Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?
Answer:
51112.5 ft^3
Explanation:
Determine the volume needed along the road when we assume a 6% shrinkage
shrinkage factor = 1 - shrinkage = 1 - 0.06 = 0.94
first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method
Volume between A1 - A2
= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]
= 125 ft * 135 ft^2
= 16875 ft^3
Volume between A2 - A3
= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]
= 125 ft * (200 ft^2 / 2)
= 12500 ft^3
Volume between A3 - A4
= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]
= 125 ft * (170 ft^2 / 2)
= 10625 ft^3
Volume between A4 - A5
(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]
= 125 ft * 115 ft^2
= 14375 ft3
Hence the total volume along the 500 ft road
= ∑ volumes between cross sectional areas
= 16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3
Finally the volume needed along this road is calculated as
Total volume * shrinkage factor
= 54375 * 0.94 = 51112.5 ft^3
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.
(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)
(b) What percentage of the hexane entering the condenser is recovered as a liquid?
Answer:
A. 72.34mol/min
B. 76.0%
Explanation:
A.
We start by converting to molar flow rate. Using density and molecular weight of hexane
= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17
= 988.5/86.17
= 11.47mol/min
n1 = n2+n3
n1 = n2 + 11.47mol/min
We have a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.47(1.00)
To get n2
(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)
0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min
0.18n2-0.05n2 = 11.47-2.0646
= 0.13n2 = 9.4054
n2 = 9.4054/0.13
n2 = 72.34 mol/min
This value is the flow rate of gas that is leaving the system.
B.
n1 = n2 + 11.47mol/min
72.34mol/min + 11.47mol/min
= 83.81 mol/min
Amount of hexane entering condenser
0.18(83.81)
= 15.1 mol/min
Then the percentage condensed =
11.47/15.1
= 7.59
~7.6
7.6x100
= 76.0%
Therefore the answers are a.) 72.34mol/min b.) 76.0%
Please refer to the attachment .
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the distance can be measured using full tape measures, what is the maxim error per tape measure allowed?
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
In beams, why is the strain energy from bending moments much bigger than the strain energy from transverse shear forces? Choose one or more of the following options.
a) The stresses due to bending moments is much more than the stresses from transverse shear.
b) The strains due to bending moments is much more than the strains from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Answer:
a) The stresses due to bending moments is much more than the stresses from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Explanation:
Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.
In an axial flow compressor air enters the compressor at stagnation conditions of 1 bar and 290 K. Air enters with an absolute velocity of 145 m/s axially into the first stage of the compressor and axial velocity remains constant through the stage. The rotational speed is 5500 rpm and stagnation temperature rise is 22 K. The radius of rotor-blade has a hub to tip ratio of 0.5. The stage work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assume for air Cp=1005 kJ/(kg·K) and γ= 1.4
Determine the followings. List your assumptions.
i. The tip radius and corresponding rotor angles at the tip, if the inlet Mach number for the relative velocity at the tip is limited to 0.96.
ii. The mass flow at compressor inlet.
iii. The stagnation pressure ratio of the stage and power required by the first stage.
iv. The rotor angles at the root section.
Answer:
i) r_t = 0.5101 m
ii) m' = 106.73 kg/s
iii) R_s = 1.26
P = 2359.8 kW
iv) β2 = 55.63°
Explanation:
We are given;
Stagnation pressure; T_01 = 290 K
Inlet velocity; C1 = 145 m/s
Cp for air = 1005 kJ/(kg·K)
Mach number; M = 0.96
Ratio of specific heats; γ = 1.4
Stagnation pressure; P_01 = 1 bar
rotational speed; N = 5500 rpm
Work done factor; τ = 0.92
Isentropic effjciency; η = 0.9
Stagnation temperature rise; ΔT_s = 22 K
i) Formula for Stagnation temperature is given as;
T_01 = T1 + C1/(2Cp)
Thus,making T1 the subject, we havw;
T1 = T_01 - C1/(2Cp)
Plugging in the relevant values, we have;
T1 = 290 - (145/(2 × 1005))
T1 = 289.93 K
Formula for the mach number relative to the tip is given by;
M = V1/√(γRT1)
Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K
Thus;
V1 = M√(γRT1)
V1 = 0.96√(1.4 × 287 × 289.93)
V1 = 0.96 × 341.312
V1 = 327.66 m/s
Now, tip speed is gotten from the velocity triangle in the image attached by the formula;
U_t = √(V1² - C1²)
U_t = √(327.66² - 145²)
U_t = √86336.0756
U_t = 293.83 m/s
Now relationship between tip speed and tip radius is given by;
U_t = (2πN/60)r_t
Where r_t is tip radius.
Thus;
r_t = (60 × U_t)/(2πN)
r_t = (60 × 293.83)/(2π × 5500)
r_t = 0.5101 m
ii) Now mean radius from derivations is; r_m = 1.5h
While relationship between mean radius and tip radius is;
r_m = r_t - h/2
Thus;
1.5h = 0.5101 - 0.5h
1.5h + 0.5h = 0.5101
2h = 0.5101
h = 0.5101/2
h = 0.2551
So, r_m = 1.5 × 0.2551
r_m = 0.3827 m
Formula for the area is;
A = 2πr_m × h
A = 2π × 0.3827 × 0.2551
A = 0.6134 m²
Isentropic relationship between pressure and temperature gives;
P1 = P_01(T1/T_01)^(γ/(γ - 1))
P1 = 1(289.93/290)^(1.4/(1.4 - 1))
P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²
Formula for density is;
ρ1 = P1/(RT1)
ρ1 = 0.9992 × 10^(5)/(287 × 289.93)
ρ1 = 1.2 kg/m³
Mass flow rate at compressor inlet is;
m' = ρ1 × A × C1
m' = 1.2 × 0.6134 × 145
m' = 106.73 kg/s
iii) stagnation pressure ratio is given as;
R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))
R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))
R_s = 1.26
Work is;
W = C_p × ΔT_s
W = 1005 × 22
W = 22110 J/Kg
Power is;
P = W × m'
P = 22110 × 106.73
P = 2359800.3 W
P = 2359.8 kW
iv) We want to find the rotor angle.
now;
Tan β1 = U_t/C1
tan β1 = 293.83/145
tan β1 = 2.0264
β1 = tan^(-1) 2.0264
β1 = 63.73°
Formula for Stagnation pressure rise is given by;
ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)
Plugging in the relevant values;
22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)
(tan 63.73 - tan β2) = 0.5641
2.0264 - 0.5641 = tan β2
tan β2 = 1.4623
β2 = tan^(-1) 1.4623
β2 = 55.63°
What is difference between a backdoor, a bot, a keylogger, and psyware,a nd a rootkit? Can they all present in the same malware?
Answer:
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
Explanation:
Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.
A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.
A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.
A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.
A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
Which method of freezing preserves the quality and taste of food?
Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
Which of the following is an example of a tax
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
I need help please thank for the help on the last one <3
A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following:
a. The amount by which this specimen will elongate in the direction of the applied stress.
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answer:
1)ΔL = 0.616 mm
2)Δd = 0.00194 mm
Explanation:
We are given;
Force; F = 52900 N
Initial length; L_o = 207 mm = 0.207 m
Diameter; d_o = 19.2 mm = 0.0192 m
Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²
Now, from Hooke's law;
E = σ/ε
Where; σ is stress = force/area = F/A
A = πd²/4 = π × 0.0192²/4
A = 0.00009216π
σ = 52900/0.00009216π
ε = ΔL/L_o
ε = ΔL/0.207
Thus,from E = σ/ε, we have;
61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)
Making ΔL the subject, we have;
ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)
ΔL = 0.616 × 10^(-3) m
ΔL = 0.616 mm
B) Poisson's ratio is given as;
υ = ε_x/ε_z
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
υ = (Δd/d_o) ÷ (ΔL/L_o)
Making Δd the subject gives;
Δd = (υ × d_o × ΔL)/L_o
We are given Poisson's ratio to be 0.34.
Thus;
Δd = (0.34 × 19.2 × 0.616)/207
Δd = 0.00194 mm
Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate
Answer:
The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.
Explanation:
If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:
60 = 4,000
95 = X
((95 x 4,000) / 60)) = X
(380,000 / 60) = X
6,333.3 = X
Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.
The pascal is actually a very small unit of pressure. To show this, convert 1 Pa = 1 N/m² to lb/ft². Atmosphere pressure at sea level is 14.7 lb/in². How many pascals is this?
Answer:
pascals is this = 101352.972 Pa
Explanation:
given data
Atmosphere pressure at sea level = 14.7 lb/in²
we convert 1 Pa = 1 N/m² to lb/ft²
so we convert here 14.7 lb/in² to pascals
we know that 1 lb/ft² = 47.990172 N/m²
so
1 lb/ft² × ft²/(12in)² = 47.990172 × 144 N/m²
it will be simplyfy
1 lb/ft² = 6894.76 N/m²
so
14.7 lb/in² = 14.7 × 6894.76 N/m²
14.7 lb/in² = 101352.972 Pa
Saturated water vapor at 150°C is compressed in a reversible steady-flow device to 1000 kPa while its specific volume remains constant. Determine the work required in kJ/kg.
Answer:
work required = 205.59 kJ/kg
Explanation:
Given data:
Temperature of water vapor = 150°c
final pressure ( P2 ) = 1000 kPa
specific volume = constant
Determine work required in kJ/kg
we apply the equation below to resolve the problem
[tex]w_{rev} = v ( P1 - P2 )[/tex] ---- ( 1 )
next we have to find the value of the specific volume and saturation pressure of water vapor at 150°c using the saturated water-temperature table
v = specific volume = 0.39248 m^3/kg
P1 = saturation pressure = 476.16 kPa
substitute values into equation 1
[tex]w_{rev} =[/tex] 0.39248 ( 476.16 - 1000 )
= -205.59 kj/kg
hence work required = 205.59 kJ/kg
Here are the commonly used Baud rate: 2400,4800,9600,19200,38400, 115200, 460800 There is an inertial measurement unit (IMU) measurement sensor that needs to update 98 bytes data (with extra 2 label bytes) every 10 ms (100Hz), what is the minimum requirement of the baud rate? (1 byte = 8 bits) Which of the above listed Baud rate you can choose to use? (please list all of them) .
Answer:
115200 and 460800
Explanation:
which of the above listed Baud rate can you choose from
Given Baud rate : 2400,4800,9600,19200,38400, 115200, 460800
The Total bytes = 98 data bytes + 2 extra label bytes for every 10 ms
= 100 bytes for every 10 ms
hence the data rate per second
= [tex]\frac{100 * 8}{10*10^{-3} }[/tex] = 80000
minimum required Baud rate = 80000
Therefore The Baud rate that can be chosen from are : 115200 and 460800
The big ben clock tower in london has clocks on all four sides. If each clock has a minute hand that is 11.5 feed in length, how far does the tip of each hand travel in 52 minutes?
Answer:
Updated question
The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?
The distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Explanation:
Let us assume the shape of the clock is circular.
the minute hand is equal to the radius = 11.5 ft
Diameter = radius x 2
Diameter = 11.5 x 2 = 23 ft
The distance traveled by the tip of the minute hand can be calculated thus;
the fraction of the circumference traveled by the minute hand would be;
52/60 = 0.8667
Circumference of the clock would be;
C = pi x d
where C is the circumference
pi is a constant
d is the diameter
C = 3.14 x 23
C = 72.22 ft
Therefore the fraction of the circumference covered by the minute hand would be;
72.22 ft x 0.8667 = 62.59 ft
Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft
is a process that is used to systematically solve problems.
design
engineering
brainstorming
O teamwork
Answer:
design
Explanation:
Design is a process used to solve problems systematically.
Human beings have specific needs and desires, which require a design process to interpret those needs and make them real from a product or service.
Design uses specific methods and techniques integrating ideals, creativity, technology and innovation to satisfy users' needs and solve problems.
What is computer programming
Answer:
Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.
Explanation:
Hope dis helps! :)
Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required, in Btu/lbm, for this compression. The gas constant of air is R.
Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and a quality of 0.2. The tank has a pressure-regulating venting valve that allows pressure to be constant. The tank is subsequently being heated until its content becomes a saturated vapor (of quality 1.0). During heating, the pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. You can neglecting the kinetic and potential energy effects.
Required:
a. Determine the total mass in the tank at the initial and final states, in kg.
b. Calculate the amount of heat (in kJ) transferred from the initial state to the final state.
Answer:
The total mass in the tank = 0.45524 kg
The amount of heat transferred = 3426.33 kJ
Explanation:
Given that:
The volume of the tank V = 0.06 m³
The pressure of the liquid and the vapor of H2O (p) = 15 bar
The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]
By applying the energy rate balance equation;
[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]
where;
[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]
Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]
If we integrate both sides; we have:
[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
We obtain the following data from the saturated water pressure tables, at p = 15 bar.
Since:
[tex]h_e =h_g[/tex]
Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]
[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]
[tex]v_g = 0.1318 \ m^3/kg[/tex]
Hence;
[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]
[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]
[tex]v_1 = 0.02728 \ m^3/kg[/tex]
Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar
[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]
Hence;
[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]
[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]
[tex]u_1 = 1193.428[/tex]
However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:
[tex]m_1 = \dfrac{V}{v_1}[/tex]
[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]
[tex]m_1 = 2.1994 \ kg[/tex]
From the question, given that the final quality; [tex]x_2 = 1[/tex]
[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]
[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]
[tex]v_2 = 0.1318 \ m^3/kg[/tex]
Also;
[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]
[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]
[tex]u_2 = 2594.5 \ kJ/kg[/tex]
Then the final mass can be calculated by using the formula:
[tex]m_2 = \dfrac{V}{v_2}[/tex]
[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]
[tex]m_2 = 0.45524 \ kg[/tex]
Thus; the total mass in the tank = 0.45524 kg
FInally; from the previous equation (1) above:
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]
Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]
Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]
Q = 3426.33 kJ
Thus, the amount of heat transferred = 3426.33 kJ
what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.
Answer:
robotic technology
Explanation:
Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.
Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.
One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.
Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.
Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.
A tube of diameter 3 cm and length 3 m has a water flow of 100 cm3/s. If the pollutant concentration in the water is constant at 2 mg/L, find the mass flux (mg/cm2-s) of pollutant through the tube due to advection.
Answer: the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
Explanation:
Given that;
Diameter of tube = 3 cm, radius r = 1.5 cm
water flow is 100 cm³/s
pollutant concentration = 2 mg/L
first we find the rate of flow of pollutant
we know that
1 L = 1000 cm³
xL = 100 cm³
100Lcm³ = xL1000cm³
xL = 100/1000
xL = 1/10 L
so 100cm³ = 1/10 L
now pollutant concentration in 100 cm³ = 1/10L × 2mg/L = 0.2 mg
Rate of flow of pollutant = 0.2 mg/s
Mass flux density is the pollutant mass per unit time per unit area
so Area of tube = πr² = 3.14 × 1.5² = 7.065 cm²
So
Mass flux = 0.2 / 7.065
Mass flux = 0.0283 mg/cm².s
Therefore, the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
Identify how the average friction and heat transfer coefficients are determined in flow over a flat plate.
A) They are determined by differentiating the local friction and heat transfer coefficients at the mid-length of the plate, and then multiplying them by the length of the plate.
B) They are determined by by integrating the local Reynolds number and Nusselt numbers over the entire plate.
C) They are determined by integrating the local friction and heat transfer coefficients over the entire plate, and then dividing them by the length of the plate.
D) They are determined by by differentiating the local Reynolds number and Nusselt numbers at the mid-length of the plate.
Answer:
C.
Explanation:
Let's have
Q = heat transfer surface
∆T = average temperature
F = area of the heat surface
Then the heat transfer coefficient = Q/∆T*F
In a flow over flat plate, the average friction and the heat transfer coefficient are determined by the integration of local friction and also great transfer coefficients over the plate entirely and then dividing by the plates length.
Therefore answer option C is the answer to this question.
please help i have no xlue
What is the Bernoulli formula?
Answer:
P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2
What overall material composition would be required to give a material made up of 50wt% mullite and 50wt% alumina at 1400°C?
Answer: overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}
Explanation:
Given that;
from the phase diagram SiO₂ - Al₂O₃
alumina at 1400°C
mullite + alumina ranges from 74 - 100% wt
so for 50% mullite and 50wt% alumina
we have;
50/100 = 100 - x / 100 - 74
0.5 = 100 - x / 26
0.5 × 26 = 100 - x
13 = 100 - x
x = 100 - 13
x = 87 wt% { AL₂O₃]
[ 100% - 87% = 13%] 13% wt SiO₂
So overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}
Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect
deadlock and kill thread, releasing all resources, (3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait, (5) resource ordering, and (6) detect deadlock and roll back thread’s actions.
a. One criterion to use in evaluating different approaches to deadlock is which
approach permits the greatest concurrency. In other words, which approach allows
the most threads to make progress without waiting when there is no deadlock?
Give a rank order from 1 to 6 for each of the ways of handling deadlock just listed,
where 1 allows the greatest degree of concurrency. Comment on your ordering.
b. Another criterion is efficiency; in other words, which requires the least processor
overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient,
assuming that deadlock is a very rare event. Comment on your ordering. Does
your ordering change if deadlocks occur frequently?
who can answer part B for me?
Answer:
b
Explanation:
The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï 7Li3 + 4He2 113Cd48 + 1 n0 ï 114Cd48 + γ[5 MeV] The microscopic thermal absorption cross sections for B-10 and Cd-113 are 3841 b and 20,600 b respectively. Which of these two materials would be the more effective radiation shield? Explain
Solution :
The nuclear reaction for boron is given as :
[tex]$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$[/tex]
And the reaction for Cadmium is :
[tex]$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$[/tex]
We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final
Answer:
74,4 litros
Explanation:
Dado que
W = nRT ln (Vf / Vi)
W = 3000J
R = 8,314 JK-1mol-1
T = 58 + 273 = 331 K
Vf = desconocido
Vi = 25 L
W / nRT = ln (Vf / Vi)
W / nRT = 2.303 log (Vf / Vi)
W / nRT * 1 / 2.303 = log (Vf / Vi)
Vf / Vi = Antilog (W / nRT * 1 / 2.303)
Vf = Antilog (W / nRT * 1 / 2.303) * Vi
Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25
Vf = 74,4 litros
Which of the following is not one of the common classifications of product liability defects? A. Manufacture B. Materials C. Packaging D. Both "Materials" and "Packaging" E. Design
Answer:
D. Both "Materials" and "Packaging"
Explanation:
Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.
There are majorly three product defects. They are :
1. Manufacturing defect
2. Design defect
3. Marketing defect