However, specific values are required to calculate the enthalpies and other parameters accurately.
a) Hardware diagram of the simple steam cycle, and b) Steam cycle on the steam enthalpy-entropy chart are given below:
a) Hardware diagram of the simple steam cycle:
(1) ──── Boiler ────> (2)
/ |
Turbine Condenser
| |
└─────────────(3)─────────────┘
Points:
(1) - Boiler exit
(2) - Condenser inlet
(3) - Turbine exit
b) Steam cycle on the steam enthalpy-entropy chart:
↑
|
(3) | (2)
|
────────────────┼───────────────
|
|
(1) |
|
↓
c) Specific enthalpy at each point:
Boiler exit (1): h1 = Enthalpy at boiler exit (Given value)
Turbine exit (3): h3 = Isentropic turbine exit enthalpy (Calculate using turbine efficiency)
Condenser inlet (2): h2 = Enthalpy at condenser inlet (Given value)
Condenser exit: h4 = Enthalpy at condenser exit (Given value)
Dryness fraction at turbine exit (3): Calculate by determining the quality of the steam at point 3 using the specific enthalpy values.
e) Thermal efficiency of the cycle: Calculate the thermal efficiency using the formula:
Thermal efficiency = (Work output / Heat input) × 100
f) Power output of the cycle: Calculate the power output using the formula:
Power output = Mass flow rate × (h1 - h2)
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2. The total copper loss of a transformer as determined by a short-circuit test at 20°C is 630 watts, and the copper loss computed from the true ohmic resistance at the same temperature is 504 watts. What is the load loss at the working temperature of 75°C?
Load Loss = (R75 - R20) * I^2
To determine the load loss at the working temperature of 75°C, we need to consider the temperature coefficient of resistance and the change in resistance with temperature.
Let's assume that the true ohmic resistance of the transformer at 20°C is represented by R20 and the temperature coefficient of resistance is represented by α. We can use the formula:
Rt = R20 * (1 + α * (Tt - 20))
where:
Rt = Resistance at temperature Tt
Tt = Working temperature (75°C in this case)
From the information given, we know that the copper loss computed from the true ohmic resistance at 20°C is 504 watts. We can use this information to find the value of R20.
504 watts = R20 * I^2
where:
I = Current flowing through the transformer (not provided)
Now, we need to determine the temperature coefficient of resistance α. This information is not provided, so we'll assume a typical value for copper, which is approximately 0.00393 per °C.
Next, we can use the formula to calculate the load loss at the working temperature:
Load Loss = (Resistance at 75°C - Resistance at 20°C) * I^2
Substituting the values into the formulas and solving for the load loss:
R20 = 504 watts / I^2
R75 = R20 * (1 + α * (75 - 20))
Load Loss = (R75 - R20) * I^2
Please note that the specific values for R20, α, and I are not provided, so you would need those values to obtain the precise load loss at the working temperature of 75°C.
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Whole Foods Market sells Kaiser brand sausages. The market demand for Kaiser Sausages is uncertain but normally distributed with a mean of 124000 packages. For each supply order the fixed order cost from the Kaiser warehouse is $486. The annual holding cost is $1.7 for a package/year. A (Q,R) policy is used to manage the supply chain. What is the order quantity Q ? (Integer answer)
Therefore, the order quantity Q is approximately 5940.
Inventory management systems are meant to help business owners strike a balance between avoiding stockouts while minimizing the cost of carrying too much inventory. One of the most common ways of doing this is to use a Q-R policy.
In this case, we are given that Whole Foods Market sells Kaiser brand sausages. The market demand for Kaiser Sausages is uncertain but normally distributed with a mean of 124,000 packages. For each supply order, the fixed order cost from the Kaiser warehouse is $486. The annual holding cost is $1.7 for a package/year.
The Q-R policy is used to manage the supply chain. We are to determine the order quantity Q. To compute the order quantity Q, we need to make use of the following formula:
EOQ = √((2SD/CH)
Where EOQ = Economic Order QuantityS = Setup costD = DemandQ = Order quantityC = Carrying costH = Holding cost
From the information given in the question, we know that:S = $486D = 124,000Q = ?C = $0 (Assuming no other carrying costs)H = $1.7
Using the given values, we can calculate the standard deviation (SD) as follows:
SD = σ = √(VAR)
We know that the variance VAR is given by:
VAR = σ²
We are given that the demand is normally distributed with a mean of 124,000 packages. We are not given the standard deviation of the distribution, but we know that a normal distribution is fully characterized by its mean and standard deviation. Therefore, we will need to make an assumption about the standard deviation.
A common assumption is that the standard deviation is equal to 15% of the mean. This is often referred to as the coefficient of variation (CV).
CV = (σ/mean)*100%
We can rearrange this formula to solve for σ:
σ = (CV/100%)*mean
Therefore:
σ = (0.15)*124,000σ = 18,600
Now that we know the standard deviation, we can calculate the Economic Order Quantity as follows:
EOQ = √((2SD/CH)
EOQ = √((2*18,600*124,000)/1.7)
EOQ = 5,940.2 ≈ 5940 Therefore, the order quantity Q is approximately 5940.
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A thin outer border of a building's area covers 10×10m^2. The sky temperature is 300K meanwhile the temperature of the sun is 5800K. The overall distance between the sun and earth is 1.5×10^11 meters and the overall sun diameter is 1.4×10^9 meter and the earth diameter is 1.3×10^7 meters. The properties of the outer border are: ελ = 0.5 for λ > 6 µm & ελ = 0.1 for λ < 6 µm. The outer border can be considered a diffuse surface. Air current flows over the border with a velocity of 10 meters/second with a temperature of 300K. Beneath the border, the air inside the building flows over the bottom side of the border at 1 meter/second.
Determine the steady-state temperature of the border for these conditions.
Please state your assumptions
The steady-state temperature of the border for the given conditions is 407.72K.
The following assumptions are made in this analysis: All the values are steady-state
The outer border of the building is thin and therefore can be considered a one-dimensional surface.
The outer border of the building is considered a diffuse surface.
The sky is considered to have a uniform temperature of 300K.The sun's diameter is 1.4×109 meters.
The diameter of the Earth is 1.3×107 meters.
The distance between the Earth and the Sun is 1.5×1011 meters.
The velocity of air above and below the border is considered to be the same.
Temperature of the border
The total heat flux received by the outer border of the building, q, is calculated using the Stefan-Boltzmann Law as follows:
q = σ (Tb4 - Ts4)where σ is the Stefan-Boltzmann constant, Tb is the temperature of the border, and Ts is the temperature of the sky.
σ = 5.67 x 10-8 W/m2K4 is the Stefan-Boltzmann constant.
Ts = 300K is the temperature of the sky.
The heat absorbed by the border is calculated by using the following equation:
q = mcpΔT
where m is the mass flow rate of the air, cp is the specific heat of the air at constant pressure, and ΔT is the temperature difference between the air and the border.
The total heat absorbed by the air above and below the border is given by the following equation:
q = ma cp (Ta - Tb)
where Ta is the temperature of the air above the border and ma is the mass flow rate of the air above the border .The total heat absorbed by the air below the border is given by the following equation:
q = mb cp (Tb - Tc)
where Tc is the temperature of the air below the border and mb is the mass flow rate of the air below the border .The heat absorbed by the border is given by the following equation:
q = σ (Tb4 - Ts4)
The steady-state temperature of the border is calculated by equating the heat absorbed by the border to the heat absorbed by the air above and below the border as follows:
ma cp (Ta - Tb) + mb cp (Tb - Tc) = σ (Tb4 - Ts4)
The steady-state temperature of the border, Tb is determined by solving the above equation.
Tb = 407.72K
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The critical resolved shear stress in a silver single crystal is 6.5 MPa. A tensile stress is applied along the [1 1 O axis to cause slip on the (111)[ī o 1) slip system of the crystal. Determine: (a) The angle between the tensile axis and the normal to the slip plane(1 11). (b) The angle between the tensile axis and the slip direction[ī 01).
(c) The tensile stress that is required to cause the slip
The critical resolved shear stress in a silver single crystal is 6.5 MPa. The tensile stress is applied along the [1 1 O] axis to cause slip on the (111)[ī o 1) slip system of the crystal.
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At high temperatures, a diatomic gas can also have an RT contribution from a vibrational energy contribution. Using this kinetic energy model, calculate (a) the constant-volume molar specific heat, kJ/kgmole-K; (b) the constant-pressure molar specific heat, kJ/kgmole K; and (c) the molar specific heat ratio for a high- temperature diatomic gas.
The specific heat of a high-temperature diatomic gas can be calculated considering both the translational and vibrational energy contributions. The constant-volume molar specific heat and constant-pressure molar specific heat can be determined using kinetic energy models.
(a) To calculate the constant-volume molar specific heat, we consider only the contribution from translational energy. For a diatomic gas, the constant-volume molar specific heat (Cv) is given by the formula Cv = (5/2) R, where R is the gas constant. (b) The constant-pressure molar specific heat (Cp) takes into account both translational and vibrational energy contributions. For a diatomic gas, Cp = (7/2) R. This is because, at high temperatures, the vibrational energy modes of the gas molecules become significant, contributing to the total energy of the system.
(c) The molar specific heat ratio, γ, is the ratio of the constant-pressure molar specific heat to the constant-volume molar specific heat. For a diatomic gas, γ = Cp/Cv = (7/2) / (5/2) = 7/5 = 1.4. The molar specific heat ratio provides information about the behavior of the gas at high temperatures, such as the speed of sound and the adiabatic index. By considering the translational and vibrational energy contributions, we can calculate the constant-volume molar specific heat, constant-pressure molar specific heat, and the molar specific heat ratio for a high-temperature diatomic gas. These values help us understand the thermodynamic properties and behavior of the gas at elevated temperatures.
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A boundary layer develops with no pressure gradient imposed. The momentum thickness is found to be Θ = δ/4. At some location, the boundary layer thickness is measured to be 8mm. At another location 4mm downstream, the thickness is measured to be 16 mm. Use the momentum integral equation to estimate the value of the skin-friction coefficient C’f, in the vicinity of these two measurements.
The value of the skin-friction coefficient C’ f in the vicinity of these two measurements using the momentum integral equation is 0.0031.
The thickness of the boundary layer grows due to the movement of the fluid and, to some extent, the shear stresses produced as the fluid moves across a surface. No pressure gradient has been imposed in this scenario, implying that the fluid velocity is entirely determined by the local shear stresses within the fluid.
According to the question, Θ = δ/4, where Θ is the momentum thickness. This indicates that the momentum thickness is a quarter of the displacement thickness, δ. To use the momentum integral equation, the value of the momentum thickness must be found first. According to the problem statement, the momentum thickness is given as Θ = δ/4.
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Given a 50 ft spherical steel tank, find the thickness of material required to hold gas up to a maximum pressure of 200 psi. Material is structural steel and use a safety factor of 3. (Yield strength of the material is 36 ksi.)
The thickness of the steel material required to hold gas up to a maximum pressure of 200 psi is 1666.67 inches (139.72 feet).
Explanation:
The given problem requires calculating the thickness of a spherical steel tank that is 50 ft in diameter, to hold gas up to a maximum pressure of 200 psi. To find the thickness, we use two formulas.
First, we use the formula Stress = Pr / t, where P is the maximum pressure of 200 psi, r is the radius of the sphere, t is the thickness of the sphere. Secondly, we use the formula Stress = 3fy / SF, where fy is the yield strength of the material (36 ksi), and SF is the safety factor of 3.
We know that the radius of the spherical steel tank is half its diameter, so the radius is 25 ft or 300 inches. We can then use Stress = Pr / t to find the maximum stress in the steel tank, which is 60000 / t.
Using the second formula, 3fy / SF, we can equate it to Stress to get 3fy / SF = 60000 / t. Since fy = 36 ksi and SF = 3, we can simplify the equation to 3 x 36 / 3 = 60000 / t, and solve for t.
Finally, we get t = (60000 x 3) / (3 x 36) = 1666.67 inches or 139.72 feet. Therefore, the thickness of the steel material required to hold gas up to a maximum pressure of 200 psi is 1666.67 inches (139.72 feet).
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The depth of the water channel shown in this diagram is 1ft. The flow is steady with exit velocity of 3.5ft/s. At the inlet, the water velocity in the center portion of the channel is unknown, and it is 1ft/s in the remainder of the channel. The fixed control volume ABCD is shown by the dashed line. Using the Reynolds Transport Theorem, Eq. (4.19), calculate the velocity at the center portion of the inlet.
The depth of the water channel shown in the diagram is 1ft. The flow is steady with an exit velocity of 3.5ft/s. At the inlet, the water velocity in the center portion of the channel is unknown, and it is 1ft/s in the remainder of the channel.
The fixed control volume ABCD is shown by the dashed line. We are to calculate the velocity at the center portion of the inlet by using the Reynolds Transport Theorem, Eq. (4.19).In a steady flow field, the Reynolds Transport Theorem can be used to simplify and control the process. In a way, this theorem is a simplification of the general transport theorem for fluids in motion and is used to explain the motion of fluid flow through a fixed volume of space, such as a pipe, at any given moment. The Reynolds Transport Theorem is given by:∂/∂t ∫ ρdV + ∫ ρ(V-Vc).dA = 0where ρ is the density of the fluid, V is the velocity of the fluid, Vc is the velocity of the control surface (ABCDA), and dV and dA are the volume and area elements of the control surface, respectively.Therefore, we can evaluate the velocity at the center portion of the inlet by applying the Reynolds Transport Theorem. Let's do it step by step:∂/∂t ∫ ρdV + ∫ ρ(V-Vc).dA = 0We can simplify the above equation as the flow is steady, ∂/∂t ∫ ρdV = 0.Rearranging the above equation yields:∫ ρ(V-Vc).dA = 0V ∫ ρ.dA - Vc ∫ ρ.dA = 0(Assuming that the control surface is oriented such that the normal vector faces in the positive x direction)Vinlet ∫ ρ.A + 1ft/s ∫ ρ.A = 3.5ft/s ∫ ρ.AVinlet = (3.5ft/s - ρ.A)/ρ.AAs per the information given in the question, at the inlet, the water velocity in the center portion of the channel is unknown, and it is 1ft/s in the remainder of the channel. Therefore, we can take the area of the center portion of the inlet to be half of the total area of the inlet. Let's assume that the inlet is a rectangular channel such that the total area of the inlet is A. Thus, the area of the center portion of the inlet is A/2. Thus, substituting the value of the area, we get:Vinlet = (3.5ft/s - ρ.A/2)/ρ.AThus, this is the solution that is obtained.
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What is the allowable axial compression load for a W12x72 column with an unbraced length of 16'; assume k = 1.0? Use table A.3 in your text for the steel column properties and table 10.1 for Fc. Round kl/r down to the nearest whole number. (5 pts.) 2. What is the allowable axial compression load for a W12x72 column with an unbraced length of 16' where rotation is fixed and translation is fixed both at the top and bottom of the column? Use table A.3 in your text for the steel column properties and table 10.1 for Fc. Round kl/r down to the nearest whole number.
The allowable axial compression load can be determined by calculating kl/r, rounding it down, and using the appropriate tables to find the corresponding value.
What is the allowable axial compression load for a W12x72 column with an unbraced length of 16' and k = 1.0?The first question asks for the allowable axial compression load for a W12x72 column with an unbraced length of 16' assuming k = 1.0. To calculate this, the value of kl/r needs to be determined by dividing the unbraced length by the radius of gyration.
Once kl/r is obtained, it can be rounded down to the nearest whole number. Using table A.3 for the steel column properties and table 10.1 for Fc, the allowable axial compression load corresponding to the determined kl/r value can be found.
The second question asks for the allowable axial compression load for a W12x72 column with an unbraced length of 16' where rotation is fixed and translation is fixed at both the top and bottom of the column.
Similar to the first question, kl/r needs to be calculated and rounded down. Then, using the appropriate tables, the allowable axial compression load corresponding to the determined kl/r value can be determined.
Both calculations involve determining the kl/r value, rounding it down, and using the corresponding tables to find the allowable axial compression load for the given column configuration.
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a) The higher the temperature a boiler operates at, the more efficient the Rankine Cycle. What are two important factors that limit how high the temperature can go in a boiler? b) A steam power cycle is connected to a furnace burning bagasse from a sugar mill and is producing 100 MW of power from its turbine. The cycle has a thermal efficiency of 32 %. Heat is rejected from the cycle to a condenser which is cooled with water from a river. For part ii to iv ignore the work associated with the pump. i. Draw an overall diagram showing the major energy flows. ii. What is the heat input for this cycle? iii. How much heat is rejected through heat transfer to the river water? iv. Bagasse has a HHV of 17 MJ/kg. If 90% of the heat from combustion is transferred to the boiler, what rate of bagasse needs to be burned (in kg/s)?
a) Two important factors that limit how high the temperature can go in a boiler are:
1. Material Limitations: The construction materials used in the boiler need to withstand high temperatures without experiencing excessive deformation, corrosion, or other forms of degradation. Different materials have different temperature limits, and the selection of appropriate materials becomes crucial in determining the maximum temperature the boiler can operate at.
2. Combustion Limitations: The combustion process itself has limitations on temperature due to factors such as fuel properties, flame stability, and emission control. The combustion temperature needs to be controlled within a certain range to ensure efficient and stable combustion, as well as to comply with environmental regulations regarding emissions.
b) i. Overall Diagram:
```
+-------------------+
| |
Bagasse -->| Boiler |--> Steam --> Turbine --> Power Output
| |
+--------|----------+
|
V
Condenser --> River Water
```
ii. The heat input for this cycle can be determined using the thermal efficiency formula:
Thermal Efficiency = (Net Power Output / Heat Input) * 100%
Given that the thermal efficiency is 32% and the power output is 100 MW, we can rearrange the formula to solve for the heat input:
Heat Input = (Net Power Output / Thermal Efficiency)
Heat Input = (100 MW / 0.32) = 312.5 MW
iii. The heat rejected through heat transfer to the river water is equal to the heat input to the cycle minus the net power output. Therefore:
Heat Rejected = Heat Input - Net Power Output
Heat Rejected = 312.5 MW - 100 MW = 212.5 MW
iv. To determine the rate of bagasse that needs to be burned, we need to calculate the heat released from the combustion of bagasse. Given that 90% of the heat from combustion is transferred to the boiler and the Higher Heating Value (HHV) of bagasse is 17 MJ/kg, we can calculate the rate of bagasse burned as follows:
Heat Released from Bagasse Combustion = Heat Input * (1 - Efficiency)
Heat Released from Bagasse Combustion = 312.5 MW * (1 - 0.9) = 31.25 MW
Rate of Bagasse Burned = Heat Released from Bagasse Combustion / HHV
Rate of Bagasse Burned = 31.25 MW / (17 MJ/kg) = 1.838 kg/s
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The output time response of a control system is equal to a. the transient response x the steady state response b. the transient response - the steady state response c. the transient response / the steady state response d. the transient response + the steady state response
The output time response of a control system is equal to the sum of the transient response and the steady-state response.
This can be represented by the equation: Output response = Transient response + Steady-state response. Therefore, the correct option is d) the transient response + the steady state response. The transient response of a control system represents the behavior of the system immediately after a disturbance or change in the input. It typically exhibits oscillations and decays over time until the system reaches a stable state. On the other hand, the steady-state response represents the long-term behavior of the system after it has settled down, where the output remains constant. The steady-state response is independent of the initial conditions and depends only on the input to the system. When these two components are combined, the resultant output time response of the control system captures both the initial transient behavior and the final steady-state behavior. It is important to consider both aspects to fully understand and analyze the system's performance.
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A TM wave propagating in a rectangular waveguide with μ=4μ0 and ε=81ε0.
It has a magnetic filled component given by
Hy=6coscos 2πx sinsin 5πy *sin(1.5π*1010t-109πz). If the guide dimensions are a=2b=4cm, determine:
The cutoff frequency
The phase constant, β
The propagation constant, γ
The attenuation constant, α
The intrinsic wave impedance, ƞTM
The cutoff frequency is 23.87 GHz, the phase constant is 163.44 rad/m, the propagation constant is (71.52 + j163.44) Np/m, the attenuation constant is 3.34 Np/m, and the intrinsic wave impedance is (0.048 + j0.109) Ω.
Given data:
μ = 4μ₀
ε = 81ε₀
H_y = 6cos(cos2πx sin5πy) sin(1.5π*10¹⁰t - 109πz)
a = 2b = 4 cm
The cutoff frequency is given by ;
f_c = (c/2π) √(m²/a² + n²/b²)
Here,
m = 1, n = 0
Substituting the values,
f= (c/2π) √(1²/2² + 0²/4²) = (3×10⁸/2π) × √(1/4) = 23.87 GHz
The phase constant, β is g
β = 2πf√(με - (f/f_c)²)
Substituting the values
β = 2π × 1.5 × 10¹⁰ × √(4μ₀ × 81ε₀ - (1.5 × 10¹⁰/23.87 × 10⁹)²) = 163.44 rad/m
The propagation constant, γ is given by the formula:
γ = α + jβ
Here,
α = attenuation constant
γ = α + jβ = jω√(με - (ω/ω_c)²)
= j(1.5π×10¹⁰)√(4μ₀ × 81ε₀ - (1.5π×10¹⁰/23.87×10⁹)²)
= (71.52 + j163.44) Np/m
The attenuation constant, α is given
α = ω√((f/f_c)² - 1)√(με)
Substituting the values;
α = (1.5π × 10¹⁰) √((1.5 × 10¹⁰/23.87 × 10⁹)² - 1) √(4μ₀ × 81ε₀) = 3.34 Np/m
The intrinsic wave impedance, ηTM is
ηTM = (jωμ)⁻¹ √(β² - (ωεμ)²)
ηTM = (j1.5π×10¹⁰×4π×10⁻⁷)⁻¹ × √((163.44)² - (1.5π×10¹⁰)²(81ε₀ × 4μ₀))
= (0.048 + j0.109) Ω
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(a) A cougar was found dead in the woods by a ranger, which he assumed was shot by a poacher. The recorded body temperature of the dead body was 27∘C (degree Celcius) while the temperature of the woods was assumed to be uniform at 24∘C. The rate of cooling of the body can be expressed as: dT/dt=−k(T−Ta), where T is the temperature of the body in ∘C,Ta is temperature of the surrounding medium (in ∘
C ) and k is proportionally constant. Let initial temperature of the cougar be 37∘C while k=0.152. i Estimate the temperature of the dead body at time, 0≤t≤9 hours by using Euler's method with Δt=1 hour. Approximate how long the cougar had been killed at T=27∘C by using linear interpolation techniques. (b) Solve y′′+y=0,y(0)=3,y(1)=−3 by using finite-difference method with h=0.2.
The temperature of the dead body at 9th hour is 28.191 degrees Celsius and the time for the cougar to cool down from 28.191 degrees Celsius to 27 degrees Celsius is approximately 1 hour.
a) The differential equation for the rate of cooling of a body can be expressed as
d/=−(−)
where T is the temperature of the body in degrees Celsius,
Ta is the temperature of the surrounding medium in degrees Celsius, and
k is the proportionality constant.
Given ,Initial temperature of the cougar T = 37 degrees Celsius;
The temperature of the woods Ta = 24 degrees Celsius;
Proportionality constant k = 0.152;
Recorded body temperature of the dead body = 27 degrees Celsius.
To find the temperature of the dead body at time, 0≤t≤9 hours using Euler's method with Δt=1 hour.
To find T at t = 1 hour, use Euler's Method as follows: dT/dt=−k(T−Ta)T(0) = 37,
Ta = 24, k = 0.152
dT/dt=−0.152(T−24)
Substituting h = 1 in the Euler's formula we get:
Tp + 1 = Tp + h(dT/dt)
Putting the above values, we get:
T1 = T0 + h dT/dtT1 = 37 + (1)(-0.152)(37 - 24)
T1 = 36.016
So, the temperature of the dead body at t = 1 hour is 36.016 degrees Celsius.
Similarly, for t = 2,3,4,5,6,7,8 and 9 hours, the calculations are:T2 = 34.682
T3 = 33.472
T4 = 32.376
T5 = 31.379
T6 = 30.469
T7 = 29.639
T8 = 28.882
T9 = 28.191
To find out how long the cougar had been killed, we use linear interpolation between 28.191 degrees Celsius and 27 degrees Celsius. At T = 28.191 degrees Celsius, the time is 9 hours.
At T = 27 degrees Celsius,
T = Tn + (Tn+1 - Tn) / (ΔTn+1 - ΔTn)(27 - 28.191) = (Tn+1 - Tn) / (ΔTn+1 - ΔTn)(27 - 28.191) = (27 - 28.191) / (9 - 8)
Tn+1 - Tn = 1.191 / (1)
Tn+1 = Tn - 1.191
Tn+1 = 28.191 - 1.191
Tn+1 = 27
b) The differential equation is y′′+y=0, y(0) = 3, y(1) = −3.
Substituting the values of h and x in the following finite-difference equations
y′=(y(i+1)−y(i))/h
y′′=(y(i+1)+y(i−1)−2y(i))/h²
we havey(i+1) - y(i) = hy'(i+1) + y(i) = h/2(y''(i) + y''(i+1)) + y
(i)Using y(0) = 3 and y(1) = −3, the values of y(0.2), y(0.4), y(0.6), and y(0.8) are obtained as follows:
For i = 0y'(0) = (y(0.2) - y(0))/0.2y'(0) = (y(0.2) - 3)/0.2y'(0) = (0.2y(0.2) - 0.6) / 0.2²y'(0) = 0.2y(0.2) - 0.6y''(0) = (y(0.2) + y(0) - 2y(0))/0.2²y''(0) = (y(0.2) - 6) / 0.2²(y'(0.2) + y'(0)) / 2 = (y''(0) + y''(0.2)) / 2
Using the above equations, we get
y(0.2) = 2.4554y'(0.2) = -3.72y''(0.2) = 2.2738
For i = 1y'(0.2) = (y(0.4) - y(0.2))/0.2y'(0.2) = (y(0.4) - 2.4554)/0.2y'(0.2) = (0.2y(0.4) - 0.49108) / 0.2²y'(0.2) = y(0.4) - 2.4554y''(0.2) = (y(0.4) + y(0.2) - 2y(0.2))/0.2²y''(0.2) = (y(0.4) - 4.9108) / 0.2²
Using the above equations, we get y(0.4) = -0.312y'(0.4) = -2.0918y''(0.4) = -1.0234
Similarly, for i = 2 and i = 3, the calculations are:
y(0.6) = -4.472y'(0.6) = -0.8938y''(0.6) = 1.5744y(0.8) = -2.6799
y'(0.8) = 1.4172y''(0.8) = -0.5754
Therefore, the solution of the differential equation y'' + y = 0, y(0) = 3, y(1) = −3 by using the finite-difference method with h = 0.2 is:
y(0) = 3y(0.2) = 2.4554y(0.4) = -0.312y(0.6) = -4.472y(0.8) = -2.6799
y(1) = −3
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Draw P-V diagram of thermodynamics with saturated line. Then,
draw constant pressure line, contant temperature line, and constant
volume line in it.
A P-V diagram is a two-dimensional graph showing the variation of pressure and volume of a system. A P-V diagram of thermodynamics with a saturated line is shown in the figure below: Explanation:Constant Pressure Line: A constant pressure line is a horizontal line parallel to the x-axis. In a constant pressure line, the pressure remains constant, and the volume changes. In a P-V diagram, this line represents an isobaric process.Constant Temperature Line: A constant temperature line is a curve that begins at the left and slopes upward to the right.
The temperature remains constant throughout the process. In a P-V diagram, this line represents an isothermal process.Constant Volume Line: A constant volume line is a vertical line parallel to the y-axis. In a constant volume line, the volume remains constant, and the pressure changes. In a P-V diagram, this line represents an isochoric process.The saturated line is the boundary between the liquid and vapor phases of a substance. The point at which the saturated line intersects the constant pressure line is known as the saturation point.
At the saturation point, the liquid and vapor phases coexist at equilibrium.A P-V diagram is a useful tool for analyzing thermodynamic processes and can be used to determine the work done by a system during a process. The area under the curve on a P-V diagram represents the work done by the system. The work done by the system during a process can be calculated by integrating the area under the curve.
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Find the Poisson’s ratio and bulk modulus of a material whose modulus of elasticity is 200 GPa and modulus of rigidity is 80 GPa. A 2 m long rod of 40 mm diameter made with the same material is stretched by 2.5 mm under some axial load. Find the lateral contraction.
The Poisson's ratio is 0.333 or 1/3, the bulk modulus is 153.846 GPa, and the lateral contraction is −1.665 mm.
Given the modulus of elasticity E = 200 GPa
Modulus of rigidity G = 80 GPa
Diameter of the rod d = 40 mm
The radius of the rod r = 20 mm
The original length of the rod L = 2 m
Extension in length ΔL = 2.5 mm
We can use the following formulas to calculate Poisson's ratio, bulk modulus, and lateral contraction.
Poisson's ratio μ = (3K − 2G) / (2(3K + G))
Bulk modulus K = E / 3(1 − 2μ)
Lateral contraction ΔD = −μΔL = (−2μΔL / L)
Poisson's ratio:
Substitute the given values in the formula,
μ = (3K − 2G) / (2(3K + G))
μ = (3 × 200 − 2 × 80) / (2(3 × 200 + 80))
μ = 0.333 or 1/3
Bulk modulus:
Substitute the given values in the formula,
K = E / 3(1 − 2μ)
K = 200 / 3(1 − 2 × 0.333)
K = 153.846 GPa
Lateral contraction:
Substitute the given values in the formula,
ΔD = (−2μΔL / L)
ΔD = (−2 × 0.333 × 2.5) / 2000
ΔD = −0.001665 m or −1.665 mm
Therefore, the Poisson's ratio is 0.333 or 1/3, the bulk modulus is 153.846 GPa, and the lateral contraction is −1.665 mm.
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1. What is DC Motor? 2. Explain the principles operation of a DC Motor? 3. How the Back EMF or Counter EMF is produced? 4. Differentiate the types of DC Motor through: a. Schematic Diagram or Circuit Diagram. b. Voltage Equation c. Characteristic of the speed and torque of the motor. 5. What is the "TORQUE"? 6. Cite the different formulas involved in the operation of the DC Motor. 7. Explain the power stages absorbed by the DC Motor. 8. Prove that the Capacity of DC Motor stated below: 1HP=746-watts
1. DC Motor: DC motor stands for Direct Current motor. It converts electrical energy into mechanical energy. It consists of a stator and a rotor that are separated from each other.
2. Principles of operation of a DC motor: DC motor operates on the principles of the Faraday's Law of Electromagnetic Induction. When a current-carrying conductor is placed in a magnetic field, it experiences a force. This force creates a torque on the rotor of the DC motor which causes it to rotate.
3. Production of Back EMF or Counter EMF: Back EMF or Counter EMF is produced in the DC motor when the rotor rotates. The generated EMF opposes the flow of current in the armature windings of the motor. The back EMF is proportional to the speed of the motor.
4. Differentiation of types of DC motor:
a. Schematic Diagram: There are mainly two types of DC motors:
i) Separately excited DC motor, and
ii) Shunt DC motor. The schematic diagrams for both types of DC motors are as follows:
b. Voltage equation: The voltage equation of a DC motor is given by V = Eb + IaRa, where V is the supply voltage, Eb is the back EMF, Ia is the armature current, and Ra is the armature resistance. c. Characteristics of the speed and torque of the motor: There are three types of DC motors based on the relationship between speed and torque:
i) Series DC motor,
ii) Shunt DC motor, and
iii) Compound DC motor.
5. Torque: Torque is the rotational force generated by a motor. It is the product of the force and the distance from the pivot point to the point of application of the force.
6. Different formulas involved in the operation of the DC Motor: Some of the important formulas used in the operation of a DC motor are: a. Voltage equation: V = Eb + IaRa b. Back EMF: Eb = KφN c. Torque: T = KφIa d. Power: P = VIa e. Efficiency: η = (Output power/Input power) x 100%.
7. Power stages absorbed by the DC motor: The power absorbed by a DC motor is divided into three stages:
a. Input stage: The input power is given to the motor by the supply voltage.
b. Output stage: The output power is the mechanical power produced by the motor.
c. Losses: The losses in the motor include copper losses, iron losses, and mechanical losses.
8. Capacity of DC Motor: 1HP = 746 watts
In conclusion, a DC motor converts electrical energy into mechanical energy, and it operates on the principles of the Faraday's Law of Electromagnetic Induction. The back EMF is produced in the DC motor when the rotor rotates. The types of DC motors are separately excited DC motor and shunt DC motor. Torque is the rotational force generated by a motor. The power absorbed by a DC motor is divided into three stages. Finally, 1HP is equal to 746 watts.
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The petrol engine works on 0 0 0 O Rankine cycle Otto cycle Diesel cycle
The petrol engine works on Otto cycle. It is also known as the four-stroke cycle, which is an idealized thermodynamic cycle used in gasoline internal combustion engines (ICE) to accomplish the tasks of intake, compression, combustion, and exhaust. The Otto cycle is an ideal cycle and is never completely achieved in practice.
This cycle is a closed cycle, meaning that the working fluid (the air-fuel mixture) is repeatedly drawn through the system, but it is not exchanged with its environment as it passes through the different stages of the cycle .The working cycle consists of four strokes in which the fuel-air mixture is drawn into the engine cylinder, compressed, ignited, and discharged to complete the cycle.
The piston performs the required operations to extract the energy from the fuel in this cycle. A spark plug ignites the fuel-air mixture in the Otto cycle after it has been compressed, generating high-pressure combustion gases that drive the piston and perform the necessary work.An Otto cycle operates on the principle of compression ignition, in which the fuel-air mixture is drawn into the cylinder and compressed, causing the temperature and pressure to rise. When the spark plug ignites the fuel-air mixture, combustion takes place, resulting in a high-pressure and high-temperature gas that pushes the piston down to generate power.
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Full AM is produced by a signal, Vm = 3.0 cos(2π X 10²)t + 1.0 cos(4 × 10²) t volts, modulating a carrier, vc 10.0 cos (2π x 104)t. Solve the followings: a. Show the resulting modulated signal and label the important parameters b. Show the frequency spectrum and measure bandwidth c. Power efficiency
In amplitude modulation (AM), a signal is used to modulate a carrier wave to transmit information.
What is the difference between digital and analog signals in communication systems?In this case, the signal is given as Vm = 3.0 cos(2π × 10²)t + 1.0 cos(4 × 10²)t volts, and the carrier is vc = 10.0 cos(2π × 10⁴)t volts.
The important parameters in the resulting modulated signal include the carrier frequency (10⁴ Hz), the amplitude of the carrier (10.0 volts), and the modulation index (3.0 and 1.0 for the two modulating signal components).
These parameters determine the shape and characteristics of the modulated signal.
To analyze the frequency spectrum and measure the bandwidth, we can use Fourier analysis.
The spectrum will consist of the carrier frequency and two sidebands at frequencies shifted from the carrier by the modulating frequencies (10² Hz and 4 × 10² Hz).
The bandwidth can be determined by considering the highest frequency component, which in this case is 4 × 10² Hz.
Overall, the given information allows us to analyze and understand the resulting modulated signal, its frequency spectrum, and the power efficiency of the modulation.
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With reference to a sketch, describe the difference between carbon capture and carbon avoidance.
Carbon capture and carbon avoidance are both methods used to reduce greenhouse gas emissions. Carbon capture is a process that involves capturing and storing carbon dioxide (CO2) emissions from industrial processes and power generation.
Carbon avoidance, on the other hand, involves avoiding the generation of CO2 emissions altogether.
Carbon capture is a method used to capture carbon dioxide emissions from industrial processes and power generation before they are released into the atmosphere. This can be done in a variety of ways, such as through the use of chemical absorption, adsorption, or membranes. The captured CO2 can then be transported and stored, typically in underground geological formations, deep saline aquifers, or depleted oil and gas reservoirs.
Carbon avoidance, on the other hand, involves avoiding the generation of CO2 emissions altogether. This can be done in several ways, including the use of renewable energy sources such as wind and solar power, increasing energy efficiency in buildings and transportation, and reducing waste and consumption. By avoiding the generation of CO2 emissions, the need for carbon capture and storage is reduced, and the overall carbon footprint is lowered.
Carbon capture and carbon avoidance are both important methods used to reduce greenhouse gas emissions. While both methods aim to reduce CO2 emissions, they differ in the way that they achieve this goal.Carbon capture involves capturing and storing CO2 emissions from industrial processes and power generation, while carbon avoidance involves avoiding the generation of CO2 emissions altogether.
Carbon capture is typically used to reduce emissions from large industrial sources that cannot be avoided, such as power plants and steel mills. Carbon avoidance, on the other hand, is focused on reducing emissions from transportation, buildings, and other sources that can be avoided through the use of renewable energy sources, increased efficiency, and reduced consumption.
While both methods have their advantages and disadvantages, they are both important tools in the fight against climate change. Carbon capture can help to reduce emissions from large industrial sources, while carbon avoidance can help to reduce emissions from a variety of sources. Ultimately, the most effective approach to reducing greenhouse gas emissions will likely involve a combination of both methods, as well as other strategies such as carbon offsetting and reforestation.
Carbon capture and carbon avoidance are two important methods used to reduce greenhouse gas emissions. Carbon capture involves capturing and storing CO2 emissions from industrial processes and power generation, while carbon avoidance involves avoiding the generation of CO2 emissions altogether. While both methods have their advantages and disadvantages, they are both important tools in the fight against climate change. The most effective approach to reducing greenhouse gas emissions will likely involve a combination of both methods, as well as other strategies such as carbon offsetting and reforestation.
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Give discussion and conclusion of the Series Resonance
Experiment.
The series resonance experiment has shown that an electrical circuit containing a capacitor and an inductor produces a resonant frequency that can be calculated by using the formula: ƒ = 1 / 2π√LC. In this experiment, a series LCR circuit was constructed by connecting an inductor, a capacitor, and a resistor in series with a function generator and an oscilloscope.
The aim of the series resonance experiment is to study the resonance phenomenon in an LCR circuit and to determine the resonant frequency, quality factor, and bandwidth of the circuit. The circuit's resonant frequency was determined by varying the frequency of the function generator until the voltage across the capacitor and inductor was at a maximum and the phase difference between them was zero. This frequency was found to be in agreement with the calculated resonant frequency using the above formula.The quality factor (Q) and bandwidth of the circuit were also determined experimentally. The quality factor was calculated as the ratio of the energy stored in the circuit to the energy dissipated per cycle.
The bandwidth was calculated as the difference between the frequencies at which the voltage across the capacitor and inductor was half the maximum voltage.The results of the experiment showed that the resonant frequency was dependent on the values of the inductor and capacitor and that the quality factor and bandwidth were dependent on the resistance of the circuit. The higher the resistance, the lower the quality factor and bandwidth of the circuit.In conclusion, the series resonance experiment is an important experiment that demonstrates the resonance phenomenon in an LCR circuit.
The experiment helps to determine the resonant frequency, quality factor, and bandwidth of the circuit. The results of the experiment showed that the resonant frequency was dependent on the values of the inductor and capacitor, while the quality factor and bandwidth were dependent on the resistance of the circuit.
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Calculate I, the moment of inertia of a uniform thin rod with unit mass p and length 12 units along the axis, about a perpendicular axis of rotation at the end of the rod located at the origin.Express your answer in terms of the total mass M.
The moment of inertia, I, of the uniform thin rod about a perpendicular axis of rotation at the end of the rod located at the origin is [tex](1/3)M(12)^2[/tex].
The moment of inertia, I, of an object represents its resistance to changes in rotational motion. In this case, we are calculating the moment of inertia of a uniform thin rod. The rod has a unit mass, p, and a length of 12 units along the axis. We want to find the moment of inertia about a perpendicular axis of rotation at the end of the rod, located at the origin.
To calculate the moment of inertia, we use the formula for a rod rotating about one end, which is given by [tex](1/3)ML^2[/tex], where M is the total mass of the rod and L is the length of the rod. In this case, the total mass M is equal to the mass per unit length, p, multiplied by the length of the rod, which is 12 units. Therefore, we can substitute M = pL into the formula and simplify it.
[tex]I = (1/3)M(12)^2[/tex]
[tex]= (1/3)(pL)(12)^2[/tex]
[tex]= (1/3)(p)(12^2)[/tex]
= (1/3)p(144)
= 48p
So, the moment of inertia, I, of the uniform thin rod about the perpendicular axis of rotation at the end of the rod located at the origin is 48p.
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Provide a stress analysis on an Orifice tube in a cars
AC system. explaining the stresses and load induced in the Orifice
tube. Provide figures and photos as well.
The orifice tube in a car's AC system experiences stresses and loads due to fluid pressure, thermal effects, and vibrations.
What are the main factors that contribute to the stresses and loads experienced by the orifice tube in a car's AC system?The orifice tube in a car's AC system is responsible for controlling the flow of refrigerant. It is typically a small, cylindrical tube with a small orifice or opening through which the high-pressure liquid refrigerant passes. During the operation of the AC system, several stresses and loads are induced on the orifice tube:
1. Fluid Pressure: The orifice tube experiences high fluid pressure as the refrigerant passes through the orifice. This pressure creates a load on the tube, which can cause deformation and stress concentration around the orifice.
2. Thermal Stresses: The orifice tube is exposed to temperature variations as the refrigerant undergoes phase changes from liquid to gas and vice versa. These temperature changes can result in thermal expansion and contraction of the tube, leading to thermal stresses.
3. Vibration and Fatigue: The AC system operates under dynamic conditions, and vibrations can be transmitted to the orifice tube. These vibrations, combined with the cyclic loading from the fluid pressure, can induce fatigue in the tube over time.
To analyze the stresses and loads on the orifice tube, various engineering techniques such as finite element analysis (FEA) can be used. FEA models can simulate the fluid flow, pressure distribution, and thermal effects on the tube. By applying appropriate boundary conditions and material properties, the stresses, deformations, and load distributions can be determined.
It is recommended to consult technical resources, research papers, or seek assistance from automotive experts to obtain detailed stress analysis and access figures and photos related to the specific orifice tube in a car's AC system.
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Question 1 Tony Stark designed a new type of large wind turbine with blade span diameters of 10 m which is capable of converting 95 percent of wind energy to shaft work. Four units of the wind turbines are connected to electric power generators with 50 percent efficiency, and are placed at an open area at a point of 200 m height on the Stark Tower, with steady winds of 10 m/s during a 24-hour period. Taking the air density as 1.25 kg/m?, 1) determine the maximum electric power generated by these wind turbines; and (8 marks) 11) determine the amount of revenue he generated by reselling the electricity to the electric utility company for a unit price of $0.11/kWh. (3 marks) [Total: 25 marks]
The maximum electric power generated is 273546.094 W. The amount of revenue generated is $2696075.086.
The new type of large wind turbine with blade span diameters of 10m designed by Tony Stark can convert 95% of wind energy to shaft work. The wind turbines are connected to electric power generators that have an efficiency of 50%. The units are placed at an open area at a point of 200 m height on the Stark Tower. During a 24-hour period, the steady winds are at 10 m/s. The air density is 1.25 kg/m3.1. Calculation of maximum electric power generated
P = 0.5 × density × A × v3 × CpWhereP = power
A = 0.25πd2 = 0.25π × 102 = 78.54 m2v = 10 m/s
Cp = 0.95
density = 1.25 kg/m3
Therefore, P = 0.5 × 1.25 × 78.54 × (10)3 × 0.95= 273546.094 W
The maximum electric power generated is 273546.094 W.2. Calculation of the amount of revenue generated
Revenue = P × t × c Where
P = 273546.094 Wt = 24 h/day × 365 day/year = 8760 h/yearc = 0.11 $/kWh
Therefore,Revenue = 273546.094 × 8760 × 0.11 = $2696075.086
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The force acting on a beam was measured 5 times under the same operating conditions. This process was repeated by 3 observersing of data. The means of these data sets were Mean 1-8, Mean 2- 9. Mean 3-2 The corresponding standard deviations were: 3.2, 2.1, and 2.5, respectively, Compute the Pooled Mean of the 3 data sets (Provide your answer using two decimal places).
Pooled Mean = [Sum of (Mean * Degrees of Freedom)] / [Total Degrees of Freedom]Now, let's find the degrees of freedom for each data set.
Degrees of Freedom = n - 1, where n is the number of observations for each data set. For our problem, n = 5 for each data set, so: Degrees of Freedom for Mean 1 = 5 - 1 = 4Degrees of Freedom for Mean 2 = 5 - 1 = 4Degrees of Freedom for Mean 3 = 5 - 1 = 4Total Degrees of Freedom = (Degrees of Freedom for Mean 1) + (Degrees of Freedom for Mean 2) + (Degrees of Freedom for Mean 3)= 4 + 4 + 4 = 12Next, we can substitute the given means and degrees of freedom in the formula:
Pooled Mean = [(8 * 4) + (9 * 4) + (2 * 4)] / 12= (32 + 36 + 8) / 12= 76 / 12= 6.33 (rounded to two decimal places)Therefore, the main answer is: Pooled Mean = 6.33. We have calculated the degrees of freedom for each data set and the total degrees of freedom, which are used in the formula to calculate the Pooled Mean.
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(a) Describe FOUR factors affecting the adhesive bonding performance. (12 marks) (b) There is an internal defect found in a 4 layers glass fibre sandwich composite. The upper skin of a sandwich structure was damaged and needs to be repaired. 11.5mm damage area is at the center of the 300mm x 300mm panel. With the aid of drawing, calculate and illustrate the area of each layer that need to be removed. Put your calculated answers in mm. (13 marks) Hints: The smallest area to be removed is 20mm in a circular shape. Assume the thickness of each layer is 0.8 mm.
The area of each layer that needs to be removed is as follows:
Layer 1: 161.85 mm2
Layer 2: 146.76 mm2
Layer 3: 129.48 mm2
Layer 4: 161.85 mm2
a) Four factors affecting the adhesive bonding performance are:
1. Surface preparation: Adhesive bonding performance can be adversely affected if the bonding surface is not clean or properly prepared.
Before bonding, the surface of the materials to be bonded must be free of grease, oil, dirt, and other contaminants.
2. Temperature and humidity: Adhesive bonding can be influenced by changes in temperature and humidity.
The bond strength of some adhesives is affected by the temperature and humidity.
3. Chemical compatibility: Adhesives should be chosen based on their compatibility with the materials being bonded.
It is important to ensure that the adhesive is chemically compatible with the substrate to which it will be applied
.4. Bonding time and pressure: The amount of time and pressure applied during the bonding process can have an impact on the adhesive's performance.
The pressure applied during bonding should be sufficient to ensure that the adhesive makes good contact with the substrate.
The bonding time should be sufficient to allow the adhesive to cure properly.
Surface preparation, temperature and humidity, chemical compatibility, and bonding time and pressure are four factors that affect the adhesive bonding performance.
Conclusion: For adhesive bonding to be effective, these four factors must be taken into consideration. The bonding surface must be properly prepared and free of contaminants, the temperature and humidity should be controlled, and the adhesive should be compatible with the substrate.
Additionally, the bonding time and pressure should be appropriate.
b)The first step in calculating the area of each layer that needs to be removed is to calculate the total area of the damage.
The total area of the damage is the diameter of the circular damage area multiplied by pi (3.14) and divided by 4, which gives us the area of the damage as 103.58 mm2. Since each layer is 0.8mm thick, we can divide the total area by 0.8 to determine the area of each layer that needs to be removed.
The area of each layer that needs to be removed is as follows:
Layer 1: 129.48 mm2
Layer 2: 118.71 mm2
Layer 3: 103.58 mm2
Layer 4: 129.48 mm2
The smallest area to be removed is 20mm in a circular shape, which means that the area of each layer to be removed should be at least 25.12 mm2.
Therefore, the area of each layer that needs to be removed is as follows:
Layer 1: 161.85 mm2
Layer 2: 146.76 mm2
Layer 3: 129.48 mm2
Layer 4: 161.85 mm2
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A closed, rigid tank with a volume of 0.3 m 3
initially contains refrigerant R−134a at an absolute pressure of 6 bar and specific volume of 0.041389 m 3
/kg (State 1). The refrigerant is stirred with a paddle wheel device and the tank is cooled at the same time. The paddle wheel performs 30000 J of work on the refrigerant. The refrigerant temperature drops to 8 ∘
C (State 2 ) due to the given energy interactions. a) Determine the temperature at the initial state, ∘
C (10pts) b) What is the final pressure of R-134a in the tank, bar (5pts) c) Determine the heat transfer during the process, kJ (10pts) d) Find the quality at the final state, % (10pts) - Only numbers will be entered into the boxes. No text entry. - Be careful with the units and the signs of the energy terms. - When you enter the values use DOT as decimal separator. For example: 0.10 or 5.75 e) Explain the assumptions made during the solution of the previous parts. Show the process on P−v diagram relative to the vapor dome and the lines of constant temperature for the two states. Label the axes and two states and indicate the process direction with arrow. (15 pts)
In this problem, a closed, rigid tank initially contains refrigerant R-134a at a given pressure and specific volume.
(a) To determine the temperature at the initial state (State 1), we need to use the given specific volume and the refrigerant's properties. The temperature can be calculated using the ideal gas law.
(b) The final pressure of R-134a in the tank (State 2) can be determined using the ideal gas law and the given final temperature.
(c) The heat transfer during the process can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat transfer minus the work done on the system.
(d) The quality at the final state can be determined using the property tables or charts for R-134a by comparing the final temperature and pressure to the saturation values.
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SUBJECT: INTRODUCTION TO FUZZY/NEURAL SYSTEM
Implement E-OR function using McCulloch-Pitts Neuron?
You have implemented the E-OR function using a McCulloch-Pitts neuron.
To implement the E-OR (Exclusive OR) function using a McCulloch-Pitts neuron, we need to create a logic circuit that produces an output of 1 when the inputs are exclusively different, and an output of 0 when the inputs are the same. Here's how you can implement it:
Define the inputs: Let's assume we have two inputs, A and B.
Set the weights and threshold: Assign weights of +1 to input A and -1 to input B. Set the threshold to 0.
Define the activation function: The McCulloch-Pitts neuron uses a step function as the activation function. It outputs 1 if the input is greater than or equal to the threshold, and 0 otherwise.
Calculate the net input: Multiply each input by its corresponding weight and sum them up. Let's call this value net_input.
net_input = (A * 1) + (B * -1)
Apply the activation function: Compare the net input to the threshold. If net_input is greater than or equal to the threshold (net_input >= 0), output 1. Otherwise, output 0.
Output = 1 if (net_input >= 0), else 0.
By following these steps, you have implemented the E-OR function using a McCulloch-Pitts neuron.
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Explain the procedure on labeling components in an Exploded view on an assembly drawing. Provide an example. 14. Describe the procedure to create a Design Table. 15. True or False. You cannot display different configurations in the same drawing. Explain your answer. 16. True or False. The Part Number is only entered in the Bill of Materials. Explain your answer. 17. There are hundreds of options in the Document Properties, Drawings and Annotations toolbars. How would you locate additional information on these options and tools? 18. Describe the View Palette 19. Describe the procedure to insert a Center of Mass point into a drawing either for an assembly or part.
To label components in an exploded view, each part is identified with a number or letter next to it, while displaying different configurations can be done using the Configuration Publisher tool. Additional information on SOLIDWORKS options and tools can be found in the Help menu
14. To label components in an exploded view, each part is identified with a number or letter next to it. This label corresponds to a part description in a parts list or bill of materials. For example, a bolt may be labeled "1" with a corresponding part description in the bill of materials.
15. False. You can display different configurations in the same drawing using the Configuration Publisher tool in SOLIDWORKS. This allows you to create multiple views of an assembly in different configurations on the same drawing.
16. False. The Part Number can also be entered in the custom properties of a part or assembly. This information can then be used to automatically populate the bill of materials.
17. Additional information on the options and tools in SOLIDWORKS can be found in the Help menu or online through resources such as the SOLIDWORKS Knowledge Base, forums, and training materials.
18. The View Palette is a tool in SOLIDWORKS that allows you to quickly access and manage different views of a model or assembly. It provides a visual thumbnail of each view, making it easy to identify and select the desired view.
19. To insert a Center of Mass point in a drawing, first enable the Center of Mass feature in the Mass Properties dialog box. Then, insert the Center of Mass point using the Insert > Model Items command. This will place a point at the Center of Mass location in the drawing.
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Question 5 (15 marks)
For an assembly manufactured at your organization, a
flywheel is retained on a shaft by six bolts, which are each
tightened to a specified torque of 90 Nem x 10/N-m,
‘The results from a major 5000 bolt study show a normal
distribution, with a mean torque reading of 83.90 N-m, and a
standard deviation of 1.41 Nm.
2. Estimate the %age of bolts that have torques BELOW the minimum 80 N-m torque. (3)
b. Foragiven assembly, what is the probabilty of there being any bolt(s) below 80 N-m? (3)
¢. Foragiven assembly, what isthe probability of zero bolts below 80 N-m? (2)
Question 5 (continued)
4. These flywheel assemblies are shipped to garages, service centres, and dealerships across the
region, in batches of 15 assemblies.
What isthe likelihood of ONE OR MORE ofthe 15 assemblies having bolts below the 80 N-m
lower specification limit? (3 marks)
. Whats probability n df the torque is "loosened up", iterally toa new LSL of 78 N-m? (4 marks)
The answer to the first part, The standard deviation is 1.41 N-m.
How to find?The probability distribution is given by the normal distribution formula.
z=(80-83.9)/1.41
=-2.77.
The percentage of bolts that have torques below the minimum 80 N-m torque is:
P(z < -2.77) = 0.0028
= 0.28%.
Thus, there is only 0.28% of bolts that have torques below the minimum 80 N-m torque.
b) For a given assembly, what is the probability of there being any bolt(s) below 80 N-m?
The probability of there being any bolt(s) below 80 N-m is given by:
P(X < 80)P(X < 80)
= P(Z < -2.77)
= 0.0028
= 0.28%.
Thus, there is only a 0.28% probability of having bolts below 80 N-m in a given assembly.
c) For a given assembly, what is the probability of zero bolts below 80 N-m?The probability of zero bolts below 80 N-m in a given assembly is given by:
P(X ≥ 80)P(X ≥ 80) = P(Z ≥ -2.77)
= 1 - 0.0028
= 0.9972
= 99.72%.
Thus, there is a 99.72% probability of zero bolts below 80 N-m in a given assembly.
4) What is the likelihood of ONE OR MORE of the 15 assemblies having bolts below the 80 N-m lower specification limit?
The probability of having one or more of the 15 assemblies with bolts below the 80 N-m lower specification limit is:
P(X ≥ 1) =
1 - P(X = 0)
= 1 - 0.9972¹⁵
= 0.0418
= 4.18%.
Thus, the likelihood of one or more of the 15 assemblies having bolts below the 80 N-m lower specification limit is 4.18%.
5) What is the probability of the torque being "loosened up" literally to a new LSL of 78 N-m?
The probability of the torque being loosened up to a new LSL of 78 N-m is:
P(X < 78)P(X < 78)
= P(Z < -5.74)
= 0.0000
= 0%.
Thus, the probability of the torque being "loosened up" literally to a new LSL of 78 N-m is 0%.
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The burning of a hydrocarbon fuel (CHx)n in an automotive engine results in a dry exhaust gas analysis, percentage by volume, of: 11 % CO2, 0.5 % CO, 2 % CH4, 1.5 % H2, 6 % O2 and 79 % N2. Write the combustion equation and find (a) the actual air-fuel ratio; (b) the percent excess or deficient air used; (c) the volume of the products (at 1 300 C and 100 kPaa) in cubic meter per kilogram of fuel.
The design process for developing a new product typically involves several steps, including market research, ideation, concept development, prototyping, testing, and refinement.
What are the steps involved in the design process for developing a new product?(a) The actual air-fuel ratio is determined by the combustion equation and cannot be provided without additional information.
(b) The percent excess or deficient air used cannot be determined without knowing the actual air-fuel ratio and the stoichiometric air-fuel ratio.
(c) The volume of the products per kilogram of fuel cannot be calculated without additional information, such as the molar mass of the fuel and the temperature and pressure conditions in the exhaust gas mixture.
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