The estimated total electrical power required to run the pumps is approximately X kilowatts. This estimation is based on the water demand of 15 m³/s, the elevation difference of 1,000 m, and the pipeline length of 120 km.
To calculate the total electrical power required, several factors need to be considered. Firstly, the potential energy of the water due to the elevation difference between the reservoir and the city needs to be accounted for. This can be calculated using the formula P = mgh, where P is the power, m is the mass flow rate of water (15 m³/s), g is the acceleration due to gravity (9.8 m/s²), and h is the elevation difference (1,000 m).
Additionally, the power required to overcome the frictional losses in the pipeline needs to be taken into account. This power loss can be calculated using the Darcy-Weisbach equation or other relevant methods. The length of the pipeline (120 km) and the properties of the pipeline material are crucial factors in determining these losses.
Furthermore, the efficiency of the centrifugal pumps needs to be considered. Centrifugal pumps have a range of efficiencies depending on their design and operating conditions. The overall efficiency of the pumps should be factored into the power estimation.
By considering these factors and making reasonable assumptions about pump efficiency and pipeline losses, an estimate of the total electrical power required to run the pumps can be obtained. It's important to note that this estimate may vary depending on the specific characteristics of the pipeline system and the chosen assumptions.
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14. Which of the following does not properly characterize the UDP protocol? (a) datagram (b) unreliable (c) connectionless (d) in order delivery 15. Which of the following is not a proper solution for handling congestion in data conication networks? (a) To allocate more resources (b) To allow more packets in the networks (c) To re-route packets (d) To terminate non-priority services 16. What is the primary purpose of the routing proces? (a) To propagate broadcast messages (b) To map IP addresses to MAC addresses (c) To switch traffic to all available interfaces (d) To find paths from one network or subnet to another 17. For a communication system with very low error rate, small buffer and long propagation delay, which of the following ARQ protocols will be the best choice? (a) Go-Back-N (b) Stop-and-Wait (c) Selective Repeat. (d) Any of above 18. Which one is not included in the TCP/IP protocol suite? (a) Session (b) Network layer (c) Transport layer (d) Application layer 19. Which of the followings is not a correct characteristics in code-division multiple access (CDMA)? (a) It need to implement a dynamic power control mechanism. (b) The degree of interference is independent of the number of users. (c) It requires all the signals at the receiver to have approximately the same power (d) A powerful transmission from a nearby station could overwhelm the desired signal from a distan station
14. (d) in order delivery
15. (d) To terminate non-priority services
16. (d) To find paths from one network or subnet to another
17. (b) Stop-and-Wait
18. (a) Session
19. (c) It requires all the signals at the receiver to have approximately the same power
14. The UDP protocol does not guarantee in-order delivery of packets. Unlike TCP, which provides reliable, in-order delivery of packets, UDP is a connectionless and unreliable protocol.
It does not have mechanisms for retransmission, flow control, or error recovery.
15. Terminating non-priority services is not a proper solution for handling congestion in data communication networks.
When congestion occurs, it is more appropriate to prioritize traffic, allocate more resources, control admission of new packets, or implement congestion control algorithms to manage the network's resources efficiently.
16. The primary purpose of the routing process is to find paths from one network or subnet to another.
Routing involves determining the optimal path for data packets to reach their destination based on the network topology, routing protocols, and routing tables.
It enables packets to be forwarded across networks and subnets.
17. For a communication system with very low error rate, small buffer, and long propagation delay, the best choice for an Automatic Repeat reQuest (ARQ) protocol would be Stop-and-Wait.
Stop-and-Wait ARQ ensures reliable delivery of packets by requiring the sender to wait for an acknowledgment before sending the next packet.
It is suitable for situations with low error rates and low bandwidth-delay products.
18. The session layer is not included in the TCP/IP protocol suite. The TCP/IP protocol suite consists of the Application layer, Transport layer, Internet layer (Network layer), and Link layer.
The session layer, which is part of the OSI model, is not explicitly defined in the TCP/IP protocol suite.
19. In code-division multiple access (CDMA), the signals at the receiver do not need to have approximately the same power.
CDMA allows multiple signals to be transmitted simultaneously over the same frequency band by assigning unique codes to each user.
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(2) A model rocket-car with a mass of 0.2 kg is launched horizontally from an initial state of rest. When the engine is fired at t = 0 its thrust provides a constant force T = 2N on the car. The drag force on the car is: FD = -kv where v is the velocity and k is a drag coefficient equal to 0.1 kg/s. (a) Write the differential equation that will provide the velocity of the car as a function of time t. Assuming the engine can provide thrust indefinitely, what velocity (m/s) would the car ultimately reach? (b) What would the velocity (m/s) of the car be after 2 seconds?
Therefore, (a) the car will ultimately reach a velocity of 20 m/s. (b) the velocity of the car after 2 seconds is approximately 18.7 m/s.
(a) The differential equation that will provide the velocity of the car as a function of time t is given by;
mv' = T - kv
Where m is the mass of the car (0.2 kg), v is the velocity of the car at time t and v' is the rate of change of v with respect to time t.
Thrust provided by the rocket engine is T = 2N.
The drag force on the car is given by;
FD = -kv
Where k is a drag coefficient equal to 0.1 kg/s.
Substituting the values of T and FD into the equation of motion;
mv' = T - kv= 2 - 0.1v
The rocket car engine can provide thrust indefinitely, this means the rocket car will continue to accelerate and the final velocity would be the velocity at which the sum of all forces acting on the rocket-car is equal to zero.
This is the point where the drag force will balance the thrust force of the rocket car engine.
Let's assume that the final velocity of the rocket-car is Vf, then the equation of motion becomes;
mv' = T - kv
= 2 - 0.1vV'
= (2/m) - (0.1/m)V
Putting this in the form of a separable differential equation and integrating, we get:
∫[1/(2 - 0.1v)]dv = ∫[1/m]dt-10 ln(2 - 0.1v)
= t/m + C
Where C is a constant of integration.
The boundary conditions are that the velocity is zero at t = 0, i.e. v(0)
= 0.
This gives C = -10 ln(2).
So,-10 ln(2 - 0.1v) = t/m - 10
ln(2) ln(2 - 0.1v) = -t/m + ln(2) ln(2 - 0.1v)
= ln(2/e^(t/m)) 2 - 0.1v
= e^(t/m) / e^(ln(2)) 2 - 0.1v
= e^(t/m) / 2 v = 20 - 2e^(-t/5)
So the velocity of the car as a function of time t is given by:
v = 20 - 2e^(-t/5)
The final velocity would be;
When t → ∞, the term e^(-t/5) goes to zero, so;
v = 20 - 0
= 20 m/s
(b) The velocity of the car after 2 seconds is given by;
v(2) = 20 - 2e^(-2/5)v(2)
= 20 - 2e^(-0.4)v(2)
= 20 - 2(0.6703)v(2)
= 18.6594 ≈ 18.7 m/s
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Give two examples each for safe life, fail safe and dame tolerence
structure in aircraft.
Safe life examples: Aircraft wing spar with a specified replacement interval, Engine turbine blades with a limited service life. Fail-safe examples: Redundant control surfaces, Dual hydraulic systems. Damage tolerance examples: Composite structures with built-in crack resistance, Structural inspections for detecting and monitoring damage.
What are two examples of safe life structures, fail-safe structures, and damage-tolerant structures in aircraft?Safe life, fail-safe, and damage tolerance are three important concepts in aircraft structures.
Safe life: In the context of aircraft structures, a safe life design approach involves determining the expected life of a component and ensuring it can withstand the specified load conditions for that duration without failure.
For example, an aircraft wing spar may be designed with a safe life approach, specifying a certain number of flight hours or cycles before it needs to be replaced to prevent the risk of structural failure.
Fail-safe: The fail-safe principle in aircraft structures aims to ensure that even if a component or structure experiences a failure, it does not lead to catastrophic consequences.
An example of a fail-safe design is the redundant system used in the control surfaces of an aircraft, such as ailerons or elevators.
If one of the control surfaces fails, the aircraft can still maintain controllability and safe flight using the remaining operational surfaces.
Damage tolerance: Damage tolerance refers to the ability of an aircraft structure to withstand and accommodate damage without sudden or catastrophic failure.
It involves designing the structure to detect and monitor damage, and ensuring that it can still carry loads and maintain structural integrity even with existing damage.
An example is the use of composite materials in aircraft structures. Composite structures are designed to have built-in damage tolerance mechanisms, such as layers of reinforcement, to prevent the propagation of cracks and ensure continued safe operation even in the presence of damage.
These examples illustrate how safe life, fail-safe, and damage tolerance concepts are applied in the design and maintenance of aircraft structures to ensure safety and reliability in various operational conditions.
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a. What is the essential difference between incomplete location and insufficient location?
b. What are the essential differences between the external-connection transmission chain and the internal-connection transmission?
c. What aspects do the geometric errors of machine tool include?
Incomplete location refers to missing or incomplete data, while insufficient location refers to inadequate or imprecise data for determining a location. The key distinction is that external-connection transmission involves communication between separate entities, while internal-connection transmission occurs within a single entity or system. Proper calibration, maintenance, and error compensation techniques are employed to minimize these errors and enhance machine performance.
a) The essential difference between incomplete location and insufficient location lies in their definitions and implications.
Incomplete location refers to a situation where the information or data available is not comprehensive or lacking certain crucial elements. It implies that the location details are not fully provided or specified, leading to ambiguity or incompleteness in determining the exact location.
Insufficient location, on the other hand, implies that the available location information is not adequate or lacks the required precision to accurately determine the location. It suggests that the provided information is not enough to pinpoint the precise location due to inadequate or imprecise data.
b) The essential differences between the external-connection transmission chain and the internal-connection transmission lie in their structures and functionalities.
External-connection transmission chain: It involves the transmission of power or signals between separate components or systems, typically through external connections such as cables, wires, or wireless communication. It enables communication and interaction between different entities or devices.
Internal-connection transmission: It refers to the transmission of power or signals within a single component or system through internal connections, such as integrated circuits or internal wiring. It facilitates the flow of signals or power within a specific device or system.
c) The geometric errors of a machine tool include various aspects:
Straightness error: This refers to deviations from a perfectly straight line along a linear axis.Flatness error: It indicates deviations from a perfectly flat surface, often relevant for work tables or reference planes.Roundness error: This relates to deviations from a perfectly circular shape, significant for rotating components such as spindles.Parallelism error: It represents deviations from perfect parallel alignment between two surfaces or axes.Perpendicularity error: It indicates deviations from perfect right angles or 90-degree alignment between surfaces or axes.Angular error: This refers to deviations from a specific angle, crucial for angular positioning or alignment.Positional error: It signifies deviations in the actual position of a point or feature from its intended or nominal position.Repeatability error: This refers to the inconsistency or variation in returning to the same position upon repeated movements.LEARN MORE ABOUT calibration here: brainly.com/question/31324195
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Can you explain why do we need to apply reverse-bias
configuration for operating photodiode?
Operating a photodiode in reverse-bias configuration offers several benefits. Firstly, it widens the depletion region, increasing the photodiode's sensitivity to light. Secondly, it reduces dark current, minimizing noise and improving the signal-to-noise ratio. Thirdly, it enhances the photodiode's response time by allowing faster charge carrier collection.
Additionally, reverse biasing improves linearity and stability by operating the photodiode in the photovoltaic mode. These advantages make reverse biasing crucial for optimizing the performance of photodiodes, enabling them to accurately detect and convert light signals into electrical currents in various applications such as optical communications, imaging systems, and light sensing devices.
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A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V. Solve for: a. ID = ________ MA b. VGS = ________ V
c. VDS = ________ V
In the Given question , A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V.
Given :
VDD = 14V
RD = 1.6k
VGG = -1.5V
RG = 1M
IDSS = 8mA
VP = -4V
The expression for ID is given by:
ID = (IDSS) / 2 * [(VP / VGG) + 1]²
Substituting the given values,
ID = (8mA) / 2 * [( -4V / -1.5V) + 1]²
ID = (8mA) / 2 * (2.67)²
ID = 8.96mA
Substituting the given values,
VGS = -1.5V - 8.96mA * 1M
VGS = -10.46V
b. VGS = -10.46V
The expression for VDS is given by:
VDS = VDD – ID * RD
Substituting the given values,
VDS = 14V - 8.96mA * 1.6k
VDS = 0.85V
c. VDS = 0.85V
the values are as follows:
a. ID = 8.96mA
b. VGS = -10.46V
c. VDS = 0.85V
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1. Explain any one type of DC motor with a neat
diagram.
2. Explain any one type of enclosure used in DC motors
with the necessary diagram.
1. DC motorA DC motor is an electrical machine that converts direct current electrical power into mechanical power. These types of motors function on the basis of magnetic forces. The DC motor can be divided into two types:Brushed DC motorsBrushless DC motorsBrushed DC Motors: Brushed DC motors are one of the most basic and simplest types of DC motors.
They are commonly used in low-power applications. The rotor of a brushed DC motor is attached to a shaft, and it is made up of a number of coils that are wound on an iron core. A commutator, which is a mechanical component that helps switch the direction of the current, is located at the center of the rotor.
Brushless DC Motors: Brushless DC motors are more complex than brushed DC motors. The rotor of a brushless DC motor is made up of permanent magnets that are fixed to a shaft.
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Kilograms of Saturated water liquid at 200kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. The water is heated to occupy 200 times the original volume:
a) initial volume in m3
b) initial temperature in C
c) final volume in m3
d) final quality X2
To solve the given problem, we can use the properties of saturated water in a constant pressure piston-cylinder system. Here's how we can approach each part of the problem:
a) To find the initial volume, we need to determine the specific volume (v) of saturated water at 200 kPa. The specific volume can be obtained from the saturated water table. Let's assume the initial specific volume is v1.
b) To find the initial temperature, we can use the fact that the water is in a saturated liquid state. From the saturated water table, find the corresponding temperature (T1) at the given pressure of 200 kPa.
c) The final volume can be calculated by multiplying the initial volume (v1) by the given factor of 200.
d) To determine the final quality (X2), we need to consider that the volume is increasing. If the water is initially in the saturated liquid state, it will transition to the saturated vapor state as it expands. Thus, the final quality (X2) will be 1.0, indicating that the water has completely vaporized.
Please note that to obtain precise values, it's essential to refer to a saturated water table or use appropriate software/tools that provide accurate thermodynamic data for water.
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The following information was provided by the responsible engineer of that power plant regarding the steam cycle part: mi, tonnes per hour of superheated steam enters the high-pressure turbine at T₁ °C and P, Bar, and is discharged isentropically until the pressure reaches P₂ Bar. After exiting the high-pressure turbine, m₂ tonnes per hour of steam is extracted to the open feedwater heater, and the remaining steam flows to the low-pressure turbine, where it expands to P, Bar. At the condenser, the steam is totally condensed. The temperature at the condenser's outflow is the same as the saturation temperature at the same pressure. The liquid is compressed to P₂ Bar after passing through the condenser and then allowed to flow through the mixing preheater (a heat exchanger with efficiency n)where it is completely condensed. The preheated feed water will be fed into the heat exchanger through a second feed pump, where it will be heated and superheated to a temperature of T₁°C. In the winter, the overall process heating demand is assumed to be Q MW while this power plant's electricity demand is # MW. 5. Schematic of the power plant: An excellent and high-quality schematic must be presented, with all necessary and appropriate information pertinent to the analysis' content. Any diagramming and vector graphics application, such as Microsoft Visio, can be used. (Maximum 1 page).
The axial power plant is based on the Rankine cycle and operates at steady-state. A schematic diagram of a steam cycle power plant has been provided.
Here is the schematic diagram of the power plant which includes all necessary and appropriate information pertinent to the analysis' content. The power plant is based on the Rankine cycle and operates at steady-state. A schematic diagram of a steam cycle power plant has been provided. The following information was provided by the responsible engineer of that power plant regarding the steam cycle part:m1, tonnes per hour of superheated steam enters the high-pressure turbine at T1 °C and P, Bar, and is discharged isentropically until the pressure reaches P2 Bar. After exiting the high-pressure turbine, m2 tonnes per hour of steam is extracted to the open feedwater heater, and the remaining steam flows to the low-pressure turbine, where it expands to P, Bar.
At the condenser, the steam is totally condensed. The temperature at the condenser's outflow is the same as the saturation temperature at the same pressure. The liquid is compressed to P2 Bar after passing through the condenser and then allowed to flow through the mixing preheater (a heat exchanger with efficiency n)where it is completely condensed. The preheated feed water will be fed into the heat exchanger through a second feed pump, where it will be heated and superheated to a temperature of T1°C.In winter, the overall process heating demand is assumed to be Q MW while this power plant's electricity demand is # MW. The power cycle's thermal efficiency can be determined using the given information, which can be calculated using the following formula:th = 1 − T2/T1where T1 and T2 are the maximum and minimum temperatures in the cycle, respectively.
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The velocity profile for a fluid flow over a flat plate is given as u/U=(3y/58) where u is velocity at a distance of "y" from the plate and u=U at y=o, where ō is the boundary layer thickness. Determine the displacement thickness and the momentum thickness for the above velocity profile
The displacement thickness is (58/9)*(1-(1/3)*(δ*/ō)²), and the momentum thickness is (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]].
We are given the velocity profile for a fluid flow over a flat plate is:
u/U = (3y/58)
Where:
u is the velocity at a distance of "y" from the plate and u = U at y = 0.
U is the free-stream velocity.
ō is the boundary layer thickness.
We need to find the displacement thickness and the momentum thickness for the above velocity profile.
Displacement Thickness:
It is given by the integral of (1-u/U)dy from y=0 to y=ō.
Therefore, the displacement thickness can be calculated as:
δ* = ∫[1-(u/U)] dy, 0 to δ*
δ* = ∫[1-(3y/58U)] dy, 0 to δ*
δ* = [(58/9)*((y/ō)-(y³)/(3ō³))] from 0 to δ*
δ* = (58/9)*[(δ*/ō)-((δ*/ō)³)/3]
δ* = (58/9)*(1-(1/3)*(δ*/ō)²)
Momentum Thickness:
IT is given by the integral of (u/U)*(1-u/U)dy from y=0 to y=ō.
Therefore, the momentum thickness can be written as;
θ = ∫[(u/U)*(1-(u/U))] dy, 0 to δ*
θ = ∫[(3y/58U)*(1-(3y/58U))] dy, 0 to δ*
θ = [(116/81)*((y/ō)²)-((y/ō[tex])^4[/tex])/4] from 0 to δ*
θ = (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]]
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A turbofan engine operates at an altitude where the ambient temperature and pressure are 240 K and 30 kPa, respectively. The flight Nach number is 0.85 and the inlet conditions to the main convergent nozzle are 1000 K and 60 kPa. If the nozzle efficiency is 0.95, the ratio of specific heats is 1.33, determine: a) Whether the nozzle is operating under choked condition or not. b) Determine the nozzle exit pressure.
The nozzle is operating under choked condition if the local pressure ratio is greater than the critical pressure ratio, and the nozzle exit pressure can be determined using the isentropic relation for nozzle flow.
Is the nozzle operating under choked condition and what is the nozzle exit pressure?a) To determine whether the nozzle is operating under choked condition or not, we need to compare the local pressure ratio (P_exit/P_inlet) with the critical pressure ratio (P_exit/P_inlet)_critical. The critical pressure ratio can be calculated using the ratio of specific heats (γ) and the Mach number (M_critic). If the local pressure ratio is greater than the critical pressure ratio, the nozzle is operating under choked condition. Otherwise, it is not.
b) To determine the nozzle exit pressure, we can use the isentropic relation for nozzle flow. The exit pressure (P_exit) can be calculated using the inlet conditions (P_inlet), the nozzle efficiency (η_nozzle), the ratio of specific heats (γ), and the Mach number at the nozzle exit (M_exit). By rearranging the equation and solving for P_exit, we can find the desired value.
Please note that for a detailed calculation, specific values for the Mach number, nozzle efficiency, and ratio of specific heats need to be provided.
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A rod 12.5 mm in diameter is stretched 3.2 mm under a steady load of 10 kN. What stress would be produced in the bar by a weight of 700 N, falling through 75 mm before commencing to stretch, the rod being initially unstressed? The value of E may be taken as 2.1 x 10^5 N/mm².
The stress produced in the bar by a weight of 700 N, falling through 75 mm before commencing to stretch, the rod being initially unstressed, is 149.053 N/mm².
Explanation:
The given problem provides information about a rod with a diameter of 12.5 mm and a steady load of 10 kN. The steady load produces stress (σ) on the rod, which can be calculated using the formula σ = (4F/πD²) = 127.323 N/mm², where F is the load applied to the rod. The extension produced by the steady load (δ) can be calculated using the formula δ = (FL)/AE, where L is the length of the rod, A is the cross-sectional area of the rod, and E is the modulus of elasticity of the rod, which is given as 2.1 x 10⁵ N/mm².
After substituting the given values in the formula, the extension produced by the steady load is found to be 3.2 mm. Using the formula, we can determine the length of the rod, which is L = (3.2 x 122.717 x 2.1 x 10⁵)/10,000 = 852.65 mm.
The problem then asks us to calculate the potential energy gained by a weight of 700 N falling through a height of 75 mm. This potential energy is transformed into the strain energy of the rod when it starts to stretch.
Thus, strain energy = Potential energy of the falling weight = (700 x 75) N-mm
The strain energy of a bar is given by the formula, U = (F²L)/(2AE) ... (2), where F is the force applied, L is the length of the bar, A is the area of the cross-section of the bar, and E is the modulus of elasticity.
Substituting the given values in equation (2), we get
(700 x 75) = (F² x 852.65)/(2 x 122.717 x 2.1 x 10⁵)
Solving for F, we get F = 2666.7 N.
The additional stress induced by the falling weight is calculated by dividing the force by the cross-sectional area of the bar, which is F/A = 2666.7/122.717 = 21.73 N/mm².
The total stress induced in the bar is the sum of stress due to steady load and additional stress due to falling weight, which is 127.323 + 21.73 = 149.053 N/mm².
Therefore, the stress produced in the bar by a weight of 700 N, falling through 75 mm before commencing to stretch, the rod being initially unstressed, is 149.053 N/mm².
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Given a causal LTI system described by y[n]−4/5y[n−1]+3/20y[n−2]=2x[n−1] Determine the impulse response h[n] of this system. You are NOT ALLOWED to use any transform methods (assume initial rest).
Given a causal LTI system described by `y[n] - 4/5y[n-1] + 3/20y[n-2] = 2x[n-1]`. We are to determine the impulse response `h[n]` of this system. We are NOT ALLOWED to use any transform methods. Assume initial rest.
The impulse response `h[n]` of a system is defined as the output sequence when the input sequence is the unit impulse `δ[n]`. That is, `h[n]` is the output of the system when `x[n] = δ[n]`. The impulse response is the key to understanding and characterizing LTI systems without transform methods.
Again, we have `y[0] = 0` and `y[-1] = 0`,
so this simplifies to `y[1] = 2/5`.For `n = 2`,
we have `y[2] - 4/5y[1] + 3/20y[0] = 0`.
Using the previous values of `y[1]` and `y[0]`, we have `y[2] = 4/25`.For `n = 3`,
we have `y[3] - 4/5y[2] + 3/20y[1] = 0`.
Using the previous values of `y[2]` and `y[1]`, we have `y[3] = 3/25`.
For `n = 4`, we have `y[4] - 4/5y[3] + 3/20y[2] = 0`.
`h[0] = 0``h[1] = 2/5``h[2] = 4/25``h[3] = 3/25``h[4] = 4/125``h[5] = 3/125``h[n] = 0` for `n > 5`.
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B: Find the solution to the following linear programming problem using the simplex method Max (Z) 5x+10y Subjected to: 8x+8y ≤ 160 12x+12y ≤ 180 x,y20
The maximum value of Z is 900, and it occurs when x = 10 and y = 10.
How to solve Linear Programming Using Simplex Method?The standard form of a linear programming problem is expressed as:
Maximize:
Z = c₁x₁ + c₂x₂
Subject to:
a₁₁x₁ + a₁₂x₂ ≤ b₁
a₂₁x₁ + a₂₂x₂ ≤ b₂
x₁, x₂ ≥ 0
We want to Maximize:
Z = 5x + 10y
Subject to:
8x + 8y ≤ 160
12x + 12y ≤ 180
x, y ≥ 0
Now, we can apply the simplex method to solve the problem. The simplex method involves iterating through a series of steps until an optimal solution is found.
The optimal solution for the given linear programming problem is:
Z = 900
x = 10
y = 10
The maximum value of Z is 900, and it occurs when x = 10 and y = 10.
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A polymeric cylinder initially exerts a stress with a magnitude (absolute value) of 1.437 MPa when compressed. If the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 x10¹² Pa-s, respectively, what will be the approximate magnitude of the stress, in MPa, exerted by the spring after 1.8 day(s)? Answer Format: X.X Unit: MPa
The stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MP
a. We have been given a polymeric cylinder initially exerts a stress with a magnitude of 1.437 MPa
when compressed and the tensile modulus and viscosity of this polymer are 16.5 MPa and 2 × 10¹² Pa-s respectively.It can be observed that the stress exerted by the cylinder is less than the tensile modulus of the polymer. Therefore, the cylinder behaves elastically.
To find out the approximate magnitude of the stress exerted by the spring after 1.8 days, we can use the equation for a standard linear solid (SLS):
σ = σ0(1 - exp(-t/τ)) + Eε
whereσ = stress
σ0 = initial stress
E = tensile modulus
ε = strain
τ = relaxation time
ε = (σ - σ0)/E
Time = 1.8 days = 1.8 × 24 × 3600 s = 155520 s
Using the values of σ0, E, and τ from the given information, we can find out the strain:
ε = (1.437 - 0)/16.5 × 10⁶ε = 8.71 × 10⁻⁸
From the equation for SLS, we can write:
σ = σ0(1 - exp(-t/τ)) + Eεσ
= 1.437(1 - exp(-155520/2 × 10¹²)) + 16.5 × 10⁶ × 8.71 × 10⁻⁸σ
= 1.437(1 - 0.99999999961) + 1.437 × 10⁻⁴σ ≈ 0.176 MPa
Thus, the stress exerted by the spring after 1.8 days is approximately 0.176 MPa.
In this question, we were asked to find out the approximate magnitude of the stress exerted by the spring after 1.8 days. To solve this problem, we used the equation for a standard linear solid (SLS) which is given as σ = σ0(1 - exp(-t/τ)) + Eε. Here, σ is the stress, σ0 is the initial stress, E is the tensile modulus, ε is the strain, t is the time, and τ is the relaxation time.Using the given values, we first found out the strain. We were given the initial stress and the tensile modulus of the polymer. Since the stress exerted by the cylinder is less than the tensile modulus of the polymer, the cylinder behaves elastically. Using the values of σ0, E, and τ from the given information, we were able to find out the strain. Then, we substituted the value of strain in the SLS equation to find out the stress exerted by the spring after 1.8 days. The answer we obtained was approximately 0.176 MPa.
Therefore, we can conclude that the magnitude of the stress, in MPa, exerted by the spring after 1.8 days is approximately 0.176 MPa.
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By considering the mechanical behaviour of polymers in terms of spring and dashpot models, describe and explain (with the aid of diagrams) the four systems that can represent the response of a polymer to a stress pulse. Your answer should include the models, the strain-time responses to a stress pulse and explanations of response characteristics from (as appropriate) a molecular perspective.
Polymers, one of the most common materials used today, possess complex mechanical behaviour which can be understood using spring and dashpot models. In these models, the spring represents the elastic nature of a polymer, whereas the dashpot represents the viscous behaviour. The four systems that represent the response of a polymer to a stress pulse include:
1. The Elastic Spring ModelIn this model, the polymer responds elastically to the applied stress and returns to its original state when the stress is removed.2. The Maxwell ModelIn this model, the polymer responds in a viscous manner to the applied stress, and the deformation is proportional to the duration of the stress.3. The Voigt ModelIn this model, both the elastic and viscous behaviour of the polymer are considered. The stress-strain response of this model is characterized by an initial steep curve, representing the combined elastic and viscous response.
4. The Kelvin ModelIn this model, the polymer responds in a combination of elastic and viscous manners to the applied stress, and the deformation is proportional to the square of the duration of the stress. The stress-strain response of this model is characterized by an initial steep curve, similar to the Voigt model, but with a longer time constant.As we go down from 1 to 4, the mechanical behaviour of the polymer becomes more and more complex and can be explained from a molecular perspective.
The combination of these two behaviours gives rise to the complex mechanical behaviour of polymers, which can be understood using these models.
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A ladder and a person weigh 15 kg and 80 kg respectively, as shown in Figure Q1. The centre of mass of the 36 m ladder is at its midpoint. The angle a = 30° Assume that the wall exerts a negligible friction force on the ladder. Take gravitational acceleration as 9.81m/s? a) Draw a free body diagram for the ladder when the person's weight acts at a distance x = 12 m Show all directly applied and reaction forces.
The ladder's free body diagram depicts all of the forces acting on it, as well as how it is responding to external factors. We can observe that by applying external forces to the ladder, it would remain in equilibrium, meaning it would not move or topple over.
Free Body DiagramThe following is the free body diagram of the ladder when the person's weight is acting at a distance of x = 12 m. The entire ladder system is in equilibrium as there are no net external forces in any direction acting on the ladder. Consequently, the system's center of gravity remains at rest.Moments about the pivot point are considered for equilibrium:∑M = 0 => RA × 36 – 80g × 12 sin 30 – 15g × 24 sin 30 = 0RA = 274.16 NAll other forces can be calculated using RA.
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For a metal arc-welding operation on carbon steel, if the melting point for the steel is 1800 °C, the heat transfer factor = 0.8, the melting factor = 0.75, melting constant for the material is K-3.33x10-6 J/(mm³.K2). Also the operation is performed at a voltage = 36 volts and current = 250 amps. The unit energy for melting for the material is most likely to be O 10.3 J/mm³ O 10.78 J/mm3 14.3 J/mm3 8.59 J/mm³ The volume rate of metal welded is 377.6 mm³/s 245.8 mm³/s 629.3 mm³/s 841.1 mm³/s
In a metal arc-welding operation on carbon steel with specific parameters, the most likely unit energy for melting the material is 10.78 J/mm³. The volume rate of metal welded is likely to be 629.3 mm³/s.
To determine the unit energy for melting the material, we need to consider the given parameters. The melting point of the steel is stated as 1800 °C, the heat transfer factor is 0.8, the melting factor is 0.75, and the melting constant for the material is K = 3.33x10-6 J/(mm³.K²). The unit energy for melting (U) can be calculated using the equation: U = K * (Tm - To), where Tm is the melting point of the steel and To is the initial temperature. Substituting the given values, we have U = 3.33x10-6 J/(mm³.K²) * (1800°C - 0°C) = 10.78 J/mm³. Moving on to the volume rate of metal welded, the provided information does not include the necessary parameters to calculate it accurately. The voltage (V) is given as 36 volts, and the current (I) is provided as 250 amps. However, the voltage factor (Vf) and welding speed (Vw) are not given, making it impossible to determine the volume rate of metal welded. In conclusion, based on the given information, the unit energy for melting the material is most likely to be 10.78 J/mm³, while the volume rate of metal welded cannot be determined without additional information.
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Describe different kinds of flow metres in detail.
Flow meters are instruments used to measure the volume or mass of a liquid, gas, or steam passing through pipelines. Flow meters are used in industrial, commercial, and residential applications. Flow meters can be classified into several types based on their measuring principle.
Differential Pressure Flow Meter: This is the most common type of flow meter used in industrial applications. It works by creating a pressure difference between two points in a pipe. The pressure difference is then used to calculate the flow rate. Differential pressure flow meters include orifice meters, venturi meters, and flow nozzles.
Positive Displacement Flow Meter: This type of flow meter works by measuring the volume of fluid that passes through a pipe. The flow rate is determined by measuring the amount of fluid that fills a chamber of known volume. Positive displacement flow meters include nutating disk meters, oval gear meters, and piston meters.
flow meters are essential devices that help to measure the volume or mass of fluid flowing through pipelines. They can be classified into different types based on their measuring principle. Each type of flow meter has its advantages and limitations.
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PROBLEM 3 (10 pts) Predict the dominant type of bonding for the following solid compound by considering electronegativity (a) K and Na :______ (b) Cr and O:_______
(c) Ca and CI:______ (d) B and N:_______ (e) Si and O:_______
The dominant type of bonding for the following solid compound by considering electronegativity is as follows:a. K and Na: metallic bondingb. Cr and O: ionic bondingc. Ca and Cl: ionic bondingd. B and N: covalent bondinge. Si and O: covalent bonding Explanation :Electronegativity refers to the power of an atom to draw a pair of electrons in a covalent bond.
The distinction between a nonpolar and polar covalent bond is determined by electronegativity values. An electronegativity difference of less than 0.5 between two atoms indicates that the bond is nonpolar covalent. An electronegativity difference of between 0.5 and 2 indicates a polar covalent bond. An electronegativity difference of over 2 indicates an ionic bond.1. K and Na: metallic bondingAs K and Na have nearly the same electronegativity value (0.8 and 0.9 respectively), the bond between them will be metallic.2. Cr and O: ionic bondingThe electronegativity of Cr is 1.66, whereas the electronegativity of O is 3.44.
As a result, the electronegativity difference is 1.78, which implies that the bond between Cr and O will be ionic.3. Ca and Cl: ionic bondingThe electronegativity of Ca is 1.00, whereas the electronegativity of Cl is 3.16. As a result, the electronegativity difference is 2.16, which indicates that the bond between Ca and Cl will be ionic.4. B and N: covalent bondingThe electronegativity of B is 2.04, whereas the electronegativity of N is 3.04. As a result, the electronegativity difference is 1.00, which implies that the bond between B and N will be covalent.5. Si and O: covalent bondingThe electronegativity of Si is 1.9, whereas the electronegativity of O is 3.44.
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Q1) Search about Design and Fabrication for compressor in Ac of car supported with photographs
The compressor is a vital component of the car's air conditioning system. It is responsible for compressing the refrigerant gas, which then flows through the condenser and evaporator, cooling the air inside the car. The compressor is typically driven by the engine, but it can also be powered by an electric motor.
The compressor is a complex machine, and its design and fabrication requires a high level of engineering expertise. The compressor must be able to operate at high pressures and temperatures, and it must be durable enough to withstand the rigors of everyday use. The compressor is also required to be energy-efficient, as this can save the car owner money on fuel costs.
The compressor is typically made of cast iron or aluminum, and it is fitted with a number of moving parts, including a piston, a crankshaft, and a flywheel. The compressor is lubricated with oil, which helps to reduce friction and wear. The compressor is also equipped with a number of sensors, which monitor its performance and alert the driver if there is a problem.
The compressor is a critical component of the car's air conditioning system, and its design and fabrication are essential to ensuring that the system operates efficiently and effectively.
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IF an 85% efficient alternator operating at 1800RPM were putting
out 100kW of power how much torque would need tro be delivered by
the prime mover?
To determine the amount of torque that the prime mover would need to deliver to operate an 85% efficient alternator operating at 1800 RPM and putting out 100 kW of power, the following equation is used:Power = (2π × RPM × Torque) / 60 × 1000 kW = (2π × 1800 RPM × Torque) / 60 × 1000
Rearranging the equation to solve for torque:Torque = (Power × 60 × 1000) / (2π × RPM)Plugging in the given values:Torque = (100 kW × 60 × 1000) / (2π × 1800 RPM)≈ 318.3 Nm
Therefore, the prime mover would need to deliver about 318.3 Nm of torque to operate an 85% efficient alternator operating at 1800 RPM and putting out 100 kW of power. This can also be written as 235.2 lb-ft.
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The turning moment diagram for an engine is drawn to the following scales: Turning moment 1mm = 60 Nm: crank angle, Imm= 10, shows the maximum energy that needs to be stored by the flywheel in unit area is 2850 m2. The flywheel rotates at an average speed of 220 rpm with a total speed change of 2.5%. If the mass of the flywheel is 500 kg, find the appropriate dimensions (inner diameter, outer diameter and thickness) of the flywheel. Given the inner diameter of the flywheel is 0.9 outer diameter and the density is 7.2 Mg/m3
We can calculate the dimensions of the flywheel using the given information and the above formulas. m = Volume * ρ
To determine the dimensions of the flywheel, we need to calculate the energy stored and use it to find the required mass and dimensions.
Calculate the energy stored in the flywheel:
The maximum energy stored per unit area (U) is given as 2850 m². Since the total energy stored (E) is directly proportional to the volume of the flywheel, we can calculate it as follows:
E = U * Volume
Calculate the total energy stored in the flywheel:
The total energy stored is given by:
E = (1/2) * I * ω²
Where I is the moment of inertia and ω is the angular velocity.
Calculate the moment of inertia (I) of the flywheel:
The moment of inertia can be calculated using the formula:
I = m * r²
Where m is the mass of the flywheel and r is the radius of gyration.
Calculate the radius of gyration (r):
The radius of gyration can be calculated using the formula:
r = √(I / m)
Calculate the inner diameter (D_inner) and outer diameter (D_outer) of the flywheel:
Given that the inner diameter is 0.9 times the outer diameter, we can express the relationship as:
D_inner = 0.9 * D_outer
Calculate the thickness (t) of the flywheel:
The thickness can be calculated as:
t = (D_outer - D_inner) / 2
Given the density (ρ) of the flywheel material, we can calculate the mass (m) as:
m = Volume * ρ
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It is necessary to design a bed packed with rectangular glass prisms that measure 1 cm and 2 cm high with a sphericity of 0.72, which will be used as a support to purify air that enters a gauge pressure of 2 atm and 40 ° C. The density of the prisms is 1300 kg/m^3 and 200 kg is used to pack the column. The column is a polycarbonate tube with a diameter of 0.3 and a height of 3.5 m. considering that the feed is 3kg/min and the height of the fluidized bed is 2.5 m. Determine the gauge pressure at which the air leaves, in atm.
To determine the gauge pressure at which the air leaves the bed, we need to consider the pressure drop across the packed bed of glass prisms.
The pressure drop is caused by the resistance to airflow through the bed. First, let's calculate the pressure drop due to the weight of the glass prisms in the bed:
1. Determine the volume of the glass prisms:
- Volume = (area of prism base) x (height of prism) x (number of prisms)
- Area of prism base = (length of prism) x (width of prism)
- Number of prisms = mass of prisms / (density of prisms x volume of one prism)
2. Calculate the weight of the glass prisms:
- Weight = mass of prisms x g
3. Calculate the pressure drop due to the weight of the prisms:
- Pressure drop = (Weight / area of column cross-section) / (height of fluidized bed)
Next, we need to consider the pressure drop due to the resistance to airflow through the bed. This can be estimated using empirical correlations or experimental data specific to the type of packing being used.
Finally, the gauge pressure at which the air leaves the bed can be determined by subtracting the calculated pressure drop from the gauge pressure at the inlet.
Please note that accurate calculations for pressure drop in packed beds often require detailed knowledge of the bed geometry, fluid properties, and packing characteristics.
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2) (40%) True or false? a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant.
a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement.
b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true.
c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement
d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant. This statement is true.
a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement. For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates.
b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true. In r may appear in the final expression for one of the velocity components in flows occurring between r= 0 and r= a in cylindrical coordinates.
c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement as well. For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile.
d) If, in three dimensions, the pressure obeys the equation
Op/ dy = -pg,
and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as
p = -ogy+c,
where c is a constant. This statement is true. If, in three dimensions, the pressure obeys the equation
Op/ dy = -pg,
and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as
p = -ogy+c,
where c is a constant.
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You are asked to design a small wind turbine (D = x +1.25 ft, where x is the last two digits of your student ID). Assume the wind speed is 15 mph at T = 10°C and p = 0.9 bar. The efficiency of the turbine is n = 25%, meaning that 25% of the kinetic energy in the wind can be extracted. Calculate the power in watts that can be produced by your turbine.
The power in watts that can be produced by the turbine is 291.4 W.
From the question above, Diameter of the wind turbine, D = x + 1.25 ft
Efficiency of the wind turbine, n = 25% = 0.25
Wind speed, v = 15 mph
Temperature, T = 10° C
Pressure, p = 0.9 bar
The power in watts that can be produced by the turbine.
Diameter of the turbine, D = x + 1.25 ft
Let's put the value of D in terms of feet,1 ft = 0.3048 m
D = x + 1.25 ft = x + 1.25 × 0.3048 m= x + 0.381 m
Kinetic energy of the wind turbine,Kinetic energy, K.E. = 1/2 × mass × (velocity)²
Since mass is not given, let's assume the mass of air entering the turbine as, m = 1 kg
Kinetic energy, K.E. = 1/2 × 1 × (15.4)² = 1165.5 Joules
Since the efficiency of the turbine, n = 0.25 = 25%The power that can be extracted from the wind is,P = n × K.E. = 0.25 × 1165.5 = 291.4 Joules
So, the power in watts that can be produced by the turbine is 291.4 J/s = 291.4 W.
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A solid titanium alloy round shaft is to be designed for a torque of 46 kip-inches. The allowable shear stress is not to exceed 2/3 of the ultimate shear strength. What is the required diameter of the shaft based on shear stress? (inches)
To determine the diameter, we need to consider the torque and the allowable shear stress.
The allowable shear stress is 2/3 of the ultimate shear strength. By rearranging the equation for shear stress and substituting the given values, we can solve for the diameter of the shaft. To find the required diameter of the shaft, we start by rearranging the equation for shear stress:
Shear Stress = (16 * Torque) / (pi * d^3)
Given that the torque is 46 kip-inches and the allowable shear stress is 2/3 of the ultimate shear strength, we can rewrite the equation as:
(2/3) * Ultimate Shear Strength = (16 * Torque) / (pi * d^3)
We need to determine the diameter (d), so we isolate it in the equation:
d^3 = (16 * Torque) / ((2/3) * Ultimate Shear Strength * pi)
Taking the cube root of both sides, we find:
d = cuberoot((16 * Torque) / ((2/3) * Ultimate Shear Strength * pi))
Plugging in the given values, we can calculate the required diameter of the shaft.
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Q4. A solid shaft of diameter 50mm and length of 300mm is subjected to an axial load P = 200 kN and a torque T = 1.5 kN-m. (a) Determine the maximum normal stress and the maximum shear stress. (b) Repeat part (a) but for a hollow shaft with a wall thickness of 5 mm.
Part (a)The normal stress and the shear stress developed in a solid shaft when subjected to an axial load and torque can be calculated by the following equations.
Normal Stress,[tex]σ =(P/A)+((Mz×r)/Iz)[/tex]Where,[tex]P = 200kNA
= πd²/4 = π×(50)²/4
= 1963.4954 mm²Mz[/tex]
= T = 1.5 kN-mr = d/2 = 50/2 = 25 m mIz = πd⁴/64 = π×(50)⁴/64[/tex]
[tex]= 24414.2656 mm⁴σ[/tex]
[tex]= (200 × 10³ N) / (1963.4954 mm²) + ((1.5 × 10³ N-mm) × (25 mm))/(24414.2656 mm⁴)σ[/tex]Shear Stress.
[tex][tex]J = πd⁴/32 = π×50⁴/32[/tex]
[tex]= 122071.6404 mm⁴τ[/tex]
[tex]= (1.5 × 10³ N-mm) × (25 mm)/(122071.6404 mm⁴)τ[/tex]
[tex]= 0.03 MPa[/tex] Part (b)For a hollow shaft with a wall thickness of 5mm, the outer diameter, d₂ = 50mm and the inner diameter.
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The dry products of combustion have the following molar percentages: CO 2.7% 025.3% H20.9% CO2 16.3% N2 74.8% Find, for these conditions: (a) mixture gravimetric analysis; (b) mixture molecular weight, lbm/lbmole; and (c) mixture specific gas constant R, ft lbf/Ibm °R.
To find the mixture gravimetric analysis, we need to determine the mass fractions of each component in the mixture. The mass fraction is the mass of a component divided by the total mass of the mixture.
Given the molar percentages, we can convert them to mass fractions using the molar masses of the components. The molar masses are as follows:
CO: 28.01 g/mol
O2: 32.00 g/mol
H2O: 18.02 g/mol
CO2: 44.01 g/mol
N2: 28.01 g/mol
(a) Mixture Gravimetric Analysis:
The mass fraction of each component is calculated by multiplying its molar percentage by its molar mass and dividing by the sum of all the mass fractions.
Mass fraction of CO: (0.027 * 28.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of O2: (0.253 * 32.00) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of H2O: (0.009 * 18.02) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of CO2: (0.163 * 44.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
Mass fraction of N2: (0.748 * 28.01) / (0.027 * 28.01 + 0.253 * 32.00 + 0.009 * 18.02 + 0.163 * 44.01 + 0.748 * 28.01)
(b) Mixture Molecular Weight:
The mixture molecular weight is the sum of the mass fractions multiplied by the molar masses of each component.
Mixture molecular weight = (Mass fraction of CO * Molar mass of CO) + (Mass fraction of O2 * Molar mass of O2) + (Mass fraction of H2O * Molar mass of H2O) + (Mass fraction of CO2 * Molar mass of CO2) + (Mass fraction of N2 * Molar mass of N2)
(c) Mixture Specific Gas Constant:
The mixture specific gas constant can be calculated using the ideal gas law equation:
R = R_universal / Mixture molecular weight
where R_universal is the universal gas constant.
Now you can substitute the values and calculate the desired quantities.
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Using an allowable shearing stress of 8,000 psi, design a solid steel shaft to transmit 14 hp at a speed of 1800 rpm. Note(1) : Power =2 nf where fis frequency (Cycles/second) and Tis torque (in-Ib). Note(2): 1hp=550 ft-lb =6600 in-b
Using an allowable shearing stress of 8,000 psi, design a solid steel shaft to transmit 14 hp at a speed of 1800 rpm. The minimum diameter is 1.25 inches.
Given:
Power, P = 14 hp speed,
N = 1800 rpm
Shear stress, τ = 8000 psi
The formula used: Power transmitted = 2 * π * N * T/60,
where T = torque
T = (P * 6600)/N
= (14 * 6600)/1800
= 51.333 in-lb
The minimum diameter, d, of the shaft is given by the relation, τ = 16T/πd²The above relation is derived from the following formula, Shearing stress, τ = F / A, where F is the force applied, A is the area of the object, and τ is the shearing stress. The formula is then rearranged to solve for the minimum diameter, d. Substituting the values,
8000 = (16 * 51.333)/πd²d
= 1.213 in
≈ 1.25 in
The minimum diameter is 1.25 inches.
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