Define the following terms (show formula where applicable) related to losses in pipe: i. Major losses
Major losses refer to the pressure losses that occur due to friction in a pipe or conduit. These losses are primarily caused by the viscous effects of the fluid flowing through the pipe. Major losses are influenced by factors such as the pipe length, diameter, roughness, and the flow rate. The major loss can be calculated using the Darcy-Weisbach formula.
ii. Minor losses:
Minor losses, also known as local losses or secondary losses, are pressure losses that occur at specific locations in a piping system, such as fittings, valves, bends, expansions, contractions, and other flow disturbances. These losses are caused by changes in flow direction, flow separation, turbulence, and other factors. Minor losses are typically expressed as a loss coefficient (K) multiplied by the dynamic pressure of the fluid. The total minor loss in a system can be calculated by summing the individual minor losses.
iii. Darcy-Weisbach formula:
The Darcy-Weisbach formula is an empirical equation used to calculate the major losses (pressure losses due to friction) in a pipe. It relates the pressure loss (ΔP) to the fluid flow rate (Q), pipe length (L), pipe diameter (D), fluid density (ρ), and a friction factor (f). The formula is as follows:
ΔP = f * (L / D) * (ρ * (Q^2) / 2)
The friction factor (f) depends on the pipe roughness, Reynolds number, and flow regime. It can be determined using charts, tables, or empirical correlations.
iv. Hagen-Poiseuille equation for laminar flow:
The Hagen-Poiseuille equation describes the flow of a viscous, incompressible fluid through a cylindrical pipe under laminar flow conditions. It relates the volume flow rate (Q) to the pressure difference (ΔP), pipe length (L), pipe radius (r), fluid viscosity (μ), and pipe resistance. The equation is as follows:
Q = (π * ΔP * r^4) / (8 * μ * L)
The Hagen-Poiseuille equation applies only to laminar flow, where the flow velocity is low, and the fluid flows in smooth, straight pipes. It does not account for the effects of turbulence.
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A square key is to be used in 40 mm diameter shaft and that will developed a 2 KN-m torque. If bearing stress of the key is 400 Mpa, determine the cross sectional dimension of square key to be used if key length is 30 mm. Answer: D
A. 324.80 mm2
B. 246.80 mm2
C. 446.80 mm2
D. 277.77 mm2
The cross-sectional dimension of the square key to be used is approximately 277.77 mm². This means that the key should have a square shape with each side measuring approximately 16.68 mm (sqrt(277.77)).
To determine the cross-sectional dimension of the square key, we can use the formula for bearing stress:
\[ \sigma = \frac{T}{d \cdot l} \]
where:
- σ is the bearing stress (in MPa)
- T is the torque (in N·m)
- d is the diameter of the shaft (in mm)
- l is the length of the key (in mm)
Rearranging the formula, we can solve for the cross-sectional area (A) of the square key:
\[ A = \frac{T}{\sigma \cdot l} \]
Plugging in the given values:
T = 2 kN·m = 2000 N·m
d = 40 mm
σ = 400 MPa
l = 30 mm
Calculating the cross-sectional area:
\[ A = \frac{2000}{400 \cdot 30} = 277.77 mm².
Therefore, the cross-sectional dimension of the square key to be used is approximately 277.77 mm². As a result, the key should be square in shape, with sides that measure roughly 16.68 mm (sqrt(277.77)).
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You have probably noticed warning signs on the highways stating that bridges may be icy even when the roads are not. Explain how this can happen. If the distance between the sun and the earth was the half of what it is L=0.5 x 1.496 x 1011 m, what would the solar constant be? The sun is a nearly spherical body that has a diameter of D = 1.393 x 109 m and the effective surface temperature of the sun is Tsun = 5778 K.
Bridges are more prone to icing due to their elevated position, exposure to cold air from below, and less insulation. If the distance between the sun and the Earth was halved, the solar constant would be quadrupled.
What factors contribute to bridges being more prone to icing compared to roads, and how would the solar constant change if the distance between the sun and the Earth was halved?Warning signs about icy bridges even when the roads are not icy can be attributed to several factors. Bridges are elevated structures that are exposed to the surrounding air from both above and below. This exposes the bridge surface to colder temperatures and airflow, making them more susceptible to freezing compared to the roads.
Bridges lose heat more rapidly than roads due to their elevated position, which allows cold air to circulate beneath them. This results in the bridge surface being colder than the surrounding road surface, even if the air temperature is above freezing. Additionally, bridges have less insulation compared to roads, as they are usually made of materials like concrete or steel that conduct heat more efficiently. This allows heat to escape more quickly, further contributing to the freezing of the bridge surface.
Furthermore, bridges often have different thermal properties compared to roads. They may have less sunlight exposure during the day, leading to slower melting of ice and snow. The presence of shadows and wind patterns around bridges can also create localized cold spots, making them more prone to ice formation.
Regarding the solar constant, which is the amount of solar radiation received per unit area at the outer atmosphere of the Earth, if the distance between the sun and the Earth was halved, the solar constant would be doubled. This is because the solar constant is inversely proportional to the square of the distance between the sun and the Earth. Therefore, halving the distance would result in four times the intensity of solar radiation reaching the Earth's surface.
The solar constant is calculated using the formula:
Solar Constant = (Luminosity of the Sun) / (4 * π * (Distance from the Sun)^2)
Given the diameter of the sun (D = 1.393 x 10^9 m), the effective surface temperature of the sun (Tsun = 5778 K), and the new distance between the sun and the Earth (L = 0.5 x 1.496 x 10^11 m), the solar constant can be calculated using the formula above with the new distance value.
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Now we're going to design another "equalizer". Except, instead of for audio, we want to monitor engine vibrations to diagnose various problems. Suppose we have a four-cylinder engine with a single camshaft. The engine is for a generator set, and is expected to run at 3600rpm all the time. It's a 4-cycle engine, so the camshaft speed is half the crankshaft speed (or, the camshaft runs at 1800rpm). We want to measure the following things... • Vibrations caused by crankshaft imbalance. • Vibrations caused by camshaft imbalance. • Vibrations caused by the exhaust wave. The exhaust wave pulses whenever an exhaust valve opens. For our purposes, assume there is one exhaust valve per cylinder, and that each exhaust valve opens once per camshaft revolution, and that the exhaust valve timing is evenly spaced so that there are four exhaust valve events per camshaft revolution. 1. Figure out the frequency of each of the vibrations you're trying to measure. 2. Set the cutoff frequencies for each of your bandpass filters.
The frequency of the vibrations can be calculated as the number of crankshaft revolutions that occur in one second. Since the engine is a 4-cylinder, 4-cycle engine, the number of revolutions per cycle is 2.
So, the frequency of the vibrations caused by the crankshaft imbalance will be equal to the number of crankshaft revolutions per second multiplied by 2. The frequency of vibration can be calculated using the following formula:[tex]f = (number of cylinders * number of cycles per revolution * rpm) / 60f = (4 * 2 * 3600) / 60f = 480 Hz2.[/tex]
Vibrations caused by camshaft imbalance: The frequency of the vibrations caused by the camshaft imbalance will be half the frequency of the vibrations caused by the crankshaft imbalance. This is because the camshaft speed is half the crankshaft speed. Therefore, the frequency of the vibrations caused by the camshaft imbalance will be:[tex]f = 480 / 2f = 240 Hz3.[/tex]
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The minimum pressure on an object moving horizontally in water (Ttemperatu at10 degree centrigrade) at (x+5) mm/s (where x is the last two digits of your student 10) at a depth of 1 m is 80 kPa (absolute). Calculate the velocity that will initiate cavitation. Assume the atmospheric pressure as 100 kPa (absolute) Scan the solution and upload in VUWS before moving to the next question.
Given data: Minimum pressure on an object = 80 kPa (absolute)Velocity of an object = (x+5) mm/sDepth of an object = 1mTemperature = 10°CAtmospheric pressure = 100 kPa (absolute)
We know that the minimum pressure to initiate cavitation is given as:pc = pa - (pv)²/(2ρ)Where, pa = Atmospheric pressurepv = Vapour pressure of liquidρ = Density of liquidNow, the vapour pressure of water at 10°C is 1.223 kPa (absolute) and density of water at this temperature is 999.7 kg/m³.Substituting the values in the above equation, we get:80 = 100 - (pv)²/(2×999.7) => (pv)² = 39.706
Now, the velocity that will initiate cavitation is given as:pv = 0.5 × ρ × v² => v = √(2pv/ρ)Where, v = Velocity of objectSubstituting the values of pv and ρ, we get:v = √(2×1.223/999.7) => v = 1.110 m/sTherefore, the velocity that will initiate cavitation is 1.110 m/s.
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A basketball has a 300-mm outer diameter and a 3-mm wall thickness. It is inflated to a 120 kPa gauge pressure. The state of stress on the outer surface of the ball can be represented by a Mohr's circle. Which of the following options is true? Choose only one option. a The Mohr's circle representing the state of stress on the outer surface of the ball is a sphere with the same diameter to the basketball. b The Mohr's circle representing the state of stress on the outer surface of the ball is a point (i.e. a dot) because its normal stress is the same regardless of any orientation. c The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot. The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses are having the same magnitude but opposite sign. This is because the ball has spherical symmetry. d The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot. The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses do not have the same magnitude but they have the same positive sign. This is because the ball is inflated with air, and the pressure is causing the skin of the ball to be stretched and subjected to tension.
The main answer for the question is option (c) The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot.
The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses are having the same magnitude but opposite sign. This is because the ball has spherical symmetry. Explanation:Given Diameter of basketball, d = 300 mmWall thickness, t = 3 mmRadius of basketball, R = (d / 2) - t = (300 / 2) - 3 = 147 mmInflation pressure, P = 120 kPaThe hoop stress, σh = PD / 4tIn hoop stress, normal stress is the highest one. It is equal to the hoop stress.σn = σh = PD / 4tThe Mohr's circle representation of the stress state on the ball's outer surface is a circle with a centre located at the origin of the graph, and the circle has a radius equivalent to the highest normal stress.
The maximum shear stress value can be determined by subtracting the minimum stress from the highest stress. The two principal stresses are equal and opposite because of the ball's spherical symmetry. Thus, option (c) is correct.
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Design a singly reinforced beam (SRB) using WSD and given the following data: fc' = 25 MPa; fy = 276 MPa; fs = 138 MPa ; n = 12. Use 28 mm diameter main bars and 12 mm diameter stirrups. Solve only the following: 1. k, j, (don't round-off) and R (rounded to 3 decimal places) 2. Designing maximum moment due to applied loads.
3. Trial b.d, and t. (Round - off d value to next whole higher number that is divisible by 25.) 4. Weight of the beam (2 decimal places).
5. Maximum moment in addition to weight of the beam. 6. Number of 28 mm diameter main bars. 7. Check for shear 8. Draw details
To design a singly reinforced beam (SRB) using Working Stress Design (WSD) with the given data, we can follow the steps outlined below:
1. Determine k, j, and R:
k is the lever arm factor, given by k = 0.85.j is the depth factor, given by j = 0.90.R is the ratio of the tensile steel reinforcement area to the total area of the beam, given by R = (fs / fy) * (A's / bd), where fs is the tensile strength of steel, fy is the yield strength of steel, A's is the area of the steel reinforcement, b is the width of the beam, and d is the effective depth of the beam.2. Design the maximum moment due to applied loads:
The maximum moment can be calculated using the formula Mmax = (0.85 * fy * A's * (d - 0.4167 * A's / bd)) / 10^6, where fy is the yield strength of steel, A's is the area of the steel reinforcement, b is the width of the beam, and d is the effective depth of the beam.
3. Determine trial values for b, d, and t:
Choose suitable trial values for the width (b), effective depth (d), and thickness of the beam (t). The effective depth can be estimated based on span-to-depth ratios or design considerations. Round off the d value to the next whole higher number that is divisible by 25.
4. Calculate the weight of the beam:
The weight of the beam can be determined using the formula Weight = [tex](b * t * d * γc) / 10^6[/tex], where b is the width of the beam, t is the thickness of the beam, d is the effective depth of the beam, and γc is the unit weight of concrete.
5. Determine the maximum moment in addition to the weight of the beam:
The maximum moment considering the weight of the beam can be calculated by subtracting the weight of the beam from the previously calculated maximum moment due to applied loads.
6. Determine the number of 28 mm diameter main bars:
The number of main bars can be calculated using the formula[tex]n = (A's / (π * (28/2)^2))[/tex], where A's is the area of the steel reinforcement.
7. Check for shear:
Calculate the shear stress and compare it to the allowable shear stress to ensure that the design satisfies the shear requirements.
8. Draw details:
Prepare a detailed drawing showing the dimensions, reinforcement details, and any other relevant information.
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Consider a substance that boils at -34°C (negative thirty four degrees Celsius) at 98 kPa. At that temperature and pressure, one kg of liquid occupies 0.0015 m³ and one kg of vapor occupies 1.16 m². At 80 kPa, this stuff boils at -38°C (negative thirty eight degrees Celsius). Using just this information: a. Estimate the enthalpy of vaporization of this substance at 98 kPa. (Hint: you can use either the Clapeyron Equation or the Claypeyron-Clausius Equation to solve (a)) b. Estimate the molar mass of the substance.
a. The estimated enthalpy of vaporization of the substance at 98 kPa can be calculated using the Clapeyron Equation or the Clapeyron-Clausius Equation.
b. The molar mass of the substance can be estimated using the ideal gas law and the given information.
a. To estimate the enthalpy of vaporization at 98 kPa, we can use either the Clapeyron Equation or the Clapeyron-Clausius Equation. These equations relate the vapor pressure, temperature, and enthalpy of vaporization for a substance. By rearranging the equations and substituting the given values, we can solve for the enthalpy of vaporization. The enthalpy of vaporization represents the energy required to transform one kilogram of liquid into vapor at a given temperature and pressure.
b. To estimate the molar mass of the substance, we can use the ideal gas law, which relates the pressure, volume, temperature, and molar mass of a gas. Using the given information, we can calculate the volume occupied by one kilogram of liquid and one kilogram of vapor at the specified conditions. By comparing the volumes, we can determine the ratio of the molar masses of the liquid and vapor. Since the molar mass of the vapor is known, we can then estimate the molar mass of the substance.
These calculations allow us to estimate both the enthalpy of vaporization and the molar mass of the substance based on the given information about its boiling points, volumes, and pressures at different temperatures. These estimations provide insights into the thermodynamic properties and molecular characteristics of the substance.
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Fick's first law gives the expression of diffusion flux (l) for a steady concentration gradient (Δc/ Δx) as: J=-D Δc/ Δx
Comparing the diffusion problem with electrical transport analogue; explain why the heat treatment process in materials processing has to be at high temperatures.
Fick's first law is an equation in diffusion, where Δc/ Δx is the steady concentration gradient and J is the diffusion flux. The equation is J=-D Δc/ Δx. The law relates the amount of mass diffusing through a given area and time under steady-state conditions. Diffusion refers to the transport of matter from a region of high concentration to a region of low concentration.
The driving force for diffusion is the concentration gradient. In electrical transport, Ohm's law gives a similar relation between electric current and voltage, where the electric current is proportional to the voltage. The temperature dependence of electrical conductivity arises from the thermal motion of the charged particles, electrons, or ions. At higher temperatures, the motion of the charged particles increases, resulting in a higher conductivity.
Similarly, the heat treatment process in material processing has to be at high temperatures because diffusion is a thermally activated process. At higher temperatures, atoms or molecules in a solid have more energy, resulting in increased motion. The increased motion, in turn, increases the rate of diffusion. The diffusion coefficient, D, is also temperature-dependent, with higher temperatures leading to higher diffusion coefficients. Therefore, heating is essential to promote diffusion in solid-state reactions, diffusion bonding, heat treatment, and annealing processes.
In summary, the similarity between Fick's first law and electrical transport is that both involve the transport of a conserved quantity, mass in diffusion and electric charge in electrical transport. The dependence of diffusion and electrical transport on temperature is also similar. Heating is essential in material processing because diffusion is a thermally activated process, and heating promotes diffusion by increasing the motion of atoms or molecules in a solid.
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The open-loop transfer function of a unit-negative-feedback system has the form of
G(s)H(s) = 1 / s(s+1).
Please determine the following transient specifications when the reference input is a unit step function:
(1) Percentage overshoot σ%;
(2) Peak time tp;
(3) 2% Settling time t.
For the given open-loop transfer function 1 / (s(s+1)), the transient specifications when the reference input is a unit step function can be determined by calculating the percentage overshoot, peak time, and 2% settling time using appropriate formulas for a second-order system.
What is the percentage overshoot?To determine the transient specifications for the given open-loop transfer function G(s)H(s) = 1 / (s(s+1)) with a unit step reference input, we need to analyze the corresponding closed-loop system.
1) Percentage overshoot (σ%):
The percentage overshoot is a measure of how much the response exceeds the final steady-state value. For a second-order system like this, the percentage overshoot can be approximated using the formula: σ% ≈ exp((-ζπ) / √(1-ζ^2)) * 100, where ζ is the damping ratio. In this case, ζ = 1 / (2√2), so substituting this value into the formula will give the percentage overshoot.
2) Peak time (tp):
The peak time is the time it takes for the response to reach its maximum value. For a second-order system, the peak time can be approximated using the formula: tp ≈ π / (ωd√(1-ζ^2)), where ωd is the undamped natural frequency. In this case, ωd = 1, so substituting this value into the formula will give the peak time.
3) 2% settling time (ts):
The settling time is the time it takes for the response to reach and stay within 2% of the final steady-state value. For a second-order system, the settling time can be approximated using the formula: ts ≈ 4 / (ζωn), where ωn is the natural frequency. In this case, ωn = 1, so substituting this value into the formula will give the 2% settling time.
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A Wheatstone bridge requires a change of 7 ohm in the unknown arm of the bridge to produce a deflection of three millimeter at the galvanometer scale. Determine the sensitivity and the deflection factor. [E 2.1]
A Wheatstone bridge is a device used for measuring the resistance of an unknown electrical conductor by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
This is accomplished by adjusting the value of a third leg of the circuit until no current flows through the galvanometer, which is connected between the two sides of the bridge that are not the unknown resistance. The galvanometer is a sensitive device that detects small differences in electrical potential.
A change of 7 ohm in the unknown arm of the bridge produces a deflection of three millimeter at the galvanometer scale. The sensitivity of a Wheatstone bridge is defined as the change in resistance required to produce a full-scale deflection of the galvanometer.
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From the technical literature and/or open sources, present the RCS of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation).
The radar cross section (RCS) of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation) can be found from the technical literature and/or open sources.
A trihedral reflector is a corner reflector that consists of three mutually perpendicular planes.
Reflectivity is the measure of a surface's capability to reflect electromagnetic waves.
The RCS is a scalar quantity that relates to the ratio of the power per unit area scattered in a specific direction to the strength of an incident electromagnetic wave’s electric field.
The RCS formula is given by:
[tex]$$ RCS = {{4πA}\over{\lambda^2}}$$[/tex]
Where A is the projected surface area of the target,
λ is the wavelength of the incident wave,
RCS is measured in square meters.
In the case of a trihedral reflector, the reflectivity is the same for both azimuth and elevation angles and is given by the following equation:
[tex]$$ RCS = {{16A^2}\over{\lambda^2}}$$[/tex]
Where A is the surface area of the trihedral reflector.
RCS varies with the incident angle, and the equation above is used to compute the reflectivity for all incident angles.
Therefore, it can be concluded that the RCS of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation) can be determined using the RCS formula and is given by the equation :
[tex]$$ RCS = {{16A^2}\over{\lambda^2}}$$.[/tex]
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A steel block [E = 29 x 103 ksi and v = 0.33] has initial side lengths all equal to 56 inches. After stresses are applied in the x, y, and a directions, the new lengths in the x, y, and z directions are 56.06 in., 56.10 in., and 55.95 in., respectively. Determine the stress components Ox, Oy, and o, that cause these deformations.
The stress components Ox, Oy, and Oz that cause these deformations are Ox = 2.07 ksi, Oy = 3.59 ksi, and Oz = -2.06 ksi, respectively.
Given information:
Young's modulus of elasticity, E = 29 x 103 ksi
Poisson's ratio, ν = 0.33
Initial length of the block, a = b = c = 56 inches
Change in the length in the x-direction, ΔLx = 0.06 inches
Change in the length in the y-direction, ΔLy = 0.10 inches
Change in the length in the z-direction, ΔLz = -0.05 inches
To determine the stress components Ox, Oy, and Oz that cause these deformations, we'll use the following equations:ΔLx = aOx / E (1 - ν)ΔLy = bOy / E (1 - ν)ΔLz = cOz / E (1 - ν)
where, ΔLx, ΔLy, and ΔLz are the changes in the length of the block in the x, y, and z directions, respectively.
ΔLx = 0.06 in.= a
Ox / E (1 - ν)56.06 - 56 = 56
Ox / (29 x 103)(1 - 0.33)
Ox = 2.07 ksi
ΔLy = 0.10 in.= b
Oy / E (1 - ν)56.10 - 56 = 56
Oy / (29 x 103)(1 - 0.33)
Oy = 3.59 ksi
ΔLz = -0.05 in.= c
Oz / E (1 - ν)55.95 - 56 = 56
Oz / (29 x 103)(1 - 0.33)
Oz = -2.06 ksi
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Determine the weight in newton's of a woman whose weight in pounds is 130. Also, find her mass in slugs and in kilograms. Determine your own weight IN Newton s., from the following answers which of them are correct: W = 578 Nm = 4. 04 slugs and m = 58. 9 kg W = 578 Nm = 4. 04 slugs and m = 68.9 kg W= 578 N, m = 8. 04 slugs and m = 78. 9 kg W= 578 N, m = 8. 04 slugs and m = 48. 9 kg
Out of the given options, the correct answer is: W = 578 N, m = 8.04 slugs and m = 78.9 kg
Given, Weight of the woman in pounds = 130. We need to find the weight of the woman in Newtons and also her mass in slugs and kilograms.
Weight in Newtons: We know that, 1 pound (lb) = 4.45 Newton (N)
Weight of the woman in Newtons = 130 lb × 4.45 N/lb = 578.5 N
Thus, the weight of the woman is 578.5 N.
Mass in Slugs: We know that, 1 slug = 14.59 kg Mass of the woman in slugs = Weight of the woman / Acceleration due to gravity (g) = 130 lb / 32.17 ft/s² x 12 in/ft x 1 slug / 14.59 lb = 4.04 slugs
Thus, the mass of the woman is 4.04 slugs.
Mass in Kilograms: We know that, 1 kg = 2.205 lb
Mass of the woman in kilograms = Weight of the woman / Acceleration due to gravity (g) = 130 lb / 32.17 ft/s² x 12 in/ft x 0.0254 m/in x 1 kg / 2.205 lb = 58.9 kg
Thus, the mass of the woman is 58.9 kg.
My weight in Newtons: We know that, 1 kg = 9.81 NMy weight is 65 kg
Weight in Newtons = 65 kg × 9.81 N/kg = 637.65 N
Thus, my weight is 637.65 N. Out of the given options, the correct answer is: W = 578 N, m = 8.04 slugs and m = 78.9 kg
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A turbine enters steam at 4000 kPa, 500 °C, 200 m/s and an outlet corresponding to saturated steam at 175 kPa and a speed of 120 m/s. If the mass flow is 2000 kg/min, and the power output is 15000 kW. Determine (a) the magnitude of the heat transferred. (b) Draw this process on the P-v diagram. (place the saturation lines)
A turbine enters steam at 4000 kPa, 500°C, 200 m/s and an outlet corresponding to saturated steam at 175 kPa and a speed of 120 m/s. If the mass flow is 2000 kg/min, and the power output is 15000 kW, we can determine
The magnitude of the heat transferred In order to calculate the magnitude of the heat transferred, we need to find the difference in enthalpy at the inlet and outlet of the turbine using the formula: Q = (m × (h2 - h1))WhereQ is the magnitude of heat transferred m is the mass flowh1 is the enthalpy of steam at the turbine inleth2 is the enthalpy of steam at the turbine outlet
We can calculate the enthalpy values using steam tables at the given pressures and temperatures. We get:
[tex]h1 = 3485.7 kJ/kgh2 = 2534.2 kJ/kg[/tex]Now, we can substitute the values to find the magnitude of heat transferred:
[tex]Q = (2000 kg/min × (2534.2 - 3485.7) kJ/kg/min) = -1.903 × 10^7 kJ/min[/tex]
Therefore, the magnitude of heat transferred is -1.903 × 10^7 kJ/min.
Initially, the steam enters the turbine at state 1 and undergoes an adiabatic (isentropic) expansion to state 2, corresponding to saturated steam at 175 kPa. This process is represented by the blue line on the diagram. The area under the curve represents the work output of the turbine, which is equal to 15000 kW in this case.
The saturation lines are represented by the red lines.
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A single start square threaded power screw is 50mm in diameter with a pitch of 8mm. The coefficient of friction is 0.08 for the collar and the threads. The frictional diameter of the collar is 1.25 times the major diameter of the screw. Determine the maximum load that can be borne by the power screw if the factor of safety of the power screw using von Mises failure theory is to be 2. The yield stress of the material of the screw is 240MPa.
Problem 3 A single start square threaded power screw is 50mm in diameter with a pitch of 8mm. The coefficient of friction is 0.08 for the collar and the threads. The frictional diameter of the collar is 1.25 times the major diameter of the screw. Determine the maximum load that can be borne by the power screw if the factor of safety of the power screw using von Mises failure theory is to be 2. The yield stress of the material of the screw is 240MPa.
A single square-thread screw is a type of screw with a square-shaped thread profile. It is used to convert rotational motion into linear motion or vice versa with high efficiency and load-bearing capabilities.
To determine the maximum load that can be borne by the power screw, we can follow these steps:
Calculate the major diameter (D) of the screw:
The major diameter is the outer diameter of the screw. In this case, it is given as 50mm.
Calculate the frictional diameter (Df) of the collar:
The frictional diameter of the collar is 1.25 times the major diameter of the screw.
Df = 1.25 * D
Calculate the mean diameter (dm) of the screw:
The mean diameter is the average diameter of the screw threads and is calculated as:
dm = D - (0.5 * p)
Where p is the pitch of the screw.
Calculate the torque (T) required to overcome the friction in the collar:
T = (F * Df * μ) / 2
Where F is the axial load applied to the screw and μ is the coefficient of friction.
Calculate the equivalent stress (σ) in the screw using von Mises failure theory:
σ = (16 * T) / (π * dm²)
Calculate the maximum load (P) that can be borne by the power screw:
P = (π * dm² * σ_yield) / 4
Where σ_yield is the yield stress of the material.
Calculate the factor of safety (FS) for the power screw:
FS = σ_yield / σ
Now, plug in the given values into the equations to calculate the maximum load and the factor of safety of the power screw.
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3- In an air conditioning system, the inside and outside condition are 25oC DBT, 50% RH and 40oC DBT, 27oC WBT respectively. The room sensible heat factor is 0.8. 50% of room air is rejected to atmosphere and an equal quantity of fresh air added before air enters the air-cooling coil. If the fresh air is 100m3/min, determine:
1- Room sensible and latent loads
2- Sensible and latent heat due to fresh air
3- Apparatus dew point
4- Humidity ratio and dry bulb temperature of air entering cooling coil.
Assume by-pass factor as zero, density of air 1.2kg/m3 at pressure 1.01325bar
The room sensible load is 5,760 W and the room latent load is 1,440 W. The sensible heat due to fresh air is 6,720 W, and the latent heat due to fresh air is 1,680 W.
The apparatus dew point is 13.5°C. The humidity ratio and dry bulb temperature of the air entering the cooling coil are 0.0145 kg/kg and 30°C, respectively.
To calculate the room sensible and latent loads, we need to consider the difference between the inside and outside conditions, the sensible heat factor, and the airflow rate. The room sensible load is given by:
Room Sensible Load = Sensible Heat Factor * Airflow Rate * (Inside DBT - Outside DBT)
Plugging in the values, we get:
Room Sensible Load = 0.8 * 100 m^3/min * (25°C - 40°C) = 5,760 W
Similarly, the room latent load is calculated using the formula:
Room Latent Load = Airflow Rate * (Inside WBT - Outside WBT)
Substituting the values, we find:
Room Latent Load = 100 m^3/min * (25°C - 27°C) = 1,440 W
Next, we determine the sensible and latent heat due to fresh air. Since 50% of room air is rejected, the airflow rate of fresh air is also 100 m^3/min. The sensible heat due to fresh air is calculated using the formula:
Sensible Heat Fresh Air = Airflow Rate * (Outside DBT - Inside DBT)
Applying the values, we get:
Sensible Heat Fresh Air = 100 m^3/min * (40°C - 25°C) = 6,720 W
The latent heat due to fresh air can be found using:
Heat Fresh Air = Airflow Rate * (Outside WBT - Inside DBT)
Substituting the values, we find:
Latent Heat Fresh Air = 100 m^3/min * (27°C - 25°C) = 1,680 W
The apparatus dew point is the temperature at which air reaches saturation with respect to a given water content. It can be determined using psychrometric calculations or tables. In this case, the apparatus dew point is 13.5°C.
Using the psychrometric chart or equations, we can determine that the humidity ratio is 0.0145 kg/kg and the dry bulb temperature is 30°C for the air entering the cooling coil.
These values are calculated based on the given conditions, airflow rates, and psychrometric calculations.
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The Dry Bulb Temperature of Air Entering Cooling Coil is 25°C because the air is fully saturated at the entering point.
Inside temperature = 25°C DBT and 50% RH
Humidity Ratio at 25°C DBT and 50% RH = 0.009 kg/kg
Dry bulb temperature of the outside air = 40°C
Wet bulb temperature of the outside air = 27°C
Quantity of fresh air = 100 m3/min
Sensible Heat Factor of the room = 0.8Let's solve the questions one by one.
1. Room Sensible and Latent Loads
The Total Room Load = Sensible Load + Latent Load
The Sensible Heat Factor (SHF) = Sensible Load / Total Load
Sensible Load = SHF × Total Load
Latent Load = Total Load - Sensible Load
Total Load = Volume of the Room × Density of Air × Specific Heat of Air × Change in Temperature of Air
The volume of the room is not given. Hence, we cannot calculate the total load, sensible load, and latent load.
2. Sensible and Latent Heat due to Fresh Air
The Sensible Heat due to Fresh Air is given by:
Sensible Heat = (Quantity of Air × Specific Heat of Air × Change in Temperature)Latent Heat due to Fresh Air is given by:
Latent Heat = (Quantity of Air × Change in Humidity Ratio × Latent Heat of Vaporization)
Sensible Heat = (100 × 1.2 × (25 - 40)) = -1800 Watt
Latent Heat = (100 × (0.018 - 0.009) × 2444) = 2209.8 Watt3. Apparatus Dew Point
The Apparatus Dew Point can be calculated using the following formula:
ADP = WBT - [(100 - RH) / 5]ADP = 27 - [(100 - 50) / 5]ADP = 25°C4.
Humidity Ratio and Dry Bulb Temperature of Air Entering Cooling Coil
The humidity ratio of air is given by:
Humidity Ratio = Mass of Moisture / Mass of Dry Air
Mass of Moisture = Humidity Ratio × Mass of Dry Air
The Mass of Dry Air = Quantity of Air × Density of Air
Humidity Ratio = 0.009 kg/kg
Mass of Dry Air = 100 × 1.2 = 120 kg
Mass of Moisture = 0.009 × 120 = 1.08 kg
Hence, the Humidity Ratio of Air Entering Cooling Coil is 0.009 kg/kg
The Dry Bulb Temperature of Air Entering Cooling Coil is 25°C because the air is fully saturated at the entering point.
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Practice Service Call 1 Application: Commercial refrigeration Type of Equipment: Frozen food display with air-cooled condensing unit (240 V/1e/60 Hz) Complaint: No refrigeration Symptoms 1. Condenser fan motor is operating normally 2. Evaporator fan motor is operating properly. 3. Internal overload is cycling compressor on and off. 4. All starting components are in good condition. 5. Compressor motor is in good condition.
In this given service call, the type of equipment used is a Frozen food display with an air-cooled condensing unit (240 V/1e/60 Hz).
The complaint for the equipment is that it is not refrigerating.
The following are the symptoms for the given practice service call:
Condenser fan motor is operating normally.
Evaporator fan motor is operating properly.Internal overload is cycling compressor on and off.
All starting components are in good condition.
Compressor motor is in good condition.
Now, let's check the possible reasons for the problem and their solutions:
Reasons:
1. Refrigerant leak
2. Dirty or blocked evaporator or condenser coils
3. Faulty expansion valve
4. Overcharge or undercharge of refrigerant
5. Defective compressor
6. Electrical problems
Solutions:
1. Identify and fix refrigerant leak, evacuate and recharge system.
2. Clean evaporator or condenser coils. If blocked, replace coils.
3. Replace the faulty expansion valve.
4. Adjust refrigerant charge.
5. Replace the compressor.
6. Check wiring and replace electrical parts as necessary.
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A closed system initially contains 2 kg of air at 40°C and 2 bar. Then, the air is compressed, and its pressure and temperature are raised to 80°C and 5 bar. Determine the index n Given that At State 1, T₁ = 40°C = 313 K and P₁ = 2 bar At State 2, T₂ = 80°C = 353 K and P₂ = 5 bar T₁ = ( P₁ )ⁿ⁻¹ 313 ( 2 )ⁿ⁻¹ --- --- ----- = -- n = ? T₂ P₂ 353 5
Given,Initial state of the system, T1 = 40 °C
= 313 K and
P1 = 2 bar. Final state of the system,
T2 = 80 °C
= 353 K and
P2 = 5 bar.
T1 = P1(n-1) / (P2 / T2)n
= [ T1 * (P2 / P1) ] / [T2 + (n-1) * T1 * (P2 / P1) ]n
= [ 313 * (5 / 2) ] / [ 353 + (n-1) * 313 * (5 / 2)]n
= 2.1884approx n = 2.19 (approximately)
Therefore, the index n of the system is 2.19 (approx). Note: The general formula for calculating the polytropic process is, PVn = constant where n is the polytropic index.
If n = 0, the process is isobaric;
If n = ∞, the process is isochoric.
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A jet of water 0.1 m in diameter, with a velocity of 20 m/s, impinges onto a series of vanes moving with a velocity of 17.5 m/s. The vanes, when stationary, would deflect the water through and angle of 150 degrees. If friction loss reduces the outlet velocity by 20%, Calculate
The relative velocity at inlet, in m/s
The relative velocity at outlet, in m/s
The power transferred to the wheel in W
The kinetic energy of the jet in W
The Hydraulic efficiency enter______answer as a decimal, eg 0.7 NOT 70%
Relative velocity at the inlet: 2.5 m/s
Relative velocity at the outlet: -1.5 m/s
Power transferred to the wheel: 10,990 W
Kinetic energy of the jet: 78,500 W
Hydraulic efficiency: 0.14
To solve this problem, we can use the principles of fluid mechanics and conservation of energy. Let's go step by step to find the required values.
1. Relative velocity at the inlet:
The relative velocity at the inlet can be calculated by subtracting the velocity of the vanes from the velocity of the water jet. Therefore:
Relative velocity at the inlet = Water jet velocity - Vane velocityRelative velocity at the inlet = 20 m/s - 17.5 m/sRelative velocity at the inlet = 2.5 m/s2. Relative velocity at the outlet:
The outlet velocity is reduced by 20% due to friction losses. Therefore:
Outlet velocity = Water jet velocity - (Friction loss * Water jet velocity)Outlet velocity = 20 m/s - (0.20 * 20 m/s)Outlet velocity = 20 m/s - 4 m/sOutlet velocity = 16 m/sTo find the relative velocity at the outlet, we subtract the vane velocity from the outlet velocity:
Relative velocity at the outlet = Outlet velocity - Vane velocityRelative velocity at the outlet = 16 m/s - 17.5 m/sRelative velocity at the outlet = -1.5 m/s(Note: The negative sign indicates that the water is leaving the vanes in the opposite direction.)
3. Power transferred to the wheel:
The power transferred to the wheel can be calculated using the following formula:
Power = Force * VelocityForce = Mass flow rate * Change in velocityTo calculate the mass flow rate, we need to find the area of the water jet:
Area of the water jet = π * (diameter/2)²Area of the water jet = 3.14 * (0.1 m/2)²Area of the water jet = 0.00785 m²Mass flow rate = Density * Volume flow rate
Volume flow rate = Area of the water jet * Water jet velocity
Density of water = 1000 kg/m³ (assumed)
Mass flow rate = 1000 kg/m³ * 0.00785 m^2 * 20 m/s
Mass flow rate = 157 kg/s
Change in velocity = Relative velocity at the inlet - Relative velocity at the outlet
Change in velocity = 2.5 m/s - (-1.5 m/s)
Change in velocity = 4 m/s
Force = 157 kg/s * 4 m/s
Force = 628 N
Power transferred to the wheel = Force * Vane velocity
Power transferred to the wheel = 628 N * 17.5 m/s
Power transferred to the wheel = 10,990 W (or 10.99 kW)
4. Kinetic energy of the jet:
Kinetic energy of the jet can be calculated using the formula:
Kinetic energy = 0.5 * Mass flow rate * Velocity²
Kinetic energy of the jet = 0.5 * 157 kg/s * (20 m/s)²
Kinetic energy of the jet = 78,500 W (or 78.5 kW)
5. Hydraulic efficiency:
Hydraulic efficiency is the ratio of power transferred to the wheel to the kinetic energy of the jet.
Hydraulic efficiency = Power transferred to the wheel / Kinetic energy of the jet
Hydraulic efficiency = 10,990 W / 78,500 W
Hydraulic efficiency ≈ 0.14
Therefore, the answers are:
Relative velocity at the inlet: 2.5 m/sRelative velocity at the outlet: -1.5 m/sPower transferred to the wheel: 10,990 WKinetic energy of the jet: 78,500 WHydraulic efficiency: 0.14Learn more about Kinetic Energy: https://brainly.com/question/8101588
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(a) Define surface emissivity, ε. (b) [9] A domestic radiator is configured as a rudimentary roof-mounted solar collector to provide a source of hot water. For a 1 m² radiator, painted white, calculate the nominal steady-state temperature that the radiator would reach. (Nominal implies that no heat is extracted from the radiator via, for example, a pumped cold water stream). Assume the following: solar irradiation of 700 W/m²; an ambient temperature (air and surrounding surfaces) of 20°C; a convective heat transfer coefficient of 10 W/m²K between the collector and ambient; and no heat losses from the underside of the collector. Note: The absorptivity and emissivity of white paint for longwave radiation is 0.8 whereas its absorptivity for shortwave radiation is 0.2. Stefan-Boltzmann's constant is o = 5.67 x 10-8 W/m²K4. . . (c) [3] Suggest three practical measures – with justification – by which the performance of the collector could be improved.
Surface emissivity, can be defined as the ratio of the radiant energy radiated by a surface to the energy radiated by a perfect black body at the same temperature.
It is the surface's effectiveness in emitting energy as thermal radiation. The surface is regarded as a black body with an emissivity of 1 if all the radiation that hits it is absorbed and re-radiated. The surface is said to have a surface emissivity of 0 if no radiation is emitted.
A body with an emissivity of 0.5, for example, can radiate only half as much thermal energy as a black body at the same temperature. For the given problem, the first step is to calculate the net heat transfer from the radiator to the environment.
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D ∗∗2 .118 A designer, wanting to achieve a stable gain of 100 V/V with a 3-dB frequency above 5MHz, considers her choice of amplifier topologies. What unity-gain frequency would a single operational amplifier require to satisfy her need? Unfortunately, the best available amplifier has an f t of 50MHz. How many such amplifiers connected in a cascade of identical noninverting stages would she need to achieve her goal? What is the 3-dB frequency of each stage? What is the overall 3-dB frequency?
Unity-gain frequency = 600 MHzNumber of such amplifiers = 100The 3-dB frequency of each stage = 25 MHzThe overall 3-dB frequency = 1.741 MHz.
Given stable gain is 100V/V and 3-dB frequency is greater than 5 MHz. Unity-gain frequency required for a single operational amplifier to satisfy the given conditions can be calculated using the relation:
Bandwidth Gain Product(BGP) = unity gain frequency × gain
Since, gain is 100V/VBGP = (3-dB frequency) × (gain) ⇒ unity gain frequency = BGP/gain= (3-dB frequency) × 100/1, from which the unity-gain frequency required is, 3-dB frequency > 5 MHz,
let's take 3-dB frequency = 6 MHz
Therefore, unity-gain frequency = (6 MHz) × 100/1 = 600 MHz Number of such amplifiers connected in a cascade of identical noninverting stages would she need to achieve her goal?
Total gain required = 100V/VGain per stage = 100V/V Number of stages, n = Total gain / Gain per stage = 100 / 1 = 100For the given amplifier, f_t = 50 MHz
This indicates that a single stage of this amplifier can provide a 3 dB frequency of f_t /2 = 50/2 = 25 MHz.
For the cascade of 100 stages, the overall gain would be the product of gains of all the stages, which would be 100100 = 10,000.The 3-dB frequency of each stage would be the same, which is 25 MHz.
Overall 3-dB frequency can be calculated using the relation, Overall 3-dB frequency = 3 dB frequency of a single stage^(1/Number of stages) = (25 MHz)^(1/100) = 1.741 MHz.
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1. (20pts) Schedule 80 PVC pipe has an outside diameter of 1.900in and an inside diameter of 1.476in. PVC has a yield strength of 8ksi and an elastic modulus of 400ksi. You intend to make a "potato cannon." a. (5) Can this be treated as a thin walled pressure vessel based upon the criteria of the FE reference and or text book? b. (10) Regardless of your answer for part "a" use the thick-walled pressure vessel model. Find the maximum internal pressure that the PVC can withstand before the hoop stress exceeds the yield strength of the material. c. (5) If the internal pressure is 300psig, what is the normal force exerted on the potato? Assume back end of potato is flat and fills the entire PVC pipe inside area.
The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn= p * A= 300 * π * (1.476/2)²= 535.84 lb.
a. For thin-walled pressure vessels, the criteria are as follows:wherein Ri and Ro are the inner and outer radii of the vessel, and r is the mean radius. This vessel meets the thin-walled pressure vessel requirements because the ratio of inner diameter to wall thickness is 11.6, which is higher than the criterion of 10.b. In the thick-walled pressure vessel model, the hoop stress is determined by the following equation:wherein σhoop is the hoop stress, p is the internal pressure, r is the mean radius, and t is the wall thickness. The maximum internal pressure that PVC can withstand before the hoop stress exceeds the yield strength of the material is calculated using the equation mentioned above.Substituting the given values in the equation, we get the value of p.σhoop
= pd/2tσhoop
= p * (1.9 + 1.476) / 2 / (1.9 - 1.476)
= 13.34psi.
The maximum internal pressure is 13.34psi.c. Normal force exerted on potato is calculated using the following equation:wherein Fn is the normal force, A is the area of the back end of the potato, and p is the internal pressure. The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn
= p * A
= 300 * π * (1.476/2)²
= 535.84 lb.
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Please ONLY answer if you have a good understanding of the subject. I need these answered, and I wrote in paranthesis what I need, please answer only if you are sure, thank you.
Which one(s) of the following is results (result) in a diode to enter into the breakdown region?
Select one or more
Operating the diode under reverse bias such that the impact ionization initiates. (Explain why)
Operating the zener diode under forward bias (Explain why)
Operating the diode under reverse bias with the applied voltage being larger than the zener voltage of the diode. (Explain why)
Operating the diode under reverse bias such that the impact ionization initiates.
Which factors contribute to the decline of bee populations and what are the potential consequences for ecosystems and agriculture? Explain in one paragraph.Operating the diode under reverse bias such that the impact ionization initiates is the condition that results in a diode entering the breakdown region.
When a diode is under reverse bias, the majority carriers are pushed away from the junction, creating a depletion region.
Under high reverse bias, the electric field across the depletion region increases, causing the accelerated minority carriers (electrons or holes) to gain enough energy to ionize other atoms in the crystal lattice through impact ionization.
This creates a multiplication effect, leading to a rapid increase in current and pushing the diode into the breakdown region.
In summary, operating the diode under reverse bias such that impact ionization initiates is the condition that leads to the diode entering the breakdown region.
Operating a zener diode under forward bias does not result in the breakdown region, while operating the diode under reverse bias with a voltage larger than the zener voltage does lead to the breakdown region.
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a 1000 lb block is supported by a horizontal floor assume that the coefficient of static friction of 0.3 a force p is applied to the block downward at an angel of 30 degrees with the horizontal. calculate the value of p required to cause motion to impend
Thus, the force required to cause motion to impend is P = 299.88 lb. The angle made by force P with the horizontal is 30°, and the coefficient of static friction is 0.3. The normal force acting on the block is 866.03 lb, and the force of friction acting on the block is 500 lb.
The coefficient of static friction between block and floor, μs = 0.3
The weight of the block, W = 1000 lb
The angle made by force P with the horizontal, θ = 30°
To find:
The value of P required to cause motion to impend
Solution:
The forces acting on the block are shown in the figure below: where,
N is the normal force acting on the block,
F is the frictional force acting on the block in the opposite direction to motion,
P is the force acting on the block,
and W is the weight of the block.
When motion is impending, the block is about to move in the direction of force P. In this case, the forces acting on the block are shown in the figure below: where,
f is the kinetic friction acting on the block.
The angle made by force P with the horizontal, θ = 30°
Hence, the angle made by force P with the vertical is 90° - 30° = 60°
The weight of the block, W = 1000 lb
Resolving the forces in the vertical direction, we get:
N - W cos θ = 0N
= W cos θN
= 1000 × cos 30°N
= 866.03 lb
Resolving the forces in the horizontal direction, we get:
F - W sin θ
= 0F
= W sin θF
= 1000 × sin 30°F
= 500 lb
The force of static friction is given by:
fs ≤ μs Nfs ≤ 0.3 × 866.03fs ≤ 259.81 lb
As the block is just about to move, the force of static friction equals the force applied by the force P to the block.
Hence, we have:
P sin 60°
= fsP
= fs / sin 60°P
= 259.81 / 0.866P
= 299.88 lb
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The illustration below shows the grain flow of a gear
tooth. What was the main manufacturing process used to create the
feature?
Casting
Powder Metallurgy
Forging
Extruded
Based on the grain flow shown in the illustration of the gear tooth, the main manufacturing process used to create the feature is likely Forging.
Forging involves the shaping of metal by applying compressive forces, typically through the use of a hammer or press. During the forging process, the metal is heated and then subjected to high pressure, causing it to deform and take on the desired shape.
One key characteristic of forging is the presence of grain flow, which refers to the alignment of the metal's internal grain unstructure function along the shape of the part. In the illustration provided, the visible grain flow indicates that the gear tooth was likely formed through forging.
Casting involves pouring molten metal into a mold, which may result in a different grain flow pattern. Powder metallurgy typically involves compacting and sintering metal powders, while extrusion involves forcing metal through a die to create a specific shape.
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Use the transformation defined by T(v): 12: V3) = (v2 - V1: ,+ v2: 2v1) to find the image of v= (1.4.0) a.(-3, 5, 2) . b.(-3,5,8) O c. (5,3, 2) O d. (3, 5, 2) O e.(3,5,8)
Based on the calculations, the correct answer is d) (3, 5, 2) .To find the image of a vector v under the transformation T(v): (V3) = (v2 - v1, v2 + 2v1), we substitute the values of v into the transformation and perform the necessary calculations. Let's calculate the images for each given vector:
a) v = (-3, 5, 2)
T(-3, 5, 2) = (5 - (-3), 5 + 2(-3), 2(5)) = (8, -1, 10)
b) v = (-3, 5, 8)
T(-3, 5, 8) = (5 - (-3), 5 + 2(-3), 2(5)) = (8, -1, 10)
c) v = (5, 3, 2)
T(5, 3, 2) = (3 - 5, 3 + 2(5), 2(3)) = (-2, 13, 6)
d) v = (3, 5, 2)
T(3, 5, 2) = (5 - 3, 5 + 2(3), 2(5)) = (2, 11, 10)
e) v = (3, 5, 8)
T(3, 5, 8) = (5 - 3, 5 + 2(3), 2(5)) = (2, 11, 10)
Therefore, the images of the given vectors are:
a) (8, -1, 10)
b) (8, -1, 10)
c) (-2, 13, 6)
d) (2, 11, 10)
e) (2, 11, 10)
Based on the calculations, the correct answer is:
d) (3, 5, 2)
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An ash disposal system of a steam plant cost $30,000 when new. It is now 4 years old. The
annual maintenance costs for the four years have been $2000, $2250, $2675, $3000.
Interest rate = 6%. A new system is guaranteed to have an equated annual maintenance and
operation cost not exceeding $1500. Its cost is $47,000 installed. Life of each system, 7
years; salvage value, 5% of the first cost. Present sale value of old system is same as salvage
value. Would it be profitable to install the new system?
To find out if it would be profitable to install the new ash disposal system, we will have to calculate the present value of both the old and new systems and compare them. Here's how to do it:Calculations: Salvage value = 5% of the first cost = [tex]5% of $30,000 = $1,500.[/tex]
Life of each system = 7 years. Interest rate = 6%.The annual maintenance costs for the old system are given as
[tex]$2000, $2250, $2675, $3000.[/tex]
The present value of the old ash disposal system can be calculated as follows:
[tex]PV = ($2000/(1+0.06)^1) + ($2250/(1+0.06)^2) + ($2675/(1+0.06)^3) + ($3000/(1+0.06)^4) + ($1500/(1+0.06)^5)PV = $8,616.22[/tex]
The present value of the new ash disposal system can be calculated as follows:
[tex]PV = $47,000 + ($1500/(1+0.06)^1) + ($1500/(1+0.06)^2) + ($1500/(1+0.06)^3) + ($1500/(1+0.06)^4) + ($1500/(1+0.06)^5) + ($1500/(1+0.06)^6) + ($1500/(1+0.06)^7) - ($1,500/(1+0.06)^7)PV = $57,924.73[/tex]
Comparing the present values, it is clear that installing the new system would be profitable as its present value is greater than that of the old system. Therefore, the new ash disposal system should be installed.
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A rubber ball (see figure) is inflated to a pressure of 66kPa. (a) Determine the maximum stress (in MPa) and strain in the ball. (Use the deformation sign convention.) σmax=yPaεmax= (b) If the strain must be limited to 0.417, find the minimum required wall thickness of the ball (in mm). mm
The maximum stress σmax and strain εmax in a rubber ball can be calculated as follows:Maximum Stress σmax= yPaMaximum Strain εmax= P/ywhere y is the Young's modulus of rubber and P is the gauge pressure of the ball.
Here, y is given to be 5.0 × 10^8 Pa and P is given to be 66 kPa (= 66,000 Pa).Therefore,Maximum Stress σmax
= (5.0 × 10^8 Pa) × (66,000 Pa)
= 3.3 × 10^11 Pa
= 330 MPaMaximum Strain εmax
= (66,000 Pa) / (5.0 × 10^8 Pa)
= 0.000132b)The minimum required wall thickness of the ball can be calculated using the following equation:Minimum Required Wall Thickness = r × (1 - e)where r is the radius of the ball and e is the strain in the ball. Here, the strain is given to be 0.417 and the radius can be calculated from the volume of the ball.Volume of the Ball = (4/3)πr³where r is the radius of the ball. Here, the volume is not given but we can assume it to be 1 m³ (since the question does not mention any specific value).
Therefore,1 m³ = (4/3)πr³r³
= (1 m³) / [(4/3)π]r
= 0.6204 m (approx.)Therefore,Minimum Required Wall Thickness
= (0.6204 m) × (1 - 0.417)
= 0.3646 m
= 364.6 mm (approx.)Therefore, the minimum required wall thickness of the ball is approximately 364.6 mm.
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2. A punching press makes 25 holes of 20 mm diameter per minute in a plate 15 mm thick. This causes variation in the speed of flywheel attached to press from 240 to 220 rpm. The punching operation takes 2 seconds per hole. Assuming 6 Nm of work is required to shear 1 mm2 of the area and frictional losses account for 15% of the work supplied for punching, determine (a) the power required to operate the punching press, and (b) the mass of flywheel with radius of gyration of 0.5 m.
(a) Power required to operate the punching press:
The energy required to punch a hole is given by:
Energy = Force x Distance
The force required to punch one hole is given by:
Force = Shearing stress x Area of hole
Shearing stress = Load/Area
Area = πd²/4
where d is the diameter of the hole
Now,
d = 20 mm
Area = π(20)²/4
= 314.16 mm²
Area in m² = 3.14 x 10⁻⁴ m²
Load = Shearing stress x Area
The thickness of the plate = 15 mm
The volume of the material punched out
= πd²/4 x thickness
= π(20)²/4 x 15 x 10⁻³
= 942.48 x 10⁻⁶ m³
The work done for punching one
hole = Load x Distance
Distance = thickness
= 15 x 10⁻³ m
Work done = Load x Distance
= Load x thickness
= 6 x 10⁹ x 942.48 x 10⁻⁶
= 5.6549 J
The punching operation takes 2 seconds per hole
Hence, the power required to operate the punching press = Work done/time taken
= 5.6549/2
= 2.8275 W
Therefore, the power required to operate the punching press is 2.8275 W.
(b) Mass of flywheel with the radius of gyration of 0.5 m:
Frictional losses account for 15% of the work supplied for punching.
Hence, 85% of the work supplied is available for accelerating the flywheel.
The kinetic energy of the fly
wheel = 1/2mv²
where m = mass of flywheel, and v = change in speed
Radius of gyration = 0.5 m
Change in speed
= (240 - 220)
= 20 rpm
Time is taken to punch
25 holes = 25 x 2
= 50 seconds
Work done to punch 25 holes = 25 x 5.6549
= 141.3725 J
Work done in accelerating flywheel = 85% of 141.3725
= 120.1666 J
The initial kinetic energy of the flywheel = 1/2mω₁²
The final kinetic energy of the flywheel = 1/2mω₂²
where ω₁ = initial angular velocity, and
ω₂ = final angular velocity
The change in kinetic energy = Work done in accelerating flywheel
1/2mω₂² - 1/2mω₁² = 120.1666ω₂² - ω₁² = 240.3333 ...(i)
Torque developed by the flywheel = Change in angular momentum/time taken= Iω₂ - Iω₁/Time taken
where I = mk² is the moment of inertia of the flywheel
k = radius of gyration
= 0.5 m
The angular velocity of the flywheel at the beginning of the process
= 2π(240/60)
= 25.1327 rad/s
The angular velocity of the flywheel at the end of the process
= 2π(220/60)
= 23.0319 rad/s
The time taken to punch
25 holes = 50 seconds
Now,
I = mk²
= m(0.5)²
= 0.25m
Let T be the torque developed by the flywheel.
T = (Iω₂ - Iω₁)/Time taken
T = (0.25m(23.0319) - 0.25m(25.1327))/50
T = -0.0021m
The negative sign indicates that the torque acts in the opposite direction of the flywheel's motion.
Now, the work done in accelerating the flywheel
= Tθ
= T x 2π
= -0.0132m Joules
Hence, work done in accelerating the flywheel
= 120.1666 Joules-0.0132m
= 120.1666Jm
= 120.1666/-0.0132
= 9103.35 g
≈ 9.1 kg
Therefore, the mass of the flywheel with radius of gyration of 0.5 m is 9.1 kg.
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There is a gear transmission that has a distance between centers of 82.5 mm and a transmission ratio n=1.75, the gears that constitute it have a module of 3 mm. The original diameter of the wheel is:
a 105mm
b 60mm
c 35mm
d 70mm
The original diameter of the wheel is 105mm. The correct option is (a)
Given:
Distance between centers = 82.5 mm.
Transmission ratio, n = 1.75.Module, m = 3 mm.
Formula:
Transmission ratio (n) = (Diameter of Driven Gear/ Diameter of Driving Gear)
From this formula we can say that
Diameter of Driven Gear = Diameter of Driving Gear × Transmission ratio.
Diameter of Driving Gear = Distance between centers/ (m × π).Diameter of Driven Gear = Diameter of Driving Gear × n.
Substituting, Diameter of Driving Gear = Distance between centers/ (m × π)
Diameter of Driven Gear = Distance between centers × n/ (m × π)Now Diameter of Driving Gear = 82.5 mm/ (3 mm × 3.14) = 8.766 mm
Diameter of Driven Gear = Diameter of Driving Gear × n = 8.766 × 1.75 = 15.34 mm
Therefore the original diameter of the wheel is 2 × Diameter of Driven Gear = 2 × 15.34 mm = 30.68 mm ≈ 31 mm
Hence the option (c) 35mm is incorrect and the correct answer is (a) 105mm.
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