Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its terminals. True False
Answer:
false
Explanation:
the changing of a prisoner sentence or another penalty to another less severe
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50
Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
Advantages
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
Disadvantages
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]
Where:
[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]
Where:
[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.
[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]
[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]
If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:
[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]
[tex]\%V_{gr} = 10.197\,\%V[/tex]
The volume percentage of graphite is 10.197 per cent.
Following are the solution to the given points:
[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.
[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]
[tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]
Calculating the weight fraction of graphite:
[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]
[tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]
Calculating the volume percent of graphite:
[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]
[tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]
Therefore, the final answer is "0.964, 0.036, and 11.368%"
Learn more Graphite:
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