The output DC voltage after using a full-wave rectifier on an AC source with a Vrms of 60V, considering a forward voltage drop of 0.7V and negligible ripple, is approximately 53.7V.
Explanation:To convert an AC voltage to DC voltage using a full-wave rectifier, you first need to understand the relationship between the root mean square (RMS) value of the AC voltage and its peak value. The RMS value is given as Vrms = 60V. For a full-wave rectified sine wave, the peak voltage (Vp) is Vrms × √2. Therefore, Vp = 60V × √2 ≈ 84.85V. However, because of the rectifier's forward voltage drop (Vf), we have to subtract this from the peak voltage to get the peak DC voltage (Vdc_peak). Therefore, Vdc_peak = Vp - Vf ≈ 84.85V - 0.7V ≈ 84.15V.
However, to find the average or output DC voltage (Vdc_avg) from a full-wave rectifier, you'd multiply the peak DC voltage by 2/π (since it's a sinusoidal wave), so Vdc_avg ≈ (2/π) × 84.15V ≈ 53.7V. Thus, the output DC voltage after using a full-wave rectifier directly on the line voltage, assuming negligible ripple, is approximately 53.7V.
An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface crack length if the surface energy of aluminum oxide is 0.90 J/m2. The modulus of elasticity of this material is 393 GPa.
Answer:
1.44 mm
Explanation:
Compute the maximum allowable surface crack length using
[tex]C=\frac {2E\gamma}{\pi \sigma_c^{2}}[/tex] where E is the modulus of elasticity, [tex]\gamma[/tex] is surface energy and [tex]\sigma_c[/tex] is tensile stress
Substituting the given values
[tex]C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm[/tex]
The maximum allowable surface crack is 1.44 mm
In a p + − njunction, the n side has a donor concentration of 1 × 1016 cm−3 . If ni = 1 × 1010 cm−3 , the relative dielectric constant Pr = 12, Dn = 50 cm2/s, Dp = 20 cm2/s, and the electron and hole minority carriers have lifetimes of τ = 100 ns and 50 ns respectively, and a forward bias of 0.6 V, calculate the hole diffusion current density 2µm away from the depletion edge on the n-side. If we double the p + doping, what effect will it have on the hole diffusion current?
Answer:
0.3
Explanation:
Please see attachment.
A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0.0254 m thick layer of an insulation (k = 0.2423 W/m∙K). The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and the temperature at the interface between the metal and the insulation.
Answer:
Q=339.5W
T2=805.3K
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.
3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature
There is a proposal in Brooklyn to construct a new mid-rise apartment building on a vacant lot at the intersection of Avenue A and 48th Street. The property is square, providing flexibility for the location of the building and an associated playground. The developer wishes to locate a small park and playground adjacent to the quietist street. The traffic volumes for Avenue A are: cars – 496; medium trucks – 52, heavy trucks – 19; and buses - 10. The traffic volumes for 48th Street are: cars – 822; medium trucks – 22, heavy trucks – 8; and buses - 3. Following his thinking, which street should the park be adjacent to? (3 points) Assuming the setbacks are the same, what is the difference in noise levels adjacent to the 2 streets? (4 points) Based upon people’s perceptions of noise differences, are the developer’s concerns valid? Why? (3 points)
Answer:
a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes
Explanation:
Data given for Avenue A
Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10
Data given for 48th Street
Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3
Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18
a. Noise Level = Number of vehicles x PCE
For Avenue A
Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa
For 48th Street
Noise level = (822 x 1) + (22 x 13 + (8 x 47) + (3 x 18) = 1538dBa
The park should adjacent to 48th street as it is quieter than Avenue A
b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence
For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa
Noise at Setback distance = 2245 - 4 = 2241dBa
Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa
The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa
c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.
Assign deliveryCost with the cost (in dollars) to deliver a piece of baggage weighing baggageWeight. The baggage delivery service charges twenty dollars for the first 50 pounds and one dolllar for each additional pound. The baggage delivery service calculates delivery charge by rounding to the next pound. Assume baggageWeight is always greater than 50 pounds.
To determine the delivery cost for baggage weighing more than 50 pounds, charge twenty dollars for the first 50 pounds and one dollar for each additional pound, rounding the weight to the nearest whole number.
Explanation:To calculate the cost to deliver a piece of baggage weighing more than 50 pounds, we must understand the pricing structure of the baggage delivery service. The service charges twenty dollars for the first 50 pounds and one dollar for each additional pound. Since the weight is rounded to the next pound, if the baggage weight is greater than 50 pounds but not a whole number, we round up to the nearest whole number before calculating the additional cost.
For example, if the baggage weighs 53.2 pounds, we round the weight to 54 pounds. The first 50 pounds cost twenty dollars, so we only need to calculate the cost for the additional weight over 50 pounds, which in this case is 4 pounds (54 - 50 = 4). Thus, the additional cost is 4 dollars (4 additional pounds × $1 per additional pound = $4). The total delivery cost would be $24 ($20 for the first 50 pounds + $4 for the additional 4 pounds).
Suppose the Bookstore is processing an input file containing the titles of books in order to remove duplicates from their list. Write a program that reads all of the titles from an input file called bookTitles.txt and writes them to an output file called noDuplicates.txt. When complete, the output files should contain all unique titles found in the input file.
Answer:
books = []
fp = open("bookTitles.txt")
for line in fp.readlines():
title = line.strip()
if title not in books:
books.append(title)
fp.close()
fout = open("noDuplicates.txt", "w")
for title in books:
print(tile, file=fout)
fout.close()
except FileNotFoundError:
print("Unable to open bookTitles.txt")
. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and side slope of 1:2 (v:h) with a discharge of 10 (m3/s). The slope is 0.0004 and Manning's roughness is 0.015.
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as
[tex]v =1/n R^{2/3} s^{1/2}[/tex]
where R is hydraulic radius
S = bed slope
[tex]Q = Av =A 1/n R^{2/3} s^{1/2}[/tex]
[tex]A = 1/2 \times (B+B+4y) \times y =(B+2y) y[/tex]
[tex]R =\frac{A}{P}[/tex]
P is perimeter [tex]= (B+2\sqrt{5} y)[/tex]
[tex]R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}[/tex]
[tex]Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}[/tex]
solving for y[tex]100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}[/tex]
solving for y value by using iteration method ,we get
y ≈ 2.5
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0mm (0.20 in.); the specimen fractured at a load of 3000 N (675 Ibf) when the distance between the support points was 40mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross sectionof 15 mm (0.6 in.) length on each edge. At what load would you expect this specimen to fracture ifthe support pointseparation is maintained at 40 mm (16 in)?
Answer:
Load to fracture = 17212.5 N
Explanation:
Firstly we need to find the flexural strength of the material. We do this by using the date obtained by the test of the circular sample and the equation shown below
σ (flexural strength) = F x L/ (π x r³), where F is the load at fracture, L is the distance between the two load points, r is the radius of the circular cross section. Substituting the values we get
σ (flexural strength) = 3000 x 40 x 10⁻³/ (π x 5 x 10⁻³)³ = 306 x 10⁶ N/m²
Now, the flexural stress for a square sample is written as
σ (flexural strength) = 3 x F x L/ 2 x b x d² , and since in a square sample breadth = width the equation becomes
σ (flexural strength) = 3 x F x L/ 2 x d³
Solving for F, we get
F = σ (flexural strength) x 2 x d³ /3 x L
For same material we will use the σ (flexural strength) as calculated above, furthermore L remains the same and d = 15 x 10⁻³. Solving for F
F = 306 x 10⁶ x 2 x (15 x 10⁻³)³ /3 x 40 x 10⁻³ = 17212.5 N
The load required for this specimen to fracture would be 17212.5 N.
A 1 mm thick sheet of metal is bent to a radius of 12 mm. After springback, it is observed that the radius is 13 mm. If this sheet is to be bent to achieve a bend angle of 90o after springback, to what angle should it be bent before springback
To solve this problem it is necessary to apply the concepts related to Arc Length.
From practical terms we know that the Angle can be calculated based on the arc length and the radius of the circle, in other words
[tex]\theta = \frac{S}{r}[/tex] or
[tex]S = \theta r[/tex]
Where,
S = Length of the arc
r = Radius
From the information given, the object to bending has a millimeter thick and a radius of 12 mm. This way your net radio is given by
[tex]R_1 = r_B+\frac{t}{2}[/tex]
After Springback the radius increases and the thickness ratio is therefore maintained.
[tex]R_2 = r_S+\frac{t}{2}[/tex]
For both cases we have different angles but that maintains the electron-magnetic proportions, therefore the arc length is maintained:
[tex]S_B = S_S[/tex]
[tex]\theta_B R_1 = \theta_S R_2[/tex]
[tex]\theta_B (r_B+\frac{t}{2})=\theta_S(r_S+\frac{t}{2})[/tex]
Re-arrange to find [tex]\theta_S[/tex]
[tex]\theta_S = \frac{\theta_B (r_B+\frac{t}{2})}{(r_S+\frac{t}{2})}[/tex]
[tex]\theta_S= \frac{90\°*(12+\frac{1}{2})}{13+\frac{1}{2}}[/tex]
[tex]\theta_S= 83.33\°[/tex]
Therefore the angle that should be bent before springback is 83.33°
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. Determine the volume of the tank, in m3, and the final pressure, in bar, using the:
(a) ideal gas equation of state.
(b) Redlich–Kwong equation.
Answer:
a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar
Explanation:
Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol
Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation
a. PV=mRT
V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3
Since it is a rigid tank the volume of the tank must remain constant and hnece we can say
T2/T1 = P2/P1, solving for P2
P2 = (150/180) x 35 = 29.17bar
b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}
where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol
solving for v1
35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}
35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}
Using Trial method to find v1
for v1 = 0.5
Right hand side becomes = {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side
for v1 = 0.4
Right hand side becomes = {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side
for v1 = 0.45
Right hand side becomes = {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35
Specific Volume = 35 m³/kmol
V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³
For Pressure P2, we know that v2= v1
P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar
A water tunnel has a circular cross-section with a radius of 1.8 m in its main section. In a test section of the tunnel the radius constricts to a value of 0.60 m. If the speed of water flow is 3.0 m/s in the main section, determine the speed of water flow in the test section.
Answer:
speed of water flow in the test is 27 m/s
Explanation:
given data
cross section radius r1 = 1.8 m
radius constricts r2 = 0.60 m
speed of water flow = 3.0 m/s
to find out
speed of water flow in the test section
solution
we using the continuity equation here
A1 × V1 = A2 × V2 ...............1
put here value we get speed
[tex](\frac{\pi }{4} d1^2) v1 = (\frac{\pi }{4} d2^2) v2[/tex]
1.8²×3 = 0.6² × v2
v2 = 27 m/s
speed of water flow in the test is 27 m/s
A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 750 N (169 lbf), the flexural strength is 105 MPa (15,000 psi), and the separation between load points is 50.0 mm (1.97 in.).
Answer:
[tex]R_{min} = 4.84\times 10^{-3} m[/tex]
Explanation:
Given data:
Applied force 750 N
Flexural strength is 105 MPa
separation is 50 mm = 0.05 m
flexural strength is given as
[tex]\sigma_f = \frac{FL}{\pi R_{min}^3}[/tex]
solving for R so we have
[tex]R_{min} = [\frac{FL}{\pi \sigma_f}]^{1/3}[/tex]
plugging all value to get minimum radius
[tex]R_{min} = [\frac{750 \times 0.05}{\pi 105 \times 10^6}]^{1/3}[/tex]
[tex]R_{min} = 4.84\times 10^{-3} m[/tex]
Darin cheers at a football game when his hometown team scores, and his Dad gives him a high five. Darin later begins cheering with his Dad at basketball, baseball, and hockey games when his hometown teams score. This is an example of Darin’s cheering behavior ____________ to other situations.
Answer:
generalizing
Explanation:
We all have a generalization system that operates as an autopilot, allowing us to be fast and consistent with our own identity. Thanks to these we package and label all the information with which we are bombed every second, to immediately think and act.
Otherwise, if we pay attention to each data individually, however tiny, every minute of our lives would become an exhausting and extremely slow process of analyzing and digesting, leaving us so overloaded to the point of collapse and not being able to function more mentally.
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?
Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation
[tex]Y = 1 - e^{-kt^n}[/tex]
here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,
[tex]0.50 = 1 - e^{-k^146^2.1}[/tex]
[tex]0.50 = e^{-k306.6}[/tex]
taking natural log on both side
ln(0.5) = -k(306.6)
[tex]k = 2.26\times 10^{-3}[/tex]
for 86 % completion
[tex]0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}[/tex]
[tex]e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14[/tex]
[tex]-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)[/tex]
[tex]t^{2.1} = 869.96[/tex]
t = 25.10 sec
Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per 10 m of pipe, do you think this pipe is
a) a new pipe,
b) an old pipe with somewhat increased roughness due to aging or
c) a very old pipe that is partially clogged by deposits. Justify your answer.
Answer:
pipe is old one with increased roughness
Explanation:
discharge is given as
[tex]V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}[/tex]
V = 7.07 m/s
from bernou;ii's theorem we have
[tex]\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l[/tex]
as we know pipe is horizontal and with constant velocity so we have
[tex]\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}[/tex]
[tex]P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma[/tex]
[tex]135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81[/tex]
solving for friction factor f
f = 0.0324
fro galvanized iron pipe we have [tex]\epsilon = 0.15 mm[/tex]
[tex]\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025[/tex]
reynold number is
[tex]Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}[/tex]
Re = 378750
from moody chart
[tex]For Re = 378750 and \frac{\epsilon}{d} = 0.0025[/tex]
[tex]f_{new} = 0.025[/tex]
therefore new friction factor is less than old friction factoer hence pipe is not new one
now for Re = 378750 and f = 0.0324
from moody chart
we have [tex]\frac{\epsilon}{d} =0.006[/tex]
[tex]\epsilon = 0.006 \times 60[/tex]
[tex]\epsilon = 0.36 mm[/tex]
thus pipe is old one with increased roughness
An air-standard cycle with constant specific heats at room temperature is executed in a closed system and is composed of the following four processes:
1–2 v = Constant heat addition from 14.7 psia and 80°F in the amount of 300 Btu/lbm
2–3 P = Constant heat addition to 3150 R
3–4 Isentropic expansion to 14.7 psia
4–1 P = Constant heat rejection to initial state
The properties of air at room temperature are cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4.
a. Show the cycle on P-v and T-s diagrams. (Please upload your response/solution using the controls provided below.) (You must provide an answer before moving on to the next part.)
b. Calculate the total heat input per unit mass
c. Determine the thermal efficiency
Answer:
a. Please see attachment .
b. 612.377
c. 24.22%
Explanation:
Please see attachment.
Compute the longitudinal modulus of elasticity (in GPa) for a continuous and aligned hybrid composite consisting of aramid and glass fibers in volume fractions of 0.22 and 0.30, respectively, within a polyester resin matrix, given the following data:
Modulus of Elasticity
(GPa)
Glass fibers
72.5
Aramid fibers
131
Polyester
2.5
=_______GPa?
Answer:
51.77 GPa
Explanation:
For elasticity for a continous and aligned hybrid composite
[tex]E_{ci}=E_{poly} (1-V_{A} -V_{G} )+E(V_{A})+E(V_{G} )\\=(2.5GPa)(1-0.22-0.3)+(131GPa)(0.22)+(72.5GPa)(0.3)\\=1.2+28.82+21.75\\=51.77GPa[/tex]
An air conditioner takes outside air at 35°C and cools it to 15°C at a rate of 1 kg/s. Estimate the amount of power needed to do this. Recalculate the needed power if it is put in recirculation mode using inside air at 24°C and cools that to 15°C.
Answer:
Q=20.1kW
recirculation mode=Q=9.045kW
Explanation:
Hi!
To solve this problem you must use the first law of thermodynamics that establishes the energy that enters a system is the same that must come out. Given the above we have the following equation
Q = mCp (T2-T1)
WHERE
Q = Heat removed to air
m = mass air flow = 1kg / s
Cp = specific heat of air = 1.005KJ / kg °C
T2 = air inlet temperature
T1 = air outlet temperature=15°C.
for the first case
Q=(1kg/s)(1.005KJ/kg°C)(35°C-15°C)
Q=20.1kW
for the second case( recirculation mode)
Q=(1kg/s)(1.005KJ/kg°C)(24°C-15°C)
Q=9.045kW
The flow rate of liquid metal into the downsprue of a mold is 500 cm3/s. The cross-sectional area at the top of the sprue is 8 cm2, and the height of the sprue is 15 cm. What should the cross-sectional area be at the base of the sprue in order to avoid separation during pouring of the molten metal
Answer:
The area at the base should be 2.91cm²
Explanation:
The separation of the molten metal may occur when the pressure of the flowing liquid falls below the atmospheric pressure and the air entrapped inside the mold tries to escape but rather gets rapped inside the casting (which is solidifying) causing the liquid to separate and the casting to be porous.
The velocity at the bottom of the sprue is written as V = √(2gh), where g is the acceleration due to gravity and h is the height of the sprue
V = √(2 x 9.81 x 100 x 15) = 171.6 cm² (Multiplying by 100 to convert 9.81 m/s to cm/s)
The expression for flow rate is Q = VA where V is the velocity as calculated above and A is the area
500 = 171.6 x A, solving for A we get A = 500/171.6 = 2.91cm²
The cross - sectional Area at the base of the sprue should be 2.91cm² to avoid separation of the molten metal.
Snow, ice, rain, and flooding all _____, making it harder to steer and harder to stop in time
Answer:
reduce traction
Explanation:
Snow, ice, rain, and flooding all reduce traction, making it harder to steer and harder to stop in time. In these conditions, you might even experience hydroplaning.
A direct shear test was performed on a specimen of dry sand. The specimen was 50 mm by 50 mm square and 25 mm thick (height). A normal stress of 192 kPa was applied to the specimen and the shear stress at failure () was 120 kPa.
a. What is the effective friction angle of the sand?
b. For a normal stress of 200 kPa, what shear force (F) will be required to cause failure of the specimen? (Report answer in N).
To solve this problem it is necessary to apply the concepts related to the normal effort and the shear effort due to failure. The shear stress can be defined based on normal stress and effective friction angle. Mathematically it can be defined as
[tex]\tau_f = \sigma_n tan\phi[/tex]
Where
[tex]\tau_f =[/tex] Shear stress at failure
[tex]\sigma_n =[/tex] Normal stress
[tex]\phi =[/tex] Effective friction angle
PART A) Our values are given as ,
[tex]\sigma_n = 192kPa[/tex]
[tex]\tau_f = 120kPa[/tex]
Replacing at the previous equation we have,
[tex]\tau_f = \sigma_n tan\phi[/tex]
[tex]120 = 192tan\phi[/tex]
[tex]\phi = tan^{-1}(\frac{120}{192})[/tex]
[tex]\phi = 32.27°[/tex]
Therefore the effective friction angle of the sand is 32.27°
B) Using the maximum possible effort and the angle previously given we can calculate the maximum shear force, from which from its definition it is possible to find the force.
[tex]\tau_f = \sigma_n tan\phi[/tex]
[tex]\tau_f = 200tan(32\°)[/tex]
[tex]\tau_f = 124.97kPa[/tex]
With the value of the shear stress from its basic definition we can find the force. By definition the shear stress is given by
[tex]\tau_f = \frac{F}{A}[/tex]
Re-arrante to find the Force
[tex]F = A\tau_f[/tex]
The shear stress is a function of the Force and the Area, therefore, the area would be the square of the sides (50mm)
Replacing in the equation we have to,
[tex]F = 124.97*10^{-3}*(50*50)[/tex]
[tex]F = 312.425N[/tex]
Therefore the shear force required to cause failure of the specimen is 312.425N
Aerial photography was taken at a flying height H ft above average terrain.
If the camera focal plane dimensions are 9 x 9 in, the focal length is fand the spacing between adjacent flight lines is X ft, what is the percent sidelap for the data given in Problem?
H 7000; f 88.90 mm; X 13,500
Answer:
% od sidelap is [tex]P_w = 0.251\ or\ 25%[/tex]
Explanation:
Given data:
flying height is 7000 ft
focal dimension is 9× 9
focal length f = 88.90 mm
side width is x = 13,500 ft
we know side width is given as
[tex]x = (1- P_w) \frac{w}{s}[/tex]
where s is scale and given as [tex]s = \frac{f}{h}[/tex]
[tex]13,500 = (1- P_w) \times \frac{\frac{9}{12}}{\frac{0.29}{7000}}[/tex]
[tex](1- P_w) = 0.748[/tex]
[tex]P_w = 0.251\ or\ 25%[/tex]
It is well established that the thermal efficiency of a heat engine increases as the temperature TL at which heat is rejected from the heat engine decreases. In a effort to increase the efficiency of a power plant, somebody suggests refrigerating the cooling water before it enters the condenser, where heat rejection takes place. Would you be in favor of this idea?Why?
Answer: No, not in favor of the idea.
This is because, an increase in the refrigerator's work output is equal to work input. In real live application, the work input of the refrigerator is always greater than the additional work output. This will result to reducing the efficiency of the power plant.
Refrigerating cooling water before the condenser in a power plant is impractical due to high energy costs, added complexity, and inefficiencies, outweighing the potential efficiency gains.
While the idea of refrigerating the cooling water before it enters the condenser in a power plant to increase efficiency might seem theoretically sound, in practice it is not advisable due to several reasons:
1.Energy Cost of Refrigeration: Refrigeration itself is an energy-intensive process. The energy required to cool the water to a significantly lower temperature would likely be greater than the energy savings achieved by the increased efficiency of the heat engine. This means that the net energy gain would be negative.
2.Second Law of Thermodynamics: According to the second law of thermodynamics, any energy conversion process, including refrigeration, involves irreversibilities and losses. The additional step of refrigeration would introduce more inefficiencies and would not be 100% efficient. Thus, the overall efficiency of the system would decrease.
3.Complexity and Cost: Adding a refrigeration system to cool the condenser water would significantly increase the complexity and cost of the power plant. This includes the initial capital cost, operational costs, and maintenance costs. The benefits gained from a marginal increase in thermal efficiency would likely not justify these additional expenses.
4. Practical Alternatives: There are more practical and cost-effective ways to improve the efficiency of a power plant. These include optimizing the thermodynamic cycle, improving the insulation, using better quality fuels, and upgrading to more efficient machinery and technology. Improving the heat exchange process itself, rather than cooling the water, is generally more practical.
5.Environmental Considerations: The energy used for refrigeration would typically come from the power plant itself or the grid, potentially leading to increased fuel consumption and higher greenhouse gas emissions. This is counterproductive to the goal of improving efficiency and sustainability.
In summary, while lowering the temperature at which heat is rejected can theoretically improve the thermal efficiency of a heat engine, refrigerating the cooling water is not a practical or economical solution due to the high energy cost, increased complexity, and overall inefficiency of the refrigeration process itself. Instead, other methods to enhance efficiency should be explored.
Consider a pump operating adiabatically at steady state. Liquid water enters at 20◦C, 100 kPa with a mass flow rate of 53 kg/min. The pressure at the pump exit is 5 MPa. The pump isentropic efficiency is 75%. Assumer negligible changes in kinetic and potential energy and the water behaves as an incompressible substance. Determine the power required by the pump, in kW.
Answer:
5778.86W
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2. find the specific weights of the water and its density using T = 20C, using thermodynamic tables
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. use the Bernoulli equation to find the height of the pump's power, taking into account that the potential and kinetic energy changes are insignificant
4. find the ideal pump power
5. find the real power of the pump using the efficiency equation
A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line througha valve. The valve is opened, and air is allowed to enter the tankuntil the density in the tank rises to 7.20 kg/m3.
Determine themass of air that has entered the tank.
To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.
Density can be defined as
[tex]\rho = \frac{m}{V}[/tex]
Where
m = Mass
V = Volume
For state one we know that
[tex]\rho_1 = \frac{m_1}{V}[/tex]
[tex]m_1 = \rho_1 V[/tex]
[tex]m_1 = 1.18*1[/tex]
[tex]m_1 = 1.18Kg[/tex]
For state two we have to
[tex]\rho_2 = \frac{m_2}{V}[/tex]
[tex]m_2 = \rho_2 V[/tex]
[tex]m_1 = 7.2*1[/tex]
[tex]m_1 = 7.2Kg[/tex]
Therefore the total change of mass would be
[tex]\Delta m = m_2-m_1[/tex]
[tex]\Delta m = 7.2-1.18[/tex]
[tex]\Delta m = 6.02Kg[/tex]
Therefore the mass of air that has entered to the tank is 6.02Kg
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are indigenous faults. What is the Mill's estimate of the number of indigenous faults remaining undetected in the program?
Answer:
Estimated number of indigenous faults remaining undetected is 6
Explanation:
The maximum likelihood estimate of indigenous faults is given by,
[tex]N_F=n_F\times \frac{N_S}{n_S}[/tex] here,
[tex]n_F[/tex] = the number of unseeded faults = 6
[tex]N_S[/tex] = number of seeded faults = 30
[tex]n_s[/tex] = number of seeded faults found = 15
So NF will be calculated as,
[tex]N_F=6\times \frac{30}{15}=12[/tex]
And the estimate of faults remaining is [tex]N_F-n_F[/tex] = 12 - 6 = 6
A game rolls a die 10 times and counts how many times the number 5 or 6 shows up.
(a) If there are a total of seven or more times, the win is 2 dollars.
(b) a total of four or more times but less than seven, the win is 1 dollars.
(c) a total of three or less, the win is 0 dollars.
Write a function called payOut that has an input row array of 10 integers representing the ten rolls of dice, and calculates the pay out.
The question asks for writing a function named 'payOut' that calculates the player's winnings based on the outcomes of rolling a die 10 times. A Python function is provided as a solution, which counts the occurrences of numbers 5 and 6 in the input array and determines the payout according to the specified rules.
Explanation:The question involves writing a function called payOut that calculates the winnings based on the outcomes of rolling a die 10 times, with specific payouts for different totals of rolling a 5 or a 6. The function takes as input an array of 10 integers, which represent the results of the die rolls, and outputs the payout based on the criteria provided.
To address this, we create a function in Python since it's a commonly used programming language for such tasks. The function will iterate through the input array, count the occurrences of the numbers 5 and 6, and then determine the payout based on the specified rules. Here's a basic implementation of this logic:
def payOut(rolls):This function works by first counting how many times a 5 or a 6 appears in the input array using a generator expression inside the sum() function. It then uses conditional statements to determine the payout based on the count, according to the game's rules.
A square aluminum plate 5 mm thick and 200 mm on a side is heated while vertically suspended in quiescent air at 40C. (A) Determine the average heat transfer coefficient for the plate when its temperature is 15C by two methods: using results from the similarity solution to the boundary layer equations, and using results from an empirical correlation. (B) Calculate the convective heat transfer loss to the air from both surfaces (C) Calculate the net radiation heat loss to a very large environment at 35C. Assume it is a two surface enclosure and all surfaces are gray and diffuse. e=1.0. Do you think you can neglect the radiation heat loss?
https://brainly.com/question/14000001
The average heat transfer coefficient would be; 4.42 W/m gK
The convective heat transfer loss to the air from both surfaces would be 4.87 W/m K.
Net radiation to a a very large environment at 35°C would be; -10.06 W
How to determine the average heat trnasferTo determine the average heat transfer, the Rayleighs number would be used. According to the formula;
Ra = gβΔTL³/Vα
= 1.827 * 10⁷
GrL = RaL/Pr
= 0.501
The correlation for estimating hL
Nu L= hL*L/k
= 4.42 W/m gK
Net radiation using the Stefan-Boltzmann equation of Q_rad = ε * σ * A * (Tplate₄ - Tsurroundings⁴)
= Q_rad_total = 2 * (-5.03 W)
= -10.06 W
Learn more about heat transfer here:
https://brainly.com/question/16055406
#SPJ2
Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel length, determine which rectangular channel, b/a = 1.0, 1.43, or 2.0, will provide the highest heat transfer rate. Is this heat transfer rate greater than, equal to, or less than the heat transfer rate associated with a circular tube? answer: b/a=2
Answer:
Please see attachment.
Explanation:
A stream of air enters a 7.00-cm ID pipe at a velocity of 30.0 m/s at 27.0°C and 1.80 bar (gauge). At a point downstrream, the air flows through a 5.50-cm ID pipe at 60.0°C and 1.63 bar (gauge). What is the velocity of the gas at the downstream point
To calculate the downstream velocity of air in a pipe, use the principle of conservation of mass and the continuity equation, which relates the product of cross-sectional area and velocity at two points in a system. The cross-sectional areas of the pipe at both points are computed, and using the continuity equation, the downstream velocity can be found. Convert diameters to meters, calculate areas, and apply the equation.
Explanation:The student's question involves calculating the velocity of air at a downstream location in a pipe system, using the principles of fluid dynamics. First, to find the velocity of gas at the downstream point, we can invoke the principle of conservation of mass, specifically the continuity equation for an incompressible fluid. This equation states that the product of the cross-sectional area (A) and velocity (V) at one point must equal the product of A and V at any other point in the system, assuming the density of the fluid remains constant.
First, we must calculate the cross-sectional areas at both points:
Initial area, A1 = rac{Using the given velocities and the continuity equation A1 x V1 = A2 x V2, we can solve for the downstream velocity, V2.
Given:
Initial velocity, V1 = 30.0 m/sDownstream temperature = 60.0°C (not necessary for solving this problem)Initial pipe diameter = 7.00 cmDownstream pipe diameter = 5.50 cmDownstream pressure = 1.63 bar (gauge) (also not necessary for solving this problem)We can convert these diameters to meters, find the areas, and then apply the continuity equation to find the downstream velocity. However, it seems the student is not required to solve for changes in air density due to temperature and pressure variations, hence we're treating the air as incompressible.
The velocity of the gas at the downstream point is approximately 57.3 m/s, considering the conservation of mass flow rate and the given conditions of temperature and pressure.
To find the velocity of the gas at the downstream point, we need to apply the principle of conservation of mass for a compressible flow, which states that the mass flow rate must be constant throughout the pipe. The mass flow rate [tex](\(\dot{m}\))[/tex] can be expressed using the following equation:
[tex]\[\dot{m} = \rho \cdot A \cdot v\][/tex]
where:
- [tex]\(\rho\)[/tex] is the density of the air,
- A is the cross-sectional area of the pipe,
- v is the velocity of the air.
First, we'll calculate the density of the air at both the upstream and downstream points using the ideal gas law:
[tex]\[\rho = \frac{P}{R \cdot T}\][/tex]
where:
- P is the absolute pressure,
- R is the specific gas constant for air [tex](\(R = 287.05 \, \text{J/(kg·K)}\))[/tex],
- T is the temperature in Kelvin.
Let's start by converting the gauge pressures to absolute pressures. Since gauge pressure is the pressure relative to atmospheric pressure, we need to add atmospheric pressure (1.01325 bar) to each gauge pressure:
Upstream Conditions:
- Gauge pressure: [tex]\(1.80 \, \text{bar}\)[/tex]
- Absolute pressure: [tex]\(P_1 = 1.80 \, \text{bar} + 1.01325 \, \text{bar} = 2.81325 \, \text{bar} = 281325 \, \text{Pa}\)[/tex]
- Temperature: [tex]\(27.0^\circ \text{C} = 27.0 + 273.15 = 300.15 \, \text{K}\)[/tex]
- Pipe diameter: [tex]\(7.00 \, \text{cm} = 0.07 \, \text{m}\)[/tex]
- Velocity: [tex]\(v_1 = 30.0 \, \text{m/s}\)[/tex]
Downstream Conditions:
- Gauge pressure: [tex]\(1.63 \, \text{bar}\)[/tex]
- Absolute pressure: [tex]\(P_2 = 1.63 \, \text{bar} + 1.01325 \, \text{bar} = 2.64325 \, \text{bar} = 264325 \, \text{Pa}\)[/tex]
- Temperature: [tex]\(60.0^\circ \text{C} = 60.0 + 273.15 = 333.15 \, \text{K}\)[/tex]
- Pipe diameter: [tex]\(5.50 \, \text{cm} = 0.055 \, \text{m}\)[/tex]
Calculate the density at both points:
[tex]\[\rho_1 = \frac{P_1}{R \cdot T_1} = \frac{281325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 300.15 \, \text{K}} \approx 3.268 \, \text{kg/m}^3\][/tex]
[tex]\[\rho_2 = \frac{P_2}{R \cdot T_2} = \frac{264325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 333.15 \, \text{K}} \approx 2.769 \, \text{kg/m}^3\][/tex]
Calculate the cross-sectional area of the pipes:
[tex]\[A_1 = \pi \left(\frac{0.07 \, \text{m}}{2}\right)^2 \approx 0.00385 \, \text{m}^2\][/tex]
[tex]\[A_2 = \pi \left(\frac{0.055 \, \text{m}}{2}\right)^2 \approx 0.00238 \, \text{m}^2\][/tex]
Mass flow rate at the upstream point:
[tex]\[\dot{m} = \rho_1 \cdot A_1 \cdot v_1 = 3.268 \, \text{kg/m}^3 \times 0.00385 \, \text{m}^2 \times 30.0 \, \text{m/s} \approx 0.378 \, \text{kg/s}\][/tex]
Velocity at the downstream point:
Using the conservation of mass flow rate:
[tex]\[\dot{m} = \rho_2 \cdot A_2 \cdot v_2\][/tex]
Solving for [tex]\(v_2\):[/tex]
[tex]\[v_2 = \frac{\dot{m}}{\rho_2 \cdot A_2} = \frac{0.378 \, \text{kg/s}}{2.769 \, \text{kg/m}^3 \times 0.00238 \, \text{m}^2} \approx 57.3 \, \text{m/s}\][/tex]
So, the velocity of the gas at the downstream point is approximately [tex]\(57.3 \, \text{m/s}\).[/tex]