Answer:
0.0198 mm/sec
Explanation:
Given:
d = 8 cm
L = 10 m = 1000 cm
v = 200 m/min
E = 100%
density = 3500 A.m^-2
Using Faraday's law of electrolysis, we have:
Q = n(e) * F
Where,
Q = electric charge
F = Faraday's constant=96500Cmol
n(e) = electron charge = 2
Therefore
Q= 96500 * 2 = 193000
Let's find the area of pipe, A = πr²(L)
But radius, r = [tex] \frac{d}{2} = \frac{4}{2} = 4cm [/tex]
Therefore,
A = 3.142 * 4² * 1000
A = 50272
To find the thickness of the coating deposited, we have :
[tex] = \frac{193000 * 7.31 * 118.7 * 200}{96500 * 50272 * 3500} = 1.98 * 10^-^3[/tex]
= 0.0198 mm/sec
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.
Find the complete solution in the given attachments.
Note: The complete Question is attached in the first attachment as the provided question was incomplete
A phase angle in the frequency domain corresponds to
a. the initial slope of the sinusoidal signal.
b. the angle at which measurements should be taken for the most accurate results.
c. a delay or advance in time as compared to a pure cosine wave.
d. the angle of the approach vector of the voltage.
Answer:
c. a delay or advance in time as compared to a pure cosine wave.
Explanation:
Electrical phase is measured in degrees, with 360° corresponding to a complete cycle. A sinusoidal voltage is proportional to the cosine or sine of the phase. Phase difference , also called phase angle , in degrees is conventionally defined as a number greater than -180, and less than or equal to +180.
The phase angle corresponds to delay or advance in time as compared to a pure cosine wave.
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?
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
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid R-134a at 26 °C. The valve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.5 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer.
Answer:
x2 = 0.5056
Qin = 22.62Kj
Explanation:
Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically to the exit, where the pressure is 40 psia. If the isentropic nozzle efficiency is 95%, determine for the nozzle,
(a) the exit velocity of the steam in ft./sec, and
(b) the amount of entropy produced in BTU/ lbm R.
Answer:
a)exit velocity of the steam, V2 = 2016.8 ft/s
b) the amount of entropy produced is 0.006 Btu/Ibm.R
Explanation:
Given:
P1 = 100 psi
V1 = 100 ft./sec
T1 = 500f
P2 = 40 psi
n = 95% = 0.95
a) for nozzle:
Let's apply steady gas equation.
[tex] h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2} [/tex]
h1 and h2 = inlet and exit enthalpy respectively.
At T1 = 500f and P1 = 100 psi,
h1 = 1278.8 Btu/Ibm
s1 = 1.708 Btu/Ibm.R
At P2 = 40psi and s1 = 1.708 Btu/Ibm.R
1193.5 Btu/Ibm
Let's find the actual h2 using the formula :
[tex] n = \frac{h_1 - h_2*}{h_1 - h_2} [/tex]
[tex] n = \frac{1278.8 - h_2*}{1278.8 - 1193.5} [/tex]
solving for h2, we have
[tex] h_2 = 1197.77 Btu/Ibm [/tex]
Take Btu/Ibm = 25037 ft²/s²
Using the first equation, exit velocity of the steam =
[tex] (1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}[/tex]
Solving for V2, we have
V2 = 2016.8 ft/s
b) The amount of entropy produced in BTU/ lbm R will be calculated using :
Δs = s2 - s1
Where s1 = 1.708 Btu/Ibm.R
At h2 = 1197.77 Btu/Ibm and P2 =40 psi,
S2 = 1.714 Btu/Ibm.R
Therefore, amount of entropy produced will be:
Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R
= 0.006 Btu/Ibm.R
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?
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 the following circuit, V"#$=120∠30ºV.Redraw the circuit in your solution.a.(4) Calculate the total input impedance seen by the source. Express in rectangular form.b.(3) Calculate the input phasor current(express answer in polar form).c.(6) Using the voltage division, calculate the phasor voltages across each component. Express final answers in polar form.d.(6) Using current divider, calculate phasor currents through L1, and C1.Show all steps. Express final answers in polar form
Answer:
Check the explanation
Explanation:
Kindly check the attached images for the step by step explanation to the question
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.
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
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
Answer:
Find the attachment for the answer
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).
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
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.
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.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to get the step by step explanation to the question above.
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
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.
Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The condensing temperature is 48 oc, and the evaporating temperature is —16 oc. If the power input to the cycle is 2.5 kW, determine: (a) the work done by the compressor in kJ/kg, (b) the heat rejected from the condenser in kJ/kg, (c) the heat absorbed by the evaporator in kJ/kg, (d) the coefficient of performance, and (e) the refrigerating efficiency.
Answer:
a. The work done by the compressor is 447.81 Kj/kg
b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg
c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg
d. The coefficient of performance is 2.746
e. The refrigerating efficiency is 71.14%
Explanation:
According to the given data we would need first the conversion of temperaturte from C to K as follows:
Temperature at evaporator inlet= Te=-16+273=257 K
Temperatue at condenser exit=Te=48+273=321 K
Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg
b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:
w=i4-i3
W/M=i4-i3
i4=W/M + i3
i4=2.5/0.05 + 397.81
i4=447.81 Kj/kg
a. Enthalpy at the compressor exit=447.81 Kj/kg
Therefore, the heat rejected from the condenser in kJ/kg=i4-i1
the heat rejected from the condenser in kJ/kg=447.81-260.51
the heat rejected from the condenser in kJ/kg=187.3 kJ/kg
c. Temperature at evaporator inlet= Te=-16+273=257 K
The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg
d. To calculate the coefficient of performance we use the following formula:
coefficient of performance=Refrigerating effect/Energy input
coefficient of performance=137.3/50
coefficient of performance=2.746
the coefficient of performance is 2.746
e. The refrigerating efficiency = COP/COPc
COPc=Te/(Tc-Te)
COPc=255/(321-255)
COPc=3.86
refrigerating efficiency=2.746/3.86
refrigerating efficiency=0.7114=71.14%
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.
Answer:
See attached image for diagrams and solution
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.
Answer:
P=8.44 kw
Explanation:
[Find the given attachment for 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.
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:
https://brainly.com/question/29475799
#SPJ5
Water at 15°C is to be heated to 65°C by passing it over a bundle of 7-m-long, 1-cm-diameter resistance heater rods maintained at 90°C. Water approaches the heater rod bundle in normal direction at a mean velocity of 0.8 m/s. The rods are arranged in-line with longitudinal and transverse pitches of SL = 4 cm and ST = 3 cm. Determine the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise.
Answer:
NL = 207
Explanation:
Solution
Now,
The mean temperature is measured as:
Tm = (T₁ - T₀)/2
= (15 + 65)/2
= 40°C
So,
we find all the thermo-physical properties of water from the table, that is properties of saturated water at T =40°C
Thermo conductivity, k = 0.631 W/m . K
Specific heat Cp = 4179 J/kg . K
Density р = 992. 1 kg/m³
The dynamic viscosity, μ = 0.653 * 10 ^⁻3 kg/m *s
Prandtl number, Pr = 4.32
At T = 15°C
рi = 992.1 kg/m³
At T = 90°C
Prandtl number, Prs = 1.96
Thus,
The maximum flow of velocity is known from the equation stated as:
Vmax = ST/ST - D *V
Here,
ST is refereed to as the transverse pitch for inline arrangements of the rods
so,
Vmax = 3/3-1 * 0.8
= 1.2 m/s
Now
The Reynolds number is determined from the equation given below
ReD =ρVmax D/μ
= 922.1 * 1.2 *(1 *10^⁻²)/ 0.653 * 10^⁻³
= 18231.55
From the table, The Nusselt number correlations fro cross flow over the tube banks for inline arrangement over the range of ReD is shown as
1000 - 2 * 10⁵
Now, the Nusselt number is determined by
NuD = 0.27ReD ^0.63 Pr^ 0.36 (Pr/Prs)^0.25
= 0.27 * (18231.55)^0.63 (4.32)^0.36 * (4.32/1.96)^0.25
=269.32
Then,
The convective transfer of heat water coefficient is determined from the equation shown by Diametral Nusselt Number
NuD =hD/k
So,
we re-write and solve for h
h = NuD * k/D
=269.32 * 0.631/(1 * 10 ^⁻2)
=16993.9 W/m² .K
Now,
The heat transfer surface area for a tube in a row is NT = 1
As = NT NLπDL
= 1*NL* π * (1 * 10^⁻2) * 4
= 0.1257NL
The logarithmic mean temperature of water is represented as
ΔTlm = Te - Ti/ln (Ts - Ti/Ts- Te)
= 65- 15/ln (90 -15/ 90 -65) = 45.51°C
Thus,
The rate of the heat transfer is determined from the equation shown below,
Q =hAsΔTlm
=16993.9 *0.1257 * NL* 45.51.......equation (1)
The mas flow rate of water is determined by the equation below
m =ρiAcV
= ρi * (STL) * V
= 999.1 8 ( 3* 10^⁻2 * 4) * 0.8
= 95.91 kg/s
The rate of heat transfer of water is determined by the equation below
Q = mcp (Te- Ti)
= 95.91 * 4179 * (65-15)
=20041146.72 W..........(Equation 2)
Now,
The number of tube rows in the direction flow is determined by measuring both equations 1 and 2 as
97219.61 NL = 20041146.72
NL =206.14
NL = 207
Therefore, the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise is NL = 207
=
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
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
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.
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.
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.
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.
A cylinder of length L would be made to carry a torque T with an angle of twist ɸ. There are two options considered: 1) hollowed cylinder with an inner radius that is equal to 0.9 of the outer radius, and 2) solid cylinder (with a different radius). If both options would be made from the same material and must have the same angle of twist ɸ under the torque T, find the ratio of the weight between the cylinder designed for option 1 and option 2.
Answer:
See explaination
Explanation:
To compare the hollow and solid cylinder we need ro use torsional formula.
And since same material and length are given.
For same torque and angle of twist there will be same polar moment of area of the section for both the cylinder.
Please kindly check attachment for further solution
g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005" The thickness of plate B is .750 +/- .005" The position tolerance of the clearance holes to one another is .014" The position tolerance of the precision holes to one another is .028" What is an appropriate MMC clearance hole diameter to allow the blocks to assemble?
Answer:
nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
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
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
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
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].
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
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.
Derive the stress strain relationship for each model A and model B. (5 points) b) Is model B equivalent to model A (same stress strain general equation)? If your answer is yes, express E1, E2, η1, and η2 in terms of E 0 1 , E 0 2 , η 0 1 , and η 0 2 . (5 points) c) If applicable, repeat the derivation in part (b) and express E 0 1 , E 0 2 , η 0 1 , and η 0 2 in terms of E1, E2, η1, and η2. (5 points) d) Derive the expression for stress relaxation for both these models and compare the expressions. How are they similar? How are they different? (15 points)
Answer:
See attached images
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.
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.
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?
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
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,
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