Natural Convection from an Oven Wall. The oven wall in Example 4.7-1 is insulated so that the surface temperature is 366.5 K instead of 505.4 K. Calculate the natural convection heat-transfer coefficient and the heat-transfer rate per m of width. Use both Eq. (4.7-4) and the simplified equation. (Note: Radiation is being neglected in this calculation.) Use both SI and English units.
Answers
Answer:
i) Heat transfer coefficient (h) = 7 w/m²k
ii) Heat transfer per meter width of wall
= h x L x 1 x (Ts - T₆₀)
= 7 x 0.3048 x (505.4 - 322) = 414.747 w/m
Explanation:
see attached image
Related Questions
A. ¿Qué opinión te merecen las palabras del n.° 138 de la carta encíclica? ¿Será real que todo está conectado? Da algún ejemplo de ello a partir de los textos leídos.
Answers
The words of No. 138 of the encyclical letter express a profound vision of interconnectedness in the world.
This idea reflects the reality that everything in life is linked in some way. For example, by studying the water cycle, we see how evaporation in one place can lead to precipitation in another, thus affecting the flora and fauna of both places.
Likewise, changes in global temperature impact terrestrial and marine ecosystems, showing how everything is connected in a complex and interdependent system.
The Question in English
A. What is your opinion of the words of No. 138 of the encyclical letter? Is it real that everything is connected? Give some example of this from the texts read.
what is the most important source of error in the Brinell test
Answers
Answer:
The measurement of the indentation.
Explanation:
Due to disparities in operators making the measurements, the results will vary even under perfect conditions. Less than perfect conditions can cause the variation to increase greatly.
A specific internal combustion engine has a displacement volume VD of 5.6 liters. The processes within each cylinder of the engine are modeled as a cold air-standard Diesel cycle with a cutoff ratio rc = 2.5. The pressure, temperature, and volume of the air at the beginning of compression are p1 = 95 kPa, T1 = 26◦C, and V1 = 6.0 liters. Use values of cv = 0.72 kJ/kg·K and γ = 1.38 for air. Determine: (a) the net work per cycle, in kJ. (b) the thermal efficiency η
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine how the valve will operate. Both the model and prototype will use water as the fluid. The model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve). If the prototype flow rate is to be 700 ft3 /s, determine the model flow rate. Use Reynolds scaling for the velocity.
Answers
Answer:
7.94 ft^3/ s.
Explanation:
So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.
Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.
Therefore; kp/ks = 1/6.
Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.
Reynolds scaling==> Hp/ 700 = (1/6)^2.5.
= 7.94 ft^3/ s
Engine oil flows through a 25-mm-diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while the tube surface temperature is maintained at 100°C. (a) Determine the oil outlet temperature for a 5-m and for a 100-m long tube. For each case, compare the log mean temperature difference to the arithmetic mean temperature difference.
Answers
The question involves calculating the outlet temperature of oil flowing through a tube with known inlet and surface temperatures for two different tube lengths, and comparing the log mean temperature difference to the arithmetic mean temperature difference in an Engineering context.
Explanation:The subject of the question is Engineering, specifically related to the thermodynamics and fluid mechanics domain. The student is given information about an oil flowing through a tube at a specific rate, with given inlet and surface temperatures, and is asked to find the oil's outlet temperature for tubes of two different lengths. The log mean temperature difference (LMTD) and the arithmetic mean temperature difference (AMTD) should be calculated and compared for both cases. This question involves the principles of heat transfer as well as fluid dynamics, which are typical topics covered in an undergraduate engineering curriculum. Additionally, the student may need to apply the concept of the thermal energy balance to determine the outlet temperature of the oil.
6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 ksi,σ2= 25 ksi,1= 0.00105, and2= 0.000195.(a) This is a plane stress state, meaning the principal stress normal to this plane is zero. Is theprincipal strain3acting normal to this plane also zero? Show why or why not. Draw the 3DMohr’s circle for both stress and strain states.(b) Determine the Modulus of Elasticity,E.(c) Determine Poisson’s ratio,ν.
Answers
Answer:
See attached image for diagrams and solution
Jane puts an unknown substance into a beaker. This substance takes the shape of the beaker. While sitting on the lab bench, untouched, the substance does not leave the open beaker. This substance is MOST likely a A) gas. B) solid. C) liquid. D) plasma.
Answers
Answer: Liquid
Explanation: This substance is liquid. Liquids are free to move and take the shape of their container, but do not expand to completely fill it.
This substance is MOST likely a liquid. Thus option C is correct.
What is beaker?Liquids can freely move and conform to the structure of their container, but they cannot enlarge to fill it entirely. A beaker is a circular container for holding liquids that is made of glass or plastic.
It is an utilitarian piece of apparatus used to measure liquids, heat them over a Bunsen burner's flame, and contain biochemical processes. It is liquid in nature. Liquids can freely move and conform to the shape of the container, but they cannot enlarge to fill it entirely. Jane fills a beaker with an unidentified chemical.
This material adopts the beaker's form. The chemical doesn't spill out of the open beaker when it's unattended on the lab table. Given that they are flexible and taking on the structure of the container in which they are in volume. Therefore, option C is the correct option.
Learn more about beaker, Here:
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A spherical seed of 1 cm diameter is buried at a depth of 1 cm inside soil (thermal conductivity of 1 Wm-1K-1) in a sufficiently large planter. There is a 1 cm thick layer of mulch (thermal conductivity of 2 Wm-1K-1) on top of the soil. The planter has top surface dimensions of 10 cm by 10 cm and is exposed to 200 Wm-2 of heat. You find the top surface temperature of the mulch (Ts,1) to be uniform at 50˚C. What is the surface temperature of the seed (Ts,2) in ˚C
Answers
Answer:
Find the attachment for the answer
Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also noisy and transmit their vibrations to the shelves and counters. A better design of the paint can shaker is possible using a balanced fourbar linkage. Design such a portable device to sit on the floor (not bolted down) and minimize the shaking forces and vibrations while still effectively mixing the paint.
Answers
Answer:
A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:
The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.
Explanation:
First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.
A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 10 in2 and that approximately 2.6 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. (convert the time from hours to years). The density of steel is 7.9 g/cm3. K
Answers
The time of submersion of the steel plate in years is; 10 years
Corrosion Penetration Rate Calculation
We are given;
Corrosion penetration rate; CPR = 200 mpyWeight of the corroded portion of plate; W = 2.6 kg = 2.6 × 10⁶ mgExposed surface area; A = 10 in²Density of the metal plate; ρ = 7.9 g/cm³The time of submersion of the steel plate is given by the formula;
t = KW/(ρA*CPR)
Now K is a constant and is equal to 534 provided CPR is in mpy and
A is in square inches.
Thus;
t = (534 * 2.6 × 10⁶)/(7.9 * 10 * 200)
t = 8.8 × 10⁴ hours
Now, 24 hours makes one day and there are 365 days in a year. Thus;
number of hours in a year = 24 * 365
Thus;
t in years = (8.8 × 10⁴)/(24 * 365)
t ≈ 10 years
Read more about corrosion at; https://brainly.com/question/5168322
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.Calculate roughly by what factor would doubling the base width of the BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor?
Answers
Complete Question:
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.
Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!
Answer & Explanation:
[Find the attachments]
Step 1 :
Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.
Emitter and base, collector and base are reverse biased then BJT in cut off region.
Three sketches one below the other is shown in Figure 1.
[find the figure in attachment]
Step 2:
Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.
Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.
[For Steps 3 4 5 6 and 7 find attachments]
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 1.1. If, after 129 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?
Answers
Answer:
3.305 * 10 ^ ⁻4
Explanation:
Solution
Given:
We calculate the value of k for which is the dependent variable in Avrami equation
y = 1 - exp (-kt^n)
exp (-kt^n) =1-y
-kt^n = ln (1-y)
so,
k =ln(1-y)/t^n
Now,
we substitute 1.1 for n, 0.50 for y, and 129 s for t
k = ln (1-0.50)/129^1.1
k= 3.305 * 10 ^ ⁻4
Now, suppose that you have a balanced stereo signal in which the left and right channels have the same voltage amplitude, 500 mVpp. This time, however, you want to be able to mix these two channels into a single inverted output while independently varying the gain of the two channels. Design and build an op amp circuit with potentiometers so that you can independently vary the gain of the left and right channels. Choose resistors so that the overall output of your circuit ranges between 0.1Vpp (when both channels are set to minimum gain) and 20Vpp (when both channels are set to maximum gain).
Answers
Answer:
R₁ = 32kΩ
Explanation:
See attached image
An aluminum cylinder bar ( 70 GPa E m = ) is instrumented with strain gauges and is subject to a tensile force of 5 kN. The diameter of the bar is 10 cm. The Poisson’s ratio of the bar is 0.33. A Wheatstone bridge is constructed to measure the axial strain. Gauge 1 measures the axial strain and gauge 2 measures the lateral strain.
Answers
Find the complete solution in the given attachments.
Note: The complete Question is attached in the first attachment as the provided question was incomplete
Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtain an expression for the view factor of the groove with respect to the surroundings outside the groove. (b) For the V groove, obtain an expression for the view factor F12, where A1 and A2 are opposite surfaces. (c) If H
Answers
The heat transfer while the reference information pertains to thermal expansion and physics-related work. Thermal expansion affects the volume, cross-sectional area, and height of objects, and these changes can be calculated using the coefficient of thermal expansion, initial dimensions, and temperature change.
Explanation:The view factor calculations in heat transfer, specifically related to grooves machined from a solid block of material. While the initial question seems to relate to this topic, the provided reference information does not align with the question and seems to cover thermal expansion and work done by forces, which are different aspects of Physics.
However, to answer the student's question regarding thermal expansion, we can consider that when temperature changes, all dimensions of an object change. The volume change ΔV can be calculated using the formula ΔV = α·V·ΔT, where α is the coefficient of thermal expansion, V is the original volume, and ΔT is the temperature change. For block A with volume L·2L·L and block B with volume 2L·2L·2L, the change in volume will be proportional to each block's respective original volume.
The change in cross-sectional area, typically lw for block A and 2Lw for block B, and change in height h or 2L for each block, will be affected in a similar manner and can be calculated using α, the original dimensions, and the temperature change ΔT.
An air standard cycle with constant specific heats is executed in a closed pistoncylinder system and is composed of the following four processes: 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion with a volume ratio, r1=V4/V3 4-1 Constant pressure heat rejection with a volume ratio, r2=V4/V1 (a) Sketch the P-v and T-s diagrams for this cycle. (b)Find out T2/T1 as a function of k, r1, r2 only. (c) Find out T4/T1 as a function of k, r1, r2 only. (d)Find out T3/T4 as a function of k, r1, r2 only. (e) Find out T3/T2 as a function of k, r1, r2 only. (f) Obtain an expression for the back work ratio for a fixed minimum-tomaximum temperature ratio T1/T3. The expression should be of a function of T1/T3, k, r1, r2 only. (g)Obtain an expression for the cycle thermal efficiency as a function of k, r1, r2 only.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to get the step by step explanation to the question above.
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given as 1:2:3 for cement, sand, and gravel with w/c of 0.50. The specific gravities of sand and gravel are 2.60 and 2.70 respectively. Entrained air content is 7.5%. How many pounds of cement, water, sand, and gravel are needed for the driveway?
Answers
Answer:
Weight of cement = 10968 lb
Weight of sand = 18105.9 lb
Weight of gravel = 28203.55 lb
Weight of water = 5484 lb
Explanation:
Given:
Entrained air = 7.5%
Length, L = 40 ft
Width,w = 12 ft
thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft
Specific gravity of sand = 2.60
Specific gravity of gravel = 2.70
The volume will be:
40 * 12 * 0.5 = 240 ft³
We need to find the dry volume of concrete.
Dry volume = wet volume * 1.54 (concrete)
Dry volume will be = 240 * 1.54 = 360ft³
Due to the 7% entarained air content, the required volume will be:
V = 360 * (1 - 0.07)
V = 334.8 ft³
At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:
Total of ratio = 1+2+3 = 6
Their respective volume will be =
Volume of cement = [tex] \frac{1}{6}*334.8 = 55.8 ft^3 [/tex]
Volume of sand = [tex] \frac{2}{6}*334.8 = 111.6 ft^3 [/tex]
Volume of gravel = [tex] \frac{3}{6}*334.8 = 167.4 ft^3 [/tex]
To find the pounds needed the driveway, we have:
Weight = volume *specific gravity * density of water
Specific gravity of cement = 3.15
Weight of cement =
55.8 * 3.15 * 62.4 = 10968 pounds
Weight of sand =
111.6 * 2.60 * 62.4 = 18105.9 lb
Weight of gravel =
167.4 * 2.7 * 62.4 = 28203.55 lb
Given water to cement ratio of 0.50
Weight of water = 0.5 of weight of cement
= 1/2 * 10968 = 5484 lb
Diborane is used in silicon chip manufacture. One facility uses a 500-lb bottle. If the entire bottle is released continuously during a 20-min period, determine the location of the 5 mg/m3 ground-level isopleth. It is a clear, sunny day with a 5 mph wind. Assume that the release is at ground level. Assume now that the bottle ruptures and that the entire contents of diborane are released instantaneously. Determine, at 10 min after the release,
Answers
Answer:
We want to determine the location after 10mins
Explanation:
The release of diborane is continuous
It is release at a rate of 500lb per 20mins
Then, let find the rate in mg/s
1 lb = 453592.37 mg
So, the mass rate Q is
Q = 500lb / 20mins
Q = 500 × 453592.37 mg / 20 × 60sec
Q = 188,996.82 mg/s
Given that mass concentration of
m~ = 5mg/m³
Then,
Rate of volume is
V~ = Q / m~
V~ = 188,996.82 / 5
V~ = 37,799.364 m³/s
The wind speed is
V = 5mph
Let convert to m/s
1 mph = 0.447 m/s
Then, 5mph = 2.235 m/s
From Pasquill Gilford, the cloud atmosphere characteristic is class A
To know the location, we will divide the velocity by heat rate
X = V / Q
X = 2.235 / 188,996.82 mg/s
X = 2.235 / 0.188996kg/s
X = 11.83 m / kg
The location is 0.000001183 m per mg of diborane
A 5% upgrade on a six-lane freeway (three lanes in each direction) is 1.25 mi long. On thissegment of freeway, there is 3% SUTs and 7% TTs, and the peak hour factor is 0.9. The lanes are 12ft wide, there is no lateral obstructions within 6ft from the roadway, and the total ramp density is 1.0 ramps per mile. What is the maximum directional peak-hour volume that can be accommodated without exceeding LOS C operating conditions
Answers
Answer:
3.586.543veh/hr
Explanation:
A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED will light up when f exceeds 50 Hz. The LED has a forward built-in voltage of 2 V. It is okay if the LED flickers when it’s ON, but it should not light up at all when OFF (Hint: use an "ideal" filter along with other components).
Answers
Answer:
See explaination
Explanation:
LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.
We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons
See attached file for detailed solution of the given problem.
Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has emissivity of 0.90 and is maintained at 950 K. The side surface is black and is maintained at 450 K. Heat is supplied from the base surface at a rate of 340 kW. Determine the emissivity of the top surface and the net rates of heat transfer between the top and bottom surfaces, and between the bottom and side surfaces. Answers: 0.44; 54.4 kW; 285.6 kW
Answers
Answer:
Check the explanation
Explanation:
Assumptions.
1. The surfaces are diffuse, may and opaque
2. steady operating conditions exist
3. Heat transfer from and to the surfaces is only due to Radiation
Consider the base surface to be surface 2 the top surface to be surface and the side surfaces to surface 3 1. cubical furnace can be considered to be three-surface enclosure. the areas and black body emissive powers of surfaces can be calculated as seen in the attached images below.
a robot arm moves so that p travels in a circle about point b which is not moving. knowing that p starts from rest, and its speed increases at a constant rate of 10mm/s, determine (a) the magnitude of the acceleration when t=4s, (b) the time for the magnitude of the acceleration to be 80 mm/s^2
Answers
Answer:
(a)10.20 mm/s² (b) 403200 s⁴
Explanation:
Solution
Recall that,
The tangible acceleration is a₁ = 10mm/s
The speed = a₁t
Normal acceleration = aₙ = v₂ /р = a₁²t₂/ р
where р = 0.8m = 800 mm
Now,
When t = 4s
v = (10) (4) = 40 mm/s
Thus,
aₙ = (40)² /800 = 2 mm/s²
Then
The acceleration is,
a = √a₁² + aₙ² = √ (10)² + (2)²
a = 10.20 mm/s²
(b) The time for he magnitude of the acceleration to be 80 mm/s^2
a² = aₙ² +a₁²
(80)² + [ (10)²t²/800]² + 10²
so,
t⁴ = 403200 s⁴
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a pressure ratio of 8, and the temperature of the gases at the turbine inlet is 2200 F. Utilizing the air-standard assumptions, determine (a) the temperature and pressure of the gases at every point of the cycle, (b) the velocity of the gases at the nozzle exit
Answers
Answer:
Pressure = 115.6 psia
Explanation:
Given:
v=800ft/s
Air temperature = 10 psia
Air pressure = 20F
Compression pressure ratio = 8
temperature at turbine inlet = 2200F
Conversion:
1 Btu =775.5 ft lbf, [tex]g_{c}[/tex] = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²
Air standard assumptions:
[tex]c_{p}[/tex]= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R
k= 1.4
Energy balance:
[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\[/tex]
As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible
hence [tex]v_{a} ^{2} = 0[/tex]
[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }[/tex]
[tex]T_{1}[/tex] = 20+460 = 480°R
[tex]T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}[/tex]= 533.25°R
Pressure at the inlet of compressor at isentropic condition
[tex]P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}[/tex]
[tex]P_{a}[/tex] = [tex](10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}[/tex]= 14.45 psia
[tex]P_{2}= 8P_{a} = 8(14.45) = 115.6 psia[/tex]
Answer:
a) The temperature and pressure of the gases at every point of the cycle are
T = 38.23 K
P = 2.91 kpa
Respectively
b) The velocity V of the gasses at the nozzle exit = 3590 m/s
Explanation: Please find the attached files for the solutions
Computer-controlled instrument panel dimming is being discussed. Technician A says the body computer dims the illumination lamps by varying resistance through a rheostat that is wired in series to the lights. Technician B says the body computer can use inputs from the panel dimming control and photo cell to determine the illumination level of the instrument panel lights on certain systems.Who is correct?A. A onlyB. B only.C. Both A and B.D. Neither A nor B.
Answers
Answer:
Answer : C ( Both A&B)
Explanation:
The level of illumination in the instrument panel lights will be determined by the panel dimming control and photocell. This panel dimming control consist of a potentiometer. The diameter positions existing in the diameter control acts like variable resistor. Based on its voltage drop BCM (Body Control Module) selects the intensity level by comparing the signal captured from photocell.
Another type is body control module receives the signal from head light rheostat which will be sent to instrument cluster. The instrument cluster controls the lamp intensity. Therefore, the statements said by both the technicians are correct.
Then, the correct option is C
You want to amplify a bio-potential signal that varies between 2.5 V and 2.6 V. Design an amplifier circuit for this signal such that the output spans 0 V to +10 V. The signal cannot be inverted. You can use any number of op amps and any number of resistors (with any values). But you can use only one +10 V DC voltage source (for powering the op amps as well as for any other needs). Clearly draw the complete circuit and show all component values.
Answers
Answer:
See attachment
Explanation:
Gain= Vo/Vin
If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7
Using above value of gain, let's design non-inverting op-amp configuration
Gain= 1+Rf/Rin
3.7= 1= Rf/Rin
2.7= Rf/Rin
If Rin=100Ω then Rf= 270Ω
Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rate of 1.8 kg/s. Water is heated by steam condensing at 120°C (hfg = 2203 kJ/kg) in the shell. If the overall heat transfer coefficient of the heat exchanger is 700 W/m2 ·°C, determine the length of the tube required in order to heat the water to 80°C using (a) the LMTD method, and (b) the ????????–NTU method. Answers: 129.5 m; 129.6 m
Answers
Answer:
Length = 129.55m, 129.55m
Explanation:
Given:
cp of water = 4180 J/kg·°C
Diameter, D = 2.5 cm
Temperature of water in = 17°C
Temperature of water out = 80°C
mass rate of water =1.8 kg/s.
Steam condensing at 120°C
Temperature at saturation = 120°C
hfg of steam at 120°C = 2203 kJ/kg
overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C
U = 700 W/m2 ·°C
Since Temperature of steam is at saturation,
temperature of steam going in = temperature of steam out = 120°C
Energy balance:
Heat gained by water = Heat loss by steam
Let specific capacity of steam = 2010kJ/Kg .°C
Find attached the full solution to the question.
g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is estimated to be 3300 kJ/h, which should be transferred out of the food department by the refrigerator. The heat rejection from the refrigerator to the environment is 4800 kJ/h. Determine the power input required by the refrigerator, in kW and the COP of the refrigerator. Is the refrigeration cycle reversible, irreversible, or impossible
Answers
Answer:
a) [tex]\dot W = 0.417\,kW[/tex], b) [tex]COP_{R} = 2.198[/tex], c) Irreversible.
Explanation:
a) The power input required by the refrigerator is:
[tex]\dot W = \dot Q_{H} - \dot Q_{L}[/tex]
[tex]\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)[/tex]
[tex]\dot W = 0.417\,kW[/tex]
b) The Coefficient of Performance of the refrigerator is:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W}[/tex]
[tex]COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]
[tex]COP_{R} = 2.198[/tex]
c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:
[tex]COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}[/tex]
[tex]COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}[/tex]
[tex]COP_{R,ideal} = 6.218[/tex]
The refrigeration cycle is irreversible, as [tex]COP_{R} < COP_{R,ideal}[/tex].
Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and requires one worker at $ 15.00 per hour. Fixed cost of the manual method is $ 5,000 per year. The automated method produces 25 pc per hour, has a fixed cost of $ 55,000 per year, and a variable cost of $ 4.50 per hour. Determine the Break – Even Point (BEP) for the two methods; that is, determine the annual production quantity at which the two methods have the same annual cost. Ignore the cost of material used in the two methods
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The break-even point is approximately 27,778 pieces per year for both manual and automated methods.
To determine the break-even point (BEP) for the two production methods, we need to equate the total annual costs of the manual method with the total annual costs of the automated method. Let's denote:
- [tex]\( Q \)[/tex] as the annual production quantity (in pieces)
- [tex]\( C_{\text{manual}} \)[/tex] as the total annual cost of the manual method
- [tex]\( C_{\text{automated}} \)[/tex] as the total annual cost of the automated method
For the manual method:
[tex]\[ C_{\text{manual}} = \text{Fixed cost} + (\text{Hourly wage} \times \text{Hours per year}) \][/tex]
[tex]\[ C_{\text{manual}} = 5000 + (15 \times Q/10 \times 24 \times 365) \][/tex]
For the automated method:
[tex]\[ C_{\text{automated}} = \text{Fixed cost} + (\text{Variable cost per hour} \times \text{Hours per year}) \][/tex]
[tex]\[ C_{\text{automated}} = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]
To find the break-even point, we set [tex]\( C_{\text{manual}} = C_{\text{automated}} \) and solve for \( Q \):[/tex]
[tex]\[ 5000 + (15 \times Q/10 \times 24 \times 365) = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]
Let's solve this equation for [tex]\( Q \)[/tex]:
[tex]\[ 5000 + 54 \times Q = 55000 + 52.20 \times Q \][/tex]
[tex]\[ 54 \times Q - 52.20 \times Q = 55000 - 5000 \][/tex]
[tex]\[ 1.80 \times Q = 50000 \][/tex]
[tex]\[ Q = \frac{50000}{1.80} \][/tex]
[tex]\[ Q \approx 27777.78 \][/tex]
So, the break-even point for the two methods is approximately 27778 pieces per year.
A water treatment plant processes 30,000 cubic meters of water each day. A square rapid-mix tank with vertical baffles and flat impeller blades will be used. The design detention time and velocity gradient are 30 seconds and 900 s-1 Determine the power input, if the temperature of the water is 20°C. µ = 1 x 10-3 kg/m•s. 1kW = 1000 J/s, 1J = 1N•m = 1 kg•m2/s2. Note: you can use the equation in its current version.
Answers
Answer:
P=8.44 kw
Explanation:
[Find the given attachment for solution]
Look at the home page of the Internet Society (www.internetsociety.org) and read about one of the designers of the original ARPANET—Larry Roberts, Leonard Kleinrock, Vinton Cerf, Robert Kahn, John Postel, or others. Learn about the early days of networking and the contributions that these individuals made to the ultimate development of the Internet. The home page of the Internet Society has links to many other places that provide a wealth of fascinating information about networks in general and the Internet and the web in particular.
Answers
Answer:
ARPANET is the direct precedent for the Internet, a network that became operational in October 1969 after several years of planning.
Its promoter was DARPA (Defense Advanced Research Projects Agency), a US government agency, dependent on the Department of Defense of that country, which still exists.
Originally, it connected research centers and academic centers to facilitate the exchange of information between them in order to promote research. Yes, being an undertaking of the Department of Defense, it is understood that weapons research also entered into this exchange of information.
It is also explained, without being without foundation, that the design of ARPANET was carried out thinking that it could withstand a nuclear attack by the USSR and, hence, probably the great resistance that the network of networks has shown in the face of major disasters and attacks.
It was the first network in which a packet communication protocol was put into use that did not require central computers, but rather was - as the current Internet is - totally decentralized.
Explanation:
Below I present as a summary some of the most relevant aspects exposed on the requested website about the origin and authors of ARPANET:
1. Licklider from MIT in August 1962 thinking about the concept of a "Galactic Network". He envisioned a set of globally interconnected computers through which everyone could quickly access data and programs from anywhere. In spirit, the concept was very much like today's Internet. He became the first head of the computer research program at DARPA, and from October 1962. While at DARPA he convinced his successors at DARPA, Ivan Sutherland, Bob Taylor and MIT researcher Lawrence G. Roberts, of the importance of this network concept.
2.Leonard Kleinrock of MIT published the first article on packet-switching theory in July 1961 and the first book on the subject in 1964. Kleinrock convinced Roberts of the theoretical feasibility of communications using packets rather than circuits, That was an important step on the road to computer networking. The other key step was to get the computers to talk together. To explore this, in 1965, working with Thomas Merrill, Roberts connected the TX-2 computer in Mass. To the Q-32 in California with a low-speed phone line creating the first wide-area (albeit small) computer network built . The result of this experiment was the understanding that timeshare computers could work well together, running programs and retrieving data as needed on the remote machine, but that the circuitry switching system of the phone was totally unsuitable for the job. Kleinrock's conviction of the need to change packages was confirmed.
3.In late 1966 Roberts went to DARPA to develop the concept of a computer network and quickly developed his plan for "ARPANET", and published it in 1967. At the conference where he presented the document, there was also a document on a concept of UK packet network by Donald Davies and Roger Scantlebury of NPL. Scantlebury told Roberts about NPL's work, as well as that of Paul Baran and others at RAND. The RAND group had written a document on packet switched networks for secure voice in the military in 1964. It happened that work at MIT (1961-1967), in RAND (1962-1965) and in NPL (1964-1967) all they proceeded in parallel without any of the investigators knowing about the other work. The word "packet" was adopted from the work in NPL and the proposed line speed to be used in the ARPANET design was updated from 2.4 kbps to 50 kbps.