Answer:
Net heat transfers: during summer = 52.253 W/m^2 during winter = 119.375 w/m^2
Explanation:
Given data :
Room temperature throughout the year = 20⁰c = 293 k
Room temperature during summer = 27⁰c = 300 k
Room temperature during winter = 14⁰c = 287 k
surface temperature of a person throughout the year = 32⁰c = 305 k
coefficient of heat transfer by natural convection (h)= 2 w /m^2 k
emissivity = 0.9
An explanation to the condition of feeling chilled during the winter and comfortable during summer can be explained with the calculation below
The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as
q = hΔt = 2 * ( 305 - 293) = 2 * 12 = 24 w /m^2
also calculate heat transfer through radiation using this formula
[tex]q_{rad} =[/tex] εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]
ε = 0.90,(emissivity) σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)
during summer :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2
therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2
During winter :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3
therefore the net heat transfer during the winter
= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2
What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]
So, the potential energy of the mass is 2940 J.
A steam power plant operates on the reheat Rankine cycle. Steam enters the highpressure turbine at 12.5 MPa and 550°C at a rate of 7.7 kg/s and leaves at 2 MPa. Steam is then reheated at constant pressure to 450 °C before it expands in the low-pressure turbine. The isentropic efficiencies of the turbine and the pump are 85 percent and 90 percent, respectively. Steam leaves the condenser as a saturated liquid. If the moisture content of the steam at the exit of the turbine is not to exceed 5 percent, determine (a) the condenser pressure, (b) the net power output, and (c) the thermal efficiency
Answer:
A) condenser pressure = 9.73 kPa,
B) 10242 kw
C) 36.9%
Explanation:
given data
entrance pressure of steam = 12.5 MPa
temperature of steam = 550⁰c
flow rate of steam = 7.7 kg/s
outer pressure = 2 MPa
reheated steam temperature = 450⁰c
isentropic efficiency of turbine( nt ) = 85% = 0.85
isentropic efficiency of pump = 90% = 0.90
From steam tables
at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK
also for an Isentropic expansion
S4s = S3 .
therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa
h4s = 2948.1 kJ/kg
nt = 0.85
nt (0.85) = [tex]\frac{h3-h4}{h3-h4s}[/tex] = [tex]\frac{3476.5 - h4}{3476.5 - 2948.1}[/tex]
making h4 subject of the equation
h4 = 3476.5 - 0.85 (3476.5 - 2948.1)
h4 = 3027.3 kj/kg
at P5 = 2 MPa and T5 = 450⁰c
h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk
at P6 , x6 = 0.95 and s5 = s6
using nt = 0.85 we can calculate for h6 and h6s
from the chart attached below we can see that
p6 = 9.73 kPa, h6 = 2463.3 kj/kg
B) the net power output
solution is attached below
c) thermal efficiency
thermal efficiency = 1 - [tex]\frac{qout}{qin}[/tex] = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a 1000 ft ramps at a high speed vo and travels 600ft in 7 s at constant deceleration before its speed is reduced to uo/2. Assuming the same constant deceleration.
Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.
Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.
Searches related to Probability questions - A person frequents one of the two restaurants KARIM or NAZEER, choosing Chicken’s item 70% of the time and fish’s item 30% of the time. Regardless of where he goes , he orders Afghani Chicken 60% of his visits. (a) The next time he goes into a restaurants, what is the probability that he goes to KARIM and orders Afghani Chicken. (b) Are the two events in part a independent? Explain. (c) If he goes into a restaurants and orders Afghani Chicken, what is the probability that he is at NAZEER. (d) What is the probability that he goes to KARIM or orders Afghani Chicken or both?
Answer:
a) 0.42
b) Independent
c) 30%
d) 0.88
Explanation:
Person chooses Chicken's item : 70% = 0.7
Person chooses fish's item : 30% = 0.3
Visits in which he orders Afghani Chicken = 60% = 0.6
a)
Probability that he goes to KARIM and orders Afghani Chicken:
P = 0.7 * 0.6 = 0.42
b)
Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.
c)
P = 0.30 because he orders Afghani Chicken regardless of where he visits.
d)
Let A be the probability that he goes to KARIM:
P(A) = 0.7 * ( 1 - 0.6 ) = 0.28
Let A be the probability that he orders Afghani Chicken:
P(B) = 0.3 * 0.6 = 0.18
Let C be the probability that he goes to KARIM and orders Afghani chicken:
= 0.7 * 0.6 = 0.42
So probability that he goes to KARIM or orders Afghani Chicken or both:
P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88
A four-cylinder four-stroke engine is modelled using the cold air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), maximum cycle temperature (T3), and engine speed in RPM, determine the efficiency and other values listed below. The specific heats for air are given as Cp 1.0045 kJ/kg-K and Cv-0.7175 kJ/kg-K.
--Given Values--
T1 (K) 325
P1 (kPa)= 185
V1 (cm^3) = 410
r=8
T3 (K) 3420
Speed (RPM) 4800
Answer:
56.47%
Explanation:
Determine the efficiency of the Engine
Given data : T1 (k) = 325, P1 (kpa) = 185,
V1 (cm^3) = 410 , r = 8, T3(k) = 3420
speed ( RPM) = 4800
USING THIS FORMULA
efficiency ( n ) = [tex]1 - (\frac{1}{(rp)^{r-1} })[/tex]
= 1 - [tex](\frac{1}{(8)^{r-1} })[/tex] = 1 - (1/8^1.4-1 )
= 0.5647 = 56.47%
Q/For the circuit showm bellow:
a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,
b) sketch vc and ic.
Answer:
hello your question is incomplete attached is the complete question
A) Vc = 15 ( 1 -[tex]e^{-t/0.15s}[/tex] ) , ic = [tex]1.5 mAe^{-t/0.15s}[/tex]
B) attached is the relevant sketch
Explanation:
applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch
[tex]R_{th}[/tex] = 8 k ohms || 24 k ohms = 6 k ohms
[tex]E_{th}[/tex] = [tex]\frac{20 k ohms(20 v)}{24 k ohms + 8 k ohms}[/tex] = 15 v
t = RC = (10 k ohms( 15 uF) = 0.15 s
Also; Vc = [tex]E( 1 - e^{-t/t} )[/tex]
hence Vc = 15 ( 1 - [tex]e^{-t/0.15}[/tex] )
ic = [tex]\frac{E}{R} e^{-t/t}[/tex] = [tex]\frac{15}{10} e^{-t/t}[/tex] = [tex]1.5 mAe^{-t/0.15s}[/tex]
attached
Given the unity feedback system
G(s)= K(s+4)/s(s+1.2)(s+2)
Find:
a. The range of K that keeps the system stable
b. The value of K that makes the system oscillate
c. The frequency of oscillation when K is set to the value that makes the system oscillate
Answer:
A.) 0 > K > 9.6
B.) K = 9.6
C.) w = +/- 2 sqrt (3)
Explanation:
G(s)= K(s+4)/s(s+1.2)(s+2)
For a closed loop stability, we can analyse by using Routh - Horwitz analysis.
To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.
Please find the attached files for the solution
Searches related to Probability questions - A person frequents one of the two restaurants KARIM or NAZEER, choosing Chicken's item 70% of the time and fish's item 30% of the time. Regardless of where he goes , he orders Afghani Chicken 60% of his visits. (a) The next time he goes into a restaurants, what is the probability that he goes to KARIM and orders Afghani Chicken. (b) Are the two events in part a independent? Explain. (c) If he goes into a restaurants and orders Afghani Chicken, what is the probability that he is at NAZEER. (d) What is the probability that he goes to KARIM or orders Afghani Chicken or both?
Answer:
a) 0.42
b) Independent
c) 30%
d) 0.88
Explanation:
Person chooses Chicken's item : 70% = 0.7
Person chooses fish's item : 30% = 0.3
Visits in which he orders Afghani Chicken = 60% = 0.6
a) Probability that he goes to KARIM and orders Afghani Chicken:
P = 0.7 * 0.6 = 0.42
b) Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.
c) P = 0.30 because he orders Afghani Chicken regardless of where he visits.
d) Let A be the probability that he goes to KARIM:
P(A) = 0.7 * ( 1 - 0.6 ) = 0.28
Let A be the probability that he orders Afghani Chicken:
P(B) = 0.3 * 0.6 = 0.18
Let C be the probability that he goes to KARIM and orders Afghani chicken:
= 0.7 * 0.6 = 0.42
So probability that he goes to KARIM or orders Afghani Chicken or both:
P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88
the partner who can lose only what he or she has invested in business is the
Answer:
partner itself
Explanation:
Answer:
limited partner
hope this answer correct :)
The following liquids are stored in a storage vessel at 1 atm and 25°C. The vessels are vented with air. Determine whether the equilibrium vapor above the liquid will be flammable. The liquids are:________.
a. Acetone
b. Benzene
c. Cyclohexane
d. Toluene Problem
Answer:
The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )
Explanation:
Liquids stored at : 1 atm , 25⁰c and they are vented with air
Determining whether the equilibrum vapor above the liquid will be flammable
We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable
A) For acetone
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 16.6513
B = 2940.46
C = -35.93
T = 298 k input values into Antoine equation
therefore ; [tex]p^{out}[/tex] = 228.4 mg
calculate volume percentage using Dalton's law
= V% = (saturation vapor pressure / pressure ) *100
= (228.4 mmHg / 760 mmHg) * 100 = 30.1%
The liquid is not flammable because its UFL = 12.8%
B) For Benzene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A= 15.9008
B = 2788.52
C = -52.36
T = 298 k input values into the above equation
[tex]p^{out}[/tex] = 94.5 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (94.5 / 760 ) * 100 = 12.4%
Benzene is not flammable under the given conditions because its UFL =7.1%
C) For cyclohexane
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 15.7527
B = 2766.63
c = -50.50
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 96.9 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= ( 96.9 mmHg /760 mmHg) * 100 = 12.7%
cyclohexane not flammable under the given conditions because its UFL= 8%
D) For Toluene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten from appendix E ( chemical process safety (3rd edition) )
A = 16.0137
B = 3096.52
C = -53.67
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 28.2 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (28.2 mmHg / 760 mmHg) * 100 = 3.7%
Toluene is flammable under the given conditions because its UFL= 7.1%
5. Assume that you and your best friend ench have $1000 to invest. You invest your money
in a fund that pays 1096 per year compound interest Your friend invests her money at a bank
that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount
for each of you ts
(a) You have Slo more than she does (b) You have $100 more than she does
You both have the same amount of money (cdShe has $10 more than you do
Correct question reads;
Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:
(a) You have $10 more than she does
(b) You have $100 more than she does
(c) You both have the same amount of money
(d) She has $10 more than you do
Answer:
(d) She has $10 more than you do
Explanation:
Using the compound interest formula
A= P [ (1-i)^n-1
Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.
My compound interest is:
= 1000 [ (1-0.1)^1-1
= $1000
$1,000 + $1,000 invested= $2,000 total amount received.
My friend's simple interest is;
To determine the total amount accrued we use the formula:
P(1 + rt) Where:
P = Invested Amount (1000)
I = Interest Amount (10,000)
r = Rate of Interest per year (10% or 0.2)
t = Time Period (1 )
= 1000 (1 + rt)
= 1000 (1 + 0.1x1)
= $1100 + $1000 invested = $2100 total amount received.
Therefore, we observe that she (my friend) has $100 more than I do.
Some characteristics of clay products such as (a) density, (b) firing distortion, (c) strength, (d) corrosion resistance, and (e) thermal conductivity are affected by the extent of vitrification. Will they increase or decrease with increasing degree of vitrification?
1. (a) increase (b) decrease (c) increase (d) decrease (e) increase
2. (a) decrease (b) increase (c) increase (d) increase (e) decrease
3. (a) decrease (b) decrease (c) increase (d) decrease (e) decrease
4. (a) increase (b) increase (c) increase (d) increase (e) increase
5. (a) increase (b) decrease (c) decrease (d) increase (e) decrease
Explanation:
1. increase This due to increase in the pore volume.
2.increase . This is due to the fact that more liquid phase will be present at the firing.
3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.
4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.
5. Increase , glass has higher thermal conductivity than the pores it fills.
A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questions. Note: Evaluate the properties of the wall as cross grain oak at 300 K.
What is the surface temperature, in °C, after 400 s?
T (0,400 sec) =
Will the surface of the wall reach the ignition temperature of oak (400°C) after 400 s?
What is the temperature, in °C, 1 mm from the surface after 400 s? T (1 mm, 400 sec) =
Answer:
a) What is the surface temperature, in °C, after 400 s?
T (0,400 sec) = 800°C
b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s
c) What is the temperature, in °C, 1 mm from the surface after 400 s?
T (1 mm, 400 sec) = 798.35°C
Explanation:
oak initial Temperature = 25°C = 298 K
oak exposed to gas of temp = 800°C = 1073 K
h = 20 W/m².K
From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k
Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.
From energy balance: [tex]\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t[/tex]
Initial temperature wall = [tex]T_i[/tex]
Surface temperature = T
Gas exposed temperature = [tex]T_{\infty}[/tex]
1000 lb boulder B is resting on a 500 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?
a. The tension in the cord connected to the truck is 200 lb.
b. The tension in the cord connected to the truck is 1200 lb.
c. The tension in the cord connected to the truck is greater than 1200 lb.
d. The normal force between A and B is 1000 lb.
e. The normal force between A and B is 1200 lb.
f. None of the above are true.
Answer:
c. The tension in the cord connected to the truck is greater than 1200 lb
e. The normal force between A and B is 1200 lb.
Explanation:
The correct question should be
A 1000 lb boulder B is resting on a 200 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?
A free body diagram is shown below.
The normal force between the the boulder and the platform will be the sum of the force, i.e 1000 lb + 200 lb = 1200 lb
For the combination of the bodies to accelerate upwards, then the tension must be greater than the normal force, i.e T > 1200 lb
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:
[tex]\sigma_b = \frac{MC}{I}[/tex]
Where,
[tex]\sigma_b[/tex] is the compressive stress or tensile stress[tex]M[/tex] is the B.M [tex]C[/tex] is the N.A distance[tex]I[/tex] is the moment of interiorSo making this formula for the max, we have:
[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]
With all this information we can put the formula as:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
See more about stress in the beam at brainly.com/question/23637191
The outer surface temperature of a glass filled with iced water may drop below the dew-point temperature of the surrounding air, causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity.
1. True
2. False
Answer: True
Explanation:
The outer surface temperature of a glass filled with iced water will drop below dew-point temperature of the surrounding air, thereby resulting into the moisture in the vicinity of the glass to condense.
After some time, the condensate may start dripping down because of gravity. This is due to the fact that water will start condensing on glass surface below due point temperature. The adhesion force that water has with glass is low.
A state of stress that occurs at a point on the free surface of the of a solid body is = 50 MPa σ x , =10 MPa σ y , and = −15 MPa xy τ .
a. Evaluate the two principal normal stresses and the one principal shear stress that can be found by coordinate system rotations in the x-y plane and give the coordinate system rotations.
b. Determine the maximum normal stress and the maximum shear stress at this point.
c. Sketch the rotations and stress magnitudes
Answer:
A) 5 MPa , 55 MPa
B) maximum stress = 55 MPa, maximum shear stress = 25 MPa
Explanation:
using the given Data
free surface of a solid body
α[tex]_{x}[/tex] = 50 MPa, α[tex]_{y}[/tex] = 10 MPa , t[tex]_{xy}[/tex] = -15 MPa
attached below is the detailed solution to the question
Answer:
I think its A) 5 MPa , 55 MPa
B) ms = 55 MPa, mss= 25 MPa
α = 50 MPa, α = 10 MPa , t = -15 MPa
Define the coefficient of determination and discuss the impact you would expect it to have on your engineering decision-making based on whether it has a high or low value. What do high and low values tell you
Answer and Explanation:
The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.
In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.
When you shift your focus, everything you
see is still in perfect focus.
True or false
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
Problem 1 (paper) Use Gauss-Jordan elimination with partial pivoting to solve the following linear system. Show all steps. Represent all values as exact fractions, not decimal numbers.
[2 1 1 4 2 1 3 1 1] [x1 x2 x3] = [8 13 10]
Problem 2 (paper) For the matrix-matrix
A = [2 3 2 4 7 6 6 11 13]
Calculate the LU decomposition A = LU where
L = [1 0 0 l21 1 0 l31 l32 1] and U = [u11 u12 u13 0 u22 u23 0 0 u33]
As follows. Mulitply out LU to get algebraic expressions for each component of A in terms of the components of L and U.
Answer:
mano não sei mas acho que vai dar certo porque isso aí é muito top mas é isso aí mano o cara tem que ser confiar mesmo viu que negócio é desse jeito mesmo entendeu porque sabe como é que é as coisas né nada é fácil mesmo hein mas é isso aí mano continua tentando aí mano porque Rapaz tu é louco doido agora tá difícil mesmo mas é isso aí o cara tem que ir saber se ele tá ligado eu deixei isso mesmo né mas é isso aí meu truta
Explanation:
É isso aí mano Espero que tenho ajudado aí beleza manda a tua pergunta aí beleza é isso aí mano É isso aí continua hein p
A drilling operation is to be performed with a 12.7 mm diameter drill on cast iron. The hole depth is 60 mm and the drill point angle is 118∘. The cutting speed is 25 m/min and the feed is 0.30 mm/rev. Calculate:___________.
a) The cutting time (min) to complete the drilling operation
b) Material removal rate (mm3/min) during the operation, after the drill bit reaches full diameter.
Answer:
a. Tm = 0.3192min.
b. MRR = 396.91mm^{3}/s.
Explanation:
Given the following data;
Drill diameter, D = 12.7mm
Depth, L = 60mm
Cutting speed, V = 25m/min = 25,000m
Feed, F = 0.30mm/rev
To find the cutting time;
Cutting time, Tm =?
[tex]Tm = \frac{L}{Fr}[/tex] .......eqn 1
We would first solve for the feed rate (F);
[tex]Fr = NF[/tex] .......eqn 2
But we need to find the rotational speed (N);
[tex]N= \frac{V}{\pi *D}[/tex]
[tex]N= \frac{25000}{3.142*12.7}[/tex]
[tex]N= \frac{25000}{39.90}[/tex]
N = 626.57rev/min.
Substiting N into eqn 2;
[tex]Fr = NF[/tex]
Fr = 626.57 * 0.30
Fr = 187.97mm/min.
Substiting F into eqn 1;
[tex]Tm = \frac{L}{Fr}[/tex]
[tex]Tm = \frac{60}{187.97}[/tex]
Tm = 0.3192min.
Therefore, the cutting time is 0.3192 minutes.
For the material removal rate (MRR);
[tex]MRR = \frac{\pi *D^{2}Fr}{4}[/tex]
[tex]MRR = \frac{3.142*12.7^{2}*187.97}{4}[/tex]
[tex]MRR = \frac{3.142*161.29*187.97}{4}[/tex]
[tex]MRR = \frac{95258.16}{4}[/tex]
[tex]MRR = 23814.54mm^{3}/min[/tex]
Time in seconds, we divide by 60;
MRR = 23814.54/60 =396.91mm^{3}/s.
Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.
A hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation louvers or glass panels is likely a(n):___________
a) Security class door
b) Maximum security door
c) Average industrial personnel door
d) High security door
Answer:
Average industrial personnel door
Explanation:
Going by the description in which the door has ventilation louvers or glass panel, this shows that it can never be any type of security door. This is because with louvers or glass panel, it can be easier for any intruder to look or break into the house or space that is intended to be protected fully.
Hence, an hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation louvers or glass panels is likely an Average industrial personnel door
A concentric tube heat exchanger is used to cool a solution of ethyl alcohol flowing at 6.93 kg/s (Cp = 3810 J/kg-K) from 65.6 degrees C to 39.4 degrees C using water flowing at 6.30 kg/s at a temperature of 10 degrees C. Assume that the overall heat transfer coefficient is 568 W/m2-K. Use Cp = 4187 J/kg-K for water.
a. What is the exit temperature of the water?
b. Can you use a parallel flow or counterflow heat exchanger here? Explain.
c. Calculate the rate of heat flow from the alcohol solution to the water.
d. Calculate the required heat exchanger area for a parallel flow configuration
e. Calculate the required heat exchanger area for a counter flow configuration. What happens when you try to do this? What is the solution?
An exothermic reaction releases 146 kJ of heat energy and 3 mol of gas at 298 K and 1 bar pressure. Which of the following statements is correct?
A) ΔU=-138.57 kJ and ΔH=-138.57 kJ
B) ΔU=-153.43 kJ and ΔH=-153.43 kJ
C) ΔU=-138.57 kJ and ΔH=-146.00 kJ
D) ΔU=-153.43 kJ and ΔH=-146.00 kJ
Answer:
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Explanation:
Given;
heat energy released by the exothermic reaction, ΔH = -146 kJ
number of gas mol, n = 3 mol
temperature of the gas, T = 298 K
Apply first law of thermodynamic
Change in the internal energy of the system, ΔU;
ΔU = ΔH- nRT
where;
R is gas constant = 8.314 J/mol.K
ΔU = -146kJ - (3 x 8.314 x 298)
ΔU = -146kJ - 7433 J
ΔU = -146kJ - 7.433 kJ
ΔU = -153.43 kJ
Therefore, the enthalpy change of the reaction ΔH is -146 kJ and change in the internal energy of the system is -153.43 kJ
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answer:
the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Explanation:
From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
The Critical Stress for a maximum internal crack can be expressed by the formula:
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
where;
[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation
[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa
Y = dimensionless parameter
a = length of the internal crack
given that ;
the maximum internal crack length is 8.6 mm
half length of the internal crack will be 8.6 mm/2 = 4.3mm
half length of the internal crack a = 4.3 × 10⁻³ m
From :
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]
[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]
[tex]Y= \dfrac{26}{13.01748802}[/tex]
[tex]Y=1.99731315[/tex]
[tex]Y \approx 1.997[/tex]
For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
when the length of the internal crack a = 3mm
half length of the internal crack will be 3.0 mm / 2 = 1.5 mm
half length of the internal crack a =1.5 × 10⁻³ m
From;
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]
[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]
[tex]\sigma_c =189.6595506[/tex]
[tex]\sigma_c =[/tex] 189.66 MPa
Thus; the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Carbon dioxide (CO2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit state of 10 bar, 520 K. The CO2 is modeled as an ideal gas, and kinetic and potential energy effects are negligible. For the compressor, determine (a) the work input, in kJ per kg of CO2 flowing, (b) the rate of entropy production, in kJ/K per kg of CO2 flowing, and (c) the isentropic compressor efficiency.
Answer:
A.) 0.08 kJ/kg.K
B.) 207.8 KJ/Kg
C.) 0.808
Explanation:
From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.
Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.
Please find the attached files for the solution and the remaining explanation.
Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percent of graphite is 91.906 per cent.
Explanation:
The volume percent of graphite ([tex]\% V_{Gr}[/tex]) is determined by the following expression:
[tex]\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%[/tex]
[tex]\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%[/tex]
Where:
[tex]V_{Gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{Fe}[/tex] - Volume occupied by the ferrite phase, measured in cubic centimeters.
The volume of each phase can be calculated in terms of its density and mass. That is:
[tex]V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}[/tex]
[tex]V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}[/tex]
Where:
[tex]m_{Gr}[/tex], [tex]m_{Fe}[/tex] - Masses of the graphite and ferrite phases, measured in grams.
[tex]\rho_{Gr}[/tex], [tex]\rho_{Fe}[/tex] - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.
Let substitute each volume in the definition of the volume percent of graphite:
[tex]\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%[/tex]
[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%[/tex]
Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:
[tex]m_{Gr} = \frac{2.5}{100}\times (100\,g)[/tex]
[tex]m_{Gr} = 2.5\,g[/tex]
[tex]m_{Fe} = 100\,g - 2.5\,g[/tex]
[tex]m_{Fe} = 97.5\,g[/tex]
If [tex]m_{Gr} = 2.5\,g[/tex], [tex]m_{Fe} = 97.5\,g[/tex], [tex]\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}[/tex] and [tex]\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}[/tex], the volume percent of graphite is:
[tex]\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%[/tex]
[tex]\% V_{Gr} = 91.906\,\%[/tex]
The volume percent of graphite is 91.906 per cent.
how can we prevent chemical hazards in labotary
Answer:
We can prevent it by:
a) By wearing GOOGLES.
b) By wearing our Lab coat.
c) Fire extinguisher should always be present in the lab.
d) Hand Gloves must be worn.
e) No playing in the lab.
f) No touching of things/equipment's e.g bottles, in the lab.
g) No eating/snacking in the lab.
h) Always pay attention, no gisting.
i) Adult/qualified person must be present in the lab with pupils/students.
Explanation:
Hope it helps.
A car radiator is a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of 0.05 kg/s, enters the radiator at 400 K and is to leave at 330 K. The water is cooled by air that enters at 0.75 kg/s and 300 K. If the overall heat transfer coefficient is 200 W/m2-K, what is the required heat transfer surface area?
Answer:
Explanation:
Known: flow rate and inlet temperature for automobile radiator.
Overall heat transfer coefficient.
Find: Area required to achieve a prescribed outlet temperature.
Assumptions: (1) Negligible heat loss to surroundings and kinetic and
potential energy changes, (2) Constant properties.
Analysis: The required heat transfer rate is
q = (m c)h (T h,i - T h,o) = 0.05 kg/s (4209J / kg.K) 70K = 14,732 W
Using the ε-NTU method,
Cmin = Ch = 210.45 W / K
Cmax = Cc = 755.25W / K
Hence, Cmin/Cmx(Th,i - Th,o) = 210.45W / K(100K) = 21,045W
and
ε=q/qmax = 14,732W / 21,045W = 0.700
NTU≅1.5, hence
A=NTU(cmin / U) = 1.5 x 210.45W / K(200W) / m² .K) = 1.58m²
1. the air outlet is..
Tc,o = Tc,i + q / Cc = 300K + (14,732W / 755.25W / K) = 319.5K
2. using the LMTD approach ΔTlm = 51.2 K,, R=0.279 and P=0.7
hence F≅0.95 and
A = q/FUΔTlm = (14,732W) / [0.95(200W / m².K) 51.2K] = 1.51m²