Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answers
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
Related Questions
Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4
Answers
Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
Deflection angle at B
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
slope at B
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
Deflection angle at C
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
Slope at C
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?
Answers
Answer:
Not seeing any other information, the best answer I can give is 2m.
Explanation:
M = magnitude
You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 6 when 16.0% of the crack length is removed.
Answers
Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
([tex]p_t[/tex])₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2
where ([tex]p_t[/tex])₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
divide both sides by 2(σ₀)₁
[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex] = 36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]
1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂
([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁
([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
This is the top part of the picture. I need someone to put the correct letter by the word !
Answers
Answer:
J = power steering pump
A = hydraulic hoses
D = pitman arm
G = rack and pinion assembly
B = idler arm
C = center link
F = tie rod
E = tie rod end
H = rack and pinion boot
I = outer tie rod end
Explanation:
You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the following code, and visualize the results as a scatter plot: %%sql SELECT cost, price FROM product What should you do before running a cell with the %%sql magic? a. Create a new DataFrame named product from productDF.select("cost", "price") b. Persist the productDF DataFrame using productDF.createOrReplaceTempView("product") c. Filter the productDF dataframe using productDF.filter("cost == price") d. Rename the columns in the productDF DataFrame using productDF.withColumnRenamed("cost", "price")
Answers
A line of students are arranged in odd and even positions. Now the students in the odd positions are to be sorted in the descending order and the students in the even position are to be sorted in ascending order, given a 1 D array. The maximum length of the line is 20. Display – ‘Invalid Size', if the input specified is zero or negative. Write an algorithm to implement the above scenario.
Answers
Answer:
The algorithm is as follows:
0. Start
1. MyArray = []
2. Input n
3. if n <1 or n > 20:
3.1print("Invalid Size")
4. else:
4.1 for [tex]i = 0[/tex] to n - 1:
4.1.1 Input MyArray[i]
4.2 even = []; odd = []
4.3 enum = 0; onum = 0
4.4 for i = 0 to n - 1:
4.4.1 if i%2 == 0:
4.4.1.1 even[enum]=list[i]
4.4.1.2 enum = enum + 1
4.4.2 else:
4.4.2.1 odd[onum]=list[i]
4.4.2.2 onum= onum + 1
4.5 MyArray.clear()
4.6 enum=0
4.7 while even:
4.7.1 minm = even[0]
4.7.2 for x in even:
4.7.2.1 if x < minm:
4.7.2.1.1 minm = x
4.7.3 MMyArray[enum] = minm
4.7.4 even.remove(minm)
4.7.5 enum = enum + 1
4.8 while odd:
4.8.1 maxm = odd[0]
4.8.2 for x in odd:
4.8.2.1 if x > maxm:
4.8.2.1.1 maxm = x
4.8.3 MMyArray[enum] = maxm
4.8.4 odd.remove(maxm)
4.8.5 enum = enum + 1
4.9 for i = 0 to n - 1:
4.9.1 print MyArray[i]
5. Stop
Explanation:
This starts the algorithm
0. Start
This creates an empty array
1. MyArray = []
This gets input for n (the length of the array)
2. Input n
If n is is less than 1 or greater than 20, then the input is invalid
3. if n <1 or n > 20:
3.1print("Invalid Size")
For valid values of n, we have:
4. else:
The italicized gets input into the array
4.1 for [tex]i = 0[/tex] to n - 1:
4.1.1 Input MyArray[i]
This creates empty arrays for even index and for odd index
4.2 even = []; odd = []
This initializes even index and odd index to 0
4.3 enum = 0; onum = 0
This iterates through the array indices
4.4 for i = 0 to n - 1:
If index is even, add array element to even
4.4.1 if i%2 == 0:
4.4.1.1 even[enum]=list[i]
4.4.1.2 enum = enum + 1
If otherwise, add array element to odd
4.4.2 else:
4.4.2.1 odd[onum]=list[i]
4.4.2.2 onum= onum + 1
Clear elements of MyArray
4.5 MyArray.clear()
Set index to 0
4.6 enum=0
Iterate through the even array
4.7 while even:
Set minimum to the first index
4.7.1 minm = even[0]
Sort the array in ascending order
4.7.2 for x in even:
4.7.2.1 if x < minm:
4.7.2.1.1 minm = x
Add the sorted array into MyArray
4.7.3 MMyArray[enum] = minm
Delete the elements of even array
4.7.4 even.remove(minm)
4.7.5 enum = enum + 1
Iterate through the odd array
4.8 while odd:
Set maximum to the first index
4.8.1 maxm = odd[0]
Sort the array in descending order
4.8.2 for x in odd:
4.8.2.1 if x > maxm:
4.8.2.1.1 maxm = x
Add the sorted array into MyArray
4.8.3 MMyArray[enum] = maxm
Delete the elements of odd array
4.8.4 odd.remove(maxm)
4.8.5 enum = enum + 1
Iterate through the indices of the double sorted MyArray
4.9 for i = 0 to n - 1:
Print each array element
4.9.1 print MyArray[i]
End algorithm
5. Stop
See attachment for the program implemented in Python
Identify the following formulas:
1. Slope Formula
2. Slope-Intercept Form
3. Standard Form
4. Point-Slope Formula
Answers
2. y=mx+b
3. Ax+By=C (I think)
4. y-y1=m(x-x1)
Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T
Answers
Answer: hello your question is incomplete below is the missing part
question :Determine the temperature of the cooled water exiting the cooling tower
answer : T = 43.477° C
Explanation:
Temp of water at exit = 45°C
mass flow rate of cooling tower = 15,000 kg/s
Temp of makeup water = 20°C
Assuming an atmospheric pressure of = 101.3 kPa
Determine temperature of the cooled water exiting the cooling tower
Water entering cooling tower at 45°C
Given that Latent heat of water at 45°C = 43.13 KJ/mol
Cp(wet air) = 1.005+ 1.884(y1)
where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273
Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C
First step : calculate the value of Q
Q = m*Cp*(ΔT) + W(latent heat)
Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)
Q = 95935.8547 KJ/s
Given that mass rate of water = 15000 kg/s
Hence the temperature of the cooled water can be calculated using the equation below
Q = m*Cp*∆T
Cp(water) = 4.2 KJ/Kg°C
95935.8547 = (15000)*(4.2)*(45 - T)
( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C
A spherical balloon is filled with a gas. The outer diameter of the balloon is 20 in. and the thick-ness is 0.012 in. Calculate the maximum permissible pressure in the balloon if the allowable tensile stress and the allowable shear stress in the balloon are 1 ksi and 0.3 ksi, respectively.
Answers
Answer:
1.4 psi
Explanation:
Before diving into the solution to the question above, let's pick out the parameters needed in solving this problem from the question.
=> The measurement for the outer diameter of the balloon = 20 inches, the measurement for the thickness = 0.012 in, the allowable tensile stress = 1ksi and the allowable shear stress in the balloon = 0.3 ksi.
The first thing to determine is the inner diameter = 20 - 2 × 0.012 in = 19.976 in.
Therefore, the tensile stress:
1000 = k × [19.976/2]÷ 2 × 0.012 = 2.4 psi.
Also, the sheer stress which is also the maximum permissible pressure in the balloon can be calculated below as:
0.3 × 1000 = k × [19.976/2]/ 4 × 0.012 = 1.4 psi.
) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Answers
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
Help, quick please. I need help with my engineering word problem
Answers
Answer:
a
Explanation:
In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answers
Answer: hello your question lacks some data below is the missing data
Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol
H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.
H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is exposed to a tensile stress of 361 MPa (52360 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.03 for Y.
Answers
Answer:
Explanation:
From the given information:
Strain fracture toughness [tex]K_k[/tex]= 75 MPa[tex]\sqrt{m}[/tex]
Tensile stress [tex]\sigma[/tex] = 361 MPa
Value of Y = 1.03
Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:
[tex]a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}[/tex]
A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.
Answers
Answer:
LOS = A
Explanation:
Given all the parameters the level of service as seen from the attached graph
is LOS = A
To determine the LOS from the attached graph
calculate the trial value of Vp
Vp = V / PHF
= (100 + 150) / 0.95 = 263 pc/h
since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1
next we will calculate the flow rate
flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]
Fhr = 1 / 1.035 = 0.966 ≈ 1
next calculate the real value of Vp
Vp = V / ( PHF * N * Fhr * Fp )
= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )
Vp ≈ 126 pc/h/In
Next calculate the density
D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In
1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address
Answers
3. A particle is projected to the right from the position S = 0, when an initial velocity of 8 m/s. If the acceleration of the particle is defined by the relation a = -0.5 v3/2, where a in m/s2 and v in m/s. Determine a) the distance the particle will have traveled when its velocity is 5 m/s b) the time when v = 1m/s c) the time require for the particle to travel 8m
Answers
Answer:
a) 3.5 m
b) 14 secs
c) 1.4 secs
Explanation:
a) Determine the distance the particle will travel
given velocity ( final velocity ) = 5 m/s
v^2 = u^2 + 2as
s = ( v^2 - u^2 ) / 2a
= ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )
= 3.5 m
b) Determine the time when v = 1m/s
V = u + at
1 = 8 + ( -0.5 * 1^3/2 ) * t
∴ t = 14 secs
c) Determine the time required for particle to travel 8 m
we will employ both equations above
V^2 = u^2 + 2as
s = 8 m , V = unknown , u = 8 m/s back to equation
V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8
∴ V^2 + 8V^3/2 - 64 = 0
resolving the above equation
V = 3.478 m/s
now using the second equation
V = u + at
3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t
hence : t = 1.4 secs
Al ejercer una fuerza de 50N sobre un resorte elastico esto se alarga desde los 15 cm hasta los 60cm¿cual es la constante elastica del resorte?
Answers
Answer:
Constante de resorte = 1.1 N/m
Explanation:
Dados los siguientes datos;
Fuerza = 50N
Extensión = 60cm - 15cm = 45cm
Para encontrar la constante del resorte;
Matemáticamente, la fuerza ejercida para estirar un resorte viene dada por la fórmula;
Fuerza = constante de resorte * extensión
Sustituyendo en la fórmula, tenemos;
50 = constante de resorte * 45
Constante de resorte = 50/45
Constante de resorte = 1.1 N/m
what is the most common type of suspensions system used on body over frame vehicles?
Answers
Answer:
Engine
Explanation:
Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.
What is a Semi-independent suspension?Semi-independent suspension give the front wheels some individual movement.
This suspension only used in rear wheels.
Thus, the correct option is Semi-independent suspension
Learn more about Semi-independent suspension
https://brainly.com/question/23838001
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calculate force and moment reactions at bolted base O of overhead traffic signal assembly. each traffic signal has a mass 36kg, while the masses of member OC and AC are 50Kg and 55kg, respectively. The mass center of mmber AC at G.
Answers
Answer:
The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F
Explanation:
I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.
Q1) Assuming that in a room full of 13 students born in 2000 and 2004 only, calculate the probability that that two persons or more have the same birthday (same day, month and year) knowing that 2000 contains 366 days and 2004 contains 366 days.
Answers
Answer:
D
Explanation:
Got it wrong so i could answer
The boundary work is positive during an expansion process.
a.
False
b.
True
Answers
Answer:
True
Explanation:
During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system
Hence, the given statement is true
Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.
Required:
Determine the dew point temperature of the products, in K.
Answers
Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as an ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%).
Answers
Answer:
a) specific work output of the actual turbine is 73.14 Btu/lbm
b) the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c) Isentropic efficiency of the turbine is 70.76%
Explanation:
Given the data in the question;
For an adiabatic turbine; heat loss Q = 0
For Initial State;
p₁ = 120 psia
T₁ = 500°F = 959.67°R
from table; { Gas Properties of Air }
At T₁ = 959.67°R
[tex]s_1^0[/tex] = 0.74102 Btu/lbm°R
[tex]h_1[/tex] = 230.98 Btu/lbm
For Finial state;
p₂ = 15 psia
T₂ = 200°F = 659.67°R
[tex]s^0_{2a[/tex] = 0.64889 Btu/lbm°R
[tex]h_{2a[/tex] = 157.84 Btu/lbm
we know that R for air is 0.06855 Btu/lbm.R
a)
The specific work output of the actual turbine Wₐ is;
W[tex]_a[/tex] = [tex]h_1[/tex] - [tex]h_{2a[/tex]
we substitute
W[tex]_a[/tex] = 230.98 - 157.84
W[tex]_a[/tex] = 73.14 Btu/lbm
Therefore, specific work output of the actual turbine is 73.14 Btu/lbm
b)
amount of specific entropy generation during the irreversible process.
To determine the entropy generation [tex]S_{gen[/tex];
[tex]S_{gen[/tex] = ΔS = [tex]s_{2a[/tex] - [tex]s_1[/tex] = [tex]s^0_{2a[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute in our values
[tex]S_{gen[/tex] = 0.64889 - 0.74102 - 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]S_{gen[/tex] = 0.64889 - 0.74102 + 0.1425457
[tex]S_{gen[/tex] = 0.050416 Btu/lbm°R
Therefore, the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c)
Isentropic efficiency of turbine η[tex]_{is[/tex]
η[tex]_{is[/tex] = {actual work output] / [ ideal work output ] = ([tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )
Now, for an ideal turbine;
ΔS = 0 = [tex]s_{2s[/tex] - [tex]s_1[/tex]
so, [tex]s_{2s[/tex] - s₁ = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
0 = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
[tex]s^0_{2s[/tex] = [tex]s_1^0[/tex] + R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute
[tex]s^0_{2s[/tex] = 0.74102 + 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]s^0_{2s[/tex] = 0.74102 - 0.1425457
[tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R
Now, from table; { Gas Properties of Air }
At [tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R; [tex]h_{2s[/tex] = 127.614 Btu/lbm
η[tex]_{is[/tex] = [( [tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )] × 100%
we substitute
η[tex]_{is[/tex] = [( 230.98 - 157.84 ) / ( 230.98 - 127.614 )] × 100%
η[tex]_{is[/tex] = [ 73.14 / 103.366] × 100%
η[tex]_{is[/tex] = 0.70758 × 100%
η[tex]_{is[/tex] = 70.76%
Therefore, Isentropic efficiency of the turbine is 70.76%
Create a 6-bit full subtractor that uses the Borrow method to subtract two 6-bit binary numbers. You can use the proper basic sub-circuit.
Answers
A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500 and a 1KO. What is the source voltage required to
supply the current
Answers
If it took a 30m capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) water from the
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Complete question:
If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.
Answer:
the flow rate of the water from the tank is 0.05 m³/min
Explanation:
Given;
volume of water in the tank, v = 30 m³
length of the waterproof faucet, L = 2cm = 0.02 m
duration of water flow through the tank, t = 10 hours
The flow rate of the water from the tank is calculated as;
[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]
Therefore, the flow rate of the water from the tank is 0.05 m³/min
Which of the following identifies the beginning phase of the engineering design process?
structural analysis
visual analysis
recognizing specifications and limitations
brainstorming possible designs
Answers
Answer:
<❤)structural analysi(❤>
Explanation:
(♨️)BRAINLEIST PLEASE(♨️)
Answer:
Structural analysis
Explanation:
Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 92 MPa m (klein.)) and a yield strength of 900 MPa (65270 psi). The flaw size resolution limit of the flaw detection apparatus is 3 mm (0.1181 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.15, what is the critical flaw length
Answers
Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness [tex]K_{tc[/tex] = 92 Mpa√m
yield strength σ[tex]_y[/tex] = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length [tex]a_c[/tex] = 1/π( [tex]K_{tc[/tex] / Yσ )²
we substitute
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (517.5 Mpa )²
[tex]a_c[/tex] = 1/π( 0.177777 )²
[tex]a_c[/tex] = 1/π( 0.03160466 )
[tex]a_c[/tex] = 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ [tex]a_c[/tex] = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection
. Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Martensite to spheroidite (b) Spheroidite to martensite (c) Bainite to pearlite (d) Pearlite to bainite (e) Spheroidite to pearlite (f) Pearlite to spheroidite (g) Tempered martensite to martensite (h) Bainite to spheroidite
Answers
Answer:
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Explanation:
The heat treatment procedure is simply the heating of a metal to a high temperature and cooling the metal back. during this process the metal will undergo certain mechanical changes
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Please help me answer this engineering question