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An Otto cycle with air as the working fluid has a compression ratio of 8.2. The ambient temperature is 298K. The maximum temperature of the Otto cycle is 2000K. Under cold air standard conditions, the thermal efficiency of this cycle is ____
Answer:
The correct answer is "57%".
Explanation:
Given:
Temperature:
[tex]T_a = 298 \ K[/tex]
[tex]T_3 = 2000 \ K[/tex]
Compression ratio (r),
= [tex]8.2[/tex]
For Otto cycle, the thermal efficiency will be:
⇒ [tex]\eta =1-\frac{1}{(r)^{\nu-1}}[/tex]
By substituting the values, we get
[tex]=1-\frac{1}{(8.2)^{1.4-1}}[/tex]
[tex]=1-\frac{1}{(8.2)^{0.4}}[/tex]
[tex]=57[/tex] (%)
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
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?
Explanation:
40.4m
Explanation:
Pressure at depth is given as
P = P, + pgh
Final pressure at depth h= 5 Po
5Po= Po + pgh
pgh = 4Po = 4 x 1.01 x 10^5
h = (4.04×10^5)/ (1000x10)
h=40.4m
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:
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tính toán dung lượng bộ nhớ yêu cầu để lưu trữ 6 phút tiếng nói theo định dàng truyền trong mạng PSTN?
Answer:
hugdxvb huiiiuttyuj ftfrr
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
what is the term RF exiciter?
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
1. Is the team developing confidence that the solution space has been fully explored?
2. Are there alternative diagrams and alternative ways to decompose the problem?
3. Have external sources been thoroughly pursued, and everyone’s ideas been accepted and integrated in the process?
4. All of the above
Answer:
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
4. All of the above
Explanation:
The team must explore its solution space, including some external sources. Then, it must integrate its findings with the ideas of team members, ensuring the consideration of all possible ways to decompose the problem. This is because employing a structured process to concept generation enables the team to come up with creative solutions to design concepts.
The mass flow rate through a centrifugal compressor is 1 kg/s. If air enters at 1 bar and 288k and leaves at 200 kN/m² and 370k, determine the power of the compressor. Take Cp = 1.103 kJ (kg.K), R = 287 kJ (kg.k)
Answer:
Explanation:
That's the correct answer to the question , thanks
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]
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/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)
What is the formula for resultant force
Answer:
F = 3 N + 2 N = 5 N
Explanation:
Resultant force F = 3 N + 2 N = 5 N to the right. The resultant force is 5 N to the right. Two forces that act in opposite directions produce a resultant force that is smaller than either individual force. To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force.
(a) What are the usual causes of mechanical failure in the component or system? ( b ) What are the general types of mechanical failure ?
Answer:
Some types of mechanical failure mechanisms are: excessive deflection, buckling, ductile fracture, brittle fracture, impact, creep, relaxation, thermal shock, wear, corrosion, stress corrosion cracking, and various types of fatigue. ... Cascading failures, for example, are particularly complex failure causes.
These waveforms are applied to a gated D latch, which is initially RESET. Which of the areas identified on the Q waveform is incorrect?
Question Completion with Options:
A) Area a B) Area b C) Area c D) Area d
Answer:
The incorrect waveform identified on the Q waveform is the:
C) Area c.
Explanation:
Area c is the incorrect waveform because its output is not correct. The Q waveform indicates that the electrical forces project toward the negative pole of the lead axis. A gated D latch is a flip flop latch with an additional control input, which determines when to change the state of the circuit. Most times, this control unit is a clock input or an enable input.
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.
You are working for a company that creates special magnetic environments. Your new supervisor has come from the financial side of the organization rather than the technical side. He has promised a client that the company can provide a device that will create a magnetic field inside a cylindrical chamber that is directed along the cylinder axis at all points in the chamber and increases in the axial direction as the square of the value of y, where y is in the axial direction and y = 0 is at the bottom end of the cylinder. Prepare a calculation to show that the field requested by your supervisor and promised to a client is impossible.
Answer:
Following are the responses to the given question:
Explanation:
When we take the entire cylinder as a surface that is:
[tex]B=B.y^2\ j\\\\[/tex]
for the magnetic filed existance
[tex]\bigtriangledown . \underset{b}{\rightarrow}=0[/tex]
[tex]\therefore[/tex]
the flub by this cyclinder is zero implies there theaming flubs = outgoing flubs
[tex]\bigtriangledown . \underset{b}{\rightarrow}\\\\=\frac{\delta }{\delta x} B_x+\frac{\delta }{\delta y } B_y+ \frac{\delta }{\delta z} B_z\\\\=\frac{\delta }{\delta x} B_0+\frac{\delta }{\delta y } B_{y^2}+ 0\\\\=\frac{\delta }{\delta x} B_0\ {y^2}=2 B_0\ y\\\\So,\\\\\bigtriangledown . \underset{b}{\rightarrow}\neq 0[/tex]
that's why it is impossible field.
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.
