A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
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13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
Write the code using the do-while loop to force the user to enter a number in the range [20,50]
Answer:
Mark as brainlist pls hello
Question in Probability and Statistics 2
: A cyclical load of 1500 lb is to be exerted at the end of a 10 in. long aluminium beam (see Figure below). The bar must survive for at least 10° cycles. What is the minimum diameter of the bar?
Answer:
the minimum diameter of the bar is 1.634 in
The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space C. His seat belt
Answer:
answer is C. his seat belt
What is working fluid and mention five example of working fluid.
Answer:
For fluid power, a working fluid is a gas or liquid that primarily transfers force, motion, or mechanical energy. ... Examples without phase change include air or hydrogen in hot air engines such as the Stirling engine, air or gases in gas-cycle heat pumps, etc.
Explanation:
Answer:
A working fluid is a gas or liquid that primarily transfers force, motion or mechanical energy
Examples:Air, pentane, chlorofluorocarbons, butane, propane and ammonia
Explain the LWD process why is it important in drilling operations?
Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.
Explanation:
pls mark brainliest
what is the term RF exiciter?
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
Future solution for air pollution in new zealand
Answer:
New Zealand may use some of these solutions to prevent air pollution
Explanation:
Using public transports.
Recycle and Reuse
No to plastic bags
Reduction of forest fires and smoking
Use of fans instead of Air Conditioner
Use filters for chimneys
Avoid usage of crackers
What is the built-in pollution control system in an incinerator called
Explanation:
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Cho thanh có tiết diện thay đổi chịu tải trọng dọc trục (hình 1).
Biết d1 = 5 cm, d2 = 8 cm, a= 15 cm, b=10cm, P1 =400kN, P2 =200kN, E= 2.104 kN/cm2.
a) Vẽ biểu đồ lực dọc.
b) Kiểm tra bền của thanh AC, [ϭ] =10 (kN/cm2).
c) Xác định chuyển vị theo phương dọc trục của tâm tiết diện C
Answer:
saay in English language
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
A route for a proposed 8-m-wide highway crosses a region with a 4-m-thick saturated, soft, normally consolidated clay (CH) above impermeable rock. Groundwater level is 1 m below the surface. The geotechnical data available during the preliminary design stage consist of Atterberg limits (LL 5 68% and PL 5 32%) and the natural water content (w 5 56%). Based on experience, the geotechnical engineer estimated the coefficient of consolidation at 8 m2 per year. To limit settlement, a 4-m-high embankment will be constructed as a surcharge from fill of unit weight 16 kN/m3.
(a) Estimate the compression and recompression indices.
(b) Estimate the total primary consolidation settlement under the center of the embankment.
(c) Plot a time–settlement curve under the center of the embankment.
(d) How many years will it take for 50% consolidation to occur?
(e) Explain how you would speed up the consolidation.
(f) Estimate the rebound (heave) when the surcharge is removed.
Sketch a settlement profile along the base of the embankment. Would the settlement be uniform along the base? Explain your answer.A route for a proposed 8-m-wide highway crosses a
4.3
While a train is standing still, its smoke blows 12 m/s north.
What will the resulting velocity be of the smoke relative to the train if the train
is moving at 25 m/s south?
(3)
Assignment Using Perman's equation, estimate the potential evapotranspiration for the month of August in locality with the following data
latitude 20 degrees north
Elevation 200m
• Mean monthly temperature 20 degreeso
• Mean relative humidity: 75%
• Mean sunshine hour= 9hrs
Wind at 2 m height equal 85 km per day
nature of surface cover =green grass
Answer:
what do you need help with
Explanation:
A(n) ____ combines two planetary gearsets to provide more gear ratio possibilities. A)compound planetary gearset B)orifice C)detent D)lock-up torque converter
Answer:
The answer is A. Compound Planetary Gearset.
Explanation:
The Compound Planetary Gear block represents a planetary gear train with composite planet gears. Each composite planet gear is a pair of rigidly connected and longitudinally arranged gears of different radii. One of the two gears engages the centrally located sun gear while the other engages the outer ring gear.
Compound planetary gear sets have at least two planet gears attached in line to the same shaft, rotating and orbiting at the same speed while meshing with different gears. Compounded planets can have different tooth numbers, as can the gears they mesh with.
a basketball player pushes down with a force of 50N on a basketball that is indlated to a gage pressure of 8.0x10^4 Pa. What is the diameter of comtact between the ball nad the floor
Answer:
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Explanation:
The basketball experiments a normal stress ([tex]\sigma[/tex]), in pascals, due to normal force from the floor ([tex]N[/tex]). By definition of normal stress, we have the following equation:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex] (1)
Where [tex]D[/tex] is the diameter of the contact area between the ball and the floor, in meters.
Please notice that magnitude of the normal force equals the magnitude of external force given by the basketball player and weight is negligible in comparison with normal and external forces.
If we know that [tex]N = 50\,N[/tex] and [tex]\sigma = 8.0\times 10^{4}\,Pa[/tex], then the diameter of the contact area is:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex]
[tex]D^{2} = \frac{4\cdot N}{\pi\cdot \sigma}[/tex]
[tex]D = 2\cdot \sqrt{\frac{N}{\pi\cdot \sigma} }[/tex]
[tex]D = 2\cdot \sqrt{\frac{50\,N}{\pi\cdot (8\times 10^{4}\,Pa)} }[/tex]
[tex]D\approx 0.028\,m[/tex] [tex](28\,mm)[/tex]
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b. DSSS and CDMA are fundamental technologies used with WiMAX BWA c. OFDM is implemented to increase spectral efficiency and to improve noise performance d. all of the statements are correct
Answer:
d. all of the statements are correct.
Explanation:
WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heated uniformly to115 oC, determine the length and the diameter of this rod to the nearest micron at the new temperature if the linear coefficient of thermal expansion of steel is 12.5 x 10-6 m/m/oC. What is the stress on the rod at 115oC.
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape and the ground (assumed to be uniform) be if the temperature of the tape rises to 30 C. Neglect the expansion of the earth (the radius of the earth is 6.37 X 106 m)
Answer:
2102.1 m
Explanation:
Temperature at the equator = 0⁰
Radius of the earth = 6.37x10⁶
Required:
We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.
Final temperature = ∆T = 303-273 = 30
S = 11x10^-6
The clearance R = Ro*S*∆T
=6.37x10⁶x 11x10^-6x30
= 2102.1m
Or 2.102 kilometers
Thank you
Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at 2208C with a mass flow rate of 1.2 kg/s. Refrigerant exits at 7 bar, 708C. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in m3 /s, and (b) the power input to the compressor, in kW.'
Answer:
a)[tex]V_1=4.88m^2/s[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)[tex]P=-119.18kW[/tex]
Explanation:
From the question we are told that:
Steady State Saturated vapor [tex]T_1= -20C=>253k[/tex]
Mass Flow rate [tex]M=1.2kg/s[/tex]
Exit Pressure [tex]P_2=7bar[/tex]
Exit Temperature [tex]T_2=70C=>373k[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>P_1=1.399 bar[/tex]
Generally the equation for Volumetric Flow rate is mathematically given by
For Inlet
[tex]V_1=m\frac{RT_1}{P_1}[/tex]
[tex]V_1=m\frac{8314*253}{1.399*10^3}[/tex]
[tex]V_1=18.97m^2/s[/tex]
For outlet
[tex]V_2=m\frac{RT_2}{P_2}[/tex]
[tex]V_2=1.2*\frac{8314*343}{7*10^3}[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)
Generally the equation for Steady state mass and energy equation is mathematically given by
[tex]P=m(h_1-h_2)[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>h_1=24.76kJ/kg[/tex]
[tex]T_2= 70C =>h_2=124.08kJ/kg[/tex]
Therefore
[tex]P=1.2(12.76-124.08)[/tex]
[tex]P=-119.18kW[/tex]
Therefore
Power input into the compressor is
[tex]P=-119.18kW[/tex]
The value of universal gas constant is same for all gases?
a) yes
b)No
Answer:
The answer of these questions is
Explanation:
b) NO
A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage of 120 V rms. The load draws a total of 10 kW at a power factor of 0.85 (lagging). Calculate the rms value of the phase currents and the magnitude of the per-phase load impedance. Draw a phasor diagram showing all tlme voltages and currents.
Answer:
Following are the solution to the given question:
Explanation:
Line voltage:
[tex]V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v[/tex]
Power supplied to the load:
[tex]P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi[/tex]
[tex]10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A[/tex]
Check wye-connection, for the phase current:
[tex]I_{ph}=I_L= 32.68\ A[/tex]
Therefore,
Phasor currents: [tex]32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A[/tex]
Magnitude of the per-phase load impedance:
[tex]Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega[/tex]
Phase angle:
[tex]\phi = \cos^{-1} \ (0.85) =31.79^{\circ}[/tex]
Please find the phasor diagram in the attached file.
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at ft/s^2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 18 ft/s^2.
Required:
Determine the minimum distance d between the cars so as to avoid a collision.
Answer:
Explanation:
Using the kinematics equation [tex]v = v_o + a_ct[/tex] to determine the velocity of car B.
where;
[tex]v_o =[/tex] initial velocity
[tex]a_c[/tex] = constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
[tex]S = d + v_ot + \dfrac{1}{2}at^2[/tex]
Then:
[tex]v_B = 60-12t[/tex]
The distance traveled by car B in the given time (t) is expressed as:
[tex]S_B = d + 60 t - \dfrac{1}{2}(12t^2)[/tex]
For car A, the needed time (t) to come to rest is:
[tex]v_A = 60 - 18(t-0.75)[/tex]
Also, the distance traveled by car A in the given time (t) is expressed as:
[tex]S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
Relating both velocities:
[tex]v_B = v_A[/tex]
[tex]60-12t = 60 - 18(t-0.75)[/tex]
[tex]60-12t =73.5 - 18t[/tex]
[tex]60- 73.5 = - 18t+ 12t[/tex]
[tex]-13.5 =-6t[/tex]
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
[tex]S_B = S_A[/tex]
[tex]d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
[tex]d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2[/tex]
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ cấu ở trạng thái cân bằng. Tại thời điểm khâu 1 ở vị trí υ1 = 900 hãy tính áp lực tại các khớp động tại B, C và A.
What statement about the print() function is true?
print() has a variable number of parameters.
print() can have only one parameter.
print() can be used to obtain values from the keyboard.
print() does not automatically add a line break to the display.
Explanation:
print() has a variable number of parameters. this is the answer.
hope this helps you
have a nice day
Why water parameters of Buriganga river vary between wet and dry seasons?
Explain.
Compute the minimum length of vertical curve that will provide 220 m stopping sight distance for a design speed of 110 km/h at the intersection of a -3.50% grade and a +2.70% grade.
i have made notes and saved it as a pdf u can take it to answer question and make ur concept good
The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m
We are given;
Stopping sight distance; S = 220 m
Design Speed; V = 110 km/h
Intersection grade 1; G1 = +2.7
Intersection Grade 2; G2 = -3.5
From the AASHTO Table attached, we can trace the value of the radius of vertical curvature for the given stopping sight distance and design speed.From the table, at S = 220 m and V = 110 km/h, we can see that;
Radius of vertical curvature; K = 74
Now, the difference in grade given is;A = G1 - G2
A = 2.7 - (-3.5)
A = 2.7 + 3.5
A = 6.2
Formula for the minimum length of vertical curve is;L = KA
Thus;
L = 74 × 6.2
L = 458.8 m
Read more about stopping sight distance at; https://brainly.com/question/2087168
In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2 and a Y value of 1.0. An in-service stress level of 200 MPa has been calculated. What is the length of a surface crack (in mm) that will lead to fracture
Answer:
[tex]l=24mm[/tex]
Explanation:
From the question we are told that:
Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]
Y value [tex]Y=1.0[/tex]
Stress level of[tex]\sigma =200 MPa[/tex]
Generally the equation for length of a surface crack is mathematically given by
[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]
[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]
[tex]l=0.024m[/tex]
Therefore
in mm
[tex]l=24mm[/tex]
