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.
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]
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.
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:
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.
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 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
. 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
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.
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.
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 1-m long, thin symmetric airfoil is placed at a free stream. Calculate the drag force for the wind speed of 150 m/s. Assume that the average friction coefficient is C subscript D equals 0.002. A equals 3 space m squared, angle of attack is 0 to the power of degree.
A. 81
B. 162 N
C. 40.5 KN
D. 0 N
Answer:
A (81 N)
Explanation:
P = F/A, where P is the pressure on an object exerted by the force F over the area A. Using "D" to represent drag force specifically, this equation can be rearranged to D = CPA, where C is a constant of proportionality that represents the friction coefficient. The pressure exerted on an object undergoing motion through a fluid can be expressed as (1/2) ρv2, where ρ is the density of the fluid and v is the velocity by which the object is moving. Therefore, D = (1/2)(C)(ρ)(v²)(A).
V = 150 m/s, C(d) = 0.002, A = 3m², ρ(density of air) = 1.2 kg/m3
Fd = (1.2 x 3 x 0.002 x 150²)/2 = 81 kg.m/s²
The drag force exerted on the airfoil is 81 N.
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.
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
Fuel Combustion and CO2 Sequestration [2016 Midterm Problem] Long-term storage of carbon dioxide in underground aquifers or old oil fields is one method to prevent release of CO2 to the atmosphere (to help mitigate climate change). This is referred to as carbon sequestration. A power plant burns fuel oil containing 39.0 mol% C, 60.3 mol% H, and 0.70 mol% S at a rate of 290 kmol/hr. An air stream flowing at 945 kmol/hr provides oxygen for the combustion process. Conversion of the fuel is 95%. Of the C that burns, 90% goes to CO2. The gases are then separated and the carbon dioxide is sequestered underground. Assume 100% separation of the CO2 from the other gases. (a) Draw a flowchart (reactor and separation unit) and label all streams. Incorporate all of the information given above (b) Calculate the percent excess air fed. (c) Determine the molar flowrate of the O2 in the stack gas leaving the plant (d) What is the mass flowrate of CO2 into the underground aquifer?
Answer:
The answers are on the attachment.
Explanation:
Three bars, each of mass 3 kg, are welded together and are pin-connected to two links BE and CF. Neglecting the weight of the links, determine the force in each link immediately after the system is released from rest.
Answer:
FCF = 14.8998 N (Compression)
FBE = 52.7342 N (Compression)
Explanation:
We have to find the mass center of ABCD which is at G.
XCG = (0.45 m /2) = 0.225 m
YCG = ∑mi*yi / ∑mi
⇒ YCG = (3 Kg*0.225m+3 Kg*0.225m+3 Kg*0m) / (3*3Kg) = 0.15 m
Now, we can apply ∑F = m*a
where a is the tangential acceleration of the links
M = 3*m = 3*3 Kg = 9 Kg
∑F = m*a ⇒ M*g*Sin 40º = M*a ⇒ a = g*Sin 40º
⇒ a = (9.81 m/s²)*Sin 40º = 6.3057 m/s²
We apply ∑MB as follows
∑MB = (0.45)*(FCF*Sin 50º) - M*g*(0.225) = - (M*a*Sin 40º)*(0.225) - (M*a*Cos 40º)*(0.15)
⇒ (0.45)*(FCF*Sin 50º) - 9*9.81*(0.225) = - (9*6.3057*Sin 40º)*(0.225) - (9*6.3057*Cos 40º)*(0.15)
⇒ FCF = 14.8998 N (Compression)
Then
∑Fy = m*ay
⇒ FCF*Sin 50º + FBE*Sin 50º - M*g = - M*a*Sin 40º
⇒ 14.8998*Sin 50º + FBE*Sin 50º - 9*9.81 = - 9*6.3057*Sin 40º
⇒ FBE = 52.7342 N (Compression)
We can see the system in the pic shown.
The question informs a statics problem in physics regarding finding the forces in supporting links of a system of welded bars in static equilibrium. Newton's laws and free body diagrams are typically used to solve such a problem.
Explanation:The student's question pertains to determining the forces in each link that supports a system of welded bars, each with a mass of 3 kg, immediately after being released from rest. To solve this problem, one would typically employ the principles of static equilibrium for rigid bodies, ensuring that the sum of forces and the sum of moments about any pivot point equals zero. This would involve drawing free body diagrams and applying Newton's second law of motion. As the links and the bars form a static structure in this scenario, the forces in the links can be found by considering the geometry of the setup and the forces due to gravity acting on the masses of the bars. However, without a detailed figure or further information about the geometry of the setup, we cannot provide a specific answer to this problem.
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
A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 16 in.2 and that approximately 3.2 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. The density of steel is 7.9 g/cm3.
Answer:
Time of submersion in years = 7.71 years
Explanation:
Area of plate (A)= 16in²
Mass corroded away = Weight Loss (W) = 3.2 kg = 3.2 x 106
Corrosion Penetration Rate (CPR) = 200mpy
Density of steel (D) = 7.9g/cm³
Constant = 534
The expression for the corrosion penetration rate is
Corrosion Penetration Rate = Constant x Total Weight Loss/Time taken for Weight Loss x Exposed Surface Area x Density of the Metal
Re- arrange the equation for time taken
T = k x W/ A x CPR x D
T = (534 x 3.2 x 106)/(16 x 7.9 x 200)
T = 67594.93 hours
Convert hours into years by
T = 67594.93 x (1year/365 days x 24 hours x 1 day)
T = 7.71 years
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
Problem 4A titanium [E = 16,500 ksi; = 5.3×10–6/°F] bar (1) and a bronze [E = 15,200 ksi; = 12.2×10–6/°F] bar (2), each restrained at one end, are fastened at their free ends by a pin of diameter d = 0.4375 in. The length of bar (1) is L1 = 19 in. and its cross-sectional area is A1 = 0.50 in.2. The length of bar (2) is L2 = 27 in. and its cross-sectional area is A2 = 0.65 in.2. What is the average shear stress in the pin at B if the temperature changes by 40°F?
Answer:
[tex]\tau_a = 22.7 ksi[/tex]
Explanation:
Given data:
[tex]\alpha = 12.2 \times 10^{-6} [/tex]degree F
Diameter is d = 0.4375
[tex]L_1 = 19 inch[/tex]
[tex]A_1 = 0.50 in^2[/tex]
[tex]L_2 = 27 inch[/tex]
[tex]A_2 = 0.65 in^2[/tex]
force in titanium and bronze will be equal
from equilibrium condition we have
F_T = F_B = p
From the information given as bar is tightened from ends thus net deformation is assummed to be zero
so we have
[tex]\Delta_T +\Delta_B = \Delta = 0[/tex]
[tex]\frac{PL}{AE}_T + \alpha \Delta T L = \frac{PL}{AE}_B + (\alpha \Delta T L)_B[/tex]
[tex]\frac{19\times P}{0.5\times 16500} + 19 \times 5.3\times 10^{-6} \times 40 = \frac{27P}{0.65 \times 15200} + 27\times 12.2\times 10^{-6} \times 40 = 0 [/tex]
solving for P we get
P = 341 kips
average shear at B
[tex]\tau_a = \frac{P}{A}[/tex]
[tex]\tau_a = \frac{341}{\frac{\pi}{4} \times 0.4375^2}[/tex]
[tex]\tau_a = 22.7 ksi[/tex]
Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier directly on the line voltage with Vf = 0.7V, what will be your output DC voltage in Volts (assuming a negligible ripple)?
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.
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]
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:
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Calculating the sin of an Angle The numerical value of cos(x) can easily be determined using the Visual Basic sin function, provided x (in radians) is given as an argument. The value of cos(x) can also be approximated by the series The accuracy of this approximation increases as the number of terms in the series (n) increases. In principle, the summation results in an exact answer when n becomes infinite. As a practical matter, the summation is usually sufficiently accurate for modestly large values of n (say, n = 5 or n = 6). Develop a visual Basic program that evaluates cos(x) using the first n terms of the series expansion. The values of x (in degrees) and n will be input values. When evaluating the series expansion, calculate Factorial using a Sub procedure. Input and output values using message boxes.
Answer:
Visual Basic program
Sub cosApprox()
' Variable declaration
Dim x, seriesSum, exactValue As Double
' Variable for holding result at each iteration
seriesSum = 0
' Initializing x value
x = CDbl(InputBox("Enter x value: ", "Cos Approx", "0"))
n = CInt(InputBox("Enter n value: ", "Cos Approx", "0"))
' Iterating for n terms
For term = 0 To n
' Calculating current value and accumulating total result
seriesSum = seriesSum + (((-1) ^ term) ((x ^ (term 2)) / (WorksheetFunction.Fact((term * 2)))))
' Incrementing loop variable
Next term
' Storing exact value
exactValue = Cos(x)
' Displaying result
MsgBox ("Approximation Value: cos(" & x & "): " & seriesSum & vbNewLine & vbNewLine & "Exact Value: cos(" & x & "): " & exactValue)
End Sub
' Factorial function
Public Function factorial(ByVal n As Integer) As Long
' Base Case
If ((n = 0) Or (n = 1)) Then
factorial = 1
Else
' Recursive case
factorial = factorial(n - 1) * n
End If
End Function
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
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP of 2.5. How much power is required to drive the unit? For the same COP, how high a high temperature would a Carnot heat pump have, assuming the same low T?
The power required to drive the heat pump is 2 MW. For the same COP, a Carnot heat pump would achieve a high temperature of 324°C, starting from a low temperature of 85°C. The calculations involve both the COP formula and the Carnot efficiency formula.
To determine the power required to drive a heat pump with a given Coefficient of Performance (COP), we first use the formula:
COP = QH / W
Where QH is the heat delivered and W is the work input. Given that the heat delivered QH is 5 MW and the COP is 2.5, we can solve for the work W as follows:
[tex]2.5 = 5 MW / W[/tex]
Solving for W gives:
[tex]W = 5 MW / 2.5 = 2 MW[/tex]
To find the maximum temperature that a Carnot heat pump would achieve with the same low temperature of 85°C and the same COP, we use the Carnot heat pump efficiency formula:
[tex]COP_{Carnot[/tex] [tex]= TH / (TH - TC)[/tex]
We know that:
[tex]2.5 = TH / (TH - 358 K)[/tex]
Rewriting it:
[tex]2.5(TH - 358) = TH[/tex]
Solving for TH, we get:
[tex]2.5TH - 895 = TH[/tex]
[tex]1.5TH = 895[/tex]
[tex]TH = 597 K[/tex]
Converting 597 K to Celsius, [tex]TH = 597 - 273 =[/tex] 324°C
Thus, the required power to drive the unit is 2 MW, and for the same COP, a Carnot heat pump would have a high temperature of 324°C, assuming the same low temperature of 85°C.
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]
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 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]
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 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