/tex]
Where,
μ: The dynamic viscosity of fluid
L: the length of tube
V: the average velocity of fluid in tube
ρ: The density of water
- The average velocity of the fluid in the tube remains the same as the exit velocity ( V2 ) because the cross sectional area ( A ) of the tube remains constant throughout the tube. Hence, the velocity also remains constant.
- The energy balance becomes:
[tex]h + L = \frac{V_2^2}{2*g} + \frac{32*u*L*V_2}{S.G*p*g*D^2} \\\\0.42 = \frac{0.605^2}{2*9.81} + \frac{32*u*(0.3)*(0.605)}{0.95*998*9.81*0.002^2} \\\\u = 0.00257 \frac{kg}{m.s}[/tex]
- Lets check the validity of the Laminar Flow assumption to calculate the major losses:
[tex]Re = \frac{S.G*p*V_2*D}{u} \\\\Re = \frac{0.95*998*0.605*0.002}{0.00257} \\\\Re = 446 < 2100[/tex]( Laminar Flow )
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)
Answer:
Exit temperature = 32°C
Explanation:
We are given;
Initial Pressure;P1 = 100 KPa
Cp =1000 J/kg.K = 1 KJ/kg.k
R = 500 J/kg.K = 0.5 Kj/Kg.k
Initial temperature;T1 = 27°C = 273 + 27K = 300 K
volume flow rate;V' = 15 m³/s
W = 130 Kw
Q = 80 Kw
Using ideal gas equation,
PV' = m'RT
Where m' is mass flow rate.
Thus;making m' the subject, we have;
m' = PV'/RT
So at inlet,
m' = P1•V1'/(R•T1)
m' = (100 × 15)/(0.5 × 300)
m' = 10 kg/s
From steady flow energy equation, we know that;
m'•h1 + Q = m'h2 + W
Dividing through by m', we have;
h1 + Q/m' = h2 + W/m'
h = Cp•T
Thus,
Cp•T1 + Q/m' = Cp•T2 + W/m'
Plugging in the relevant values, we have;
(1*300) - (80/10) = (1*T2) - (130/10)
Q and M negative because heat is being lost.
300 - 8 + 13 = T2
T2 = 305 K = 305 - 273 °C = 32 °C
An 60-m long wire of 5-mm diameter is made of steel with E = 200 GPa and ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the allowable tension in the wire (b) the corresponding elongation of the wire
Answer:
a) 2.45 KN
b) 0.0375 m
Explanation:
[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]
[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]
[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]
(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]