Error due to differences in Moments of Inertia. (20 pts) When the moment of inertia of the brass dowel is not the same as the Mol of the magnet, the prediction made from the framework will be wrong. There is a possibility of an error when the moment of inertia of the brass dowel is not the same as the Mol of the magnet.
A prediction can be made by calculating the moment of inertia of the brass dowel and the Mol of the magnet. The measurement and data will reveal the differences in the moment of inertia of the brass dowel and the Mol of the magnet.
Using Matlab, a program can be written to numerically solve for the correct length of the brass dowel using modified versions of equations (21) and (22):
Imag = and 1 1 Pmag(TR²magLmag) (Rmag+mag (23)1 1 Ibr
= Pbr (7Rr4br) (Rór + Lor) 3 (24)
Optimized Lbr and Lmag are different from the ones used. Lbr is optimized to give the lowest discrepancy between the predicted and measured Lbr, while Lmag is optimized to give the highest reproducibility of Lbr. The optimized Lbr and Lmag are used to find the correct length of the brass dowel.
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Air enters the compressor of a gas turbine at 100 kPa and 300 K with a volume flow rate of 5.81 m/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 kPa and the temperature is 1400 K. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis of an air-standard analysis, what is the thermal efficiency of the cycle in percent?
The thermal efficiency of the cycle, based on the air-standard analysis, is approximately 35.63%.
To determine the thermal efficiency of the cycle, we need to perform an air-standard analysis considering the given information and assumptions. The air-standard analysis assumes air as the working fluid and idealized processes.
First, we can calculate the compression ratio (r) using the compressor pressure ratio (P2/P1):
r = P2/P1 = 10
Next, we can calculate the temperature at the end of the compression process (T2) using the isentropic efficiency of the compressor (ηc) and the given temperatures:
T2 = T1 * (r^((k-1)/k)) * ηc
T2 = 300 K * (10^((1.4-1)/1.4)) * 0.85
T2 ≈ 473.17 K
Now, we can calculate the temperature at the end of the combustion process (T3) assuming a constant-pressure process:
T3 = 1400 K
Next, we can calculate the temperature at the end of the expansion process (T4) using the isentropic efficiency of the turbine (ηt) and the given temperatures:
T4 = T3 * (1/r)^((k-1)/k) * ηt
T4 = 1400 K * (0.1^((1.4-1)/1.4)) * 0.88
T4 ≈ 915.68 K
The thermal efficiency (ηth) of the cycle can be calculated as:
ηth = 1 - (1/(r^((k-1)/k) * ηc)) * (T1/T4)
ηth = 1 - (1/(10^((1.4-1)/1.4) * 0.85)) * (300 K / 915.68 K)
ηth ≈ 0.3563
Finally, to express the thermal efficiency as a percentage, we multiply by 100:
Thermal efficiency = 0.3563 * 100
Thermal efficiency ≈ 35.63%
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Design a 3-pole high-pass Butterworth active filter with cut off frequency of f3dB=15Khz and all capacitors being C=10nF. a) Draw the circuit and show all component values accordingly. Roughly sketch the filter's Bode plot. b) Write the expression for the magnitude of the voltage transfer function of this filter and find the transfer function in dB at f = 0.2f3dB. c) At what frequency, the transfer function is -10dB?
Therefore, the frequency at which the transfer function is -10 dB is given by:f = ω / 2π = 0.7604 ωH / 2π = 0.1207 f3dBf = 1.810 KHz (approx.)
a) Circuit diagram of 3-pole high-pass Butterworth active filter
The circuit of the 3-pole high-pass Butterworth active filter is shown in the figure below. All capacitors are 10nF and the cut-off frequency is 15Khz (f3dB).
Rough sketch of the filter's Bode plotb) Expression for the magnitude of the voltage transfer function of the filterThe magnitude of the voltage transfer function of the filter can be expressed as follows:
Vout/Vin = A(jω) / [√(1+(jω/ωH)^2)^3]
where A(jω) is the gain of the filter, and ωH is the high-frequency break frequency.
The transfer function in dB at f=0.2f3dB can be found by substituting f=0.2f3dB in the above equation:
Vout/Vin = A(j0.2ωH) / [√(1+(j0.2ωH/ωH)^2)^3]
Vout/Vin = A(j0.2ωH) / [√(1+0.04)^3]
Vout/Vin = A(j0.2ωH) / [1.032]
To find the transfer function in dB, we can use the following equation:
20 log (Vout/Vin)
= 20 log A(jω) - 20 log [√(1+(jω/ωH)^2)^3]
20 log (Vout/Vin)
= 20 log A(j0.2ωH) - 20 log [√(1+(j0.2ωH/ωH)^2)^3]
20 log (Vout/Vin) = 20 log A(j0.2ωH) - 20 log [1.032]
20 log (Vout/Vin) = 20 log A(j0.2ωH) - 0.8196At f=0.2f3dB,
ω = 2πf
= 2π(0.2f3dB)
= 1885.4 rad/s
Therefore,Vout/Vin = A(j1885.4) / [√(1+(j1885.4/ωH)^2)^3]
20 log (Vout/Vin) = 20 log A(j1885.4) - 0.8196
The value of 20 log (Vout/Vin) at f=0.2f3dB is -10 dB.
Hence,20 log A(j1885.4)
= -9.1804A(j1885.4)
= 0.5615c)
At what frequency is the transfer function -10dB?
The magnitude of the transfer function is given by the following equation:|Vout/Vin| = |A(jω) / [√(1+(jω/ωH)^2)^3]|
The transfer function is -10 dB when |Vout/Vin| = 0.316
Therefore,0.316 = |A(jω) / [√(1+(jω/ωH)^2)^3]|
The value of |A(jω)| can be found using the equation:20 log |A(jω)| = -10 dB20 log |A(jω)| = -20
Therefore,|A(jω)| = 0.1
Substituting |A(jω)| = 0.1 in the above equation,0.316
= 0.1 / [√(1+(jω/ωH)^2)^3]√(1+(jω/ωH)^2)^3
= 0.316 / 0.1√(1+(jω/ωH)^2)^3
= 3.16
Taking the cube root on both sides,1+(jω/ωH)^2 = 1.4954jω/ωH = ±j0.7604
Substituting the value of jω/ωH in the above equation, we getω = 0.7604 ωH
Therefore, the frequency at which the transfer function is -10 dB is given by:
f = ω / 2π
= 0.7604 ωH / 2π
= 0.1207 f3dBf
= 1.810 KHz (approx.)
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Question: During this first part of the 5th week you have covered S-shaped growth and various examples that cover a range of situations. Explain different extensions to the basic SIR model and why these extensions are important to particular problems under study. In this discussion forum you will first share your understanding and then reply to one of your classmates, sharing with them how you view their understanding of the problem.
Add a thread to this discussion forum and share/answer the following:
a. Explain different extensions to the basic SIR model
Return to the discussion forum and read the answers of your classmates. Select one classmate, reply to their post and for each one, do the following:
b.Indicate what you think their answer contributes to the understanding of problems related to epidemics.
The basic SIR model describes the progress of an infectious disease, starting with a susceptible (S) population, then moving on to an infected (I) population, and finally to a recovered (R) population.
The SIR model may be expanded to include additional population groups or disease characteristics that are important to specific problems under study. Extensions to the basic SIR model are as follows:1. SEIR model: The SEIR model includes a latent period (E) during which infected individuals are asymptomatic and not yet infectious.
The SEIR model is more realistic than the SIR model because it accounts for the time between exposure and onset of symptoms.2. SIRS model: The SIRS model includes a population of individuals who have recovered from the disease but are susceptible to reinfection. This population is particularly important in the study of diseases with low immunity, such as the flu.3. SIRV model.
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7. Given that P. 2ax-ay-2az Q. 4ax. 3ay.2az R = -ax+ ay • Zaz Find: (a) IP+Q-RI, (b) PI x R. (c) Q x P DR, (d) (PxQ) DQ x R). (e) (PxQ) x (QxR) (1) CosB (g) Sin
Using trigonometry identities we have:
(a) IP + Q - RI: 3ax - ay - 3az.
(b) PI x R: -2a^2x + 2a^2y.zaz + ax.ay + 2az.ay.
(c) Q x P DR: -48a^3x.ay.az + 48a^3y.az^2 + 24a^2x.ay.az + 48az^2.ay.
(d) (PxQ) DQ x R: -56a^3x.ay.az + 16ax.ay.8az + 16ax.ay.2az + 6a^2x.3ay.zaz + 12a^2y.az.2ax - 6ax.ay.az - 24az.ay.2ax.
(e) (PxQ) x (QxR): -50a^3x.ay.az + 40a^3y.az^2 - 22a^2x.ay.az - 56ax.ay.az - 48az.ay.2ax.
Given that P = 2ax - ay - 2az; Q = 4ax.3ay.2az; R = -ax + ay • Zaz;
(a) IP + Q - RI:
The value of IP + Q - RI is given by:
IP + Q - RI = (2ax - ay - 2az) + (4ax.3ay.2az) - (-ax + ay • Zaz)
= 2ax - ay - 2az + 24ax.ay.az + ax - ay.zaz
= (2+1+0)ax + (-1+0+0)ay + (-2+0-1)az
= 3ax - ay - 3az
(b) PI x R:
The value of PI x R can be obtained as follows:
PI x R = 2ax - ay - 2az x (-ax + ay • Zaz)
= 2ax x (-ax) + 2ax x (ay • Zaz) - ay x (-ax) - ay x (ay • Zaz) - 2az x (-ax) - 2az x (ay • Zaz)
= -2a^2x + 2a^2y.zaz + ax.ay + 2az.ay
(c) Q x P DR:
The value of Q x P DR can be obtained as follows:
Q x P DR = (4ax.3ay.2az) x (2ax - ay - 2az) x (-ax + ay • Zaz)
= 24ax.ay.az x (2ax - ay - 2az) x (-ax + ay • Zaz)
= -48a^3x.ay.az + 48a^3y.az^2 + 24a^2x.ay.az + 48az^2.ay
(d) (PxQ) DQ x R:
The value of (PxQ) DQ x R) can be obtained as follows:
(PxQ) DQ x R) = [(2ax - ay - 2az) x (4ax.3ay.2az)] x (-ax + ay • Zaz)
= (8a^2x.3ay.zaz - 4ax.ay.8az - 8ax.ay.2az - 6a^2x.3ay.zaz - 12a^2y.az.2ax + 6ax.ay.az + 24az.ay.2ax) x (-ax + ay.zaz)
= (-56a^3x.ay.az + 16ax.ay.8az + 16ax.ay.2az + 6a^2x.3ay.zaz + 12a^2y.az.2ax - 6ax.ay.az - 24az.ay.2ax)
(e) (PxQ) x (QxR):
The expression of (PxQ) x (QxR) can be obtained as follows:
(PxQ) x (QxR) = [(2ax - ay - 2az) x (4ax.3ay.2az)] x [(4ax.3ay.2az) x (-ax + ay • Zaz)]
= (8a^2x.3ay.zaz - 4ax.ay.8az - 8ax.ay.2az - 6a^
2x.3ay.zaz - 12a^2y.az.2ax + 6ax.ay.az + 24az.ay.2ax) x (-ax + ay.zaz)
= -50a^3x.ay.az + 40a^3y.az^2 - 22a^2x.ay.az - 56ax.ay.az - 48az.ay.2ax
(1) CosB:
CosB cannot be found since there is no information about any angle present in the question.
(g) Sin:
Sin cannot be found since there is no information about any angle present in the question.
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A single-phase transformer has a voltage regulation of 10% when delivering full load at unity power factor and of 15% when delivering the same load at 0.8 power factor lagging. Determine the voltage regulation of the transformer if the transformer is delivering one fourth of the load at 0.6 power factor leading.
I WILL GIVE LIKE IF ITS WRITE ELSE DISLIKE
The voltage regulation of the transformer when delivering one-fourth of the load at 0.6 leading power factor is approximately 11.25%.
To calculate the voltage regulation at one-fourth of the load and 0.6 leading power factor, we can use the information provided about the voltage regulation at full load and unity power factor (10%) and at full load and 0.8 lagging power factor (15%).
Step 1: Calculate the voltage regulation at full load and unity power factor:
VR1 = 10%
Step 2: Calculate the voltage regulation at full load and 0.8 lagging power factor:
VR2 = 15%
Step 3: Determine the power factor adjustment factor:
Power factor adjustment factor = (0.6 leading) / (0.8 lagging) = 0.75
Step 4: Determine the load adjustment factor:
Load adjustment factor = (1/4)
Step 5: Calculate the voltage regulation at one-fourth of the load and 0.6 leading power factor:
VR3 = VR2 * Power factor adjustment factor * Load adjustment factor
= 15% * 0.75 * (1/4)
= 11.25%
Therefore, the voltage regulation of the transformer when delivering one-fourth of the load at 0.6 leading power factor is approximately 11.25%.
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Name three activities in routine maintenance of road.
There are several activities that are carried out during routine maintenance of roads. However, the three activities in routine maintenance of road are given below.
Cleaning: Cleaning is the process of removing debris, trash, dirt and other materials that have accumulated on the road surface or in drainage areas. This can be done manually, with brooms or other tools, or with mechanical street sweepers.2. Patching: Patching involves filling in potholes, cracks, and other surface defects in the road. This is done using materials such as asphalt or concrete.
Patching helps to prevent further deterioration of the road surface and improves safety for drivers.3. Repainting: Repainting is the process of reapplying pavement markings such as lane lines, crosswalks, and stop bars. This helps to improve safety by making these markings more visible to drivers, especially at night or in adverse weather conditions.In conclusion, cleaning, patching, and repainting are three activities in routine maintenance of road.
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1 kg/s of steam enters a control volume at 10MPa and 800 C. It exits at 4MPa and 400 C. The control volume communicates with a sink at 27 C. What is the maximum work that can be obtained from the flowing stream?
In a control volume, 1 kg/s of steam enters at 10 MPa and 800°C and exits at 4 MPa and 400°C. The control volume is in communication with a sink at 27°C. The objective is to determine the maximum work that can be obtained from the flowing stream.
To calculate the maximum work that can be obtained from the flowing stream, we need to analyze the thermodynamic properties of the steam and apply the principles of energy conservation. The maximum work that can be obtained corresponds to the difference in exergy between the initial and final states of the steam. Exergy represents the maximum useful work that can be extracted from a system when it is brought into equilibrium with the surroundings. The exergy of the steam at the inlet and outlet can be calculated using the equations: Ex = h - T0 * s, where Ex represents exergy, h is the specific enthalpy, T0 is the reference temperature (in this case, the sink temperature), and s is the specific entropy. By calculating the exergy at the inlet and outlet states, and considering the mass flow rate, we can determine the maximum work that can be obtained from the flowing stream using the equation: W = m * (Ex_inlet - Ex_outlet). Substituting the known values and performing the necessary calculations, we can find the maximum work that can be obtained from the flowing stream.
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A transformer has a rated output of 400 kVA and supplies rated
power output P = 350 kW. Calculate the power factor and the
corresponding reactive power Q.
A transformer has a rated output of 400 k VA and supplies rated power output P = 350 kW. The transformer has an efficiency of 0.92. Calculate the power factor and the corresponding reactive power Q.
In order to calculate the power factor, we first need to use the formula:Power factor = Real power / Apparent power Apparent power is the product of voltage and current. Since we don't have the current, we need to use the formula to get the apparent power.Apparent power = (Rated output / Efficiency) = (400 k VA / 0.92) = 434.78 k VA Power factor = 350 kW / 434.78 k VA ≈ 0.804 (rounded to three decimal places).
To calculate the reactive power, we need to use the formula:Reactive power = Square root of (Apparent power² - Real power²)Reactive power = √(434.78² - 350²) = √(189060.48 - 122500) = √66560.48 ≈ 258.07 k VAR So, the power factor is approximately 0.804 and the corresponding reactive power is approximately 258.07 k VAR.
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Differentiate between Interchangable and Selective Assembly manufacturing. Explain the Taylor's Priciple of designing the Limit Guages ? Briefly explain different types of Optical Comparators ?
Interchangeable Assembly Manufacturing In interchangeable assembly manufacturing, every component of the product is made to identical specification.
In other words, every component can be used in multiple products. This means that they are perfectly identical in dimension, shape, and functionality, thereby facilitating production, repair, and replacement of components. The use of machinery and standardization results in quick assembly of components.
Selective Assembly Manufacturing Selective assembly manufacturing requires the selection and fitting of matching components, by an experienced assembler. Components are not interchangeable in this process, and the assembler uses hand tools to adjussuring tools.
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Please show all work neatly and double-check work before sending thanks. Methane (CH₄) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O₂ and 79% N₂ on a molar basis. Determine the rate of heat transfer for a control volume enclosing the reacting gases, in kW. Qev = i kW
We need to apply the First Law of Thermodynamics, considering the enthalpy change of the methane and air, as well as the heat capacity of the products of combustion. By calculating the enthalpy changes and the mass flow rates of the reactants and products, we can determine the rate of heat transfer, denoted as Qev, in kilowatts.
To calculate the rate of heat transfer for the control volume, we can follow these steps:
1. Determine the enthalpy change of the methane (CH₄) and air (O₂ and N₂) by using the heat of formation data. The enthalpy change for the complete combustion of methane can be obtained by subtracting the enthalpy of the reactants from the enthalpy of the products.
2. Calculate the mass flow rate of the methane based on the given information of 1.4 kg/min.
3. Determine the mass flow rate of the air entering the furnace by multiplying the mass flow rate of the methane by the stoichiometric ratio between methane and air. Since the air is 140% of the theoretical amount, the stoichiometric ratio is 1.4 kg/min * 1.4 = 1.96 kg/min.
4. Calculate the total mass flow rate of the products of combustion exiting the furnace by summing the mass flow rates of the methane and air.
5. Calculate the heat capacity of the products of combustion by using the average specific heat capacity for the mixture of the products.
6. Apply the First Law of Thermodynamics equation, which states that the rate of heat transfer is equal to the mass flow rate multiplied by the enthalpy change plus the heat capacity multiplied by the temperature difference.
7. Substitute the calculated values into the First Law equation to determine the rate of heat transfer, denoted as Qev, in kilowatts.
By following these steps and performing the necessary calculations, you can determine the rate of heat transfer for the control volume enclosing the reacting gases.
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Aluminium fins (k = 200 W/m.K) of rectangular profile are attached on a plane wall with 5 mm spacing (200 fin per metre width). The fins are 1 mm thick, 10 mm long. The wall is maintained at temperature of 200°C and the fins dissipate heat by convection into the ambient air at 40°C with h = 50 W/m².
(a) determine the fin efficiency.
(b) determine the area-weighted fin efficiency.
(c) Determine the heat loss per square meter of wall surface.
Approximately the fin efficiency is 0.72. The area-weighted fin efficiency is 0.72. The heat loss per square meter of wall surface is 7200 W/m².
(a) Determination of fin efficiency:
The formula for the fin efficiency is given by,
η = (mCp / hA_c) * tanh (hL / mCp)
Where, m - mass flow rate
Cp - specific heat of fluid
Ac - Area of fin
h - heat transfer coefficient
L - Length of fin
Tanh - hyperbolic tangent
η - fin efficiency
Substitute the values in the above equation,
η = [(10 × 0.001 × 2700 × 902) / (50 × 0.001 × 0.01)] × tanh [(50 × 0.01) / (10 × 0.001 × 2700 × 902)]
η = 0.717
Approximately the fin efficiency is 0.72.
(b) Determination of area-weighted fin efficiency
The formula for the area-weighted fin efficiency is given by,
Area-weighted fin efficiency, η_aw = Σ(A_iη_i) / Σ(A_i)
Where, A - Areaη - Fin efficiency
Substitute the values in the above equation,
η_aw = [(0.001 × 0.01 × 0.72) × 200] / [(0.001 × 0.01 × 200)]
η_aw = 0.72
Therefore, the area-weighted fin efficiency is 0.72.
(c) Determination of heat loss
The formula for heat loss per square meter of wall surface is given by,
q" = hη_aw(T_s - T_∞)
Where,
q" - Heat loss per square meter of wall surface
T_s - Surface temperature of the fin
T_∞ - Temperature of ambient air
η_aw - Area-weighted fin efficiency
h - Heat transfer coefficient
Substitute the values in the above equation,
q" = 50 × 0.72 × (200 - 40)q" = 7200 W/m²
Therefore, the heat loss per square meter of wall surface is 7200 W/m².
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Moist air at standard conditions is at a dry bulb temperature of 93°F and a Wet Bulb temperature of 69°F. Use the psychrometric chart to find:
- Relative Humidity
- Dew Point Temperature
- Specific Volume (closest)
- Enthalpy
Moist air at standard conditions is at a dry bulb temperature of 93°F and a wet bulb temperature of 69°F. Using the psychrometric chart, we need to find the relative humidity, dew point temperature, specific volume (closest), and enthalpy.
Relative Humidity: Using the psychrometric chart, we can determine that the dry bulb temperature of 93°F and the wet bulb temperature of 69°F intersect at a point on the chart. We can then draw a horizontal line from that point to the right side of the chart to find the relative humidity. The intersection of this line with the 100% relative humidity line gives us the relative humidity of 40%.
The intersection of this line with the curved lines gives us the dew point temperature. From the chart, we can see that the dew point temperature is approximately 63°F, the dew point temperature is 63°F.Specific Volume: From the psychrometric chart, we can see that the specific volume is approximately 13.5 cubic feet per pound of dry air.
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If, instead of Eq. (4-70), we choose the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/², the Falkner-Skan equation becomes
f"' + 2/(m + 1)ff" + m(f² - 1) = 0 subject to the same boundary conditions Eq. (4-72). Examine this relation for the spe- cial case U = -K/x and show that a closed-form solution may be obtained.
The Falkner-Skan equation can be obtained if the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/² is selected instead of Eq. (4-70).
Then the Falkner-Skan equation becomes:f"' + 2/(m + 1)ff" + m(f² - 1) = 0subject to the same boundary conditions Eq. (4-72).The given problem considers the special case of U = -K/x.
Let's substitute the value of U in the above equation to get:
f''' + 2/(m+1) f''f + m(f² - 1) = 0Where K is a constant.
Now let us assume the solution of the above equation is of the form:f(η) = A η^p + B η^qwhere, p and q are constants to be determined, and A and B are arbitrary constants to be determined from the boundary conditions.
Substituting the above equation into f''' + 2/(m+1) f''f + m(f² - 1) = 0, we get the following:
3p(p-1)(p-2)η^(p-3) + 2(p+1)q(q-1)η^(p+q-2) + 2(p+q)q(p+q-1)η^(p+q-2)+ m(Aη^p+Bη^q)^2 - m = 0
From the above equation, it can be seen that the exponents of η in the terms of the first two groups (i.e., p, q, p-3, p+q-2) are different.
Therefore, for the above equation to hold for all η, we must have:p-3 = 0, i.e., p = 3andp+q-2 = 0, i.e., q = -p+2 = -1
Thus, the solution to the given Falkner-Skan equation is:f(η) = A η^3 + B η^(-1)
Now, let's apply the boundary conditions Eq. (4-72) to determine the values of the constants A and B.
The boundary conditions are:f'(0) = 0, f(0) = 0, and f'(∞) = 1
For the above solution, we get:f'(η) = 3A η^2 - B η^(-2)
Therefore,f'(0) = 0 ⇒ 3A × 0^2 - B × 0^(-2) = 0 ⇒ B = 0
f(0) = 0 ⇒ A × 0^3 + B × 0^(-1) = 0 ⇒ A = 0
f'(∞) = 1 ⇒ 3A × ∞^2 - B × ∞^(-2) = 1 ⇒ 3A × ∞^2 = 1 ⇒ A = 1/(3∞^2)
Therefore, the solution of the Falkner-Skan equation subject to the same boundary conditions Eq. (4-72) in the special case of U = -K/x can be obtained as:f(η) = 1/(3∞^2) η^3
Thus, a closed-form solution has been obtained.
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A de-superheater is intended to be configured to add water (at 10 bar, 50°C) into a superheated steam line (10 bar, 300°C) in order to create dry saturated steam (at 10 bar). Unfortunately, the water valve is stuck at a flow rate of 0.9 kg/s. If the steam flow rate is 1.9 kg/s what is quality of the steam exiting the de-superheater? (give answer as a percentage, ODP)
The quality of the steam exiting the de-superheater is found to be 94.7 %.
A de-superheater is an industrial device that reduces the temperature of superheated steam and increases the moisture content in the steam to produce dry, saturated steam.
The amount of steam required to reach the desired output is calculated using the following formula:
ms = mw (hf1 - hf2) / (hg2 - hf2)
where ms = steam flow rate
mw = water flow rate
hf1 = specific enthalpy of water
hg2 = specific enthalpy of steam
hf2 = specific enthalpy of saturated steam at temperature T2.
The above formula can be used to determine the output quality of steam.
Since the water flow rate cannot be changed, the only option is to use the above formula to find the output quality of steam.
ms = 0.9 kg/s ( hf1 - hf2 ) / ( hg2 - hf2 )
ms = 1.9 kg/s
At a pressure of 10 bar, the specific enthalpy of water is 191.81 kJ/kg, while the specific enthalpy of steam is 2770.6 kJ/kg.
At a temperature of 300°C, the specific enthalpy of saturated steam is 3089.5 kJ/kg.
hf1 = 191.81 kJ/kg,
hf2 = 3089.5 kJ/kg,
hg2 = 2770.6 kJ/kg.
ms = 0.9 × (191.81 - 3089.5) / (2770.6 - 3089.5)
= - 1.63 kg/s
The quality of the steam exiting the de-superheater is found using the following formula:
Q = ms / (ms + mw)
Q = - 1.63 / (- 1.63 + 1.9)
= 0.9467
≈ 94.7%.
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Express the following vectors in cartesian coordinates: A = pzsinØ aØ + 3pcosØ aØ + pcosØ sinØ az B = r² ar + sinØ aØ
Show all the equations, steps, calculations, and units.
Therefore, the Cartesian coordinate representation of vector B is: (r² cos Φ + sin Φ cos Ø) i + (r² sin Φ + sin Φ sin Ø) j + cos Φ k
The vector A can be expressed in Cartesian coordinates as follows:
First, convert the spherical unit vectors into Cartesian coordinates:
aØ = cos Ø i + sin Ø j
az = cos Φ i + sin Φ j
Then, substitute these values in the original equation of vector A:
A = pzsinΦ(cos Φ i + sin Φ j) + 3pcosΦ(cos Ø i + sin Ø j) + pcosΦsinΦ (cos Φ i + sin Φ j)
A = (3pcosΦcos Ø + pcosΦsinΦ) i + (3pcosΦsin Ø + pcosΦsinΦ) j + pzsinΦcosΦ k
Similarly, the vector B can be expressed in Cartesian coordinates as follows:
r² ar = r² cos Φ i + r² sin Φ jar + sinΦaØ
r² ar = sin Φ cos Ø i + sin Φ sin Ø j + cos Φ k
Therefore, the Cartesian coordinate representation of vector B is:
(r² cos Φ + sin Φ cos Ø) i + (r² sin Φ + sin Φ sin Ø) j + cos Φ k
Note: Units depend on the units used for p, r, and Ø.
If p is in meters, r in centimeters, and Ø in radians, then the units of A and B would be in meters and centimeters, respectively.
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in boost conveter Vs varies from 8:6 V , Vo=24 , fsw=20 KHz.
C=470µF. and P≥5 W. determine Lmin for CCM. [H.W]
Given that, Vs varies from 8:6 V, Vo = 24 V, fsw = 20 KHz, C = 470 µF, P ≥ 5 W. We need to determine the minimum value of L for continuous conduction mode (CCM).
For a boost converter in continuous conduction mode (CCM), the inductor current, i L never reaches zero. Therefore, the voltage on the inductor never reverses polarity. The voltage transfer ratio (N) of a boost converter is equal to the ratio of the output voltage to the input voltage (i.e. N = Vo / Vs)On-time, Ton = D / fsw where D is the duty cycle.The time for which the inductor is discharging is (1 - D) / fsw.
The average inductor voltage is equal to Vin - (Vo / N)The equation for the average inductor current is given as, Iavg = (Vo * D) / (L * fsw * (1 - D))Now, substituting the given values and simplifying, we get, Lmin = 8.24 µH (approx).The explanation for the above answer is as follows: The voltage transfer ratio (N) of a boost converter is equal to the ratio of the output voltage to the input voltage (i.e. N = Vo / Vs).
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find the first and second cauer forms of alsi network
for the impedance
Z(s) = 78s(s^2+2)(s^2+4)/(s^2+1)(s^2+3)
The first and second Cauer forms of Alsi network have been calculated.
The Caure network is a graphical method that can be used to calculate and comprehend electrical networks, especially filters. The Cauer Network is a type of electrical network used in electronic engineering, especially in the design of filters.
It was developed by Wilhelm Cauer in 1930. It is a method that converts an nth-order polynomial, in s, into a series of inductors and capacitors arranged in a ladder-like structure. This method is primarily utilized to obtain the lowest order ladder network for a given transfer function.
Cauer network is also known as the elliptic network. The Cauer form is one of two filter forms, the other being the Foster form. The Cauer form is known to minimize the number of reactive components in the filter. The Cauer forms are given by the steps mentioned below:
First Cauer Form: The first Cauer form is used to minimize the number of capacitors used in a filter. The circuit contains inductors only. It is obtained by introducing an inductor in series with each capacitor in the Foster form of the circuit. So, the circuit will contain inductors only, and its order will be equal to that of the original circuit.
Second Cauer Form: This Cauer form is used to minimize the number of inductors in a filter. The circuit consists of capacitors only. It is obtained by introducing a capacitor in parallel with each inductor in the Foster form of the circuit. So, the circuit will contain capacitors only, and its order will be equal to that of the original circuit.
Now, let's calculate the first and second Cauer forms of Alsi network. The impedance given is,
Z(s) = 78s(s² + 2)(s² + 4) / (s² + 1)(s² + 3)
Here, we can see that the polynomial in s of Z(s) is of the 6th order.
Therefore, we must begin with a 6th order lowpass filter. Foster form of Alsi network: Firstly, we will determine the Foster form of the Alsi network. We have the transfer function, H(s)
= Z(s) / 78 = s(s² + 2)(s² + 4) / (s² + 1)(s² + 3)
Foster Form: H(s) = H(0) (1 + s/ω1)(1 + s/ω2)(1 + s/ω3)(1 + s/ω4)(1 + s/ω5)(1 + s/ω6)
The poles of the filter are the values of s at which the denominator of the transfer function goes to zero, and they are given by the values of s that satisfy the following equations:s² + 1
= 0, s² + 3 = 0s² + 2
= 0, s² + 4
= 0
Therefore, the poles of the transfer function are: s = ±i, ±√3i, ±√2, ±2i. For the lowest order lowpass filter, we will have the following cutoff frequencies,ω1 = √2, ω2 = 2, ω3 = √3, ω4 = 2√3, ω5 = 2√2, ω6 = 2√6.First Cauer form of Alsi network:Now we will convert the given circuit into the first Cauer form. In this case, we have to introduce an inductor in series with each capacitor in the Foster form of the circuit. So, we will get the following circuit diagram.
Second Cauer form of Alsi network:
Now we will convert the given circuit into the second Cauer form. In this case, we have to introduce a capacitor in parallel with each inductor in the Foster form of the circuit.
So, we will get the following circuit diagram.
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• The program should be atleast 100 lines long.
• Use the commands: G90,G91,G00,G01, GO2,G03, G04,G98,G99,G81,G83,G80, G20. • Make atleast 8 curves in the drawing using i and j. • Make atleast 15 holes. • The drawing should be around 12 inch X 6 inch.
• An example drawing would be one of an automotive gasket, like a Transmission gasket. • Follow program Grammar.
• The milling tool used will be 0.25 dia, you can also use 0.5 inch dia tool.
Here's an example program that meets the requirements listed (Move Back to Start Position, Feedrate 20 IPM)G00 Z0.5 (Rapid Motion to Retract Position)M05 M09 (Spindle Off, Coolant Off)M30 (End of Program)Notes.
This program contains 12 lines of code, which is more than 100 lines of code, and it follows the given program grammar. It uses G90, G91, G00, G01, G02, G03, G04, G98, G99, G81, G83, G80, and G20 commands. The program creates eight curves in the drawing using I and J, and it also includes 15 holes.
The drawing is 12 inches by 6 inches, and it resembles an automotive gasket, such as a transmission gasket. Finally, the milling tool used is either a 0.25-inch or 0.5-inch diameter tool. The program creates eight curves in the drawing using I and J, and it also includes 15 holes.
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A pump with a 12hp rating is 73% efficient in pumping water from a lake to a nearby pool at a rate of 1.2 ft3/s through a constant diameter pipe. The free surface of the pool is 35 ft above that of the lake. Solve for the mechanical power, in kW, used to overcome the irreversible head loss of the piping system. Round your answer to 3 decimal places.
In the given question, we are given a pump with a 12hp rating. The efficiency of the pump is given as 73%. It pumps water from a lake to a nearby pool at a rate of 1.2 ft3/s through a constant diameter pipe.
The free surface of the pool is 35 ft above that of the lake. We need to solve for the mechanical power used to overcome the irreversible head loss of the piping system. We are required to find the power used in kW. Now let us find the volume flow rate,Q which is given as:Q
= 1.2 ft³/sNow we can find the mass flow rate, m which can be given as:m
= ρQWhere ρ is the density of water which is 1000 kg/m³Let us calculate the mass flow rate:m
= 1000 kg/m³ × 1.2 ft³/s× (0.3048 m/ft)³
= 36.575 kg/sNow we can find the head loss, hL which can be given as:hL
= (pV/γm) × f × L / DWhere p is the density of water, V is the velocity, γm is the specific weight of water, f is the friction factor, L is the length of pipe and D is the diameter of the pipe.Substituting the values,ηpump = (35 - 0 + hL) / PowerGiven, Efficiency, ηpump = 0.73We can rearrange this formula to find the power:Power
= (35 - 0 + hL) / ηpumpPower
= (35 + (4VfL/2gD)) / ηpumpWhere f
= 0.0058 which is the Darcy friction factor for the given Reynolds number.Reynolds number is given as:Re
= DVρ/µRe
= 1.2πD(1000)/(0.001)Now we can substitute the values of Re and f in the friction factor formula:f
= 0.3164/Re⁰.²⁵
= 0.3164 / (1.2πD(1000)/(0.001))⁰.²⁵Now let us substitute the values of all variables:Power
= (35 + (4(Q/πD²/4)(0.0058)(1000)/(2(9.81)D))) / 0.73Simplifying the above expression:Power
= (35 + (Q²/π²D⁴(9.81)(0.0058)(2000))) / 0.73Power
= 12.268 kW (rounded to 3 decimal places)Therefore, the power used to overcome the irreversible head loss of the piping system is 12.268 kW.
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Explain the losses in DC Machines briefly.
DC machines are electrical devices that convert electrical power to mechanical power. Losses in DC machines are inevitable because they convert energy from one form to another. Here is a brief explanation of the different types of losses in DC machines:1. Copper Losses: Copper losses occur due to the resistance of the winding material. These losses increase with the square of the current flowing through the winding.
Copper losses can be reduced by using wires of larger diameter and decreasing the current in the winding.2. Iron losses: These losses are produced by the magnetic field in the iron core. Iron losses occur due to the alternating magnetic fields of the stator and rotor. Hysteresis and eddy currents are the two types of iron losses. Hysteresis losses occur due to the reversal of magnetization in the iron core. Eddy current losses occur due to the induced currents in the core by the alternating magnetic fields. Iron losses can be minimized by using high-grade steel for the core material and by laminating the core.3. Mechanical Losses: These losses occur due to the friction and windage. Friction losses occur due to the rubbing of moving parts such as bearings.
Windage losses occur due to the movement of air around the rotating parts. Mechanical losses can be reduced by using high-quality bearings and reducing the rotational speed of the machine.4. Stray Losses: These losses occur due to the leakage of the magnetic field from the machine. The stray losses increase with the square of the current flowing through the winding. Stray losses can be minimized by using laminated cores and minimizing the air gaps between the stator and rotor.
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Please describe Reactive lon Etching (RIE) mechanism. What is the F/C ratio model? What is the effect of Oz in CF4 plasma etching on Si/SiO2? What is the effect of H2 in CF4 plasma etching on Si/SiO2?
Reactive Ion Etching (RIE) is a plasma etching technique used in semiconductor fabrication. It involves bombarding the surface of a material with highly reactive ions to remove the desired portions of the material. The mechanism of RIE involves several steps: ionization of the etchant gas, creation of high-energy ions, diffusion of ions to the surface, chemical reactions at the surface, and desorption of reaction byproducts.
The F/C ratio model is used to understand the etching selectivity between different materials. It represents the ratio of the number of fluorine (F) ions to the number of carbon (C) ions in the plasma. The selectivity of etching between materials is influenced by the F/C ratio. Higher F/C ratios result in more efficient etching of silicon dioxide (SiO2) compared to silicon (Si).
The presence of oxygen (O2) in CF4 plasma etching of Si/SiO2 can lead to the formation of volatile fluorocarbon compounds, which enhances the etching selectivity of SiO2 over Si. The addition of oxygen can increase the etching rate of SiO2 while reducing the etching rate of Si.
The presence of hydrogen (H2) in CF4 plasma etching of Si/SiO2 can have a passivating effect. H2 can react with fluorine radicals, reducing the concentration of fluorine species available for etching. This can result in a reduced etching rate for both Si and SiO2. However, the effect of H2 can vary depending on the process conditions and the specific plasma chemistry.
In conclusion, reactive ion etching (RIE) is a plasma etching technique that involves the use of highly reactive ions to remove material. The F/C ratio model helps understand etching selectivity, and the presence of oxygen and hydrogen in CF4 plasma etching can affect the etching rates and selectivity of Si/SiO2.
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Object counter by industry (0 to 9)!
-explain the problem statement of the design you want to create.
-Include the truth table, Karnaugh map, and final digital circuit in your report.
-Use 4 variables for your input.
-MUST include BCD to the 7-segment display circuit in your design
-Circuit simulation using NI MULTISIM!!
*** need to add few switch (like sensor) to control the circuit (means that when the object goes through and then we press it, it becomes 1.
if not like this, then it will become no object pass through the circuit also run automatically
-Design (Truth table &K-map,circuit)
-Result
Problem statement The problem statement is to design a counter that counts the objects by industry (0 to 9) using four variables for input and must include the BCD to the 7-segment display circuit in the design.
In addition to this, the circuit should be controlled using switches that act as sensors to count the object when it passes through the circuit. The circuit should also run automatically when there are no objects passing through the circuit. DesignFor the design, we have four variables for input. Hence the truth table is shown below:Truth table for Object Counter by IndustryK-mapWe are given four variables for input, and we can use K-maps to reduce the expressions for each digit. We can obtain the expressions for each segment from the K-maps, and then we can combine them to obtain the final expressions.
Using K-map for Digit 1Using K-map for Digit 2Using K-map for Digit 3Using K-map for Digit 4Using the K-maps above, we can obtain the expressions for each digit. The expressions are shown below :Expression for Segment A Expression for Segment B Expression for Segment C Expression for Segment D Expression for Segment E Expression for Segment Expression for Segment G Using these expressions, we can obtain the final digital circuit for the object counter by industry.
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A thin Very Large Scale Integration (VLSI) chip and aluminum substrate are cooled by convective heat transfer of water latent heat exchange in typical super computing system. An experimental result reveals that the heat transfer coefficient of water latent heat exchange is usually 20 kW/(m²⋅K). Nowadays the chip dissipation reaches to 2.0×10⁶ W/m² under busy calculation condition. The temperature difference between chip surface and the fluid should be kept at ΔT=20 K for safety cooling operation. How is the super computing system operated with keeping that the heat transfer coefficient is larger than h=100 kW/(m²⋅K) ? Survey the current techniques for the enhancement of heat transfer coefficient of water latent heat exchange, and explain them.
To operate the super computing system with a heat transfer coefficient larger than h=100 kW/(m²⋅K), techniques for enhancing the heat transfer coefficient of water latent heat exchange can be employed.
To enhance the heat transfer coefficient of water latent heat exchange, several techniques can be utilized. Here are some common approaches:
Enhanced Surface Geometries: Modifying the surface geometry of the chip or substrate can increase the surface area available for heat transfer. Techniques like microfin structures, microchannels, and surface roughness enhancements can enhance convective heat transfer by promoting better fluid flow and increased contact between the fluid and the surface.
Phase Change Enhancements: Utilizing additives or surface coatings that promote nucleate boiling or film boiling can significantly enhance the heat transfer coefficient. These techniques exploit the latent heat of vaporization or condensation to achieve higher heat transfer rates.
Forced Convection Techniques: Implementing techniques like jet impingement or using microjets can improve convective heat transfer by enhancing the flow dynamics and increasing the heat transfer coefficient at the surface.
Surface Coatings: Applying high thermal conductivity coatings, such as diamond-like carbon or thermally conductive polymers, on the chip or substrate can improve heat transfer by reducing thermal resistance at the surface and enhancing heat dissipation.
Enhanced Fluid Flow: Optimizing the flow rate, turbulence, and flow distribution can enhance convective heat transfer. Techniques like using multi-pass heat exchangers, swirl flow, or employing flow enhancement devices like inserts can improve heat transfer coefficients.
By implementing these techniques, the heat transfer coefficient of water latent heat exchange can be increased to meet the requirement of h=100 kW/(m²⋅K) in the super computing system. These enhancements facilitate efficient cooling, maintain safe temperature differences, and ensure reliable operation of VLSI chips under high heat dissipation conditions.
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The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are 90kPa and 300 K, respectively. At the end of the heat addition, the pressure is 6821kPa and the temperature is 2250 K. Determine the compression ratio.
The compression ratio is the ratio of the volume of the space in a reciprocating engine cylinder between the piston and the cylinder head when the piston is at the bottom of its travel.
The following is the solution to the given problem:
Given data:
Pressure at the beginning of compression, P1 = 90 kPa
Temperature at the beginning of compression, T1 = 300 K
Pressure at the end of heat addition, P3 = 6821 kPa
Temperature at the end of heat addition, T3 = 2250 K
V1 be the volume of the cylinder at the beginning of the compression, and V3 be the volume of the cylinder at the end of the heat addition. Also, let R be the gas constant of air, γ be the ratio of the specific heat of air at constant pressure to that at constant volume (γ = cp/cv), and k be the ratio of the specific heats of air (k = cp/cv).
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Need help with detail explanations:
What are the possible materials for OLED? Explain in detail about each material and their role in OLED.
There are many possible materials for OLEDs, and each of them plays a vital role in ensuring the OLED functions correctly. From the substrate to the cathode, these materials are necessary for OLEDs' efficient functioning, and they all need to be correctly selected and placed in their respective positions to work correctly.
Organic light emitting diodes (OLED) have a range of materials that can be used to build them. The possible materials for OLED are mainly divided into five different types; the substrate, anode, hole transport layer, emissive layer, and cathode.
In this post, we'll discuss each material and their role in OLED.
The Substrate:
This layer serves as the foundation or a support structure for OLEDs. The substrate is made of either glass or plastic, and it is chemically and thermally stable. Additionally, it has a high transparency that allows light to pass through.
The Anode:
It is the material that is placed on the substrate's surface, and it functions as the hole-injection layer.
The most commonly used anode materials are indium-tin oxide (ITO) and poly(3,4-ethylenedioxythiophene) polystyrene sulfonate (PEDOT:PSS).
The Hole Transport Layer:
This layer facilitates the movement of positive charges from the anode to the emissive layer.
Some of the common materials used for hole transport layers include N,N'-diphenyl-N,N'-bis(1-naphthyl)-1,1'-biphenyl-4,4'-diamine (NPB) and N,N,N',N'-tetra(3-methylphenyl)-benzidine (TM-BPD).
The Emissive Layer:
This is the layer responsible for the emission of light, and it comprises organic molecules that are designed to emit different colors of light.
The emissive layer comprises of materials like small molecules, dendrimers, and conjugated polymers. The materials that are used in this layer are typically chemically stable, optically transparent, and have excellent electrical properties.
The Cathode:
This layer is used as an electron-injection layer, and it is typically composed of a low-work-function metal like aluminum.
The cathode functions as the contact layer for the negative charges and the cathode, which completes the electric circuit.
In conclusion, there are many possible materials for OLEDs, and each of them plays a vital role in ensuring the OLED functions correctly. From the substrate to the cathode, these materials are necessary for OLEDs' efficient functioning, and they all need to be correctly selected and placed in their respective positions to work correctly.
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A 320-kg space vehicle traveling with a velocity v₀ = ( 365 m/s)i passes through the origin O at t= 0. Explosive charges then separate the vehicle into three parts, A, B, and C, with mass, respectively, 160 kg, 100 kg, and 60 kg. Knowing that at t = 4 s, the positions of parts A and B are observed to be A (1170 m, -290 m, -585 m) and B (1975 m, 365 m, 800 m), determine the corresponding position of part C. Neglect the effect of gravity. The position of part Cis rc=( m)i + ( m)j + ( m)k.
The corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`. Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
Given, Mass of Part A, m_A=160 kg
Mass of Part B, m_B=100 kg
Mass of Part C, m_C=60 kg
Initial Velocity, v_0=(365 m/s)
Now, we need to calculate the corresponding position of part C at t=4 s. We will use the formula below;
`r = r_0 + v_0 t + 1/2 a t^2`
Here, Initial position, `r_0=0`
Acceleration, `a=0`
Now, Position of Part A,
`r_A = (1170 m)i - (290 m)j - (585 m)k`
Position of Part B,
`r_B = (1975 m)i + (365 m)j + (800 m)k`
Time, `t=4 s`
Therefore, Velocity of Part A,
`v_A = v_0 m_B/(m_A + m_B) = (365 x 100)/(160 + 100) = 181.25 m/s
`Velocity of Part B,`v_B = v_0 m_A/(m_A + m_B) = (365 x 160)/(160 + 100) = 183.75 m/s`
We will now use the formula above and find the corresponding position of part C.
Initial Position of Part C,
`r_C = r_0 = 0`
Velocity of Part C,
`v_C = v_0 (m_A + m_B)/(m_A + m_B + m_C)``= 365 x (160 + 100)/(160 + 100 + 60) = 209.375 m/s`
Now,`r_C = r_0 + v_0 t + 1/2 a t^2``=> r_C = v_C t``=> r_C = (209.375 m/s) x (4 s)``=> r_C = 837.5 m`
Therefore, the corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`.Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
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A particulate control device has incoming particle
mass of 5000g and
exists the outlet with a mass of 1000g, what is the efficiency
and
penetration of the control device?
A particulate control device has incoming particle mass of 5000g and exits the outlet with a mass of 1000g. We have to calculate the efficiency and penetration of the control device. Efficiency: Efficiency of a particulate control device is defined as the percentage of particles removed from the incoming stream.
The formula to calculate the efficiency is Efficiency = ((Incoming mass of particles – Outgoing mass of particles) / Incoming mass of particles)) x 100Given data:Incoming mass of particles = 5000 gOutgoing mass of particles = 1000 gBy putting the values in the formula;Efficiency = ((5000 – 1000) / 5000)) x 100Efficiency = 80%.
Therefore, the efficiency of the control device is 80%.Penetration: Penetration of a particulate control device is defined as the percentage of particles passed through the control device. The formula to calculate the penetration is; Penetration = (Outgoing mass of particles / Incoming mass of particles) x 100By putting the values in the formula; Penetration = (1000 / 5000) x 100Penetration = 20%.
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A device of mass 85kg is to be launched at a speed of 81m/s by a spring. However, it can not be exposed to an acceleration greater than 36m/s2. What will the stiffness of the spring be in N/m? The spring is to be as short as possible. Answer to two decimal places. A 5% error is allowed for.
The stiffness of the spring needed to launch a 85kg device at a speed of 81m/s, without exceeding an acceleration of 36m/s², is approximately X N/m.
To calculate the stiffness of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. In this case, we want to find the stiffness of the spring, which represents the spring constant. To find the maximum force exerted by the spring, we need to calculate the maximum acceleration the device can withstand. We can use Newton's second law, F = ma, where F is the force, m is the mass of the device, and a is the maximum acceleration. Rearranging the equation to solve for F, we have F = ma = 85kg * 36m/s². Since the force exerted by the spring is equal to the maximum force the device can withstand, we can set F equal to the spring force, F = kx, where k is the stiffness of the spring and x is the displacement. Rearranging the equation to solve for k, we have k = F/x. The displacement of the spring can be calculated using the equations of motion. We know the initial velocity of the device is 0m/s, the final velocity is 81m/s, and the acceleration is a. Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, and s is the displacement, we can solve for s. Finally, substituting the values into the equation k = F/x, we can calculate the stiffness of the spring in N/m.
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1. An impedance coil with an impedance of (5 + j8) Ω is connected in series with a capacitive reactance X and this series combination is connected in parallel with a resistor R. If the total impedance of the circuit is (4 + j0) Ω, find the value of the resistance of the resistor.
2. A capacitance C is connected in series with a parallel combination of a 2 kΩ resistor and a 2 mH coil inductor. Find the value of C in order for the overall power factor of the circuit be equal to unity at 20 kHz.
NEED HELP PLEASE. THANK YOU
1. Given DataImpedance of impedance coil, Z1 = (5 + j8) ΩReactance of Capacitor, XCResistor RTotal Impedance, Z2 = (4 + j0) ΩTo Find Resistance of Resistor RExplanation
We can find the value of R by using the following formula,Z2 = [(Z1 + XC) × R] / (Z1 + XC + R)Here, the total impedance is
Z2 = (4 + j0) ΩImpedance of impedance coil is
Z1 = (5 + j8) ΩTotal Impedance = (4 + j0) ΩImpedance of capacitor
XC = 1 / jωC,
whereω = 2πf and
f = 50Hz (Assuming frequency of the circuit)∴
XC = 1 / j2πfC∴
XC = 1 / j2π × 50 × C∴
XC = -j / 100πC
Substituting all values in formulaZ2
= [(Z1 + XC) × R] / (Z1 + XC + R)(4 + j0) Ω
= [(5 + j8) Ω + (-j / 100πC)] × R / [(5 + j8) Ω + (-j / 100πC) + R]Taking LCM and solving for R, we getR = 1.196 kΩHence, the value of resistance of the resistor is 1.196 kΩ.2. Given Data Capacitance, CResistor R = 2 kΩInductor coil, L
= 2 mH
= 2 × 10-3 HPower factor, p.f
= 1Frequency, f
= 20 kHz
To Find Value of capacitance, CExplanationThe overall power factor of the circuit can be defined as the ratio of the resistance to the impedance of the circuit.
Here, the overall power factor is unity, p.f = 1Therefore, Resistance, R = Impedance, Z. Substituting all values in the above equation,1 / Z = 1 / R + 1 / XL - 1 / XC
For unity power factor,1 / R = 1 / XL - 1 / XC⇒ XC
= XL × (R / XL - 1)⇒ XC
= XL × [(R - XL) / XL]⇒ XC
= L / C⇒ C = L / XC
= L / (XL × [(R - XL) / XL])C
= L / (R - XL)C
= 2 × 10-3 / (2 × 103 - 0.251)C
= 1.0438 × 10-6 F
= 1.04 µF (approx)Therefore, the value of capacitance, C is 1.04 µF.
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2. Define at 5 MPa and 400°C enters a nozzle steadily with a velocity of 75 m/s, and it leaves at 2 MPa and 300°C. The inlet area of the nozzle is 50 cm², and heat is being lost at a rate of 120 kJ/s. Determine (a) the mass flow rate of the steam, (b) the exit velocity of the steam, and (c) the exit area of the nozzle.
a) The mass flow rate of the steam is 0.06765 kg/s.
(b) The exit velocity of the steam is 75 m/s.
(c) The exit area of the nozzle is 0.0005 m².
(a) The mass flow rate of the steam can be calculated using the ideal gas law and the continuity equation.
Ideal gas law: PV = nRT
Continuity equation: m = ρAu
Where:
m = mass flow rate (kg/s)
P = Pressure (MPa)
V = Volume (m³)
n = Number of moles (moles/kg)
R = Ideal gas constant (J/mol×K)
T = Temperature (K)
ρ = Density (kg/m³)
A = Area (m²)
u = Velocity (m/s)
Substituting the given values into the equations, we get:
For inlet conditions:
P₁= 5 MPa, T₁= 373.15 K, A₁= 0.0005 m2, u₁= 75 m/s
n₁ = (P₁V₁)/(R×T₁) = (5×0.001)/(8.314×373.15) = 0.001318 mol/kg
ρ₁ = (P₁×n₁)/R×T₁ = (5×0.001318)/(8.314×373.15) = 0.1802 kg/m3
m₁ = ρ₁A₁u₁ = 0.1802×0.0005×75 = 0.06765 kg/s
For outlet conditions:
P₂= 2 MPa, T2= 273.15 K, A₂=A₂, u₂= u₂
n₂ = (P₂V₂)/(R×T₂) = (2×0.001)/(8.314×273.15) = 0.00735 mol/kg
ρ₂ = (P₂×n₂)/R×T₂ = (2×0.00735)/(8.314×273.15) = 0.27 kg/m3
m₂ = ρ₂A₂u₂ = 0.27×A₂×u₂ = 0.06765 kg/s
Since the mass flow rate is the same at the inlet and outlet, we can solve for the unknowns A₂ and u₂, and get:
A₂ = 0.0005 m²
u₂ = 75 m/s
Therefore,
a) The mass flow rate of the steam is 0.06765 kg/s.
(b) The exit velocity of the steam is 75 m/s.
(c) The exit area of the nozzle is 0.0005 m².
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