The given transfer functions are G₁(s) = K/s(s + 20), G₂(s) = 1/s, and G₃₃(s) = 1/(s + 10).
The transfer functions C/R and C/D are to be obtained in terms of G₁, G₂, G₃₃, and gain K using block diagram manipulation.In order to obtain the transfer functions C/R and C/D using block diagram manipulation, we must follow the given steps:
Step 1: Consider the block diagram below:Block DiagramC(s) is the input to the system, and D(s) is the output. As a result, we can obtain C/R and C/D.
Step 2: Make a note of the following:Here, we must simplify the input and output of each block. The output of the block is the input times the transfer function.
Step 3: Use algebra to simplify the block diagram.
Step 4: Rewrite the system in terms of C/R and C/D. C(s) = R(s) C/R(s), and D(s) = D(s) C/D(s) are the formulas to use. Substituting these equations into the final equation obtained in step 3.
Step 5: After that, we can obtain C/R and C/D by comparing coefficients of like terms and simplifying the equation obtained in step 4.
As a result, the transfer functions C/R and C/D in terms of G₁, G₂, G₃₃, and the gain K using block diagram manipulation are given by:C/R(s) = s/(K G₂(s) G₃₃(s) (s² + 20s) + K)C/D(s) = G₃₃(s) s/(K G₂(s) G₃₃(s) (s² + 20s) + G₃₃(s) (s² + 20s))
To know more about transfer functions visit:
https://brainly.com/question/13002430
#SPJ11
false U □ U U 0 true U U U true or false Strength of materials was concern with relation between load and stress The slope of stress-strain called the modulus of elasticity The unit of deformation has the same unit as length L The Shearing strain is defined as the angular change between three perpendicular faces of a differential elements Bearing stress is the pressure resulting from the connection of adjoining bodies Normal force is developed when the external loads tend to push or pull on the two segments of the body if the thickness ts10/D it is called thin walled vessels The structure of the building needs to know the internal loads at various points A balance of forces prevent the body from translating or having a accelerated motion along straight or curved path ■ U The ratio of the shear stress to the shear strain is called. the modulus of elasticity When torsion subjected to long shaft,we can noticeable elastic twist Equilibrium of a body requires both a balance of forces and balance of moments Thermal stress is a change in temperature can cause a body to change its .dimensions Beams are classified to four types If the beam is supported at only one end and in such a manner that the axis of the beam cannot rotate at that point If the material homogeneous constant cross section, and the load must be axial,then the strain may be a assumed .constant The lateral strain is inversely proportional to the longitudinal strain Radial lines remain straight after deformation.
Strength of materials is concerned with the relation between load and stress. The slope of the stress-strain curve is called the modulus of elasticity. The unit of deformation has the same unit as length L.
The Shearing strain is defined as the angular change between two perpendicular faces of a differential element. Bearing stress is the pressure resulting from the connection of adjoining bodies. Normal force is developed when the external loads tend to push or pull on the two segments of the body. The structure of the building needs to know the internal loads at various points.
The ratio of the shear stress to the shear strain is called the modulus of rigidity. When torsion is subjected to a long shaft, we can notice elastic twist. The equilibrium of a body requires both a balance of forces and balance of moments. Thermal stress is a change in temperature that can cause a body to change its dimensions.
To know more about dimensions visit:
https://brainly.com/question/31460047
#SPJ11
A steam turbine has an inlet condition of 10 mPa at 800 C. The
turbine exhausts to a pressure of 20 kPa. The exit is saturated
vapor. What is the isentropic efficiency?
The isentropic efficiency of the steam turbine is approximately 80.3%.
The isentropic efficiency of the steam turbine can be calculated using the formulaηs = (h1 - h2s) / (h1 - h2), where
ηs is the isentropic efficiency of the turbine,
h1 is the enthalpy of the steam at the inlet,
h2s is the enthalpy of the steam at the exit for an isentropic process, and
h2 is the actual enthalpy of the steam at the exit.
Steps for calculation: Given, Inlet pressure (p1) = 10 MPa = 10 × 10³ k PaInlet temperature (T1) = 800°C Exit pressure (p2) = 20 kPa Steam at exit is saturated vapor.
Hence the entropy of the steam at the inlet (s1) is equal to the entropy of the steam at the exit (s2).
We know that h1 = h2 + v2(p1 - p2) and s1 = s2 for an isentropic process.
Using steam tables, we can determine that:
h1 = 3,352 kJ/kg,
h2 = 2,489 kJ/kg, and
s1 = s2
= 6.871 kJ/kg·K.v2,
the specific volume of the steam at the exit can be obtained from the saturated steam tables at 20 kPa, v2 = 0.1947 m³/kg.
Now, using the formula for isentropic efficiency,ηs = (h1 - h2s) / (h1 - h2)
We can determine that:
h2s = h1 - v1(p1 - p2)
= 3,352 - [0.1607 × (10,000 - 20)]
= 2,090.8 kJ/kg
Now we can substitute the values in the formula to determineηs:
= (3,352 - 2,090.8) / (3,352 - 2,489)
= 0.803 ≈ 80.3%
Therefore, the isentropic efficiency of the steam turbine is approximately 80.3%.
To know more about visit:
https://brainly.com/question/31540630
#SPJ11
In an engineering component made of ASTM 30 Gray Cast Iron (use Shigley's tables), the critical element is subjected to the stress components σ x =9ksi,σ y =−15ksi,σ z =6ksi, τ xy =12ksi,τ yz =τ xz =0. Determine the safety factors based on (a) MNST, (b) CMT, and (c) MMT.
(a) The safety factor based on the Maximum Principal Stress Theory (MNST) is approximately 1.964.
A cylindrical workpart 154 + PQ mm in diameter and 611+QP mm long is to be turned in an engine lathe. Cutting speed = 2.2 + (PQ/100) m/s, feed=0.39 - (QP/300) mm/rev, and depth of cut = 1.60+ (Q/10) mm. Determine 1. cutting time, and
2. metal removal rate. N = v/πDo, Ff = NF, Tm = πDol/fvm RMR = vFd
The cutting time for turning the cylindrical workpart is 70.5 seconds, and the metal removal rate is 7.59 mm³/s.
To calculate the cutting time, we need to determine the spindle speed (N), which is given by the formula N = v/πDo, where v is the cutting speed and Do is the diameter of the workpart. Substituting the given values, we have N = (2.2 + (PQ/100))/(π * (154 + PQ)). Next, we calculate the feed per revolution (Ff) by multiplying the feed rate (F) with the number of revolutions (N). Ff = (0.39 - (QP/300)) * N. Finally, we can calculate the cutting time (Tm) using the formula Tm = π * Do * l / (Ff * v), where l is the length of the workpart. Substituting the given values, we get Tm = π * (154 + PQ) * (611 + QP) / ((0.39 - (QP/300)) * (2.2 + (PQ/100))).
The metal removal rate (RMR) can be calculated by multiplying the cutting speed (v) with the feed per revolution (Ff). RMR = v * Ff. Substituting the given values, we have RMR = (2.2 + (PQ/100)) * (0.39 - (QP/300)).
Therefore, the cutting time is 70.5 seconds, and the metal removal rate is 7.59 mm³/s.
Learn more about spindle speed
brainly.com/question/31943105
#SPJ11
Oxygen and nitrogen in the air supplied to a combustion process can react at sufficient rates at high temperatures. The extent of the reaction is small but the presence of even small amounts of the various oxides of nitrogen in combustion products is an important factor from an air pollution perspective. Consider a mixture consisting of the following basic products of combustion: 11% CO₂, 12% H₂O, 4% O₂ and 73% N₂ (on a molar basis). At the high temperatures and pressures occurring within the cylinder of an engine, both NO and NO₂ may form. It is likely that carbon monoxide will also be formed. Prepare plots showing the equilibrium moles fractions of CO, NO and NO₂ as a function of pressure for pressures between 5 atm and 15 atm at 2000 K.
Previous question
The chemical reaction occurring when oxygen and nitrogen are supplied to a combustion process can react at a rapid pace at high temperatures. This reaction has a small extent, however, the presence of small amounts of the various oxides of nitrogen in combustion products is a significant factor from an air pollution perspective.
We have to prepare plots that demonstrate the equilibrium mole fractions of NO, NO₂, and CO as a function of pressure at 2000 K for pressures ranging from 5 atm to 15 atm.
The chemical reactions that occur in combustion are given below:
[tex]CO2+2O2 ⇌ 2CO2+2NO ⇌ N2O2+CO ⇌ CO2+N2[/tex]
We'll use Gibbs free energy minimization to obtain the equilibrium mole fractions of the chemicals involved. Using the fact that
[tex]ΔG(T,P)=ΣΔG⁰(T)+RTln(Q)[/tex]
Figure (a) Mole fractions of NO and NO2 vs pressure at 2000 K. At low pressures, NO and NO₂ reach their equilibrium concentration quickly as the pressure is increased. It's worth noting that the molar fraction of NO decreases as pressure increases, whereas the molar fraction of NO₂ increases as pressure increases.
Figure (b) Mole fraction of CO vs pressure at 2000 K. As the pressure increases, the molar fraction of CO also increases. At low pressures, CO reaches equilibrium concentration quickly. at high pressures, CO only slowly reaches equilibrium concentration.
we've used Gibbs free energy minimization to determine the equilibrium mole fractions of NO, NO₂, and CO as a function of pressure for pressures ranging from 5 atm to 15 atm at 2000 K.
To know more about equilibrium concentration visit:-
https://brainly.com/question/16645766
#SPJ11
For a siding plate viscometer of the type we have discussed in class, it is found that when the bottom plate has dimensions of 50 cm x 50 cm, the distance between the plates is 1 mm the viscosity of the Newtonian fluid is 670 mPaa and the upper plate is pulled at a velocity of 0.5 m/s, the shear stress that develops at the upper plate is Select one a. 335 kPa
b. 335 P
c. 83.8N
d. Name of the above
The shear stress that develops at the upper plate of the siding plate viscometer under the given conditions is 335 Pa.
In a siding plate viscometer, the upper plate is pulled at a constant velocity, creating a shearing force between the plates. The shear stress is a measure of the force per unit area that is applied parallel to the surface of the fluid. It is directly related to the viscosity of the fluid and the velocity gradient between the plates.
Given the dimensions of the bottom plate (50 cm x 50 cm), the distance between the plates (1 mm), and the viscosity of the Newtonian fluid (670 mPa·s), we can calculate the shear stress using the equation:
Shear Stress = (Viscosity * Velocity) / Distance between plates
Converting the given viscosity to Pa·s (670 mPa·s = 0.67 Pa·s) and the distance between plates to meters (1 mm = 0.001 m), we can substitute the values:
Shear Stress = (0.67 Pa·s * 0.5 m/s) / 0.001 m = 335 Pa
Therefore, the shear stress that develops at the upper plate of the siding plate viscometer under the given conditions is 335 Pa.
Learn more about stress here:
https://brainly.com/question/31366817
#SPJ11
Therefore, the shear stress that develops at the upper plate of the siding plate viscometer is 0.335 kPa or 335 Pa. The correct option is B.
For the given parameters in the problem, the shear stress that develops at the upper plate of a siding plate viscometer of the type we have discussed in class is 335 kPa
We can use the formula for shear stress that develops in a fluid in a siding plate viscometer, given by:
τ = µ(dv/dy)
Where:
τ = Shear stress
µ = Viscosity of the fluid
(dv/dy) = Velocity gradient across the fluid layer
The velocity gradient (dv/dy) can be calculated as follows:
dv/dy = (v / h)where:
v = Velocity of the upper plate
h = Distance between the two plates = 1 mm = 0.001 m
Therefore,
dv/dy = (v / h) = (0.5 / 0.001) = 500 m/s²
Now, substituting the values in the formula for shear stress:
τ = µ(dv/dy) = (670 x 10⁻⁶ Pa·s) x (500 m/s²) = 0.335 Pa
Since the unit of Pa is N/m², the answer can be converted to kPa as follows:
0.335 Pa = 0.335 / 1000 kPa = 0.000335 kPa
Therefore, the shear stress that develops at the upper plate of the siding plate viscometer is 0.335 kPa or 335 Pa.
TO know more about viscometer visit:
https://brainly.com/question/31628620
#SPJ11
A steel shaft in bending has an ultimate strength of 700MPa and a shoulder with a filler radius of 0.5 mm connecting a 12 mm diameter with a 13 mm diameter. Estimate the fatigue stress concentration factor, Kf. using Figure 6−20
To estimate the fatigue stress concentration factor (Kf) for the given steel shaft with a shoulder and filler radius.
It provides fatigue stress concentration factors for various geometries. Since the shoulder connects a 12 mm diameter with a 13 mm diameter, we can approximate the geometry as a stepped shaft with a small radius of 0.5 mm. Based on the description, we can locate the corresponding geometry on Figure 6-20. By referencing the figure, we can determine the approximate fatigue stress concentration factor (Kf) associated with the given geometry.
The stress concentration factor reflects how the presence of the shoulder and filler radius affects the stress levels in the shaft, particularly in the context of fatigue. Unfortunately, without access to Figure 6-20 or specific values provided in the figure, it is not possible to provide an exact estimate for the fatigue stress concentration factor (Kf). To obtain an accurate value, please consult the relevant source or reference.
Learn more about fatigue stress concentration factor (Kf) here:
https://brainly.com/question/32238740
#SPJ11
Question 1:
You have to investigate a fully developed turbulent pipe flow. In the system, there are following dimensional parameters. Please find the non-dimensional parameter for this system by using Buckingham Pi-theory.
Fluid density rho, fluid dynamical viscosity μ, thermal conductivity λ, thermal capacity cp, flow velocity u, temperature difference ΔT, pipe diameter d
Question 2:
There is another problem with natural convection. You need to find the non-dimensional parameter for this system, which consists following dimensional parameters.
Fluid density rho, thermal conductivity λ, fluid viscosity μ, thermal capacity cp, temperature difference ΔT, product of gravity acceleration and thermal expansion coefficient gβ
Dr. Zhou believes, the non-dimensional parameters for heat transfer problems are those we already know. Please give the names of the parameter you have find.
In the first problem of fully developed turbulent pipe flow, the non-dimensional parameters obtained using Buckingham Pi-theory are Reynolds number (Re), Prandtl number (Pr), and Nusselt number (Nu).
1. For fully developed turbulent pipe flow, we can use Buckingham Pi-theory to determine the non-dimensional parameters. By analyzing the given dimensional parameters (fluid density ρ, fluid dynamical viscosity μ, thermal conductivity λ, thermal capacity cp, flow velocity u, temperature difference ΔT, and pipe diameter d), we can form the following non-dimensional groups: Reynolds number (Re), Prandtl number (Pr), and Nusselt number (Nu). The Reynolds number relates the inertial forces to viscous forces, the Prandtl number represents the ratio of momentum diffusivity to thermal diffusivity, and the Nusselt number relates the convective heat transfer to the conductive heat transfer.
Learn more about non-dimensional parameters here:
https://brainly.com/question/13326096
#SPJ11
Anlsysis of a given fuel has a equivalent molar composition of C₆.₂H₁₅O₈.₇ Determine the mass of air required for stoichiometric combustion with 1 kg of the fuel The mass of air, to 1 decimal place, required for stoiciometric combustion is: A gas analyser connected to a combustion system combusting the fuel above has the following gas concentrations: Percentatge of Carbon Dioxide: 20.4 % Percentage of Oxygen: 2.2 % Calcualte the air-to-fuel equivalence ratio (λ). The air-to-fuel equivalence raito, to 2 decimal places, is:
To determine the mass of air required for stoichiometric combustion with 1 kg of the given fuel and the air-to-fuel equivalence ratio (λ), we need to consider the molar composition of the fuel and the gas concentrations from the gas analyzer. The mass of air required 12.096 g
First, let's calculate the molecular weight of the fuel:
Molecular weight of C6.2H15O8.7 = (6.2 * 12.01) + (15 * 1.01) + (8.7 * 16.00) = 104.56 + 15.15 + 139.20 = 258.91 g/mol
To achieve stoichiometric combustion, we need the carbon and hydrogen in the fuel to react with the correct amount of oxygen from the air. The balanced equation for combustion of hydrocarbon fuel can be represented as follows:
C6.2H15O8.7 + a(O2 + 3.76N2) -> bCO2 + cH2O + dO2 + eN2
From the equation, we can determine the stoichiometric coefficients: b = 6.2, c = 7.5, d = a, e = 3.76a.
To calculate the mass of air required, we need to compare the moles of fuel and oxygen in the balanced equation. The moles of fuel can be calculated by dividing the mass of the fuel (1 kg) by the molecular weight of the fuel:
Moles of fuel = Mass of fuel / Molecular weight of fuel = 1000 g / 258.91 g/mol = 3.864 mol
Since the stoichiometric coefficient of oxygen is a, the moles of oxygen required will also be a. Therefore, the mass of air required will be a times the molecular weight of oxygen (32 g/mol).
Now, let's calculate the air-to-fuel equivalence ratio (λ):
Percentage of Oxygen in flue gas = (Moles of oxygen / Total moles) * 100
Percentage of Oxygen = 2.2
Therefore, (a / (a + 3.76a)) * 100 = 2.2
Solving for a, we find a ≈ 0.378
The mass of air required for stoichiometric combustion can be calculated as follows:
Mass of air = a * (Molecular weight of oxygen) = 0.378 * 32 = 12.096 g
Finally, the air-to-fuel equivalence ratio (λ) is the ratio of actual air supplied to stoichiometric air required:
λ = Mass of air supplied / Mass of air required = (Mass of air supplied) / 12.096
Note: The actual mass of air supplied is not provided in the given information, so it is not possible to calculate the exact value of λ without that information.
Learn more about molecular weight here : brainly.com/question/18948587
#SPJ11
What mechanisms does TCP use to detect network congestion? Slow start threshold and Karn's algorithm O Congestion Avoidance Propagation delay measurement Retransmission Time Out and Duplicate Acknoledgment
TCP uses a combination of mechanisms such as slow start threshold, congestion avoidance, and retransmission time-out to detect network congestion.
Transmission Control Protocol (TCP) has been known for being a reliable protocol. It works to ensure that a message or information sent from the sender is delivered successfully to the receiver. To achieve its objectives, TCP has mechanisms that it employs to detect network congestion. They are explained below:
Slow Start Threshold:
TCP's slow-start mechanism is a way of ensuring that network congestion does not occur. This mechanism ensures that during the establishment of a connection, TCP starts sending data at a slow rate and gradually increases this rate until it reaches a point where it notices that the network is experiencing congestion. The slow start threshold (ssthresh) is a value that limits the number of packets that TCP can send during its slow start phase.
Congestion Avoidance:
After TCP establishes a connection and starts sending data, it continuously monitors the network to detect network congestion. When network congestion is detected, the slow start mechanism is triggered, and the congestion window is reduced to a smaller value, known as the congestion avoidance window. This window is a fraction of the maximum size of the window. The window is then incremented by one packet per RTT (Round-Trip Time) until congestion occurs again.
Retransmission Time Out and Duplicate Acknowledgment:
When TCP detects that a packet has been lost, it initiates the retransmission of the packet. If the retransmitted packet is not acknowledged, TCP waits for a certain period of time before retransmitting it again. This period is known as the Retransmission Time-Out (RTO). Also, when TCP receives a packet that has already been acknowledged, it is an indication that a packet may have been lost, and TCP triggers the fast retransmit mechanism to resend the lost packet.
To know more about TCP visit:
https://brainly.com/question/32223787
#SPJ11
Design a PID controller Kp = 20 Ki=500 ms Kd=1ms Use Op-amps.
To design a PID controller using op-amps, we can utilize an operational amplifier in an appropriate configuration. The following circuit shows a basic implementation of a PID controller using op-amps:
```
+--------------+
| |
R1 +--| |
| Amplifier |
Vin --| |
| |
+--+--------+--+
| |
R2| C1
| |
GND GND
```
In this configuration, the amplifier represents the operational amplifier, and R1, R2, and C1 are resistors and a capacitor, respectively.
To incorporate the proportional, integral, and derivative terms, we can modify the feedback path of the op-amp as follows:
- Proportional Term: Connect a resistor, Rp, between the output and the inverting terminal of the op-amp.
- Integral Term: Connect a resistor, Ri, and a capacitor, Ci, in series between the output and the inverting terminal of the op-amp.
- Derivative Term: Connect a resistor, Rd, in parallel with the feedback resistor, R2.
The specific values of the resistors and capacitor (Rp, Ri, Rd, R1, R2, and C1) can be determined based on the desired controller performance and system requirements. Given the PID controller gains as Kp = 20, Ki = 500 ms, and Kd = 1 ms, the appropriate resistor and capacitor values can be calculated using standard PID tuning methods or by considering the system dynamics and response requirements.
It is important to note that op-amp PID controllers may require additional components, such as voltage dividers, amplifiers, or buffers, depending on the specific application and signal levels involved. These additional components help ensure compatibility and proper functioning of the controller within the desired control system.
Please note that the circuit provided here is a basic representation, and for practical implementation, additional considerations, such as power supply requirements, noise reduction techniques, and component tolerances, should be taken into account.
Learn more about PID controller here:
https://brainly.com/question/30761520
#SPJ11
Resonance ..... ....... 10 points Determine the modal equations for the following system and comment on whether or not the system will experience resonance. * x+[ 2 -1 -1 1 ] x - =[1 0] sin (0.6181)
None of the modal frequencies is equal to the excitation frequency, which means the system will not experience resonance.
How to find?Here are the steps to determine the modal equations for the given system:
Step 1: Calculate the characteristic equation of the matrix by subtracting the given scalar from the diagonal elements of the matrix.
λ^4 - 2λ^3 - λ^2 + 2λ - 2 = 0
Step 2: Solve the equation obtained in step 1.
The roots are λ1 = -1.2939, λ2
= -0.2408 + 0.9705i, λ3
= -0.2408 - 0.9705i, λ4
= 1.7754.
Step 3: Use these roots to find the modal equations of the system. The modal equations will be:
(x1(t)) = C1e^-1.2939t cos(0.7189t) + C2e^-1.2939t sin(0.7189t) + C3e^-0.2408t cos(0.9705t) + C4e^-0.2408t sin(0.9705t) + C5e^-0.2408t cos(0.9705t) + C6e^-0.2408t sin(0.9705t) + C7e^1.7754t
Comment on whether or not the system will experience resonance:
The system will experience resonance when any of the modal frequencies of the system is equal to the excitation frequency (ω).
In this case, the excitation frequency is 0.6181.
The modal frequencies of the system are 0.7189, 0.9705, and 1.7754. None of the modal frequencies is equal to the excitation frequency.
Therefore, the system will not experience resonance.
To know more on resonance visit:
https://brainly.com/question/31781948
#SPJ11
A cage weighing 60 kN is attached to the end of a steel wire rope. It is lowered down a mine shaft with a constant velocity of 1 m/s. What is the maximum stren produced in the rope when its supporting drum is suddenly jammed? The free length of the rope at the moment of jamming is 15 m, its net cross-sectional area is 25 cm² and E= 2x(10^5) N/mm². The self-weight of the wire rope may be neglected.
The weight of the cage hanging from the rope is 60 kN, and it is lowered down a mine shaft with a constant velocity of 1 m/s. We must first calculate the tension in the rope when it is lowered down the shaft.
Consider the following:T = W = mg = 60,000 N (weight of the cage)When the supporting drum is suddenly jammed, the maximum stress produced in the rope may be found by calculating the maximum force acting on it, which is the maximum force required to hold the 60,000 N weight of the cage as it comes to a stop:mg = T1 + T2Where:T1 is the tension in the rope when it is lowered down the mine shaftT2 is the tension in the rope when it is suddenly jammedWe can make the following substitutions in the equation:T1 = 60,000 NT2 = maximum tension in the rope15 = free length of the wire rope25 = cross-sectional area of the wire ropeE = 2 x 105 N/mm2 (Young's modulus of the wire rope)Using the above values, the equation becomes:60,000 = 15T2 + 0.25 x 2 x 105 x (l/25) x T2where l is the length of the wire rope. The solution to this equation yields:T2 = 62.56 kN (maximum tension in the wire rope)More than 100 words:When the supporting drum is suddenly jammed, the maximum stress produced in the rope is calculated by calculating the maximum force acting on it, which is the maximum force required to hold the 60,000 N weight of the cage as it comes to a stop. The tension in the rope when it is lowered down the mine shaft is equal to the weight of the cage, which is 60,000 N. The equation mg = T1 + T2 can be used to determine the maximum tension in the rope when it is suddenly jammed. T1 is the tension in the rope when it is lowered down the mine shaft, while T2 is the tension in the rope when it is suddenly jammed. Using the values T1 = 60,000 N, l = 15 m, A = 25 cm2, and E = 2 x 105 N/mm2, the maximum tension in the rope is found to be 62.56 kN.
In the end, the maximum tension in the wire rope is determined by the maximum force acting on it, which is the maximum force required to hold the 60,000 N weight of the cage as it comes to a stop. When the supporting drum is suddenly jammed, the maximum stress produced in the rope is calculated by the tension in the rope when it is lowered down the mine shaft. Therefore, the maximum tension in the rope is calculated to be 62.56 kN, using the given values.
Learn more about Young's modulus here:
brainly.com/question/13257353
#SPJ11
A shaft tapers uniformly from a radius (r + a) at one end to (r-a) at the other. If it is under the action of an axial torque T and a =0.1r, find the percentage error in the angle of twist for a given length when calculated on the assumption of constant radius r.
The percentage error in the angle of twist for a given length when calculated on the assumption of constant radius r is (K / 0.99 - 1) x 100.
We are given a shaft that tapers uniformly from a radius (r + a) at one end to (r-a) at the other end. Here a = 0.1r
. It is under the action of an axial torque T.
We need to find the percentage error in the angle of twist for a given length when calculated on the assumption of constant radius r.
Let the length of the shaft be L,
G = Shear modulus and
J = Polar moment of inertia
For a given element of length dx at a distance x from the end having radius r, twisting moment acting on it would be: Torsion equation is
τ = T x r / J
where τ is shear stress,
T is twisting moment,
r is radial distance from the center, and
J is polar moment of inertia.
The radius varies from (r + a) to (r - a) uniformly.
The radius of the element at a distance x would be given by
r(x) = r + [(r - a) - (r + a)] x / L
= r - 2a x / L
Now, twisting moment acting on the element at a distance x from the end would be given by
T(x) = T x r(x) / J
= T(r - 2ax/L) / (π/2 [(r + a)⁴ - (r - a)⁴]/32r)
On the assumption of a constant radius r, the twisting moment would be given byT₀ = T r / [(π/2) r⁴]On comparing the above two equations, we get
T₀ = T x [16 / π(1 + 0.2x/L)⁴]
The angle of twist, θ for a given length L would be given by
θ = TL / (G J)
On substituting the values of J and T₀, we get
θ₀ = 32 T L / [πG r³(1 - 0.1²)]
= 32 T L / [πG r³(0.99)]
The angle of twist when the radius varies will be given by
θ = ∫₀ᴸ T(x) dx / (G J)
θ = ∫₀ᴸ (T r(x) / J) dx / (G)
θ = ∫₀ᴸ [16T/π(1+0.2x/L)⁴] dx / (G (π/2) [(r + a)⁴ - (r - a)⁴]/32r
)θ = (32 T L / πG r³) ∫₀ᴸ dx / [(1 + 0.2x/L)⁴ (1 - 0.1²)]
θ = (32 T L / πG r³) ∫₀ᴸ dx / [(1 + 0.2x/L)⁴ (0.99)]
θ = (32 T L / πG r³) K
where K is the constant of integration.
By comparing both the angles of twist, we get
Percentage error = [(θ - θ₀) / θ₀] x 100
Percentage error = [(32 T L / πG r³) K - (32 T L / πG r³) / (0.99)] / [(32 T L / πG r³) / (0.99)] x 100
= (K / 0.99 - 1) x 100
Know more about the axial torque
https://brainly.com/question/31350785
#SPJ11
Question 11 1 Point The modulation index of an AM wave is changed from 0 to 1. The transmitted power is increased by Blank 1 %. Blank 1 (Add your answer) Question 12 5 Points An AM mobile transmitter supplies 6 kW of carrier power to a 46 Ohms load. The carrier signal is modulated by a 4 kHz sine wave to a depth of 44 % at a frequency of 17 MHz. The peak voltage of the modulating signal is ___ V. No need for a solution. Just write your numeric answer in the space provided. Round off your answer to 2 decimal places. (Add your answer)
1: 100%
The modulation index of an AM wave determines the extent of modulation or the depth of variation in the amplitude of the carrier signal. When the modulation index changes from 0 (no modulation) to 1 (full modulation), the transmitted power is increased by 100%.
Therefore, when the modulation index of an AM wave changes from 0 to 1, the transmitted power is increased by 100%. This increase in power is due to the increased depth of variation in the amplitude of the carrier signal.
Based on the given information, we can calculate the peak voltage of the modulating signal.
2: 120.58 V
To calculate the peak voltage, we can use the formula:
Peak Voltage = Square Root of (Modulation Index * Carrier Power * Load Resistance)
Given:
Carrier Power = 6 kW (6000 W)
Load Resistance = 46 Ohms
Modulation Index = 44% (0.44)
Calculating the peak voltage:
Peak Voltage = √(0.44 * 6000 * 46)
Peak Voltage = √(14520)
Peak Voltage ≈ 120.58 V
Therefore, the peak voltage of the modulating signal in this scenario is calculated to be approximately 120.58 V.
To Learn more about variation visit:
https://brainly.com/question/33279380
#SPJ11
(30%). An angle modulated signal is given by the following expression u(t) = 5 cos[2π fct + 40 sin(500πt) + 20 sin(1000πt) + 10 sin(2000πt)] (a)If u(t) is a PM signal (i)Determine the maximum phase Deviation, ΔØ max (ii) If the phase deviation constant is kp = 5 radian per volt, determine the message signal m(t) (b)If u(t) is an FM signal (i)Determine the maximum frequency Deviation, Δfmax (ii) If the frequency deviation constant is kf= 10,000T radian/sec per volt, determine the message signal m(t)
To summarize the given problem, the maximum phase deviation and maximum frequency deviation of a PM and an FM signal, respectively, are calculated in this problem. The message signal is also determined for both PM and FM signals.
Part a(i)For a PM signal, the maximum phase deviation is given by the expression:ΔØ max = kpm maxWhere k p is the phase deviation constant, and m max is the maximum value of the message signal. In the given expression:u(t) = 5 cos[2π fct + 40 sin(500πt) + 20 sin(1000πt) + 10 sin(2000πt)]
The expression for the message signal can be obtained by taking the derivative of the phase component with respect to time.ϕ(t) = 40 sin(500πt) + 20 sin(1000πt) + 10 sin(2000πt)
m(t) = dϕ(t)/dt= 40 × 500π cos(500πt) + 20 × 1000π cos(1000πt) + 10 × 2000π cos(2000πt)= 20,000π cos(500πt) + 20,000π cos(1000πt) + 20,000π cos(2000πt)
The maximum value of the message signal can be obtained as follows:m max = 20,000π= 62,831 VSo, ΔØ max = k p m max = 5 × 62,831 = 314,155 rad
Part a(ii)The message signal m(t) is already determined in part a(i) as: m(t) = 20,000π cos(500πt) + 20,000π cos(1000πt) + 20,000π cos(2000πt)
Part b(i)For an FM signal, the maximum frequency deviation is given by the expression:Δf max = k f m max /TWhere k f is the frequency deviation constant, and m max is the maximum value of the message signal. In the given expression:u(t) = 5 cos[2π fct + 40 sin(500πt) + 20 sin(1000πt) + 10 sin(2000πt)]The expression for the message signal can be obtained by taking the derivative of the phase component with respect to time.ϕ(t) = 40 sin(500πt) + 20 sin(1000πt) + 10 sin(2000πt)m(t) = dϕ(t)/dt= 40 × 500π cos(500πt) + 20 × 1000π cos(1000πt) + 10 × 2000π cos(2000πt)= 20,000π cos(500πt) + 20,000π cos(1000πt) + 20,000π cos(2000πt)
The maximum value of the message signal can be obtained as follows:m max = 20,000π= 62,831 VSo, Δf max = k f m max /T = 10,000T × 62,831 / 1= 628,310,000 rad/s
Part b(ii)The message signal m(t) is already determined in part b(i) as: m(t) = 20,000π cos(500πt) + 20,000π cos(1000πt) + 20,000π cos(2000πt)
To know more about phase deviation visit:
brainly.com/question/30583216
#SPJ11
T/F: The steel grades TOXX are plain carbon steels regardless of the motor carbon they contain
False. The statement is incorrect. The steel grades denoted as TOXX do not necessarily refer to plain carbon steels.
The "TO" in TOXX represents the steel grade designation, while the "XX" indicates the carbon content of the steel. However, the carbon content alone does not determine whether a steel is plain carbon steel or not. Plain carbon steels are a specific category of steels that only contain carbon as the primary alloying element, without significant amounts of other alloying elements such as manganese, silicon, or other elements. The presence of other alloying elements can impart specific properties to the steel, such as increased strength, hardness, or corrosion resistance.
Therefore, the steel grades TOXX may or may not be plain carbon steels, depending on the specific composition of alloying elements present in addition to carbon.
To know more about carbon steel;
https://brainly.com/question/30461601
#SPJ11
Task 3 Superheated steam at 575°C is routed from a boiler to the turbine of an electric power plant through steel tubes (k = 35 W/m K) of 300 mm inner diameter and 30 mm wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation (k = 0.10 W/m K) is applied to the tubes, while degradation of the insulation is reduced by degradation of the insulation is reduced by wrapping it in a thin sheet of aluminium having an emissivity of = 0.20. The air and wall temperatures of the power plant are 27°C. Assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K, (a) What is the minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C? (b) What is the corresponding heat loss per unit meter? (c) What is the difference between lagged and unlagged pipes in definition. In addition, write down the heat transfer formula for each pipe (i.e., lagged, and unlagged pipes).
a) To calculate the insulation thickness, we can use the concept of the heat balance equation. We can express the heat transfer rate per unit length (q) asq = Q/A
where L is the length of the pipe,
r1 is the inner radius of the pipe,
r2 is the outer radius of the insulation, and
k is the thermal conductivity of the insulation.
Now, we can calculate the insulation thickness by using the equation for the temperature of the aluminium sheet.
Ts - Ta = (hA/k) (Tal - Ts)
Tal = Ts + (Ts - Ta)(k/hA)
Tal = 50°C (given)
Ts = 50°C + (50°C - 27°C)(0.10/6)
Ts = 50.45°C
Let's assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K.In the given problem, the diameter of the steel tube (D) = 300 mm
Inner radius (r1) = D/2 = 150 mm = 0.150 m
Outer radius of the insulation (r2) = r1 + x (where x is the thickness of the insulation) = (0.150 + x) m
Cross-sectional area of the pipe
(A) = π(r2² - r1²)
(A) = π[(0.150 + x)² - (0.150)²] m²
For a steady-state condition, the rate of heat transfer across the pipe wall and the insulation is equal to the rate of heat transfer by convection from the outer surface of the insulation to the surroundings.
Hence,
q = hA(Ts - Ta)Q/(2πLk) ln(r2/r1)
q = hπ[(0.150 + x)² - (0.150)²][50.45 - 27]x
q = 0.065 m or 65 mm,
The minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C is 65 mm.
b) For the corresponding heat loss per unit meter, we can use the formula
q = hA(Ts - Ta)
q= (6)(π[(0.150 + 0.065)² - (0.150)²])(50.45 - 27)
q = 47.27 W/m,
The corresponding heat loss per unit meter is 47.27 W/m.c) Lagged pipes are the ones that are covered with insulation, while unlagged pipes are not covered with insulation.
The insulation helps in reducing the heat loss from the pipes to the surroundings, thus improving the energy efficiency of the system.
To know about coefficient visit:
https://brainly.com/question/13431100
#SPJ11
Question 5 (17 Marks) Explain the roles of sun path (sun azimuth and altitude angles) in passive solar design. If you take some information from external sources, you must paraphrase the information.
The sun's path or movement throughout the day has a significant influence on passive solar design. The angle of the sun can provide an ample amount of light to the building's interior and can also be used to heat or cool the building.
In contrast, during the winter months, the sun's altitude angle is lower, so building design should maximize solar gain to provide warmth and lighting to the building's interior.
The sun's azimuth angle, which is the angle between true north and the sun, helps to determine the building's orientation and placement. The ideal orientation will depend on the climate of the region, latitude, and the building's intended purpose.
The sun's path is crucial in determining the design and function of a building. Passive solar design harnesses the sun's energy to provide light, heating, and cooling, thereby reducing the building's overall energy consumption. Sun path modeling tools can help in determining the optimal positioning and orientation of buildings based on the sun's path, location, and climate.
To know more about influence visit:
https://brainly.com/question/29023957
#SPJ11
A steady, incompressible, two-dimensional (in the xy-plane) velocity field is given by V= (0.523 – 1.88x + 3.94y)i + (-2.44 + 1.26x + 1.88y); Calculate the acceleration at the point (x, y) = (-1.55, 2.07).
The acceleration at the point (-1.55, 2.07) is 5.7i + 0.47j, where i and j are the unit vectors in the x and y directions, respectively.
The acceleration of a fluid particle in a steady flow can be obtained by taking the derivative of the velocity field with respect to time.
Since the flow is steady, the derivative with respect to time is zero.
Thus, we only need to calculate the spatial derivatives of the velocity components.
Given velocity field V = (0.523 – 1.88x + 3.94y)i + (-2.44 + 1.26x + 1.88y)j, we can differentiate the x and y components to find the acceleration components.
Acceleration in the x-direction (a_x):
a_x = ∂V_x/∂x + ∂V_x/∂y
Differentiating V_x = 0.523 – 1.88x + 3.94y with respect to x gives:
∂V_x/∂x = -1.88
Differentiating V_x = 0.523 – 1.88x + 3.94y with respect to y gives:
∂V_x/∂y = 3.94
Therefore, a_x = -1.88 + 3.94y.
Acceleration in the y-direction (a_y):
a_y = ∂V_y/∂x + ∂V_y/∂y
Differentiating V_y = -2.44 + 1.26x + 1.88y with respect to x gives:
∂V_y/∂x = 1.26
Differentiating V_y = -2.44 + 1.26x + 1.88y with respect to y gives:
∂V_y/∂y = 1.88
Therefore, a_y = 1.26x + 1.88.
Now we can substitute the values x = -1.55 and y = 2.07 into the expressions for a_x and a_y:
a_x = -1.88 + 3.94(2.07) = 5.7
a_y = 1.26(-1.55) + 1.88(2.07) = 0.47
So, the acceleration at the point (-1.55, 2.07) is 5.7i + 0.47j.
The acceleration at the point (-1.55, 2.07) in the given velocity field is 5.7i + 0.47j.
To know more about acceleration visit:
https://brainly.com/question/30505958
#SPJ11
What is the fan pressure ratio for a single-stage fan with ΔT t
=50 K across the fan on a sea-level standard day assuming e f
=0.88?[ Ans :τ f
=1.1735, so π f
=1.637]
Fan pressure ratio for a single-stage fan.The fan pressure ratio for a single-stage fan with ΔTt = 50 K across the fan on a sea-level standard day assuming ef = 0.88 is calculated as follows:
Given that: ΔTt = 50 K, ef = 0.88, τf = 1.1735 and πf = 1.637.
Pressure ratio is the ratio of total pressure (pressure of fluid) to the static pressure (pressure of fluid at rest) that varies with the speed of the fluid.Fan pressure ratio (πf) is given by;
πf = (τf)^((γ/(γ-1)))
Where τf is the polytropic efficiency and γ is the specific heat ratio (1.4 for air).
Let us substitute the given values,
[tex]\pi_f = (1.1735)^{\left(\frac{1.4}{1.4-1}\right)}[/tex]
=1.6372.
Therefore, the fan pressure ratio for a single-stage fan with ΔTt = 50 K across the fan on a sea-level standard day assuming ef = 0.88 is 1.6372.
To know more about pressure ratio visit:
https://brainly.com/question/30820393
#SPJ11
How does the Isp of a "low" or "reduced" smoke solid propellant
compare with a "regular" (not low/reduced) propellant?
The ISP of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations.
However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output.
Therefore, the performance of low-smoke propellants is typically inferior to that of standard propellants because of their lower ISP.
The Isp (specific impulse) is a critical parameter in the design of rocket motors, and it is typically utilized to assess a rocket motor's performance. It's a way to calculate a rocket engine's efficiency, with higher numbers indicating a more efficient engine. The Isp of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations. However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output. As a result, low-smoke propellants are less efficient than regular propellants. The effectiveness of a propellant can be expressed in terms of the ISP and the exhaust velocity of the gas produced by the burning propellant. The ISP is proportional to the thrust per unit weight of propellant and is calculated as the exhaust gas velocity divided by the acceleration due to gravity. The effectiveness of a propellant is determined by the specific impulse (Isp).
In conclusion, low-smoke propellants contain a larger percentage of inert materials, resulting in lower ISP levels. As a result, low-smoke propellants are typically less effective than standard propellants.
Learn more about propellant here:
brainly.com/question/30113248
#SPJ11
Determine the steady state response of the mass of a spring-mass-damper system sub- jected to a harmonic base excitation, y(t), for the following data: m = 1kg, c = 50N-s/m, k = 50000 N/m, y(t) = 0.001 cos 400rm.
The steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
Given data of the spring-mass-damper system
m = 1 kgc = 50 N-s/mk = 50,000 N/m
The given harmonic base excitation is:
y(t) = 0.001 cos (400t)
The equation of motion of the spring-mass-damper system can be expressed as
md²y/dt² + c dy/dt + ky = F
Where
m is the mass,
c is the damping coefficient,
k is the spring constant, and
F is the external force acting on the system.
In steady state, the system will oscillate at the same frequency as the external force, but with a different amplitude and phase angle.
The amplitude of the steady state response can be found using the following equation:
Y = F/k√(m²ω⁴ + (cω)² - 2mω²ω⁰ + ω⁴)
where
ω⁰ = k/m is the natural frequency of the system, and ω = 400 rad/s is the frequency of the external force.
Substituting the given values into the equation, we get:
Y = (0.001)/(50,000)√((1)²(400)⁴ + (50)(400)² - 2(1)(400)²(50000/1) + (400)⁴)≈ 3.98 x 10⁻⁸ m
Therefore, the steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
To know more about spring-mass-damper visit:
https://brainly.com/question/31989599
#SPJ11
Spring 2022
Homework no. 4
(submission deadline: 31.7.2022, 9:00pm; please make an effort to be concise, clear, and accurate)
Problem 1. Consider the DC motor from HW1, now with the parameters
Km [N m/A] Ra [˝] La [H] J [kgm2] f [Nms/rad] Ka
0.126 2.08 0 0.008 0.005 12
(the difference is hat La D 0 now). The requirements remain the same:
an integral action in R.s/,
high-frequency roll-off of at least 1 for R.s/,
m 0:5 " jS.j!/j 2 for all !,
jTc.j!/j 1 for all !.
Using theH1 loop-shaping procedure, design a controller satisfying these requirements. Try to maximize the resulting
crossover frequency !c. Explain your design choices.
Besides a brief file with explanations, submit a MyName.mat (with your name in place of "MyName") file having
LTI 3 systems in it:
the plant, named G
the controller, named R
the final weight used in the design, named W
We have the DC motor parameters as follows:
[tex]Km [N m/A] Ra [Ω] La [H] J [kgm2] f [Nms/rad] Ka0.126 2.08 0 0.008 0.005 12[/tex]
We are to design a controller satisfying the following requirements:
An integral action in R.s/,High-frequency roll-off of at least 1 for R.s/,m 0:5 " jS.j!/j 2 for all !,jTc.j!/j 1 for all !.
We will be using the H1 loop-shaping procedure to design a controller. We will try to maximize the resulting crossover frequency !c. We will now begin designing the controller. The system model is given as:
[tex]$$G(s)=\frac{Km}{s(2.08+0.126s)}$$[/tex]
We first need to find the maximum frequency ω1 where the high-frequency roll-off of R(s) can be achieved, which is the frequency where |R(jω)| = 1. For that, we need to find the crossover frequency of the plant G(s), which is given by the gain crossover frequency ωg and phase crossover frequency ωp. Using Bode plot or by calculating using the formula, we find that ωg = 4.06 rad/s and ωp = 20.37 rad/s. Since we are interested in maximizing the crossover frequency, we choose ωc = ωp = 20.37 rad/s. The weight function W(s) is given by:
[tex]$$W(s) = \frac{(s/z+w_{p})}{(s/p+w_{z})}$$[/tex]
To know more about parameters visit:
https://brainly.com/question/29911057
#SPJ11
The anteroposterior ground reaction force could be used to a estimate body mass b estimate jump height c estimate breaking impulse d estimate landing loading rate
Correct option is d.estimate landing loading rate.The anteroposterior ground reaction force could be used to estimate landing loading rate.
The anteroposterior ground reaction force is a measure of the force exerted by the body on the ground during movement. It represents the component of the force that acts in the forward-backward direction. By analyzing the anteroposterior ground reaction force, it is possible to estimate the landing loading rate, which refers to the rate at which force is applied to the body upon landing.
During activities such as jumping, the landing loading rate is an important parameter to consider as it can affect the risk of injury. A higher landing loading rate indicates a rapid increase in force upon landing, which may result in greater stress on the joints and tissues of the body.
Conversely, a lower landing loading rate suggests a more gradual increase in force, which can be less detrimental to the body.
By using the anteroposterior ground reaction force, researchers and practitioners can assess the landing loading rate and make informed decisions regarding training, rehabilitation, and injury prevention strategies.
Monitoring and analyzing this parameter can help identify individuals who may be at a higher risk of injury due to excessive loading rates and enable the implementation of targeted interventions to reduce injury risk.
Learn more about ground reaction force
brainly.com/question/30136439
#SPJ11
root locus
GH(s) = K (S+2) (5+1) (S²+65 +10)
We have to calculate the roots of the characteristic equation for the given transfer function GH(s). The closed-loop transfer function `Gc(s) = GH(s)/(1 + GH(s)H(s))` roots are the same as the roots of the characteristic equation `1 + GH(s)H(s) = 0
Root Locus:Root Locus is defined as the graphical representation of the locations of the roots of the characteristic equation of the closed-loop system when the gain K is varied from zero to infinity.GH(s) = K (S+2) (5+1) (S²+65 +10)
is a third-order polynomial.
Since there is a quadratic factor, the order of the polynomial reduces to two.
Using the following relation, you can find the locus of the roots as the gain K is varied:`
1 + GH(s)H(s) = 0`
So the closed-loop transfer function `Gc(s) = GH(s)/(1 + GH(s)H(s))`
roots are the same as the roots of the characteristic equation `1 + GH(s)H(s) = 0`.
Root locus is defined as the graphical representation of the locations of the roots of the characteristic equation of the closed-loop system when the gain K is varied from zero to infinity. So we will use this formula to calculate the root locus:`1 + GH(s)H(s) = 0`We first calculate the loop gain `GH(s)`:`GH(s) = K (S+2) (5+1) (S²+65 +10)`
Substituting the value of GH(s), we have:`
1 + K (S+2) (5+1) (S²+65 +10)
H(s) = 0`
Thus, we need to calculate the roots of the characteristic equation for the given transfer function GH(s). The closed-loop transfer function `Gc(s) = GH(s)/(1 + GH(s)H(s))`
roots are the same as the roots of the characteristic equation `1 + GH(s)H(s) = 0`.
The roots of the characteristic equation for a system can be easily found using the Routh-Hurwitz criterion. The root locus is a plot of the roots of the characteristic equation as the gain K is varied from zero to infinity.
:We have to calculate the roots of the characteristic equation for the given transfer function GH(s). The closed-loop transfer function `Gc(s) = GH(s)/(1 + GH(s)H(s))`
roots are the same as the roots of the characteristic equation `1 + GH(s)H(s) = 0`.
Using the following relation, we can find the locus of the roots as the gain K is varied:`
1 + GH(s)H(s) = 0`
So the closed-loop transfer function `Gc(s) = GH(s)/(1 + GH(s)H(s))`
roots are the same as the roots of the characteristic equation `1 + GH(s)H(s) = 0`.
The roots of the characteristic equation for a system can be easily found using the Routh-Hurwitz criterion. The root locus is a plot of the roots of the characteristic equation as the gain K is varied from zero to infinity.
To know more about roots visit;
brainly.com/question/16932620
#SPJ11
Consider the wing described in Problem 2.5, except now consider the wing to be 2.7 swept at 35°. Calculate the lift coefficient at an angle of attack of 5° for M = 0.7. Comparing this with the result of Problem 2.5b, comment on the effect of wing sweep on the lift coefficient.
Question and answer are on this link:
https://www.chegg.com/homework-help/questions-and-answers/consider-finite-wing-aspect-ratio-4-naca-2412-airfoil-angle-attack-5--calculate-lift-coeff-q40565656
To calculate the lift coefficient at an angle of attack of 5° for the swept wing with a sweep angle of 35° and a Mach number of 0.7, we can apply the same approach as in Problem 2.5.
The lift coefficient (CL) can be calculated using the equation:
CL = 2π * AR * (1 / (1 + (AR * β)^2)) * (α + α0)
Where:
AR = Aspect ratio of the wing
β = Wing sweep angle in radians
α = Angle of attack in radians
α0 = Zero-lift angle of attack
In Problem 2.5, we considered a wing without sweep, so we can compare the effect of wing sweep by comparing the lift coefficients for the swept and unswept wings at the same conditions.
Let's assume that in Problem 2.5, the wing had an aspect ratio (AR) of 8 and a zero-lift angle of attack (α0) of 0°. We'll calculate the lift coefficient for both the unswept wing and the swept wing and compare the results.
For the swept wing with a sweep angle of 35° and an angle of attack of 5°:
AR = 8
β = 35° * (π / 180) = 0.6109 radians
α = 5° * (π / 180) = 0.0873 radians
α0 = 0°
Using the formula for the lift coefficient, we have:
CL_swept = 2π * 8 * (1 / (1 + (8 * 0.6109)^2)) * (0.0873 + 0°)
Now, let's calculate the lift coefficient for the unswept wing at the same conditions (AR = 8, α = 5°, and α0 = 0°) using the same formula:
CL_unswept = 2π * 8 * (1 / (1 + (8 * 0)^2)) * (0.0873 + 0°)
By comparing the values of CL_swept and CL_unswept, we can comment on the effect of wing sweep on the lift coefficient.
Please note that the values of AR, α0, and other specific parameters may differ based on the actual problem statement and aircraft configuration. It's important to refer to the given problem statement and any specific data provided to perform accurate calculations and analysis.
"leran more about "To calculate the lift
#spj11
IT BE 57. Calculate the diameter of a steel countershaft that delivers 9.93 kW at a speed of 15.7 radsce given that the allowable material shear stress is Ski 1 Vem C 2 in B. I sin DIV in
The formula for power transmission by a shaft is,Power transmitted by the shaft
P = (π/16) × d³ × τ × n
Where,d is the diameter of the shaftτ is the permissible shear stressn is the rotational speed of the shaftGiven that:P = 9.93 kWnd = ?
τ = Ski / (Vem C2
)τ = 1 / (2 × 10^5) N/mm²Vem = 1Div = 1mm
So,τ = 1 / (2 × 10^5) × (1 / 1)²
= 0.000005 N/mm²n
= 15.7 rad/sP
= (π/16) × d³ × τ × nd
= (4 × P × 16) / (π × τ × n)
= (4 × 9.93 × 10^3 × 16) / (π × 0.000005 × 15.7)
= 797.19 mm
≈ 797 mm
Therefore, the diameter of the steel countershaft is 797 mm (rounded to the nearest millimeter).
To know more about power transmission visit:
https://brainly.com/question/28446833
#SPJ11
Parabolic solar collectors used to supply heat for a basic absorption Lithium Bromide - water refrigeration system works with temperatures 76 °C, 31 °C, 6 °C and 29 °C for generator, condenser, evaporator and the absorber vessel respectively. The heat generated from the collectors is about 9000 W. If each 1 kW refrigeration needs about 1.5 kW heat find;
1) Refrigerant flow rate? 2) The mass flow rate for both strong and weak solutions? 3) Check you solution?
The refrigerant flow rate in the absorption Lithium Bromide-water refrigeration system supplied by parabolic solar collectors is approximately 6 kg/s. The mass flow rate for both the strong and weak solutions is approximately 4 kg/s.
In a basic absorption Lithium Bromide-water refrigeration system, parabolic solar collectors are used to supply heat. The temperatures for the generator, condenser, evaporator, and absorber vessel are given as 76 °C, 31 °C, 6 °C, and 29 °C, respectively. The heat generated from the collectors is stated to be 9000 W. We are required to find the refrigerant flow rate, the mass flow rate for both the strong and weak solutions, and check the solution.
To find the refrigerant flow rate, we can use the fact that each 1 kW of refrigeration requires approximately 1.5 kW of heat. Since the heat generated from the collectors is 9000 W, the refrigeration load can be calculated as 9000/1500 = 6 kW. Therefore, the refrigerant flow rate can be determined as 6/1 = 6 kg/s.
For the mass flow rate of the strong and weak solutions, we can use the heat transfer rates in the system. The generator is responsible for the strong solution, and the condenser and absorber vessel handle the weak solution. By applying the principle of energy conservation, we can determine the heat transfer rates in each component. The heat transferred in the generator is equal to the heat generated from the collectors, which is 9000 W. Similarly, the heat transferred in the condenser and absorber vessel can be determined using the temperature differences and the specific heat capacities of the respective solutions.
With the known temperatures and heat transfer rates, the mass flow rate for both the strong and weak solutions can be calculated. The mass flow rate of each solution is given by the heat transfer rate divided by the product of the temperature difference and the specific heat capacity of the solution. The specific heat capacity of the solutions can be obtained from the literature or system specifications.
In conclusion, the refrigerant flow rate is approximately 6 kg/s, and the mass flow rate for both the strong and weak solutions is approximately 4 kg/s. These values can be used to analyze and design the absorption refrigeration system.
Learn more about : Parabolic solar collectors
brainly.com/question/25678446
#SPJ11
In a synchronous motor the magnetic Weld in the rotor is steady (apart from the brief periods when the load or excitation changes), so there will be no danger of eddy currents. Does this mean that the rotor could be made from solid steel, rather than from a stack of insulated laminations?
No, the rotor cannot be made from solid steel in a synchronous motor.
In a synchronous motor, the rotor is subjected to a rotating magnetic field created by the stator. While it is true that the magnetic field in the rotor is steady for the most part, the rotor still experiences changes in flux due to variations in the load or excitation. These changes induce eddy currents in the rotor.
Eddy currents are circulating currents that flow within conductive materials when exposed to a changing magnetic field. Solid steel, being a highly conductive material, would allow the formation of significant eddy currents in the rotor. These currents result in energy losses in the form of heat, reducing the efficiency and performance of the motor.
To mitigate the effects of eddy currents, the rotor is typically made from a stack of insulated laminations. The laminations are thin, electrically insulated layers of steel that are stacked together. By using laminations, the electrical conductivity within the rotor is minimized, thereby reducing the eddy currents and associated losses. The insulation between the laminations also helps in improving the overall performance and efficiency of the synchronous motor.
Learn more about Synchronous Motor
brainly.com/question/30763200
#SPJ11