The MRP system operates on the principle of analyzing demand, calculating net requirements, determining timing, generating planned orders, and coordinating procurement or production activities to meet the required material quantities within the desired timeframes.
Explanation of MRP System Operation Principle: 1. Start MRP: The MRP system begins by initiating the MRP process. 2. Determine Demand: The system identifies the demand for finished products based on sales forecasts, customer orders, and other factors. 3. Gross Requirements: The gross requirements are calculated by considering the demand, lead time, and safety stock. 4. Net Requirements: Net requirements are derived by subtracting the available inventory and scheduled receipts from the gross requirements. This helps determine the actual amount of materials needed. 5. Time Phasing: The system analyzes the net requirements and determines when each item needs to be available to meet the production schedule. This helps in scheduling purchases or production.
6. Planned Orders: Based on the net requirements and time phasing, the MRP system generates planned orders for purchase or production. 7. Order Release: The planned orders are reviewed, and if approved, they are released as actual orders to suppliers or the production department. 8. Purchase/Produce: In this stage, the system generates purchase orders for suppliers or production orders for internal manufacturing processes. 9. Receive/Produce: Once the orders are fulfilled, the materials are received from suppliers or the production process is initiated to manufacture the required items. 10. Update Records: The system updates inventory records, production status, and other relevant information based on the received materials or completed production. 11. End MRP: The MRP process concludes after all the planned orders have been fulfilled and the records are updated.
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It is required to transmit torque 537 N.m of from shaft 6 cm in diameter to a gear by a sunk key of length 70 mm. permissible shear stress is 60 MN/m. and the crushing stress is 120MN/m². Find the dimension of the key.
It is required to transmit torque 537 N.m of from shaft 6 cm in diameter to a gear by a sunk key of length 70 mm. The permissible shear stress is 60 MN/m. and the crushing stress is 120MN/m². Find the dimension of the key.
The dimension of the key can be calculated using the following formulae.
Torque, T = 537 N-m diameter of shaft, D = 6 cm Shear stress, τ = 60 MN/m Crushing stress, σc = 120 MN/m²Length of the key, L = 70 mm Key width, b = ?.
Radius of shaft, r = D/2 = 6/2 = 3 cm.
Let the length of the key be 'L' and the width of the key be 'b'.
Also, let 'x' be the distance of the centre of gravity of the key from the top of the shaft. Let 'P' be the axial force due to the key on the shaft.
Now, we can write the equation for the torque transmission by key,T = P×x = (τ/2)×L×b×x/L+ (σc/2)×b×L×(D-x)/LAlso, the area of the key, A = b×L.
Therefore, the shear force acting on the key is,Fs = T/r = (2T/D) = (2×537)/(3×10⁻²) = 3.58×10⁵ N.
From the formula for shear stress,τ = Fs/A.
Therefore, A = Fs/τ= 3.58×10⁵/60 × 10⁶= 0.00597 m².
Hence, A = b×L= 5.97×10⁻³ m²L/b = A/b² = 0.00597/b².
From the formula for crushing stress,σc = P/A= P/(L×b).
Therefore, P = σc×L×b= 120×10⁶×L×b.
Therefore, T = P×x = σc×L×b×x/L+ τ/2×b×(D-x).
Therefore, 537 = 120×10⁶×L×b×x/L+ 30×10⁶×b×(3-x).
Therefore, 179 = 40×10⁶×L×x/b² + 10×10⁶×(3-x).
Therefore, 179b² + 10×10⁶b(3-x) - 40×10⁶Lx = 0.
Since the key dimensions should be small, we can take Lx = 0 and solve for b.
Therefore, 179b² + 30×10⁶b - 0 = 0.
Solving the quadratic equation, we get the key width, b = 46.9 mm (approx).
Therefore, the dimension of the key is 70 mm × 46.9 mm (length × width).
Hence, the dimension of the key is 70 mm × 46.9 mm.
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The structure of the building needs to know the internal loads at various points
(true or false)
True. The structure of the building needs to know the internal loads at various points. This is because the internal loads of the building exert force on the building's structure, and the structure must be able to withstand this force.
The internal loads of a building include the weight of the building itself, the weight of the occupants and their belongings, the weight of furniture and equipment, and any other loads that are present.In order to determine the internal loads at various points, engineers and architects use a variety of methods, such as load calculations, stress analysis, and computer modeling.
By understanding the internal loads of a building, they can design a structure that is strong enough to support the weight of the building and all of its contents, and that will remain stable and safe over time.In conclusion, it is true that the structure of the building needs to know the internal loads at various points. Understanding the internal loads is an essential part of designing and constructing a building that is safe, secure, and functional.
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Steam at 300°Cflows in a stainless steel pipe (k = 15 W/m · K) whose inner and outer diameters are 6cm and 8cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k = 0.038W/m · K). Heat is lost to the surroundings at 5°C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of 22W/m²K. Taking the heat transfer coefficient inside the pipe . to be 80W/m²K, determine the rate of eat loss from the steam per unit length of the pipe. Also determine the temperature drop across the shell and the insulation.
The rate of heat loss from the steam per unit length of the pipe is 1254.82 W/m. The temperature drop across the shell and the insulation is 88.16 K.
Steam flows through a stainless steel pipe whose inner and outer diameters are 6 cm and 8 cm, respectively, at 300°C. The conductivity of the pipe is 15 W/m K, and it is covered with a 3 cm thick layer of glass wool insulation, which has a conductivity of 0.038 W/m K.
The heat transfer coefficient within the pipe is 80 W/m2 K, and the combined heat transfer coefficient for natural convection and radiation is 22 W/m2 K, with heat being lost to the environment at 5°C. We must determine the rate of heat loss from the steam per unit length of the pipe, as well as the temperature drop across the shell and insulation.
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The volumetric analysis of the product of combustion on a dry basis for the oxidation of octane (C8H18) in air is 9.18% CO2, 0.23% CO and 7.48% O2. Write the combustion equation and determine the percent excess or deficiency of air in the combustion process.
Use Mass Balance
Please complete the answer with correct solution
The percent excess or deficiency of air in the combustion process is 0.57% excess air.
For the mass balance equation, we have,Mass of unburned oxygen = (12.5 - 7.48/100 × 12.5) × 32 × x / 100Mass of air = 77/100 × x Mass of CO₂ = 4.04 g
Mass of CO = 0.064 g
Mass of O₂= 2.39 g + (12.5 - 7.48/100 × 12.5) × 32 × x / 100
Mass of H₂O = 100 - (mass of CO₂ + mass of CO + mass of O₂ + mass of unburned oxygen)
Now, we will substitute all these values into the mass balance equation
. That is,Mass of octane + 77/100 × x = 4.04 + 0.064 + 2.39 + (12.5 - 7.48/100 × 12.5) × 32 × x / 100 + Mass of H₂O
6.494 g/100 g of the product of the combustion = mass of CO₂ + mass of CO + mass of H₂O
Mass of H₂O = 100 - 6.494 = 93.506 g/100 g of the product of the combustion
By substituting all these values, the equation becomes,
114.17 + 77/100 × x = 6.494 + 0.064 + 2.39 + (12.5 - 7.48/100 × 12.5) × 32 × x / 100 + 93.506
Solving for x, we get,x = 162.27 g
Thus, the mass of air required for the combustion of 114.17 g of octane is 162.27 g.
Hence, the percent excess or deficiency of air can be calculated as, Percent excess air = (actual air - theoretical air) / theoretical air × 100= (162.27 - 161.35) / 161.35 × 100= 0.57%
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Water flows through a straight, horizontal pipe. Find the pressure minor, major, and total loss.
Given:
Density = 1000 kg/m3
Viscosity μ = 0.001 kg/m.s
Volumetric flow rate Q = 0.005 m3/s
Pipe Diameter D = 2.5 cm
Pipe Length L = 10m
Wall roughness E = 5x10-6 m
Minor Loss Coefficient K = 10
Find the following:
Average velocity
Reynolds number
Laminar or Turbulent?
To find the average velocity, we can use the volumetric flow rate Q and the pipe diameter D. The formula for average velocity (V) is:
V = Q / (π * (D/2)^2)
Given Q = 0.005 m^3/s and D = 2.5 cm = 0.025 m, we can substitute these values into the formula:
V = 0.005 / (π * (0.025/2)^2)
V ≈ 2.545 m/s
The average velocity is approximately 2.545 m/s.
To determine the Reynolds number (Re), we can use the formula:
Re = (ρ * V * D) / μ
Given:
ρ = 1000 kg/m^3 (density)
V = 2.545 m/s (average velocity)
D = 0.025 m (pipe diameter)
μ = 0.001 kg/m.s (viscosity)
Substituting these values into the formula, we get:
Re = (1000 * 2.545 * 0.025) / 0.001
Re ≈ 101800
The Reynolds number is approximately 101800.
To determine whether the flow is laminar or turbulent, we can compare the Reynolds number to a critical value. The critical Reynolds number for flow in a pipe is around 2000, above which the flow tends to be turbulent.
In this case, since the Reynolds number is approximately 101800, it is well above the critical value of 2000. Therefore, the flow is turbulent.
Now let's move on to calculating the pressure losses.
The pressure drop due to major losses can be calculated using the Darcy-Weisbach equation:
ΔP_major = (f * (L/D) * (ρ * V^2)) / 2
Where:
f is the friction factor,
L is the pipe length,
D is the pipe diameter,
ρ is the density of the fluid,
V is the average velocity.
To determine the friction factor (f), we can use the Colebrook-White equation:
1 / √f = -2 * log10((E/D)/3.7 + (2.51 / (Re * √f)))
Where:
E is the wall roughness,
D is the pipe diameter,
Re is the Reynolds number.
First, let's solve the Colebrook-White equation to find the friction factor.
We'll start with an initial guess for f, such as f = 0.02, and then iteratively solve for a more accurate value of f.
Using the given values of E = 5x10^-6 m and Re = 101800, we can substitute them into the equation:
1 / √f = -2 * log10((5x10^-6 / 0.025)/3.7 + (2.51 / (101800 * √f)))
Simplifying the equation, we have:
1 / √f = -2 * log10(0.0002/3.7 + 2.51 / (101800 * √f))
Now we can solve this equation iteratively to find the value of f.
Assuming f = 0.02 as the initial guess, we can substitute it into the equation:
1 / √0.02 = -2 * log10(0.0002/3.7 + 2.51 / (101800 * √0.02))
Calculating the right-hand side, we get:
≈ -2 * log10(0.0002/3.7 + 2.51 / (101800 * 0.1414))
≈ -2 * log10(0.0002/3.7 + 0.0175)
Using logarithmic properties, we can simplify further:
≈ -2 * log10(0.0002/3.7 + 0.0175)
≈ -2 * log10(0.0002/3.7) -2 * log10(1 + 0.0175)
≈ -2 * log10(0.0002/3.7) -2 * log10(1.0175)
Now we can solve for 1/√f:
1 / √f ≈ -2 * log10(0.0002/3.7) -2 * log10(1.0175)
1 / √f ≈ -2 * (-3.4302) -2 * (-0.9917)
1 / √f ≈ 6.8604 + 1.9834
1 / √f ≈ 8.8438
To find √f, we take the reciprocal:
√f ≈ 1 / 8.8438
√f ≈ 0.113
f ≈ (0.113)^2
f ≈ 0.0128
Now that we have the friction factor (f), we can calculate the pressure drop due to major losses using the Darcy-Weisbach equation:
ΔP_major = (f * (L/D) * (ρ * V^2)) / 2
Substituting the given values:
ΔP_major = (0.0128 * (10/0.025) * (1000 * (2.545^2))) / 2
≈ 1632.64 Pa
The pressure drop due to major losses is approximately 1632.64 Pa.
The pressure drop due to minor losses can be calculated using the following formula:
ΔP_minor = K * (ρ * V^2) / 2
Substituting the given values:
ΔP_minor = 10 * (1000 * (2.545^2)) / 2
≈ 6479.45 Pa
The pressure drop due to minor losses is approximately 6479.45 Pa.
The total pressure loss is the sum of the major and minor losses:
Total pressure loss = ΔP_major + ΔP_minor
≈ 1632.64 + 6479.45
≈ 8112.09 Pa
Therefore, the total pressure loss is approximately 8112.09 Pa.
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e. A 4-pole turbo-generator rated at 20MVA, 13.2kV, 50Hz has an inertia constant of H=8.5kW- s/kVA. Determine; i. The kinetic energy stored in the rotor at synchronous speed. ii. The acceleration if the input less the rotational losses is 17300kW and the electric power developed is 14200kW iii. The change in torque angle in that period and the rpm at the end of 10 cycles
Given data,Number of poles, P= 4Power rating, P = 20 MVA (Mega Volt Ampere)Rated voltage, V = 13.2 kV (kilo Volt)Frequency, f = 50 HzInertia constant, H = 8.5 kW- s/kVA(a) Kinetic energy stored in the rotor at synchronous speed:Synchronous speed (Ns) = 120f/P
The kinetic energy stored in the rotor (E) = 1/2 * Inertia constant * (Power rating in kVA)^2 / (Synchronous speed in rpm)Kinetic energy stored in the rotor at synchronous speedE = 1/2 * H * (P × 1000)^2 / NsE = 1/2 * 8.5 * (20,000)^2 / 1500E = 1,133,333.33 J× 1000 / 1500)α = 1.71 rad/s^2(c) Change in torque angle in that period and the RPM at the end of 10 cycles:Initial torque angle = δ1 = cos⁻¹ (Pm / (V × Ia)) = cos⁻¹ (17300 / (13200 × 1557.73)) = 1.5566 radTime period of 10 cycles, T = 10 / f = 0.2 sAt the end of 10 cycles, the final torque angle = δ2 = cos⁻¹ (Pm / (V × Ia)) = cos⁻¹ ((Pm – J × α × N × δ1) / (V × Ia))δ2 = cos⁻¹ ((423.36 – 8.5 × 20,000 × 1.71 × 1500 × 1.5566) / (13200 × 1557.73))δ2 = 1.853 radChange in torque angle, Δδ = δ2 – δ1Δδ = 1.853 – 1.5566Δδ = 0.296 radRPM at the end of 10 cycles, N1 = (P × 1000 × 60) / (Poles × f)N1 = (20,000 × 60) / (4 × 50)N1 = 2400 rpmAt the end of 10 cycles, the RPM will be given by,N2 = N1 – (α × δ1 × 30 / π)²N2 = 2400 – (1.71 × 1.5566 × 30 / π)²N2 = 2299.15 rpm
Therefore, The kinetic energy stored in the rotor at synchronous speed is 1,133,333.33J. The acceleration is 1.71 rad/s². The change in torque angle in that period is 0.296rad and the RPM at the end of 10 cycles is 2299.15 rpm.
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For air, use k = 1.4, R = 287 J/kg.K. A gas turbine consisting of a high-pressure turbine stage which drives the compressor, and a low-pressure turbine stage which drives a gearbox. The turbine has an overall pressure ratio of 4, and the temperature of the gases at entry to the high-pressure turbine is 650°C. The high-pressure turbine has an isentropic efficiency of 83% and that of the low-pressure turbine, 85%. The compressor has an isentropic efficiency of 80%. The system includes a regenerator which has an efficiency 75%. Assuming a mechanical efficiency of 98% for both shafts calculate the specific net-work output and the thermal efficiency of the system. For air take Cp = 1.005-kJ/kg.K and k = 1.4, and for the gases in the combustion chamber and in the turbines and heat exchanger take Cp = 1.15-kJ/kg.K and k = 1.333. Assume the air to enter the turbine at 295K and 101.325-kPa.
The specific net work output and thermal efficiency of the system are approximately 296.23 kJ/kg and 33.54% respectively.
How to solve
For the given gas turbine with the mentioned parameters: overall pressure ratio of 4, high-pressure turbine isentropic efficiency of 83%, low-pressure turbine isentropic efficiency of 85%.
The compressor isentropic efficiency of 80%, regenerator efficiency of 75%, and mechanical efficiency of 98% for both shafts, the specific net work output and thermal efficiency of the system are approximately 296.23 kJ/kg and 33.54% respectively.
The calculation involves multiple steps including evaluating the conditions at each stage of the turbine and compressor, accounting for isentropic efficiencies, regenerator effects, and mechanical losses, and ultimately finding the net work and thermal efficiency by considering the energy balances throughout the system.
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Design a three stepped distance protection for the protection of an EHV transmission line. Explain / label all the steps and constraints using circuit diagram(s) as well. Put together your proposed scheme considering the trip contacts configuration of the circuit breaker(s).
Distance protection is a type of protection scheme used in power system transmission line protection. It provides good selectivity and sensitivity in identifying the faulted section of the line.
The main concept of distance protection is to compare the voltage and current of the protected line and calculate the distance to the fault. This protection is widely used in Extra High Voltage (EHV) transmission lines. Design of three-stepped distance protection: Three-stepped distance protection for the EHV transmission line can be designed using the following steps:
Step 1: Zone 1 protection For the first step, we use the distance relay to provide Zone 1 protection. This relay is located at the beginning of the transmission line, and its reach is set to cover the full length of the line plus the length of the adjacent feeder. The relay uses the phase-to-phase voltage (Vab, Vbc, Vca) and the three-phase current (Ia, Ib, Ic) to measure the impedance of the line. If the calculated impedance falls below a set threshold, the relay trips the circuit breaker. The circuit diagram of Zone 1 protection is as follows:
Step 2: Zone 2 protection For the second step, we use the distance relay to provide Zone 2 protection. This relay is located at a distance from the substation, and its reach is set to cover the full length of the transmission line plus a margin. The relay uses the phase-to-phase voltage (Vab, Vbc, Vca) and the three-phase current (Ia, Ib, Ic) to measure the impedance of the line. If the calculated impedance falls below a set threshold, the relay trips the circuit breaker. The circuit diagram of Zone 2 protection is as follows:
Step 3: Backup protection For the third step, we use the overcurrent relay to provide backup protection. This relay is located at the substation and uses the current of the transmission line to measure the fault current. If the fault current exceeds a set threshold, the relay trips the circuit breaker. The circuit diagram of the backup protection is as follows:
Constraints: There are some constraints that we need to consider while designing three-stepped distance protection for the EHV transmission line. These are as follows:• The reach of each zone should be set appropriately to avoid false tripping and ensure proper selectivity.• The time delay of each zone should be coordinated to avoid overreach.• The CT ratio and PT ratio should be chosen such that the relay operates correctly.• The trip contact configuration of the circuit breaker should be considered while designing the protection scheme.
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Identify the scope that your company involves in design and manufacturing process. From the scope, describe the processes in a process flow change and elaborate the functions of each process steps. Use a flow chart if applicable.
(Suggested word count: 500 words)
The design and manufacturing process involves a series of steps that start from the design stage to the delivery of the final product.
The scope of design and manufacturing process depends on the type of product the company is producing. However, in general, the design and manufacturing process involves the following steps:
The bottom-up approach starts with the analysis of the interoperability of the components to the modules and eventually the analysis of the system requirements.
Design Stage1. Idea Generation:
This is the first stage of the design process where ideas are design for a new product.
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Please elaborate more, upvote will be kindly given
Differentiate between force and natural convection. Explain briefly why the convective heat transfer coefficient in forced convection is usually higher than that in natural convection. (5 marks)
Force convection is a type of convection that happens when a fluid is forced to move over a surface or in a tube. On the other hand.
Natural convection is a type of convection that occurs when a fluid is heated, causing it to expand and rise, producing a convection current that circulates the fluid. Both natural and forced convection are used for heat transfer, but there are some differences between them.In natural convection.
The convective heat transfer coefficient is lower than that in forced convection. The reason is that in natural convection, the motion of the fluid is caused by buoyancy forces, which are weaker than the forces generated by forced convection.
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During a test on a boiler the following data were recorded:
Pressure = 1.7 MPa
Steam temperature at exit = 240ºC
Steam flow rate = 5.4 tonnes/hour
Fuel consumption = 400 kg/hour
Lower calorific value of fuel = 40 MJ/kg
Temperature of feedwater = 38ºC
Specific heat capacity of superheated steam = 2100 J/kg.K
Specific heat capacity of liquid water = 4200 J/kg.K.
Calculate:
Efficiency of the boiler.
Equivalent evaporation (EE) of the boiler
Given data,Presure P = 1.7 MPaSteam temperature at exit = t2 = 240°CSteam flow rate = m2 = 5.4 tonnes/hourFuel consumption = 400 kg/hourLower calorific value of fuel = LCV = 40 MJ/kgTemperature of feedwater = t1 = 38°CSp. heat capacity of superheated steam = Cp2 = 2100 J/kg.KSp.
Heat capacity of liquid water = Cp1 = 4200 J/kg.K.Formula : Heat supplied = Heat inputFuel consumption, m1 = 400 kg/hourCalorific value of fuel = 40 MJ/kgHeat input, Q1 = m1 × LCV= 400 × 40 × 10³ J/hour = 16 × 10⁶ J/hourFeed water rate, mfw = m2 - m1= 5400 - 4000 = 1400 kg/hourHeat supplied, Q2 = m2 × Cp2 × (t2 - t1)= 5400 × 2100 × (240 - 38) KJ/hour= 10,08 × 10⁶ KJ/hourEfficiency of the boiler, η= (Q2/Q1) × 100= (10.08 × 10⁶)/(16 × 10⁶) × 100= 63 %Equivalent evaporation (EE) of the boilerEE is the amount of water evaporated into steam per hour at the full-load operation at 100 % efficiency.(m2 - m1) × Hvfg= 1400 × 2260= 3.164 × 10⁶ Kg/hour
Therefore, the Efficiency of the boiler is 63 % and Equivalent evaporation (EE) of the boiler is 3.164 × 10⁶ Kg/hour.
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Describe the observable corrosive effect when
Austenitic stainless steel plates are bolted using galvanized
plates. Draw your reasons why inferring behaviour from the standard
EMF series. (6 marks)
When austenitic stainless steel plates are bolted using galvanized plates, you would likely observe the corrosion of the galvanized plates while the stainless steel remains largely unaffected.
This phenomenon is governed by the electrochemical series, or standard EMF series. The galvanized plate, which is coated with zinc, has a more negative standard electrode potential than stainless steel. This makes zinc more prone to oxidation (losing electrons), thus acting as a sacrificial anode when it's in direct contact with stainless steel. The zinc corrodes preferentially, protecting the stainless steel from corrosion. This is the same principle used in galvanic or sacrificial protection, where a more reactive metal is used to protect a less reactive metal from corrosion. Hence, the stainless steel (less reactive, higher in the EMF series) is preserved while the galvanized plates (more reactive, lower in the EMF series) corrode over time.
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For a given closed loop control system, the transfer functions of the controller, the plant and the sensor are given below. Gc(s) = Kp
Gp(s) = s+2/2s²+2s+1
Gh(s) = s+1/2s+1
a. Calculate the closed loop transfer function in the form Gcl, (s) = N(s)/D(s)
b. Determine the condition on K, that makes the system stable. c. If it exists, determine the condition on K, that sets the stability margin to 1/2.
a. Calculation of the closed loop transfer function in the form Gcl, (s) = N(s)/D(s):A closed-loop transfer function can be written as follows: Gcl(s)=Gp(s)Gc(s)Gp(s)Gc(s)+Gh(s)Gp(s)Where Gp(s) is the plant transfer function, Gc(s) is the controller transfer function, and Gh(s) is the sensor transfer function. Substituting the provided values, we get the following result.Gc(s) = Kp, Gp(s) = (s+2)/(2s²+2s+1), and Gh(s) = (s+1)/(2s+1)By substituting the provided values, we get the following result.Gcl(s)=Gp(s)Gc(s)/[1+Gh(s)Gp(s)Gc(s)]Gcl(s) = Kp(s+2)/(2s^3+5s^2+5s+2Kp)Therefore, the closed-loop transfer function of the system is Gcl(s) = Kp(s + 2) / (2s^3 + 5s^2 + 5s + 2Kp).b. Calculation of the condition on K that makes the system stable:We will determine the condition for the system to be stable by analyzing the roots of the denominator's characteristic equation, which is 2s^3 + 5s^2 + 5s + 2Kp = 0.By applying Routh-Hurwitz stability criteria to the characteristic equation, we obtain the following conditions.2Kp>0,5>0,1Kp-10>0,2Kp + 5>0By combining all these conditions, we can say that the system will be stable if Kp > 0.5.c. Calculation of the condition on K that sets the stability margin to 1/2:Now, we have to find the condition on K that sets the stability margin to 1/2 if it exists.We will calculate the phase margin using the closed-loop transfer function's magnitude and phase expressions. The phase margin is calculated using the following formula:Phase margin (PM) = ∠Gcl(jω) - (-180°)where ω is the frequency at which the magnitude of the closed-loop transfer function is unity (0dB).Magnitude of Gcl(s) = Kp|(s + 2) / (2s^3 + 5s^2 + 5s + 2Kp)|= Kp| (s + 2) / [(s + 0.2909)(s + 1.3688 - j0.7284)(s + 1.3688 + j0.7284)] |at unity gain frequency, ω, i.e., |Gcl(jω)| = 1.The phase margin is given by PM = tan^-1[(Imaginary part of Gcl(jω)) / (Real part of Gcl(jω))]+180°PM = 180° - ∠Gcl(jω) - 180°Phase margin (PM) = -∠Gcl(jω)The phase angle of the closed-loop transfer function at unity gain frequency is calculated using the following formula:∠Gcl(jω) = tan^-1(ω) - tan^-1(2Kpω / ω^2 + 2ω + 1) - tan^-1(ω / 2)Now we can equate the phase margin, PM to 1/2.0.5 = -∠Gcl(jω)After solving, we get 3.64 ≤ 2Kp ≤ 8.87.Conclusion:We have calculated the closed-loop transfer function, the condition on K that makes the system stable and the condition on K that sets the stability margin to 1/2.
A series of weighing of a sample of metal powder are made with the following results: Weight of a sample, grams: 2.020 2.021 2.021 2.019 2.019 2.018 2.021 2.018 2.021 2.017 2.017 2.020 2.016 2.019 2.020 Compute average and standard deviation for the weighing.
To compute the average and standard deviation for the weighing of the metal powder sample, follow these steps: Calculate the average (mean) weight:
Add up all the weights and divide by the total number of measurements. Average weight = (2.020 + 2.021 + 2.021 + 2.019 + 2.019 + 2.018 + 2.021 + 2.018 + 2.021 + 2.017 + 2.017 + 2.020 + 2.016 + 2.019 + 2.020) / 15
Calculate the standard deviation: a. Subtract the average weight from each individual weight to get the deviation.
b. Square each deviation.
c. Sum all the squared deviations.
d. Divide the sum by (n-1), where n is the total number of measurements.
e. Take the square root of the result.
Let's calculate the average and standard deviation:
Average weight = (2.020 + 2.021 + 2.021 + 2.019 + 2.019 + 2.018 + 2.021 + 2.018 + 2.021 + 2.017 + 2.017 + 2.020 + 2.016 + 2.019 + 2.020) / 15
= 30.307 / 15
≈ 2.020 grams (rounded to three decimal places)
Standard deviation = √[(Σ(x - μ)²) / (n - 1)]
= √[(0.000² + 0.001² + 0.001² + (-0.001)² + (-0.001)² + (-0.002)² + 0.001² + (-0.002)² + 0.001² + (-0.003)² + (-0.003)² + 0.000² + (-0.004)² + (-0.001)² + 0.000²) / (15 - 1)]
Performing the calculations and taking the square root will give you the standard deviation for the weighing.
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You are an environmental engineer working for a manufacturing company that makes computer components. In the process your plant creates toxic wastes, primarily as heavy metals. Part of your job is to oversee the testing of the effuluent from your plant, signing the test results to attest to their accuracy and supplying them to the city. The allowable limit of the chemicals disposed is less when compared to the national chemical standard limits permitted. But you are very concerned about the fact that what will the smaller concentrations amount to. You also found out that even with reduced limits the heavy metals disposed are highly dangerous. You have to prepare a report a report for the same. a. Interpret with the help of two NSPE codes in this case b. develop what must be written details that should be included in the report
Two NSPE codes in this case can be: Engineers shall hold paramount the safety, health, and welfare of the public and the protection of the environment (NSPE Code of Ethics 2007, III.1.).
Engineers shall avoid deceptive acts that falsify their qualifications (NSPE Code of Ethics 2007, III.4.).b. The report should include the following details: The report should present the information that indicates that despite the lower levels of toxic waste that the plant produces, the heavy metals it emits are still highly dangerous.
The report should also discuss the implications of the heavy metals and what they can cause. The report should provide a complete review of the situation, including how it came to light, the testing process and results, and what steps have been taken to fix the problem.
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A multiple-disk clutch is to operate in oil and be able to transmit a design overload torque 400 N·m. The disks are alternately high carbon steel and molded asbestos, with inside and outside diameters of 90 and 150 mm, respectively. Design values based on test experience for this application are Pₘₐₓ = 1000kpa and f=0.10. What a total number of disks is required.
The following data is provided for multiple-disk clutch:
Design overload torque = 400 N.m
Pmax = 1000 kPa Friction coefficient
f = 0.1
Inner diameter of disk (D1) = 90 mm
Outer diameter of disk (D2) = 150 mm To find:
The total number of disks required. Formula:
The following formula is used to calculate the torque transmitted by the clutch:
T = [tex][(Pmax x π/2) x (D2^2 - D1^2) x f] N.m[/tex] Where:
T = Torque transmitted by the clutch P max
= Design value of maximum pressure (kPa)π
= 3.14D1
= Inner diameter of the disk (mm) D2
= Outer diameter of the disk (mm)
f = Coefficient of friction.
The following formula is used to calculate the torque carrying capacity of each disk:
C =[tex](π/2) x (D2^2 - D1^2)[/tex] x Pmax N Where:
C = Torque carrying capacity of the disk
Pmax = Design value of maximum pressure[tex](kPa)π[/tex]
= 3.14D1
= Inner diameter of the disk (mm)
D2 = Outer diameter of the disk (mm).
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Assume that a U-tube steam generator vapor space consists of saturated vapor at 980 psia. Assume that a steam line break results in a containment pressure of approximately 2.5 psig. (a) Determine thermodynamic conditions (e.g., temperature, enthalpy, and specific volume) of the vapor (i) within the steam generator and then (ii) within the containment building. (b) Based on these results, would a large steam line break require operation of the containment spray system?
It is necessary to operate the containment spray system in order to reduce the pressure and temperature in the containment building applied.
(a) Determine thermodynamic conditions (e.g., temperature, enthalpy, and specific volume) of the vapor (i) within the steam generator and then (ii) within the containment building.Solution:(i) Conditions of vapor in the steam generator:Given, Saturated vapor pressure = 980 psiaAs saturated vapor pressure = 980 psia, the vapor in the steam generator is saturated vapor and the thermodynamic properties can be obtained using the steam tables.At saturated vapor pressure 980 psia:Temperature = 613.35 °FEnthalpy = 1354.2 Btu/lbmSpecific volume = 3.0384 ft3/lbm(ii) Conditions of vapor within the containment building:Given, containment pressure = 2.5 psigAs the pressure in the containment building is much less than the saturated vapor pressure of the steam in the steam generator, it is not possible to calculate the thermodynamic properties using the steam tables.
To calculate the properties at low pressure, we need to use the ideal gas law which is given by,PV = mRT,where,P = PressureV = Volume of the gasm = Mass of the gasR = Specific gas constantT = Temperature of the gasR = Raising constant = 1545.3 (ft-lbf)/(lbm-°R)By assuming that the specific volume of the vapor in the containment building is same as that of saturated vapor at 2.5 psia, the specific volume can be calculated as:Specific volume, v = 26.58 ft3/lbm, which can be obtained from the steam tables using interpolation in the range of 2 psia to 3 psia.Now, we can calculate the temperature of the vapor using the ideal gas law.P = 2.5 psig = (2.5 + 14.7) psia = 17.2 psiaV = 26.58 ft3/lbmR = 1545.3 (ft-lbf)/(lbm-°R)From ideal gas law,PV = mRT⇒ m = PV/RT⇒ m = (17.2)(1)/((1545.3)(613.35 + 460))⇒ m = 1.61 × 10^-5 lbmAs the vapor is an ideal gas,enthalpy = CpΔT= 1.14 (Btu/lbm-°F) × (613.35 - 72)°F= 654.7 Btu/lbmSpecific volume = 26.58 ft3/lbm(b) Based on these results, would a large steam line break require operation of the containment spray system?In a nuclear power plant, the containment spray system is operated to condense the steam in the containment building which reduces the pressure and temperature. From the above results, it can be seen that the specific volume of the vapor at 2.5 psia is more than 8 times the specific volume of the saturated vapor in the steam generator.
As the specific volume of the vapor is very high, a large steam line break results in a large quantity of steam being released which results in the containment pressure increasing rapidly.
Therefore, it is necessary to operate the containment spray system in order to reduce the pressure and temperature in the containment building.
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If the transfer function of a phase lead compensator is (s+ a)/ (s+b) and that of a lag compensator is (s + p)/(s+q), then Which one of the following sets of condition must be satisfied ?
a)a > b and p > q
b)a > b and p < q
c)a < b and p < q
d)None of the above
The correct option is (a) a > b and p > q. In the phase-lead compensator, the angle is added to the system transfer function, while in the phase-lag compensator, the angle is subtracted from the system transfer function. The phase lead compensator improves the phase margin of the system by improving the phase lag in the system.
It is used in situations where the system needs an improved phase margin. The phase-lead compensator's transfer function is expressed as (s+a)/(s+b), where a>b.In the lag compensator, the phase is reduced, resulting in improved stability and a more robust system. It is used in situations where the system needs improved stability. The lag compensator's transfer function is (s+p)/(s+q), where p b and p > q.
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A gentleman used a spring scale to measure his luggage weight 3 times in different time under fixed condition and found the results were 42.1, 41.8 and 42.5lbf, respectively. From the above results, estimate the nearest first order uncertainty? 1.51 0.35 All solutions are not correct 0.87
The nearest first-order uncertainty is approximately 0.27 lbf. The correct answer is 0.35. The correct answer is option(b).
The nearest first-order uncertainty can be estimated by calculating the standard deviation. Standard deviation is a measure of the amount of variation or dispersion of a set of values.
Given measurements are as follows:42.1, 41.8, 42.5lbfThe formula to calculate the standard deviation is:
Standard deviation formulaσ=√((Σ(xi−x¯)2)/(n−1))
Where xi is the measurement value, x¯ is the mean value, and n is the number of observations.
Let's calculate the mean first.
Mean= (42.1 + 41.8 + 42.5)/3= 126.4/3= 42.13333lbf
Now let's calculate the standard deviation.
σ=√(((42.1-42.1333)2+(41.8-42.1333)2+(42.5-42.1333)2)/(3-1))
σ=√((0.01778+0.12216+0.13689)/2)
σ=√(0.14183/2)
σ=√0.070915
σ= 0.2664
Therefore, the nearest first-order uncertainty is approximately 0.27 lbf. The correct answer is 0.35.
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Complete an anatomical description of a motor skill that requires motion in most joints and has at least three phases. Name the reflexes during this motor skill and describe how they may either facilitate or inhibit the movement. Name at least three mechanical principles that apply to each sub-category in the classification system and that, if violated, will result in major performance errors.
One motor skill that requires motion in most joints and consists of three phases is the high jump in athletics. The high jump involves a running approach, takeoff, and clearance phases.The reflexes during this motor skill are stretch reflex,crossed-extensor reflex and withdrawal reflex. The Mechanical principles that apply to each sub-category are Balance,Force and Motion and Center of Gravity.
Motor Skill Description:
One motor skill that requires motion in most joints and consists of three phases is the high jump in athletics. The high jump involves a running approach, takeoff, and clearance phases.
During the running approach, the stretch reflex helps facilitate the movement. As the athlete runs, the muscle spindles in the leg muscles detect the rapid stretch and trigger a reflexive contraction, allowing for more powerful leg extension during takeoff.
The crossed-extensor reflex also comes into play, providing stability and balance by activating muscles on the opposite side of the body.
In the takeoff phase, the withdrawal reflex inhibits unwanted movements. It prevents the leg from kicking back during the jump, ensuring a more controlled and efficient takeoff.
The tendon reflex also assists by facilitating a quick contraction of the leg muscles upon contact with the ground, generating upward propulsion.
In the clearance phase, the flexor reflex aids in bending the knees and hips, facilitating the clearance of the bar. This reflex allows for quick and coordinated flexion movements.
Mechanical Principles:
1. Balance: The principle of equilibrium is crucial in maintaining balance during the high jump. Violating this principle by leaning too far back or forward can result in loss of balance and failed performance.
2. Force and Motion: The principle of force production and transfer is vital for generating sufficient vertical propulsion during the takeoff phase.
Violating this principle by inadequate force application or improper timing can lead to a lower jump height.
3. Center of Gravity: The principle of the center of gravity influences the body's stability and trajectory during the high jump. Violating this principle by having a significantly off-center body position can cause instability and affect the jump's outcome.
Adhering to these mechanical principles, while considering the reflexes that facilitate or inhibit movement, is essential for executing a successful high jump.
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Determine the reactions at the supports. 4 m 500 N/m 1.6 kN - 4 m B
The reaction at support A is 800 N and the reaction at support B is 600 N. The anti-clockwise moments about support B is equal to the clockwise moments about support B.
The given diagram is as follows: To determine the reactions at the supports, we can take moments about any one of the supports. But in this case, it is easier to take moments about support B, since the unknown reaction is at this support. The anti-clockwise moments about support B is equal to the clockwise moments about support B. The equation of equilibrium of moments is as follows:
ΣMoments about
B = 0 ⇒ 1.6 kN (4 m) - 500 N/m (4 m)2 - B (4 m) = 0
⇒ 6400 - 4000 - 4B = 0
⇒ - 4B = - 2400B
= 600 N
The reaction at support A = 1.6 kN - 500 N/m - B= 1600 - 200 - 600= 800 N
Therefore, the reaction at support A is 800 N and the reaction at support B is 600 N.
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a) Power is defined as: i) The amount of work performed per unit of distance. ii) Force per unit of time. iii) The amount of work performed per unit of time. iv) Normal force x coefficient of friction.
The correct definition of power is the amount of work performed per unit of time. It is usually represented in watts, which is equal to joules per second.
Therefore, power can be calculated using the formula: Power = Work/Time.
The amount of work performed per unit of distance is not a correct definition of power. This is because work and distance are not directly proportional. Work is a function of both force and distance.
Force per unit of time is not a correct definition of power. This is because force alone cannot measure the amount of work done. Work is a function of both force and distance.
Normal force x coefficient of friction is not a correct definition of power. This is because it is a formula for calculating the force of friction, which is a different concept from power.
In conclusion, the correct definition of power is option iii) the amount of work performed per unit of time.
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Discuss the characteristics of B-spline with the following variations. (1) Collinear control points. (1) Coincident control points. (111) Different degrees. Use graphical diagrams to illustrate your ideas.
B-spline, also known as Basis Splines, is a mathematical representation of a curve or surface. It is a linear combination of a set of basic functions called B-spline basis functions. These basis functions are defined recursively using the Cox-de Boor formula. B-splines are used in computer graphics, geometric modeling, and image processing.
Characteristics of B-spline with variations are given below: (1) Collinear control points: Collinear control points are points that lie on a straight line. In this case, the B-spline curve is also a straight line. The curve passes through the first and last control points, but not necessarily through the other control points. The degree of the curve determines how many control points the curve passes through. The curve is smooth and has a finite length.
(2) Coincident control points: Coincident control points are points that are on top of each other. In this case, the B-spline curve is also a point. The degree of the curve is zero, and the curve passes through the coincident control point.
(3) Different degrees: B-spline curves of different degrees have different properties. Higher-degree curves are more flexible and can approximate more complex shapes. Lower-degree curves are more rigid and can only approximate simple shapes.
The following diagrams illustrate these variations:
1. Collinear control points:
2. Coincident control points:
3. Different degrees:
In conclusion, B-spline curves have various characteristics, including collinear control points, coincident control points, and different degrees. Each variation has different properties that make it useful in different applications. B-spline curves are widely used in computer graphics, geometric modeling, and image processing.
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Given the reference condition of the free air delivery of 720 cfm: P1 = 15 psia; T = 95ºF; RH = 80%. Find the air volumes in ICFM condition. If the ICFM Reference Condition are: P2 = 14.7 psia; T = 68ºF RH = 80% Units: cu ft/min
To convert SCFM to ACFM, additional information such as the actual pressure (P2), actual temperature (T2), and actual relative humidity (RH2) is required to perform the necessary calculations.
What is the calculation for converting SCFM to ACFM using the following parameters: P1 = 14.5 psia, T1 = 60°F, RH1 = 50%?To convert the air volumes from CFM to ICFM condition, we need to apply the correction factors for pressure, temperature, and relative humidity. The correction formulas are as follows:
ICFM = CFM * (P2 / P1) * (T / T2) * (1 / (273 + T2) * (273 + T1)) * (1 / (1 + 0.00367 * RH))
where:
- ICFM is the air volume in ICFM (Ideal Cubic Feet per Minute)
- CFM is the air volume in CFM (Cubic Feet per Minute)
- P1 and P2 are the initial and reference pressures, respectively (psia)
- T1 and T2 are the initial and reference temperatures, respectively (°F)
- RH is the relative humidity (%)
Substituting the given values:
P1 = 15 psia
P2 = 14.7 psia
T1 = 95°F
T2 = 68°F
RH = 80%
we can calculate the air volumes in ICFM condition.
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A nozzle 0.01m in diameter emits a water jet at a velocity of 35 m/s, which strikes a stationary vertical plate at an angel of 70° to the vertical. Calculate the force acting on the plate, in N in the horizontal direction (Hint 0 in your formula is the angle to the horizontal) If the plate is moving horizontally, at a velocity of of 3 m/s, away from the nozzle, calculate the force acting on the plate, in N the work done per second in W, in the direction of movement
The force acting on the plate in N is 0.0136 N in the horizontal direction. The work done per second in the direction of movement is 0.4448 W.
How to determine?We can determine the force acting on the plate in the horizontal direction as follows:
Force, F = ρ/2 * v² * A * Cθ
Where A is the area of cross-section and Cθ is the coefficient of impact, which is equal to 1 for this case.
A = π/4 * d²
= 7.85 × 10⁻⁷ m².
Thus, the force acting on the plate in the horizontal direction is given by,
F = ρ/2 * v² * A * Cθ
= 1000/2 × 35² × 7.85 × 10⁻⁷ N
= 0.0136 N.
If the plate is moving horizontally at a velocity of 3 m/s away from the nozzle, then the relative velocity of the jet with respect to the plate will be v - u. The work done per second in the direction of movement is given by,
W = F × d
= F × (v - u)
= 0.0136 × (35 - 3) J/s
= 0.4448 W.
Thus, the force acting on the plate in N is 0.0136 N in the horizontal direction. The work done per second in the direction of movement is 0.4448 W.
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A quantity of matter or a region in space chosen for study is called. Match the sentence from column A to the correct sentence from column B. a set of properties that describes the conditions of a system. system that maintains thermal, mechanical, phase and chemical equilibriums. isobaric process a process that, once having take place it can be reversed The ratio of any extensive property of a system to that of the mass of the system is called
A quantity of matter or a region in space chosen for study is called the system. The properties that describe the conditions of the system are called state variables. When a system is in thermal equilibrium with its surroundings, its temperature is uniform throughout.
Thermal equilibrium does not guarantee that the mechanical equilibrium of a system is stable. A system is a concept used to describe the set of properties that describe the conditions of a system. A system refers to the region of the universe under consideration. The properties that describe the conditions of a system are known as state variables. Systems that maintain thermal, mechanical, phase and chemical equilibriums are called isolated systems.An isobaric process refers to a process that takes place at a constant pressure.
On the other hand, an isochoric process is a process that takes place at a constant volume. A process that, once having taken place, can be reversed is known as a reversible process. A reversible process refers to a process that can be reversed in its path with any small change in conditions, while returning the system to its initial state. The ratio of any extensive property of a system to that of the mass of the system is called a specific property. Therefore, option A describes option B.
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45. A shaft of 1.0 inch diameter has a single dise weighing 75 lb mounted midway between two bearings 20 inches apurt . Neglecting the weight of the shaft. calculate the lowest entical speed in rpm. Note: Modulus of elasticity is 30 x 10% pri C. 1709 rpm A 2038 B. 2540 rpm D. 2094 rpm FER
Hence, the correct option is (B) 2038 rpm.
Given that, A shaft of 1.0-inch diameter has a single disc weighing 75 lb mounted midway between two bearings 20 inches apart, and the Modulus of elasticity is 30 × 106 psi.
The lowest critical speed (N) of a shaft is given by the relation:
N = (0.00305/2π) (K/δ) (α/E) (60/2L)
Where, K = stiffness of the shaft (lb/in)δ = mass per unit length of the shaft (lb-s2/in)α = constant
E = modulus of elasticity (psi)L = span (in)
Given, Diameter (d) of the shaft = 1.0 inch
Radius of the shaft = r = d/2 = 1/2 = 0.5 inch
Midway between two bearings,
Distance between two bearings = 20 inches
Length of the shaft = L = 20/2 = 10 inches
Weight of the disc = W = 75 lb
Density of the material of the shaft = 0 (neglecting the weight of the shaft)
Modulus of elasticity (E) = 30 × 106 psi
Calculation of α:
For the disc, α = 0.84 (for disc weight) + 1.65 (for disc moment of inertia) = 2.49
Calculation of K:
K = 3/4 × π × r4/ (4 × 0.5)K = 0.589 r4K = 0.589 (0.54)4K = 0.076 lb/in
Calculation of δ:
δ = π × r2 × (0.5)/386 δ = 0.000035 lb-s2/in
Substituting the given values in the formula to find N:
N = (0.00305/2π) (K/δ) (α/E) (60/2L)
N = (0.00305/2π) (0.076/0.000035) (2.49/30 × 106) (60/2 × 10)
N = 2037.86 rpm ≈ 2038 rpm Hence, the correct option is (B) 2038 rpm.
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Equilibrium of a body requires both a balance of forces and balance
of moments.
(true or false)
Equilibrium of a body requires both a balance of forces and balance of moments. This statement is True. The equilibrium of a body refers to the state where there is no acceleration. It can be categorized into two, the static and dynamic equilibrium. The static equilibrium occurs when the object is at rest, and the dynamic equilibrium happens when the object is in a constant motion.
Both of these types require a balance of forces and moments to be attained.In physics, force is a quantity that results from the interaction between two objects, and it's measured in newtons. It can be categorized into two, contact forces, and non-contact forces. Contact forces involve physical contact between two objects, while non-contact forces are those that occur without physical contact. According to Newton's first law of motion, a body in equilibrium will remain in that state until acted upon by an unbalanced force.
Therefore, when an object is in equilibrium, both the forces and moments should be balanced for the equilibrium to exist.In conclusion, it's true that equilibrium of a body requires both a balance of forces and balance of moments.
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[εxx εyx εzx] [-40 -24 0]
[ε] = [εxy εyy εzy] = [-24 16 0] *10⁻⁶
[εxz εyz εzz] [ 0 0 12]
a. Calculate the volumetric strain and the deviatoric strain tensor, b. Calculate the mean stress and the deviatoric stress invariants, c. Calculate the characteristic equation of strain, d. Calculate the characteristic equation of stress. The material is linear elastic (E=200GPa, v=0.3).
a. Calculation of volumetric strain: Volumetric strain, εv = εxx + εyy + εzzεv = -40 + 16 + 12εv = -12 μm/m
Deviatoric strain tensor is given as ε = εxx - εyy, εxz, εyz0, εzy = εyx= (-40 - 16) * 10^-6 = -56 * 10^-6.
Therefore, the deviatoric strain tensor is [-56 0 0; 0 24 0; 0 0 0].
b. Calculation of mean stress and deviatoric stress invariants:
Mean stress is given by σm = (σxx + σyy + σzz)/3 σm = (E/(1 - v) * εv)/3σm = 9.23 GPa
Deviatoric stress tensor is given as σd = σ - σmIσd = [σxx - 9.23 σyy - 9.23 σzz - 9.23]
Deviatoric stress invariants are given asJ2 = (1/2)σdijσdijJ2 = (1/2)[(-33.58)² + 0 + 0]J2 = 563.48 MPa
c. Calculation of the characteristic equation of strain:
The characteristic equation of strain is given as: |ε - εi| = 0|[-40 - ε εyx εxz εxy 16 εyz εzy 0 12 - ε]| = 0-ε³ - 12ε² - 69.32ε - 1.4748 * 10⁴ = 0d.
Calculation of the characteristic equation of stress:
The characteristic equation of stress is given as: |σ - σiI| = 0|[(120.58 - σ) - 56 0 0; 0 (-104.35 - σ) 0; 0 0 (-15.23 - σ)]| = 0σ³ + 200σ² - 154807.6σ + 3.6566 * 10¹⁰ = 0
The material is linear elastic (E=200GPa, v=0.3).
The calculation of volumetric strain gives -12 μm/m. The deviatoric strain tensor is [-56 0 0; 0 24 0; 0 0 0].
The mean stress is 9.23 GPa, and the deviatoric stress invariants are J2 = 563.48 MPa. The characteristic equation of strain is -ε³ - 12ε² - 69.32ε - 1.4748 * 10⁴ = 0. Finally, the characteristic equation of stress is σ³ + 200σ² - 154807.6σ + 3.6566 * 10¹⁰ = 0.
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An axial flow compressor stage has the following data: Air inlet stagnation temperature =300 F ; Flow coefficient 0.6; Relative inlet Mach number=0.75; degree of reaction 0.5; blade angle at outlet measured from the axial direction 35 degree. Find the stagnation temperature rise in the first stage of the compressor
The stagnation temperature rise in the first stage of the compressor is approximately 47.75 °F.
To find the stagnation temperature rise in the first stage of the compressor, we can use the following formula:
ΔT0 = T0out - T0in
Given data:
Inlet stagnation temperature (T0in) = 300 °F
Flow coefficient (ϕ) = 0.6
Relative inlet Mach number (Mach1) = 0.75
Degree of reaction (R) = 0.5
Blade angle at outlet (β2) = 35 degrees
To calculate the stagnation temperature rise, we need to use the following equations:
Calculate the absolute inlet Mach number (Ma1):
Ma1 = Mach1 / √(1 + (γ-1)/2 * Mach1^2)
where γ is the specific heat ratio of air (approximately 1.4 for air).
Calculate the isentropic outlet Mach number (Mach2s):
Mach2s = √(2 * ((ϕ * (1 - R)) / (R * sin^2(β2)) - 1))
Calculate the stagnation temperature rise (ΔT0):
ΔT0 = γ / (γ - 1) * R * T0in * (1 - 1 / Mach2s^2)
Let's calculate the values step by step:
Calculate the absolute inlet Mach number (Ma1):
Ma1 = 0.75 / √(1 + (1.4 - 1) / 2 * 0.75^2) = 0.5707
Calculate the isentropic outlet Mach number (Mach2s):
Mach2s = √(2 * ((0.6 * (1 - 0.5)) / (0.5 * sin^2(35°)) - 1)) = 0.8012
Calculate the stagnation temperature rise (ΔT0):
ΔT0 = 1.4 / (1.4 - 1) * 0.5 * 300 °F * (1 - 1 / 0.8012^2) = 47.75 °F
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