Answer:
Mass flow rate = 15.96kg/s
Explanation:
From
P1V1=mRT1
V1=RT1/P1
V1=0.0492m³/kg
M=A(U)/V1
Where u is the velocity
A=πd²/4
M=π(0.2²)/4×25/0.0492
M=15.96kg/s
Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-ton sluice gate of a dam. Quality of construction and maintenance (including lubrication) are good, resulting in an estimated friction coefficient of only 0.1 for the screw. Collar friction can be neglected, as ball thrust bearings are used. Assuming that, because of gate friction, each screw must provide a lifting force of 26 tons, what power is required to drive each screw when the gate is being raised at the rate of 3 ft/min
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
he desk has a weight of 80 lb and a centerof gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcomethe static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. Thecoefficients of static and kinetic friction at A andB are μs= 0.45 andμk= 0.25, respectively.
Answer:
[tex]N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb[/tex]
Explanation:
Na and Nb are the vertical reactions on each of the two legs at A and at B
For the horizontal forces:
[tex]Fcos(30)-0.5N_a-0.5N_b=0\\0.5N_a+0.5N_b= Fcos(30)\\N_a+N_b= 2Fcos(30)[/tex]
For the vertical forces:
[tex]N_a+N_b-Fsin(30)-75=0\\N_a+N_b=Fsin(30)+80[/tex]
Therefore equating both equations:
[tex]2Fcos(30)=Fsin(30)+80\\F(2cos(30)-sin(30))=80\\F=\frac{80}{2cos(30)-sin(30)} =64.93N[/tex]
After the desk star to slide:
sum of all vertical force = ma , therefore:
[tex]N_a+N_b-64.93sin(30)-80=0\\N_a+N_b=64.93sin(30)+80[/tex]
sum of all horizontal force = ma
[tex]64.93cos(30)-0.2N_a-0.2N_b=\frac{80lb}{32.2ft/s^2}a\\ 0.2(N_a+N_b)=64.93cos(30)-\frac{80lb}{32.2ft/s^2}a\\N_a+N_b=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}a}{0.2}=324.65-12.42a[/tex]
equating both equations:
[tex]324.65-12.42a=64.93sin(30)+80\\12.42a=324.65-64.93sin(30)-80\\12.42a=212.185\\a=17.08ft/s^2[/tex]
From the moment equation:
[tex]4N_b-80(2)-64.93(3)=\frac{-80}{32.2} (17.08)(2)\\N_b=67.48lb[/tex]
[tex]N_a=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}(17.08)}{0.2}-67.48 = 45.04lb[/tex]
For each leg: [tex]N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb[/tex]
A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator
Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
What is Postflow used to protect?
Answer:
The idea is to protect the puddle while it cools
Explanation:
3. A storage tank is connected to a pond (at atmospheric pressure!) by a length of 4" pipe and a gate valve. From previous operating experience, it has been found that when the tank is at a pressure of 3 atm the flow through the pipe is 35 m3/h when the gate valve is fully open. If the pressure in the tank increases to 5 atm, what will be the maximum discharge rate from the tank?
Answer:
V2= 21m³/h
Explanation:
According to Boyle's law, pressure and flow rate of gas can be calculated from the following equation:
P1V1=P2V2
3 × 35 = 5 × V2
V2= 21m³/h
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2 (b) Show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs. (c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs
Answer:
Explanation:
a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2
We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX
If AB = 00 select [tex]I_o[/tex]
If AB = 01 select [tex]I_1[/tex]
If AB = 1_(B is don't care), select [tex]I_2[/tex]
However, the truth table is attached and shown in the first file below.
Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.
b) We are show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
For four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output
This question is a multiplexer which is a topic in digital circuit.
Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.
If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.
a)
Two 2-to-1 multiplexers to form a 3-to-1 MUX.
If AB = 00, select [tex]I_o[/tex]
If AB = 01, select [tex]I_1[/tex]
If AB = 1- (B is don't care) select I
The truth table for the above scenario is in the attached document below.
Figure 1 and 2 represents the solution to this question.
b).
Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.
In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.
Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.
In the attached diagram, figure 4 shows how it is structured.
We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.
Learn more about multiplexers here;
https://brainly.com/question/25953942
The effective resistance of parallel resistors is always _____ than the lowest individual value.
a) more
b) less
c) no different than
Answer:
A
Explanation:
Answer:
the answer is a
Explanation:
it is a because thats what the answer is
A wheel tractor is operating in it is fourth gear range with full rated revolution per minute. Tractors speed is 7.00 miles per hour. Ambien air temperature is 60 Fahrenheit, and operating attitude is sea level. This tractor is towing a fill material loaded pneumatic trailer while climbing with 5 % slop. Rolling resistance is 55lb/ton. Tractor is single axle and it is operating weight is 74,946 lb. The loaded trailer weighs 55,000 lb. The weight distribution is for the combined tractor- trailer unit is 53% to the drive axle and 47% to the rear axle.
The manufacturer, for the environmental conditions as given above, rates the tractive effort of the new tractor at 330 rimpull HP. What percentage of this manufacturers rated rimpul HP actually develop?
If the symbol is on the top and bottom of the reference line, we call this __________ sides.
A. arrow
B. other
C. both
Hot oil is to be cooled by water in a 1-shell-pass and 8-tube-passes heat exchanger. The tubes are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 5 m, and the overall heat transfer coefficient is 310 W/m2.K. Water flows through the tubes at a rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/is. The water and the oil enter at temperatures of 20 C and 150 C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.
Answer:
Rate of heat transfer is 66.8°C
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
1. Deformation of a Cylindrical Pressure Vessel A thin–walled cylindrical pressure vessel of inner radius r and thickness t r is subjected to an internal pressure p. The pressure vessel is made from steel which has elastic material constants E = 200 GPa and ν = 0.30. If the vessel has a length ` = 3 m, an inner radius of r = 0.65 m, and a wall thickness t = 15 mm, determine the following items for a pressure p = 25 MPa. Assume that elongation is positive and contraction is negative. (a) Determine the hoop (circumferential) strain in the pressure vessel. (b) Determine the axial (longitudinal) strain in the pressure vessel. (c) Determine change in the circumference of the pressure vessel caused by the internal pressure. (d) Determine the change in the outer diameter of the pressure vessel caused by the internal pressure. (e) Determine the change in the length of the pressure vessel caused by the internal pressure.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Two identical bulbs are connected to a 12-volt battery in parallel. The voltage drop across the first bulb is 12 volts as measured with a voltmeter. What is the voltage drop across the other bulb?
Answer:
12 volts
Explanation:
The voltages across parallel-connected items are identical. (In fact, that's why you can measure the voltage by connecting the voltmeter in parallel with the circuit element.)
The voltage drop across each bulb is 12 volts.
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? (c) In bullet format compare and contrast the expected mechanical behavior of hypoeutectoid and hypereutectoid steels in terms of: (i) Yield strength (ii) Ductility (iii) Hardness (iv) Tensile strength (d) If you want to choose an alloy to make a knife or ax blade would you recommend a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your recommendation in 1-3 bullet points. (e) If you wanted a steel that was easy to machine to make a die to press powders or stamp a softer metal, would you choose a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your choice in 1-3 bullets.
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
(a)
Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.
Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.
(b)
The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.
Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite. The carbon concentration for both ferrites is 0.022 wt% C.
(c)
(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.
(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.
(iii) Hardness: hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.
(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.
(d)
I would recommend hypereutectoid steel alloy to make a knife or ax blade
1- Hardness is required at the surface of the blades.
2- Ductility is not needed for such application.
3- Due to constant impact, the material will not easily yield to stress.
(e)
I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.
1- hypoeutectoid steel alloys have high machinability, hence better productivity
2- It will be used on softer metals, hence its fitness for the application
3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.
Explanation:
See all together above
For what type of metal is high speed steel drill best suited?
Answer:
high speed steel I believe
A complex gear drawing done on a drawing sheet marked M-1 has many section views showing important interior details of the gear. One of the cutting-plane lines is marked at the ends with a callout in a circular bubble that says 7 above a line and M-3 below the line. To find this detail, you would
Answer:
The answer is "go to sheet M-3 and look for a detail labeled 7".
Explanation:
In the given question some information is missing, that is choices so, the correct choice can be described as follows:
In gear drawing, we use equipment that sorts a very important technical reference necessary for machinery design. If a manufacturer wants a tool in the production of a new computer, two choices are available to design the new equipment itself. To use standard features that have already been developed. In this gear drawing to find the details we go to sheet in M-3 and for the detailed labeled 7.It is desired to obtain 500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor. What should be the angle of the AC to AC converter to be used? Calculate the THD of the current drawn from the mains (consider up to the 12th harmonic)?
Answer:
14.5° ; THD % = 3.873 × 100 = 387.3%.
Explanation:
Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;
(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".
(2). Consideration of up to 12th harmonic.
So, let us delve right into the solution to the question above;
Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;
Irms = reactive power /Vrms = 500/230 = 2.174 A.
Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.
Step two: Calculate the THD.
Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.
h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.
Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.
THD % = 3.873 × 100 = 387.3%.
Step four: angle AC - Ac converter
theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.
A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountainous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF=0.90.
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remain the same but the peak-hour factor increases to 0.95?
Process: (1) determine passenger car equivalent for trucks and buses; (2) determine passenger car equivalent for recreational vehicles; (3) calculate heavy vehicle factor; (4) determine 15-min passenger equivalent flow rate for current conditions; (5) determine 15-min passenger equivalent flow rate at full capacity; (6) calculate the volume for current and capacity conditions; (7) take the difference of the two volumes to determine how many vehicles were added
Answer:
The number of vehicles added to this highway before the capacity is reached is 1,511 vehicles.
Explanation:
see attached image
A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 5.9 × 10^(-11) m^2/s, and the diffusion flux is found to be 4.7 × 10^(-7) kg/m^2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.5 kg/m^3.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
Answer:
0.544×10–³
Explanation:
Please see the attached file for the solution
The amusement park ride consists of a fixed support near O, the 6-m arm OA, which rotates about the pivot at O, and the compartment, which remains horizontal by means of a mechanism at A. At a certain instant, β=30ο, 2 0.75 rad/s, and 0.5 rad/s , all clockwise. Determine the horizontal and vertical forces (F and N) exerted by the bench on the 75-kg rider at P. Compare your results with the static values of these forces. (Use x-y coordinate system and vector equations
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
As discussed in the text, one possible performance enhancement is to do a shift and add instead of an actual multiplication. Since 9 x 6, for example, can be written (2 x 2 x 2 + 1) x 6, we can calculate 9 x 6 by shift ing 6 to the left 3 times and then adding 6 to that result. Show the best way to calculate 0 x33 x 0 x 55 using shift s and adds/subtracts. Assume both inputs are 8-bit unsigned integers.
Answer:
The best way to calculate 0 x33 x 0 x 55 using shift s and adds/subtracts and assuming both inputs are 8-bit unsigned integers is attached below
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values to explain your result. Assume the water pressure is 30 psistr
Answer:
hoop stresslongitudinal stressmaterial usedall this could led to the failure of the garden hose and the tear along the length
Explanation:
For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :
hoop stress ( which acts along the circumference of the pipe):
αh = [tex]\frac{PD}{2T}[/tex] EQUATION 1
and Longitudinal stress ( acting along the length of the pipe )
αl = [tex]\frac{PD}{4T}[/tex] EQUATION 2
where p = water pressure inside the hose
d = diameter of hose, T = thickness of hose
we can as well attribute the failure of the hose to the material used in making the hose .
assume for a thin cylindrical pipe material used to be
[tex]\frac{D}{T}[/tex] ≥ 20
insert this value into equation 1
αh = [tex]\frac{20 *30}{2}[/tex] = 60/2 = 30 psi
the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose
A converging–diverging nozzle is designed to produce a Mach number of 2.5 with air. (a) What operating pressure ratio (prec/pt inlet) will cause this nozzle to operate at the first, second, and third critical points? (b) If the inlet stagnation pressure is 150 psia, what receiver pressures represent operation at these critical points? (c) Suppose that the receiver pressure were fixed at 15 psia. What inlet pressures are necessary to cause operation at the critical points?
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
(20 points) (Assessment of Outcome 1): A plant has two identical standby generator units for emergency use. In the area of the generators, the normal noise level registers 82 dBA on the sound-level meter with the generators turned off. When one generator switches on, the SLM needle jumps to 85.8 dBA. What will the dBA reading be when the second generator also turns on (so that both generators are on)
Answer:
It wouldn't get any louder then maybe 3db more
Explanation:
There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.
8. When supplying heated air for a building, one often chooses to mix in some fresh outside air with air that has been heated from the building as it passes through the furnace. An insulated mixing chamber is used to combine two streams of air to be used in a building. One stream of air, brought from the outside, enters at 2 kg/s, at a pressure of 120 kPa, and a temperature of 5oC. The second stream of air, coming from the building’s furnace, has a mass flow rate of 8 kg/s, a pressure of 120 kPa, and a temperature of 35oC. The combined stream is then delivered to the warm space at 120 kPa. Determine the rate of entropy generation for this mixing chamber.
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Tech A says that when checking tire pressure, the tire should be " cold." Tech B says that the tires should be driven more than 3 miles before checking tire presure. Who is correct?
Answer: Technician A is correct.
Explanation:
Technician A is correct because temperature of a tire will affect its pressure reading.
Tires attract heat because of their dark colour and then motion on the road generates heat. A car owner or technician should know that tire pressure is most accurate when the tire is cold (especially when atmospheric temperature is cool too).
If the tires are driven three miles first, their temperature will be high (due to the rubbing of the tires on the surface of the road). This higher temperature will result in higher per square inch (psi) readings.
Temperature has a great influence on the tire pressure.
Even if the tire is driven up to or more than 3 miles, it should still be left to cool, before tire pressure is checked.
The tire manufacturer's rating should be the maximum possible tire pressure. If an abnormal reading is gotten, the gauge should be properly checked.
Steam enters and leaves a horizontal 1D pipe steadily at a constant speed with the specific enthalpy to be 4000 kJ/kg and 1500 kJ/kg respectively. Assuming a mass flow rate of 0.5 kg/sec and If there is no significant change in potential energy from inlet to exist, determine the rate of heat transfer between the pipe and its surroundings.
Answer:
The rate of heat transfer between pipe and its surrounding is found to be 1250 KW
Explanation:
Assuming no significant change in potential energy. When we apply the first law of thermodynamics to the given system of 1-D pipe, we get the following expression:
Q = m(h₁ - h₂)
where,
Q = Heat Transfer between pipe and surrounding = ?
m = mass flow rate of steam = 0.5 kg/sec
h₁ = Specific Enthalpy of steam at entrance = 4000 KJ/kg
h₂ = Specific Enthalpy of steam at exit = 1500 KJ/kg
Therefore, using these values in the equation, we get the value of heat transfer:
Q = (0.5 kg/sec)(4000 KJ/kg - 1500 KJ/kg)
Q = 1250 KW
A large truck drives down the highway at 10 m/s hauling a rectangular trailer that is 6 m long, 2 m wide, and 2 m tall. The trailer contains frozen food and is therefore temperature-controlled. Its external surface can be approximated to be a consistent 15°C, while the outside air is at 20°C. Assume the heat transfer on the front, back, and bottom of the trailer is negligible.
a) How much cooling power must be provided to maintain the temperature controlled trailer? (i.e. What is the total heat transfer rate from the air to the trailer)?b) What is the minimum local heat transfer coefficient on the surface of the trailer? Where does the minimum occur?c) What percentage of the total heat transfer is occuring over a laminar boundary layer?d) If the cooling system can only provide 5 KW of cooling, what is the fastest speed that this truck can drive while still adequately maintaining the temperature within the trailer?
Answer:
3w/m²k
Explanation:
Base on the scenario been described in the question, the solution to the given problem solve in the file attached below
Determine the drag on a small circular disk of 0.02-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.89 and a viscosity 10000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
Answer:
33.3%
Explanation:
Given that:
specific gravity (SG) = 0.89
Diameter (D) = 0.01 ft/s
Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]
Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]
Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]
the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]
Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%
Discuss the ethics of the circumstances that resulted in the Columbia shuttle disaster. Considering the predictions that were made years before the disaster, as well as the reliability of the Binomial distribution and its implications, what could or should the engineers associated with the program have done differently
Explanation:
This is not so much a mathematical issue as a case study, because the response will inevitably require us to test the special Columbic shuttle disaster scenario. I would suggest that you read this in detail and present the points accordingly. Here I give as many points as I think are relevant.
The failure of a space program is definitely a complex situation, more than a simple binomial distribution. It's definitely not as simple as repeating the flip of a coin. There are several coherent factors and situations that govern the overall coordination and execution of such an event. The problem is, those who are running a project like this are still making a trade off,It is never the case that they sealed the lid on any chance of failure between multiple parameters. You try to do something, but often, as is the case above, the potentially dangerous situation is impossible or uncontrollable. Since the root cause of failure, which is dried out tiles that can not withstand heat and water, it appears that owing to the constant use of the shuttle the head architects have not foreseen this.
a) The current that goes through a 100 mH inductor is given as
i(t) = 6 - 2e^-2t A t >= 0
Find the voltage v(t) across the inductor.
b) The voltage v(t) = 5sin(5t) V is applied across the terminals of a 200 mH inductor. The initial current through the inductor is i(0) = -10 A. Find the current i(t) through the inductor for t > 0.
Answer:
A) V(t) = 0.4e^-2t
B) i(t) = (25tsin5t+10) A for t>0
Explanation:
Formula for calculating voltage across an inductor is expressed as:
V = Ldi/dt
Given L = 100mH = 100×10^-3
If i(t) = 6 - 2e^-2t A t >= 0
di/dt = (-2)(-2)e^-2t
di/dt = 4e^-2t
If t ≥ 0
V(t) = 100×10^-3 × (4e^-2t)
V(t) = 0.1×4e^-2t
V(t) = 0.4e^-2t for t≥0
B) Applying the same formula as above
V = Ldi/dt
Vdt = Ldi
V/L dt = di
On integration
Vt/L = i + C
When t = 0, i = -10A
Substituting the values into the formula
V(0)/L = -10 + C
0 = -10+C
C = 10
To get the current i(t) through the inductor for t>0,
Since Vt/L = i + C
Given V(t) = 5sin5t Volts
L = 200mH = 200×10^-3H
C = 10
On substituting
(5sin5t)t/0.2 = i + 10
25tsin5t = i + 10
i(t) = (25tsin5t-10) A for t>0
