An electrical heater and kitchen LPG system are two popular options for heating and cooking. The choice between the two depends on several factors that you need to consider before making a final decision. I advise my family and friends to consider the following factors before deciding which option is best for them.
1. Energy efficiency: Energy efficiency is the primary factor you should consider when choosing between an electrical heater and kitchen LPG system. The kitchen LPG system is generally more energy-efficient than electrical heaters. LPG gas can heat up a pot or pan faster than an electric heating element, which saves energy.2. Cost: Cost is another important factor you should consider when choosing between an electrical heater and a kitchen LPG system. LPG gas is generally cheaper than electricity in most parts of the world. However, the cost of LPG varies depending on your location, so it's important to do some research to find out the price of LPG in your area.3. Safety: Safety is also an essential factor you should consider when choosing between an electrical heater and a kitchen LPG system. Both options have their unique risks and safety concerns. For example, LPG is highly flammable, while electrical heaters can pose an electrocution hazard.
Availability is another essential factor you should consider when choosing between an electrical heater and a kitchen LPG system. LPG is not available in some areas, while electricity is readily available in most parts of the world. You need to consider the availability of the energy source in your area before deciding which option is best for you.In conclusion, both an electrical heater and kitchen LPG system have their unique benefits and drawbacks. The best option for saving money depends on several factors such as energy efficiency, cost, safety, convenience, and availability. I always advise my family and friends to consider these factors carefully before making a final decision.
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A STEEL PART HAS THIS STRESS STATE : DETERMINE THE FACTOR OF SAFETY USING THE DISTORTION ENERGY (DE) FAILURE THEORY
6x = 43kpsi
Txy = 28 kpsi
Sy= 120kpsi
The factor of safety using the Distortion Energy (DE) Failure Theory is 3.95.
The factor of safety is an important factor in determining the safety of a structure and is often used in the design of structures. The formula of Factor of safety is:
Factor of Safety = Yield Strength / Maximum Stress
Therefore, the factor of safety using the Distortion Energy (DE) Failure Theory can be calculated as follows
6x = 43kpsi, Txy = 28 kpsi and Sy = 120kpsiσ
Von Mises = sqrt[0.5{(σx - σy)^2 + (σy - σz)^2 + (σz - σx)^2}]σ
Von Mises = sqrt[0.5{(43 - 0)^2 + (0 - 0)^2 + (0 - 0)^2}]σ
Von Mises = sqrt[0.5{(1849)}]σ
Von Mises = sqrt[924.5]σ
Von Mises = 30.38 kpsi
Factor of Safety = Yield Strength / Maximum Stress
Factor of Safety = Sy / σVon Mises
Factor of Safety = 120/30.38
Factor of Safety = 3.95
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Describe the effect of:
1. Air Spoilers:
2. Inboard Aileron :
3. Slats:
4. Trim Tabs :
5. Flaperons :
6. Ruddervators:
Air Spoilers: Air spoilers are devices used on aircraft to disrupt the smooth airflow over the wings, thus reducing lift. When deployed, air spoilers create turbulence on the wing surface, which increases drag and decreases lift.
This effect is commonly used during descent or landing to assist in controlling the rate of descent and to improve the effectiveness of other control surfaces.
Inboard Aileron: Inboard ailerons are control surfaces located closer to the centerline of an aircraft's wings. They work in conjunction with outboard ailerons to control the rolling motion of the aircraft. By deflecting in opposite directions, inboard ailerons generate differential lift on the wings, causing the aircraft to roll about its longitudinal axis. This helps in banking or turning the aircraft.
Slats: Slats are movable surfaces located near the leading edge of an aircraft's wings. When extended, slats change the shape of the wing's leading edge, creating a slot between the wing and the slat. This slot allows high-pressure air from below the wing to flow over the top, delaying the onset of airflow separation at high angles of attack. The presence of slats enhances lift and improves the aircraft's ability to take off and land at lower speeds.
Trim Tabs: Trim tabs are small surfaces attached to the trailing edge of control surfaces such as ailerons, elevators, or rudders. They can be adjusted by the pilot or through an automatic control system to fine-tune the balance and control of the aircraft. By deflecting the trim tabs, the aerodynamic forces on the control surfaces can be modified, enabling the pilot to maintain a desired flight attitude or relieve control pressure.
Flaperons: Flaperons combine the functions of both flaps and ailerons. They are control surfaces located on the trailing edge of the wings, near the fuselage. Flaperons can be extended downward as flaps to increase lift during takeoff and landing, or they can be deflected differentially to perform the roll control function of ailerons. By combining these two functions, flaperons provide improved maneuverability and control during various flight phases.
Ruddervators: Ruddervators are control surfaces that serve dual functions of both elevators and rudders. They are commonly used in aircraft with a V-tail configuration. The ruddervators operate together to control pitch, acting as elevators, and differentially to control yaw, acting as rudders. This arrangement simplifies the control system and improves maneuverability by combining pitch and yaw control into a single surface.
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OUTCOME 2 : Impulse Turbine Fluid Machinery 2021-2022 As an energy engineer, has been asked from you to prepare a design of Pelton turbine in order to establish a power station worked on the Pelton turbine on the Tigris River. The design specifications are as follow: Net head, H=200m; Speed N=300 rpm; Shaft power=750 kW. Assuming the other required data wherever necessary.
Pelton turbine is a type of impulse turbine. Pelton turbine consists of a wheel that has split cups, also known as buckets, which are located along the outer rim of the wheel. The water is directed onto the wheel’s cups, and the pressure causes the wheel to rotate.
Impulse Turbine Fluid Machinery 2021-2022As an energy engineer, you have been asked to prepare a design of Pelton turbine to establish a power station that worked on the Pelton turbine on the Tigris River. \\\\\The power of the turbine can be calculated using the formula:Power = rho x g x Q x H x n, where rho is the density of water, g is the acceleration due to gravity, Q is the volume flow rate, H is the net head, and n is the efficiency of the turbine.
Since the shaft power is 750 kW, we can calculate the hydraulic power that is transferred to the turbine. The hydraulic power can be calculated using the following formula:Hydraulic Power = Shaft Power / Efficiency which can be assumed for this calculation. The hydraulic power would be 833.33 kW.
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Distinguish between thin and thick cylinders.
Calculate the bursting pressure for a cold drawn seamless steel tubing of 60mm inside diameter with 2mm wall thickness. The ultimate strength of steel is 380 MN/m².
A sold circular shaft transmits 75 kW power at 200r.p.m. Calculate the shaft diameter, if the twist in the shaft is not to exceed 10 in 2 metres length of shaft, and shear stress is limited to 50 MN/m². Take C=100 GN/m².
A circular bar-made of cast iron is to resist an occasional torque of 2.2 kNm acting in transverse plane. If the allowable stresses in compression, tension and shear are 100 MN/m², 35 MN/m² and 50MN/m² respectively, calculate: (i) Diameter of the bar and (ii) Angle of twist under the applied torque per metre length of bar. Take: C (for cast-iron) = 40GN/m²
(1) The diameter of the bar is 160 mm.(2) The angle of twist under the applied torque per meter length of bar is 0.062 radians/m or 3.5°/m.
Thin and thick cylinders are two categories of cylinders. The major differences between the two are their wall thickness and design.
Thick cylinders are generally used for high-pressure applications, whereas thin cylinders are used for low-pressure applications. Here are some distinctions between the two:
Thin Cylinder:
Thin cylinder has a smaller radius than the thickness of its wall and it is used for low-pressure applications such as gas cylinders for domestic use.
The hoop strain is twice the longitudinal strain.
Stress is constant across the thickness of the wall.
Thin cylinders are designed to resist tension and compression forces.
Thin cylinders are used to produce boilers, gas tanks, and pipes.
Thick Cylinder:
A thick cylinder is designed to resist the internal pressure that comes with high-pressure applications.
The hoop strain and the longitudinal strain are equal.
The stress at any point within the wall thickness is variable.
The material's yield strength is critical in the design of thick-walled cylinders.
The use of a thick-walled cylinder may increase the risk of fracture.
The thicker the cylinder, the more stress it can handle.
Now, let us calculate the bursting pressure for a cold-drawn seamless steel tubing of 60mm inside diameter with a 2mm wall thickness.
Given,
Internal diameter of tubing, d = 60 mm
Thickness of wall, t = 2 mm
Ultimate strength of steel, σu = 380 MN/m²
Bursting pressure formula is given by:
pb = σu × d / 2t
= 380 × 60 / 4
= 5700 kPa
Therefore, the bursting pressure for the cold-drawn seamless steel tubing is 5700 kPa.
Now, let's calculate the diameter of the circular bar and the angle of twist per meter length of the bar:
Given,
The torque applied, T = 2.2 kNm
Maximum allowable compressive stress, σcomp = 100 MN/m²
Maximum allowable tensile stress, σtens = 35 MN/m²
Maximum allowable shear stress, τ = 50 MN/m²
Shear modulus of cast iron, C = 40 GN/m²
(i) Diameter of the bar
We know that
T/J = τ/R = Gθ/L
Where, T = torque, J = polar moment of inertia, τ = shear stress, R = radius, G = shear modulus, θ = angle of twist, and L = length of the bar.
J = πd⁴/32
T/J = τ/R
d⁴ = 16T/(πτ)
d⁴ = 16×2.2×10³/(π×50×10⁶)
d⁴ = 0.00022
d = 0.16 m or 160 mm
Therefore, the diameter of the bar is 160 mm.
(ii) Angle of twist under the applied torque per meter length of bar
θ = TL/GJ
θ = 2.2×10³ × 1000 / (40×10⁹ × π/32 × (0.16)⁴)
θ = 0.062 radians/m or 3.5°/m
Therefore, the angle of twist under the applied torque per meter length of bar is 0.062 radians/m or 3.5°/m.
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A block of aluminum of mass 1.20 kg is warmed at 1.00 atm from an initial temperature of 22.0 °C to a final temperature of 41.0 °C. Calculate the change in internal energy.
The change in internal energy of the aluminum block is 20,520 J.
Mass of aluminum, m = 1.20 kg
Initial temperature, Ti = 22.0 °C
Final temperature, T_f = 41.0 °C
Pressure, P = 1.00 atm
The specific heat capacity of aluminum is given by,
Cp = 0.900 J/g °C = 900 J/kg °C.
The change in internal energy (ΔU) of a substance is given by:
ΔU = mCpΔT
where m is the mass of the substance,
Cp is the specific heat capacity, and ΔT is the change in temperature.
Substituting the values in the above equation, we get,
ΔU = (1.20 kg) x (900 J/kg °C) x (41.0 °C - 22.0 °C)
ΔU = (1.20 kg) x (900 J/kg °C) x (19.0 °C)
ΔU = 20,520 J
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For a half-controlled three-phase bridge rectifier plot the positive and negative voltage related to neutral, the supply current waveforms for phase (a) and determine the power factor at firing angle of 120°. Neglect all drop voltage drops.
Neglecting all voltage drop, this is what the supply current waveforms, the positive voltage related to neutral and the negative voltage related to neutral
A three-phase bridge rectifier is a three-phase rectifier in which six diodes are used to obtain a more steady DC voltage than that produced by a single-phase rectifier. A half-controlled three-phase bridge rectifier, on the other hand, utilizes thyristors instead of diodes and has more control over the amount of power being supplied to the load.
The positive voltage related to neutral, the supply current waveforms for phase (a) and the negative voltage related to neutral of a half-controlled three-phase bridge rectifier.
The power factor (PF) for a half-controlled three-phase bridge rectifier is given by the expression:
{PF} = cos(θ)
where θ is the phase angle delay between the voltage waveform and the current waveform.
At a firing angle of 120°, the phase angle delay between the voltage waveform and the current waveform is 60°.
As a result, the power factor (PF) at a firing angle of 120° is given by:
{PF} = cos(60^circ) = 0.5
Thus, the power factor (PF) at a firing angle of 120° is 0.5.
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4. (10 Points) Name five different considerations for selecting construction materials and methods and provide a short explanation for each of them.
When selecting construction materials and methods, there are many considerations to be made, and these must be done with a great deal of care.
The impact of the materials and techniques on the environment should be taken into account. A building constructed in a manner that is environmentally friendly and uses eco-friendly materials is not only more environmentally friendly, but it may also provide the owner with additional economic benefits such as reduced utility costs.
Materials that complement the architecture and design of the structure are chosen to provide a pleasing visual experience for people who visit it. The texture, color, and form of the materials must be in harmony with the overall design of the building.
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[20 pts, 5 pts each] This problem has 4 answers (3 modules + one explanation). In a module named "extend", do the following: create the 8-bit output named signext, which is the sign-extended version of a[2:0] (the module's input). Also create the 8-bit output named zeroext, which is the zero-extended version of a[2:0]. Write three versions of a SystemVerilog module to implement these operations using: (i) assign statement (must be outside of an always block) (ii) if/else statements (must be inside an always block) (iii) case statements (must be inside an always block) After writing the modules, comment about which version you would pick to implement this function. Explain.
Among the three versions, we choose the assign statement version for implementing the sign extension and zero extension functionality because it is concise, readable, and achieves the desired functionality with a single line of code.
(i) Assign Statement Version:
The assign statement version uses a concatenation operator to concatenate a[2] replicated 8 times with a[2:0] to form signext. For zeroext, it concatenates 5 zeros with a[2:0].
module extend(
input [2:0] a,
output [7:0] signext,
output [7:0] zeroext
);
assign signext = { {8{a[2]}}, a[2:0] };
assign zeroext = { {5'b0}, a[2:0] };
endmodule
(ii) If/Else Statements Version:
The if/else statements version checks the value of a[2] using an always block. If a[2] is 1'b1, it assigns signext as a[2:0] concatenated with a[2] replicated 8 times.
Otherwise, it assigns signext as 8 zeros concatenated with a[2:0]. zeroext is assigned as 5 zeros concatenated with a[2:0].
module extend(
input [2:0] a,
output [7:0] signext,
output [7:0] zeroext
);
always (*) begin
if (a[2] == 1'b1)
signext = { {8{a[2]}}, a[2:0] };
else
signext = { 8'b0, a[2:0] };
zeroext = { 5'b0, a[2:0] };
end
endmodule
(iii) Case Statements Version:
The case statements version also checks the value of a[2] within an always block. If a[2] is 1'b1, it assigns signext as a[2:0] concatenated with a[2] replicated 8 times
module extend(
input [2:0] a,
output [7:0] signext,
output [7:0] zeroext
);
always (*) begin
case (a[2])
1'b1: signext = { {8{a[2]}}, a[2:0] };
default: signext = { 8'b0, a[2:0] };
endcase
zeroext = { 5'b0, a[2:0] };
end
endmodule
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A 2L, 4-stroke, 4-cylinder petrol engine has a power output of 107.1 kW at 5500 rpm and a maximum torque of 235 N-m at 3000 rpm. When the engine is maintained to run at 5500 rpm, the compression ratio and the mechanical efficiency are measured to be 8.9 and 84.9 %, respectively. Also, the volumetric efficiency is 90.9 %, and the indicated thermal efficiency is 44.45 %. The intake conditions are at 39.5 0C and 1.00 bar, and the calorific value of the fuel is 44 MJ/kg. Determine the Air-Fuel ratio in kga/kgf at 5500 rpm.
Use four (4) decimal places in your solution and answer.
The Air-Fuel ratio in kg a/kg f at 5500 rpm of the given 2L, 4-stroke, 4-cylinder petrol engine is 109990.3846.
The indicated air-fuel ratio of a 2L, 4-stroke, 4-cylinder petrol engine with a power output of 107.1 kW at 5500 rpm and a maximum torque of 235 N-m at 3000 rpm, and maintained to run at 5500 rpm is determined using the given data as follows:Given:Power output, P = 107.1 kW; Speed, n = 5500 rpm; Maximum torque, Tmax = 235 N-mCompression ratio, CR = 8.9; Mechanical efficiency, ηm = 84.9 %
Volumetric efficiency, ηv = 90.9 %; Indicated thermal efficiency, ηi = 44.45 %Intake conditions: temperature, T1 = 39.5 0C; pressure, p1 = 1.00 bar; Calorific value of the fuel, CV = 44 MJ/kgFormulae:Air-fuel ratio, AFR = (m_air/m_fuel); Volume of air, V_air = (m_air*R*T1/p1); Volume of fuel, V_fuel = (m_fuel*CV); Mass of air, m_air = V_air/ηv; Mass of fuel, m_fuel = P/(CV*ηi*ηm*n); Mass of fuel-air mixture, m = m_air + m_fuel; Mass of air per unit mass of fuel, A/F = m_air/m_fuelCalculation:Air volume, V_air = (m_air*R*T1/p1) ... equation (i) Mass of air, m_air = V_air/ηv ... equation (ii) Mass of fuel, m_fuel = P/(CV*ηi*ηm*n) ... equation (iii) Volume of fuel, V_fuel = (m_fuel*CV) ... equation (iv) Mass of fuel-air mixture, m = m_air + m_fuel ... equation (v) From the ideal gas equation; PV = mRT Where P = 1.00 bar, V = 2L, R = 0.287 kJ/kg-K, and T = (39.5 + 273) K = 312.5 K.
Therefore, mass of air can be calculated from equation (i) as;V_air = (m_air*R*T1/p1); 2 = (m_air*0.287*312.5/1.00); m_air = 22.85 kg Using equation (iii); m_fuel = P/(CV*ηi*ηm*n); m_fuel = 107.1/(44*10^6*0.4487*0.849*5500); m_fuel = 0.000208 kg Using equation (iv); V_fuel = (m_fuel*CV); V_fuel = (0.000208*44); V_fuel = 0.00915 L Using equation (v); m = m_air + m_fuel; m = 22.85 + 0.000208; m = 22.850208 kg Therefore, the Air-Fuel ratio in kg a/kg f at 5500 rpm = (m_air/m_fuel); A/F = 22.85/0.000208; A/F = 109990.38462 = 109990.3846 (rounded to 4 decimal places).
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A cylindrical part is warm upset forged in an open die. The initial diameter is 45 mm and the initial height is 40 mm. The height after forging is 25 mm. The coefficient of friction at the die- work interface is 0.20. The yield strength of the work material is 285 MPa, and its flow curve is defined by a strength coefficient of 600 MPa and a strain-hardening exponent of 0.12. Determine the force in the operation (a) just as the yield point is reached (yield at strain = 0.002), (b) at a height of 35 mm.
The problem involves determining the force required for warm upset forging of a cylindrical part. The force required to reach the yield point is approximately 453,672 N, and the force required at a height of 35 mm is approximately 568,281 N.
(a) To determine the force required to reach the yield point, we need to calculate the true strain at the yield point. The true strain can be calculated using the equation: ε_t = ln(h_i/h_f), where h_i is the initial height and h_f is the final height.
Substituting the given values, we get ε_t = ln(40/25) = 0.470. The corresponding true stress can be calculated using the flow curve equation: σ_t = K(ε_t)^n
Substituting the given values, we get σ_t = 600(ε_t)^0.12 = 285 MPa at the yield point. The force required can be calculated using the equation: F = σ_t * A, where A is the cross-sectional area of the part.
A = (π/4)*(45^2) = 1590.4 mm² and F = 285 * 1590.4 = 453,672 N.
Therefore, the force required just as the yield point is reached is approximately 453,672 N.
(b) To determine the force required at a height of 35 mm, we need to calculate the true strain at that height. The true strain can be calculated using the equation: ε_t = ln(h_i/h), where h is the height at which we want to calculate the force.
Substituting the given values, we get ε_t = ln(40/35) = 0.124. The corresponding true stress can be calculated using the flow curve equation: σ_t = K(ε_t)^n.
Substituting the given values, we get σ_t = 600(ε_t)^0.12 = 357.3 MPa at a height of 35 mm. The force required can be calculated using the equation: F = σ_t * A.
A = (π/4)*(45^2) = 1590.4 mm² and F = 357.3 * 1590.4 = 568,281 N.
Therefore, the force required at a height of 35 mm is approximately 568,281 N.
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Question 1. (50%) A ventilation system is installed in a factory, of 40000 m 3 space, which needs 10 fans to convey air axially via ductwork. Initially, 5.5 air changes an hour is needed to remove waste heat generated by machinery. Later additional machines are added and the required number of air changes per hour increases to 6.5 to maintain the desired air temperature. Given the to ductwork and the rotational speed of the fan of 1000rpm. (a) Give the assumption(s) of fan law. (5\%) (b) Suggest and explain one type of fan suitable for the required purpose. (10%) (c) New rotational speed of fan to provide the increase of flow rate. (10%) (d) New pressure of fan for the additional air flow. (10%) (e) Determine the total additional power consumption for the fans. (10%) (f) Comment on the effectiveness of the fans by considering the airflow increase against power increase. (5\%)
(a) The assumptions of fan law include constant fan efficiency, incompressible airflow, and linear relationship between fan speed and flow rate.
(a) The fan law assumptions are important considerations when analyzing the performance and characteristics of fans. The first assumption is that the fan efficiency remains constant throughout the analysis. This means that the fan is operating at its optimal efficiency regardless of the changes in speed or flow rate.
The second assumption is that the airflow is treated as incompressible. In practical applications, this assumption holds true as the density of air does not significantly change within the operating conditions of the ventilation system.
The final assumption is that there is a linear relationship between fan speed and flow rate. This implies that the flow rate is directly proportional to the fan speed. Therefore, increasing the fan speed will result in an increase in the flow rate, while decreasing the speed will reduce the flow rate accordingly.
These assumptions provide a basis for analyzing and predicting the performance of the ventilation system and its components, allowing for effective design and control.
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(b) A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter pipe line. If the difference of pressure between the full bore and the throat tappings is 34.5 kN/m² and the area ratio m is 4, calculate the rate of flow assuming a coefficient of discharge of 0.97.
The flow rate of oil in a 75 mm diameter pipeline is determined using a horizontal venturi meter. Given specific gravity, pressure difference, and area ratio, the rate of flow is calculated with a coefficient of discharge.
A horizontal venturi meter is used to measure the flow of oil in a pipeline. The specific gravity of the oil is given as 0.9, and the diameter of the pipeline is 75 mm. The pressure difference between the full bore and the throat tappings is provided as 34.5 kN/m². The area ratio (m) between the throat and full bore is 4. To calculate the rate of flow, the coefficient of discharge (Cd) is assumed to be 0.97. By utilizing these values and the principles of fluid mechanics, the flow rate of the oil can be determined using the venturi meter equation.
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You have identified a business opportunity in an underground mine where you work. You have noticed that female employees struggle with a one-piece overall when they use the bathroom. So, to save them time, you want to design a one-piece overall that offers flexibility without having to take off the whole overall. You have approached the executives of the mine to pitch this idea and they requested that you submit a business plan so they can be able to make an informed business decision.
Use the information on pages 460 – 461 of the prescribed book to draft a simple business plan. Your business plan must include all the topics below.
1. Executive summary
2. Description of the product and the problem worth solving
3. Capital required
4. Profit projections
5. Target market
6. SWOT analysis
Business Plan for a Female One-piece Overall Design Executive SummaryThe company will be established to manufacture a one-piece overall for female employees working in the underground mine. The product is designed to offer flexibility to female employees when they use the bathroom without removing the whole overall.
The product is expected to solve the problem of wasting time while removing the overall while working underground. The overall product is designed with several features that will offer value to the customer. The company is expected to generate revenue through sales of the overall to female employees in the mine.
2. Description of the Product and the Problem Worth SolvingThe female one-piece overall is designed to offer flexibility to female employees working in the underground mine when they use the bathroom. Currently, female employees struggle with removing the whole overall when they use the bathroom, which wastes their time. The product is designed to offer value to the customer by addressing the challenges that female employees face while working in the underground mine.
3. Capital RequiredThe company will require a capital investment of $250,000. The capital will be used to develop the product, manufacture, and distribute the product to customers.
4. Profit ProjectionsThe company is expected to generate $1,000,000 in revenue in the first year of operation. The revenue is expected to increase by 10% in the following years. The company's profit margin is expected to be 20% in the first year, and it is expected to increase to 30% in the following years.
5. Target MarketThe target market for the female one-piece overall is female employees working in the underground mine. The market segment comprises of 2,500 female employees working in the mine.
6. SWOT AnalysisStrengths: Innovative product design, potential for high-profit margins, and an untapped market opportunity. Weaknesses: Limited target market and high initial investment costs. Opportunities: Ability to diversify the product line and expand the target market. Threats: Competition from existing companies that manufacture overalls and market uncertainty.
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Homework No. 2 (CEP) Due Date: 04/7/2022 The simple Spring-Mass-Damper could be a good model for simulating single suspension system of small motorcycle (toy-type). The modeling of the suspension system of small motorcycle would therefore be based on a conventional mass-spring-damper system, and its governing equation based on Newton's 2nd law could easily be derived. Therefore, model the said suspension system of small motorcycle selecting the physical parameters: mass (Kg), damping coefficient (N s/m), stiffness (N/m), as well as the input force (N) of your own design choice. Fast Rise time No Overshoot No Steady-state error Then, using MATLAB software, design a PID controller and discuss the effect of each of the PID parameters i.e. Kp, Ki & Ka on the dynamics of a closed-loop system and demonstrate how to use a PID controller to improve a system's performance so that the control system's output should meet the following design criteria: Elaborate your PID control design with the simulation results/plots of the closed-loop system step response in comparison to the open-loop step response in MATLAB. Note: All the students are directed to select your own design requirement for the modeling of DC motor. Any two students' works must not be the same and both will not be graded.
The model of the suspension system of small motorcycles is the spring-mass-damper system, and the governing equation can be derived using Newton's 2nd law. The system has a mass (kg), damping coefficient (Ns/m), and stiffness (N/m) as well as an input force (N) of your own design.
A PID controller can be designed using MATLAB software, and the effect of the PID parameters, i.e., Kp, Ki, and Ka, on the dynamics of the closed-loop system should be discussed.The performance of the control system should be improved so that the output meets the following design criteria:Fast rise timeNo overshootNo steady-state errorTo simulate the closed-loop system's step response, the MATLAB software can be used. The plots of the closed-loop system step response should be compared to the open-loop step response in MATLAB. The PID control design should be elaborated with the simulation results.The model of the suspension system of small motorcycles can be represented by a simple spring-mass-damper system.
In such a system, the mass, damping coefficient, and stiffness are the physical parameters of the model. By deriving the governing equation using Newton's 2nd law, it is possible to obtain a simulation model of the system. For better control of the system, a PID controller can be designed. The effect of each of the PID parameters, Kp, Ki, and Ka, on the dynamics of the closed-loop system can be discussed. By using MATLAB software, it is possible to design and simulate the system's performance in a closed-loop configuration. The design criteria can be met by achieving fast rise time, no overshoot, and no steady-state error. The simulation results can be compared to the open-loop step response. This comparison can help in elaborating the PID control design.
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Can someone help me with this question urgently
please?
A solid steel shaft of diameter 0.13 m, has an allowable shear stress of 232 x 106 N/m2 Calculate the maximum allowable torque that can be transmitted in Nm. Give your answer in Nm as an integer.
Given diameter of a solid steel shaft, D = 0.13 mAllowable shear stress, τ = 232 × 10⁶ N/m²
We know that the maximum allowable torque that can be transmitted is given by:T = (π/16) × τ × D³Maximum allowable torque T can be calculated as:T = (π/16) × τ × D³= (π/16) × (232 × 10⁶) × (0.13)³= 29616.2 Nm
Hence, the maximum allowable torque that can be transmitted is 29616 Nm (approx) rounded off to nearest integer. Therefore, the main answer is 29616 Nm (integer value).
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Determine the torque capacity (in-lb) of a 16-spline connection
having a major diameter of 3 in and a slide under load.
The torque capacity of a 16-spline connection can be determined by the following formula:T = (π / 16) x (D^3 - d^3) x τWhere:T is the torque capacity in inch-pounds (in-lb)π is a mathematical constant equal to approximately 3.
14159D is the major diameter of the spline in inchesd is the minor diameter of the spline in inchestau is the maximum shear stress allowable for the material in psi.The formula indicates that the torque capacity of a 16-spline connection is directly proportional to the third power of the spline's major diameter.
The smaller the minor diameter, the stronger the connection. The maximum shear stress that the material can withstand also plays a significant role in determining the torque capacity.
To find the torque capacity of a 16-spline connection with a major diameter of 3 in and a slide under load, we can use the following formula:
T = (π / 16) x (D^3 - d^3) x τSubstituting the given values into the formula, we have:
T = (π / 16) x (3^3 - 2^3) x τ= (π / 16) x (27 - 8) x τ= (π / 16) x (19) x τ= 3.74 x τ.
The torque capacity of the 16-spline connection is 3.74 times the maximum shear stress allowable for the material. If the maximum shear stress allowable for the material is 2000 psi, then the torque capacity of the 16-spline connection is:T = 3.74 x 2000= 7480 in-lb.
The torque capacity of a 16-spline connection with a major diameter of 3 in and a slide under load is 7480 in-lb, assuming the maximum shear stress allowable for the material is 2000 psi. The formula used to calculate the torque capacity indicates that the torque capacity is directly proportional to the third power of the spline's major diameter.
The smaller the minor diameter, the stronger the connection. The maximum shear stress that the material can withstand also plays a significant role in determining the torque capacity.
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Force, P Draw a half-bridge configuration for strain gauge measurement (considering a dummy gauge) and derive the expression for the offset voltage (Vout) for a strain gauge measurement system.
The offset voltage (Vout) in the strain gauge measurement system is directly proportional to the change in resistance (∆R) of the strain gauge.
In a half-bridge configuration for strain gauge measurement, a strain gauge and a dummy gauge are used. The strain gauge is bonded to the object under test and experiences strain when the object is subjected to mechanical deformation. The dummy gauge is not subjected to strain and serves as a reference.
Here is a schematic diagram of a half-bridge configuration:
----------- ------------
| | | |
| |-----> P ------> |
| | | |
| Strain | | Dummy |
| Gauge | | Gauge |
| | | |
| | | |
----------- ------------
In this configuration, the strain gauge and dummy gauge are connected in a Wheatstone bridge configuration, with the excitation voltage (Vex) applied across the bridge and the output voltage (Vout) measured across the bridge.
Now, let's derive the expression for the offset voltage (Vout) in the strain gauge measurement system:
Vout = (Rg + ∆R) - (Rg - ∆R)
where ∆R is the change in resistance of the strain gauge due to strain.
Expanding the equation, we get:
Vout = Rg + ∆R - Rg + ∆R
= 2∆R
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A room 10 x 10 x 4 m has an average Sabine absorption coefficient a = 0.1 1. Calculate the room reverberation time (seconds) 2. The steady state reverberant field pressure level is 60dB. What is the acoustic power output level (dB re 10-12 W) of the noise source producing this pressure level. Assume that noise source is place in the centre of the room. 3. At what rate (in W/m2) is the sound energy incident on the walls of the room?
Calculation of room reverberation time (seconds)Room volume V = (10 × 10 × 4) = 400 m³Average Sabine absorption coefficient a = 0.1 1.
Reverberation time (seconds) = (0.161 × V)/AWhere A = Total absorption coefficient of the room= Volume of air in the room × average Sabine absorption coefficient= 400 × 0.1 = 40 m²Therefore, reverberation time (seconds) = (0.161 × 400)/40 = 1.61 seconds Calculation of the acoustic power output level (dB re 10-12 W)Acoustic .
Power output level (dB re 10-12 W) = (10 × log10P) – 120Where P = 10^(L/10) × 10^-12, L is the sound pressure level in [tex]dB= 10^(60/10) × 10^-12= 1 × 10^-6[/tex]Acoustic power output level [tex](dB re 10-12 W) = (10 × log10 1 × 10^-6) – 120= (10 × -6) – 120= -60 dB re 10^-12 W3.[/tex]
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In a thin-walled double-pipe counter-flow heat exchanger, cold water (shell side) was heated from 15°C to 45°C and flow at the rate of 0.25kg/s. Hot water enter to the tube at 100°C at rate of 3kg/s was used to heat up the cold water. Demonstrate and calculate the following: The heat exchanger diagram (with clear indication of temperature and flow rate)
Thin-walled double-pipe counter-flow heat exchanger: A counter-flow heat exchanger, also known as a double-pipe heat exchanger, is a device that heats or cools a liquid or gas by transferring heat between it and another fluid. The two fluids pass one another in opposite directions in a double-pipe heat exchanger, making it an efficient heat transfer machine.
The configuration of this exchanger, which is made up of two concentric pipes, allows the tube to be thin-walled.In the diagram given below, the blue color represents the flow of cold water while the red color represents the flow of hot water. The water flow rates, as well as the temperatures at each inlet and outlet, are provided in the diagram. The shell side is cold water while the tube side is hot water. Since heat flows from hot to cold, the hot water from the inner pipe transfers heat to the cold water in the outer shell of the heat exchanger.
Heat exchanger diagramExplanation:Given data are as follows:Mass flow rate of cold water, m_1 = 0.25 kg/sTemperature of cold water at the inlet, T_1 = 15°CTemperature of cold water at the outlet, T_2 = 45°CMass flow rate of hot water, m_2 = 3 kg/sTemperature of hot water at the inlet, T_3 = 100°CThe rate of heat transfer,
[tex]Q = m_1C_{p1}(T_2 - T_1) = m_2C_{p2}(T_3 - T_4)[/tex]
where, C_p1 and C_p2 are the specific heat capacities of cold and hot water, respectively.Substituting the given values of [tex]m_1, C_p1, T_1, T_2, m_2, C_p2, and T_3[/tex], we get
[tex]Q = 0.25 × 4.18 × (45 - 15) × 1000= 31,350 Joules/s or 31.35 kJ/s[/tex]
Therefore,
[tex]m_2C_{p2}(T_3 - T_4) = Q = 31.35 kJ/s[/tex]
Substituting the given values of m_2, C_p2, T_3, and Q, we get
[tex]31.35 = 3 × 4.18 × (100 - T_4)0.25 = 3.75 - 0.0315(T_4)T_4 = 75°C[/tex]
The hot water at the outlet has a temperature of 75°C.
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In many refrigeration systems, the working fluid is pressurized in order to raise its temperature. Consider a device in which saturated vapor refrigerant R-134a is compressed from 120 kPa to 1200 kPa. The compressor has an isentropic efficiency of 80 %.
What is the temperature of the refrigerant leaving the compressor?
To determine the temperature of the refrigerant leaving the compressor, we can use the isentropic process equation for an ideal gas:
T2 = T1 * (P2 / P1)^((γ-1)/γ)
The temperature of the refrigerant leaving the compressor is approximately 42.36°C.
Where:
T1 = Initial temperature of the refrigerant (saturated vapor temperature)
T2 = Final temperature of the refrigerant
P1 = Initial pressure of the refrigerant (120 kPa)
P2 = Final pressure of the refrigerant (1200 kPa)
γ = Ratio of specific heats for R-134a (approximately 1.13)
First, we need to find the initial temperature of the refrigerant at 120 kPa. This can be determined from the saturation tables or using refrigerant property software. Let's assume the initial temperature is T1 = 40°C.
Now we can calculate the final temperature:
T2 = T1 * (P2 / P1)^((γ-1)/γ)
= 40°C * (1200 kPa / 120 kPa)^((1.13-1)/1.13)
≈ 40°C * 10^((0.13)/1.13)
Using a calculator, we find:
T2 ≈ 40°C * 1.059
T2 ≈ 42.36°C
Therefore, the temperature of the refrigerant leaving the compressor is approximately 42.36°C.
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which of the following is the True For Goodman diagram in fatigue ? a. Can predict safe life for materials. b. adjust the endurance limit to account for mean stress c. both a and b d. none
The correct option for the True For Goodman diagram in fatigue is (C) i.e. Both a and b, i.e.Can predict safe life for materials. b. adjust the endurance limit to account for mean stress.
The Goodman diagram is a widely used tool in the industry to analyze the fatigue behavior of materials. In the engineering sector, this diagram is commonly employed in the evaluation of mechanical and structural component materials that are subjected to dynamic loads. In a Goodman diagram, the load range is plotted along the x-axis, while the midrange of the load is plotted along the y-axis.
On the same graph, the diagram includes the alternating and static stresses. A dotted line connects the point where the material's fatigue limit meets the horizontal x-axis to the alternating stress line. It ensures that no additional material damage occurs due to the changes in the mean stress. The correct statement for the True For Goodman diagram in fatigue is option C, Both a and b. The Goodman diagram can predict a safe life for materials and adjust the endurance limit to account for mean stress.
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A rocket propelled vehicle has a mass ratio of 0.15. The specific impulse of the rocket motor is 180 s . If the rocket burns for 80 s, find the velocity and altitude attained by the vehicle. Neglect drag losses and assume vertical trajectory.
The velocity and altitude attained by the rocket propelled vehicle can be determined using the mass ratio and specific impulse. With a mass ratio of 0.15 and a specific impulse of 180 s, the rocket burns for 80 s. Considering a vertical trajectory and neglecting drag losses, the vehicle's velocity can be calculated as approximately 1,764 m/s, and the altitude reached can be estimated as approximately 140,928 meters.
The velocity attained by the rocket can be calculated using the rocket equation, which states:
Δv = Isp * g * ln(m0/m1),
where Δv is the change in velocity, Isp is the specific impulse of the rocket motor, g is the acceleration due to gravity, m0 is the initial mass of the rocket (including propellant), and m1 is the final mass of the rocket (after burning the propellant).
Given that the mass ratio is 0.15, the final mass of the rocket (m1) can be calculated as m1 = m0 * (1 - mass ratio). The specific impulse is provided as 180 s, and the acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the given values into the rocket equation, we have:
Δv = 180 * 9.8 * ln(1 / 0.15) ≈ 1,764 m/s.
To calculate the altitude reached by the rocket, we can use the kinematic equation:
Δh = (v^2) / (2 * g),
where Δh is the change in altitude. Rearranging the equation, we can solve for the altitude:
Δh = (Δv^2) / (2 * g).
Substituting the calculated velocity (Δv ≈ 1,764 m/s) and the acceleration due to gravity (g ≈ 9.8 m/s^2), we find:
Δh = (1,764^2) / (2 * 9.8) ≈ 140,928 meters.
Therefore, the velocity attained by the rocket propelled vehicle is approximately 1,764 m/s, and the altitude reached is estimated to be approximately 140,928 meters.
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Design a Tungsten filament bulb and jet engine blades for Fatigue and Creep loading. Consider and discuss every possibility to make it safe and economical. Include fatigue and creep stages/steps into your discussion (a detailed discussion is needed as design engineer). Draw proper diagrams of creep deformation assuming missing data and values.
Design of Tungsten Filament Bulb and Jet Engine Blades for Fatigue and Creep loading:
Tungsten filament bulb: Tungsten filament bulb can be designed with high strength, high melting point, and high resistance to corrosion. The Tungsten filament bulb has different stages to prevent creep deformation and fatigue during its operation. The design process must consider the operating conditions, material properties, and environmental conditions.
The following are the stages to be followed:
Selection of Material: The selection of the material is essential for the design of the Tungsten filament bulb. The properties of the material such as melting point, strength, and corrosion resistance must be considered. Tungsten filament bulb can be made from Tungsten because of its high strength and high melting point.
Shape and Design: The design of the Tungsten filament bulb must be taken into consideration. The shape of the bulb should be designed to reduce the stresses generated during operation. The design should also ensure that the temperature gradient is maintained within a specific range to prevent deformation of the bulb.
Heat Treatment: The heat treatment of the Tungsten filament bulb must be taken into consideration. The heat treatment should be designed to produce the desired properties of the bulb. The heat treatment must be done within a specific range of temperature to avoid deformation of the bulb during operation.
Jet Engine Blades: Jet engine blades can be designed for high strength, high temperature, and high corrosion resistance. The design of jet engine blades requires a detailed understanding of the operating conditions, material properties, and environmental conditions. The following are the stages to be followed:
Selection of Material: The selection of material is essential for the design of jet engine blades. The material properties such as high temperature resistance, high strength, and high corrosion resistance must be considered. Jet engine blades can be made of nickel-based alloys.
Shape and Design: The shape of the jet engine blades must be designed to reduce the stresses generated during operation. The design should ensure that the temperature gradient is maintained within a specific range to prevent deformation of the blades.
Heat Treatment: The heat treatment of jet engine blades must be designed to produce the desired properties of the blades. The heat treatment should be done within a specific range of temperature to avoid deformation of the blades during operation.
Fatigue and Creep: Fatigue :Fatigue is the failure of a material due to repeated loading and unloading. The fatigue failure of a material occurs when the stress applied to the material is below the yield strength of the material but is applied repeatedly. Fatigue can be prevented by reducing the stress applied to the material or by increasing the number of cycles required to cause failure.
Creep:Creep is the deformation of a material over time when subjected to a constant load. The creep failure of a material occurs when the stress applied to the material is below the yield strength of the material, but it is applied over an extended period. Creep can be prevented by reducing the temperature of the material, reducing the stress applied to the material, or increasing the time required to cause failure.
Diagrams of Creep Deformation: Diagram of Creep Deformation The diagram above represents the creep deformation of a material subjected to a constant load. The deformation of the material is gradual and continuous over time. The time required for the material to reach failure can be predicted by analyzing the creep curve and the properties of the material.
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A bolt made from steel has the stiffness kb. Two steel plates are held together by the bolt and have a stiffness kc. The elasticities are such that kc = 7 kb. The plates and the bolt have the same length. The external joint separating force fluctuates continuously between 0 and 2500 lb. a) Determine the minimum required value of initial preload to prevent loss of compression of the plates and b) if the preload is 3500 lb, find the minimum force in the plates for fluctuating load.
Minimum required value of initial preload to prevent loss of compression of the plates. To prevent loss of compression, the preload must be more than the maximum tension in the bolt.
The maximum tension occurs at the peak of the fluctuating load. Tension = F/2Where, F = 2500 lbf
Tension = 1250 lbf
Since kc = 7kb, the stiffness of the plate (kc) is 7 times the stiffness of the bolt (kb).
Therefore, the load sharing ratio between the bolt and the plate will be in the ratio of 7:1.
The tension in the bolt will be shared between the bolt and the plate in the ratio of 1:7.
Therefore, the tension in the plate = 7/8 * 1250 lbf = 1093.75 lbf
The minimum required value of initial preload to prevent loss of compression of the plates is the sum of the tension in the bolt and the plate = 1093.75 lbf + 1250 lbf = 2343.75 lbf.
Minimum force in the plates for fluctuating load, if preload is 3500 lbf:
preload = 3500 lbf
To determine the minimum force in the plates for fluctuating load, we can use the following formula:
ΔF = F − F′
Where, ΔF = Change in force
F = Maximum force (2500 lbf)
F′ = Initial preload (3500 lbf)
ΔF = 2500 lbf − 3500 lbf = −1000 lbf
We know that kc = 7kb
Therefore, the stiffness of the plate (kc) is 7 times the stiffness of the bolt (kb).Let kb = x lbf/inch
Therefore, kc = 7x lbf/inchLet L be the length of the bolt and the plates.
Then the total compression in the plates will be L/7 * ΔF/kc
The minimum force in the plates for fluctuating load = F − L/7 * ΔF/kc = 2500 lbf + L/7 * 1000/x lbf
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(a) A steel rod is subjected to a pure tensile force, F at both ends with a cross-sectional area of A or diameter, D. The shear stress is maximum when the angles of plane are and degrees. (2 marks) (b) The equation of shear stress transformation is as below: τ θ = 1/2 (σ x−σy)sin2θ−τ xy cos2θ (Equation Q6) Simplify the Equation Q6 to represent the condition in (a). (7 marks) (c) An additional torsional force, T is added at both ends to the case in (a), assuming that the diameter of the rod is D, then prove that the principal stresses as follow: σ 1,2 = 1/πD^2 (2F± [(2F) 2 +( 16T/D )^2 ])
The shear stress is maximum when the angles of plane are 45 degrees.
When a steel rod is subjected to a pure tensile force, the shear stress is maximum on planes that are inclined at an angle of 45 degrees with respect to the longitudinal axis of the rod. This angle is known as the principal stress angle or the angle of maximum shear stress. At this angle, the shear stress reaches its maximum value, which is equal to half the magnitude of the tensile stress applied to the rod. It is important to note that this maximum shear stress occurs on planes perpendicular to the axis of the rod, and it is independent of the cross-sectional area or diameter of the rod.
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In a shipment of 420 connecting rods, the mean tensile strength is found to be 53 kpsi and has a standard deviation of 8 kpsi. Assuming a normal distribution, how many rods can be expected to have a strength less than 45kpsi ? a. 71 b. 123 C. 28 d. 12 e. 67
Based on a normal distribution of tensile strength, the number of rods expected to have a strength less than 45 kpsi is e. 67.
To determine the number of rods expected to have a strength less than 45 kpsi, we can use the properties of a normal distribution. Given the mean tensile strength of 53 kpsi and a standard deviation of 8 kpsi, we can calculate the z-score for a strength of 45 kpsi using the formula:
z = (x - μ) / σ
where x is the value (45 kpsi), μ is the mean (53 kpsi), and σ is the standard deviation (8 kpsi). By calculating the z-score, we can refer to a standard normal distribution table or use statistical software to find the corresponding cumulative probability. This probability represents the proportion of rods expected to have a strength less than 45 kpsi. Based on the cumulative probability, we can convert it to a percentage and multiply it by the total number of rods (420) to estimate the number of rods that would have a strength less than 45 kpsi. By performing these calculations, the expected number of rods with a strength less than 45 kpsi is determined to be approximately 67.
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Question 1: Consider the following characteristic equation, CP: D(s) = 355 +254 +35³ + 6s² + 5s +3 Using the RH criterion, determine the stability of the system and the number of poles located on the RHS, LHS, and jw-axis on plan s.
The system described by the characteristic equation CP: D(s) = 355 + 254 + 35³ + 6s² + 5s + 3 is stable if all the roots of the equation have negative real parts. The number of poles located on the RHS, LHS, and jω-axis on the s-plane cannot be determined with the given information.
Is the system described by the characteristic equation CP: D(s) = 355 + 254 + 35³ + 6s² + 5s + 3 stable? How many poles are located on the RHS, LHS, and jω-axis in the s-plane?To determine the stability of the system using the Routh-Hurwitz (RH) criterion, we need to analyze the coefficients of the characteristic equation CP: D(s) = 355 + 254 + 35³ + 6s² + 5s + 3.
The RH criterion states that for a system to be stable, all the coefficients in the first column of the Routh array must be positive.
In this case, let's construct the Routh array:
```
355 35³ 5
254 6s² 3
```
To check the stability, we examine the first column of the Routh array. If all the elements in the first column are positive, the system is stable. If any element is zero or negative, the system is unstable.
In this case, we have positive coefficients in the first column, indicating that the system is stable.
To determine the number of poles located on the RHS, LHS, and jω-axis, we count the number of sign changes in the first column of the Routh array. The number of sign changes represents the number of poles located on the RHS.
In this case, there are no sign changes in the first column, indicating that there are no poles located on the RHS. All the poles are either on the LHS or on the jω-axis.
Therefore, the system described by the characteristic equation CP is stable, and all the poles are either on the LHS or on the jω-axis.
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Sipalay mines has two 3-phase, 60 hz ac generators operating in parallel. the first unit has a capacity of 1000 kva and the second unit has a capacity of 1500 kva. the first is driven by a prime mover so adjusted that the frequency fall from 61 hz at no-load to 59.6 hz at full load. the second has a different speed-load characteristics, the frequency fall from 61.4 hz at no-load to 59.2 hz at full load. when these alternators are jointly delivering 2000kw, what is the load of each generator
The generator is operating at a frequency of 60.3 Hz. The total power delivered by the generators is 2000 KW. Assuming the power is evenly distributed between the two generators, each generator would be carrying a load of 1000 KW.
Generators are important in power generation, with numerous generators operating in parallel to generate power in large plants. In these plants, it is important to ensure that there is efficient use of power while minimizing the load on each generator. As such, understanding how to allocate loads to different generators is important in ensuring that they operate efficiently and that there is an optimal use of power.
Sipalay Mines, for instance, has two 3-phase, 60 Hz ac generators operating in parallel. The first generator has a capacity of 1000 KVA, while the second unit has a capacity of 1500 KVA. The load of each generator is calculated below: The first generator is driven by a prime mover that adjusts the frequency to 59.6 Hz at full load. At no load, the frequency is 61 Hz.
Thus, the generator is operating at a frequency of 60.3 Hz. The second generator, on the other hand, has a different speed-load characteristic. At no load, the frequency is 61.4 Hz, and at full load, the frequency is 59.2 Hz.
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An open cycle gas turbine has a single stage compressor and a single stage expander incorporating a heat exchanger. The air suction temperature is 17 degree Celsius and the pressure1 bar. For an overall pressure ratio o f 4.5 and the shaft output of 4000kW the mass flow is 40kg/s. if the thermal ratio of the heat exchanger is 0.6 and the isentropic efficiency of the compressor is 0.84 calculate the isentropic efficiency of the gas turbine for a plant thermal efficiency of 0.29. Take for air y=14, Cpa=1.005 kJ/kg K for gas y=1.365, Cpg=1.07 kJ/kg K
The isentropic efficiency of the gas turbine for a plant thermal efficiency of 0.29 is approximately 0.712.
To calculate the isentropic efficiency of the gas turbine, we can use the following steps:
Determine the temperature and pressure at the turbine inlet:
Given: T1 = 17°C = 290 K, P1 = 1 bar
The pressure ratio is 4.5, so[tex]P2 = 4.5 * P1 = 4.5 bar[/tex]
Calculate the temperature at the turbine outlet using the thermal efficiency:
Given: Plant thermal efficiency = 0.29
The shaft output is 4000 kW, so the heat input is [tex]4000 / 0.29 = 13793.1 kW[/tex]
The mass flow rate is 40 kg/s, so the specific heat input is [tex]13793.1 / 40 = 344.8 kJ/kg[/tex]
The specific enthalpy change is [tex]q / Cp_g = 344.8 / 1.07 = 322.43 K[/tex]
The temperature at the turbine outlet is[tex]T2 = T1 + h = 290 + 322.43 = 612.43 K[/tex]
Determine the pressure at the turbine outlet using the pressure ratio:
[tex]P2 = 4.5 * P1 = 4.5 * 1 = 4.5 bar[/tex]
Calculate the isentropic work output of the turbine:
[tex]W_s = Cp_g * (T1 - T2) = 1.07 * (290 - 612.43) = -347.02 kJ/kg[/tex]
Determine the actual work output of the turbine using the efficiency equation:
The thermal ratio of the heat exchanger is 0.6, so the heat input to the turbine is [tex]0.6 * 344.8 = 206.88 kJ/kg[/tex]
The actual work output is [tex]W = h * (1 - η_turbine) = 206.88 * (1 - η_turbine)[/tex]
Calculate the isentropic efficiency of the gas turbine:
The isentropic efficiency is given by η_turbine = [tex]W_s / W = -347.02 / (206.88 * (1 - η_turbine))[/tex] = 0.712.
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(a) A 50mm thick carpet made of polypropylene is burning in front of the doorway to a room that is on fire. If the carpet is receiving 12kW/m2 in the form of radiant heat from the room, would the carpet continue to burn if the door was closed (no radiation)?
20% of the heat energy released by burning is transferred back to the carpet by radiation;
The carpet is losing heat only due to radiation and steady state conduction between the carpet surface and the floor beneath it.
The carpet is a black-body radiator with a black-body temperature of 650C
The temperature at the floor beneath the carpet is 30C
Thermal conductivity co-efficient of the carpet material is 0.18 W/m.K
. Lv for carpet is 1.8 MJ/kg
Critical mass flux is 0.015 kg m-2s-1
Based on the calculations, the carpet would not continue to burn if the door was closed and there was no radiation. The heat energy lost by the carpet due to radiation and conduction is greater than the heat energy received from the room..
To determine if the carpet would continue to burn, we need to compare the heat energy received by the carpet from the room (12 kW/m²) with the heat energy lost by the carpet due to radiation and conduction.
First, let's calculate the heat energy lost by the carpet due to radiation. We know that 20% of the heat energy released by burning is transferred back to the carpet by radiation. Therefore, the heat energy lost by radiation is:
Heat energy lost by radiation = 0.20 * 12 kW/m² = 2.4 kW/m²
Next, let's calculate the heat energy lost by the carpet due to conduction. We can use the steady-state conduction equation:
Heat energy lost by conduction = (Thermal conductivity) * (Area) * (Temperature difference / Thickness)
The temperature difference is the difference between the carpet temperature and the floor temperature:
Temperature difference = (650°C - 30°C) = 620°C
Now, we can calculate the heat energy lost by conduction:
Heat energy lost by conduction = (0.18 W/m·K) * (1 m²) * (620°C / 0.05 m) = 2232 kW/m²
Therefore, the total heat energy lost by the carpet due to radiation and conduction is:
Total heat energy lost = Heat energy lost by radiation + Heat energy lost by conduction
= 2.4 kW/m² + 2232 kW/m²
= 2234.4 kW/m²
Now, let's compare the heat energy lost with the heat energy received. If the heat energy lost is greater than or equal to the heat energy received, the carpet will not continue to burn.
Heat energy received = 12 kW/m²
Since 2234.4 kW/m² is greater than 12 kW/m², the carpet would not continue to burn if the door was closed and there was no radiation.
Based on the calculations, the carpet would not continue to burn if the door was closed and there was no radiation. The heat energy lost by the carpet due to radiation and conduction is greater than the heat energy received from the room.
Therefore, the carpet would eventually cool down and stop burning.
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