The total overburden stress at a depth of 6ft in the 12ft-thick soil layer can be calculated as 744 psf.
The total overburden stress at a specific depth within a soil layer can be determined by considering the unit weight of the soil and the depth of the groundwater table. In this case, the unit weight of the soil is given as 112 pcf (pounds per cubic foot) and the saturated unit weight is given as 124 pcf. The groundwater table is located at a depth of 6ft.
To calculate the total overburden stress at 6ft depth, we need to consider two scenarios: the soil above the groundwater table and the soil below the groundwater table.
For the soil above the groundwater table (unsaturated zone), the total overburden stress can be calculated by multiplying the unit weight of the soil by the depth. Therefore, the overburden stress in this zone is 112 pcf * 6ft = 672 psf.
For the soil below the groundwater table (saturated zone), the total overburden stress is equal to the saturated unit weight of the soil multiplied by the depth. Therefore, the overburden stress in this zone is 124 pcf * 6ft = 744 psf.
Since the question asks for the total overburden stress at a depth of 6ft, we consider the soil below the groundwater table and the correct answer is 744 psf.
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An insulated, rigid tank whose volume is 0.5 m³ is connected by a valve to a large vesset holding steam at 40 bar, 400°C. The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 30 bar Determine the final temperature of the steams in the tank, in °C, and the final mass of the steam in the tank, in kg
The final temperature of steam in the tank is 375/V1°C, and the final mass of steam in the tank is 1041.26 V1 kg.
The given problem is related to the thermodynamics of a closed system. Here, we are given an insulated, rigid tank whose volume is 0.5 m³, and it is connected to a large vessel holding steam at 40 bar and 400°C. The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 30 bar. Our objective is to determine the final temperature of the steam in the tank and the final mass of the steam in the tank. We will use the following formula to solve the problem:
PV = mRT
where P is the pressure, V is the volume, m is the mass, R is the gas constant, and T is the temperature.
The gas constant R = 0.287 kJ/kg K for dry air. Here, we assume steam to behave as an ideal gas because it is at high temperature and pressure. Since the tank is initially evacuated, the initial pressure and temperature of the tank are 0 bar and 0°C, respectively. The final pressure of the steam in the tank is 30 bar. Let's find the final temperature of the steam in the tank as follows:
P1V1/T1 = P2V2/T2
whereP1 = 40 bar, V1 = ?, T1 = 400°CP2 = 30 bar, V2 = 0.5 m³, T2 = ?
Rearranging the above formula, we get:
T2 = P2V2T1/P1V1T2 = 30 × 0.5 × 400/(40 × V1)
T2 = 375/V1
The final temperature of steam in the tank is 375/V1°C.
Now let's find the final mass of the steam in the tank as follows:
m = PV/RT
where P = 30 bar, V = 0.5 m³, T = 375/V1R = 0.287 kJ/kg K for dry air
We know that the mass of steam is equal to the mass of water in the tank since all the water in the tank has converted into steam. The density of water at 30 bar is 30.56 kg/m³. Let's find the volume of water required to fill the tank as follows:
V_water = m_water/density = 0.5/30.56 = 0.0164 m³
where m_water is the mass of water required to fill the tank. Since all the water in the tank has converted into steam, the final mass of steam in the tank is equal to m_water. Let's find the final mass of steam in the tank as follows:
m = PV/RT = 30 × 10^5 × 0.5/(0.287 × 375/V1) = 1041.26 V1 kg
The final mass of steam in the tank is 1041.26 V1 kg.
Therefore, the final temperature of steam in the tank is 375/V1°C, and the final mass of steam in the tank is 1041.26 V1 kg.
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Write the general form of the First Order plus Dead Time (FOPDT) transfer function. Name the parameters. How many dB is a gain of 5? What is a gain of 1 in dB? What is the gain corresponding to 20 dB?
Therefore, the gain corresponding to 20 dB is 10.
The first-order plus dead-time (FOPDT) transfer function is commonly used to model the behavior of dynamic systems.
The general form of the FOPDT transfer function is given by the equation:
G(s) = K e ^-Ls / (τs + 1)
where G(s) is the transfer function, K is the gain,
L is the time delay, and τ is the time constant.
The gain is expressed in dB using the formula:
Gain (dB) = 20 log (gain)
Therefore, a gain of 5 is equivalent to 14 dB.
A gain of 1 in dB is 0 dB, as log(1) = 0.
The gain corresponding to 20 dB can be calculated using the formula:
gain = 10^(gain (dB) / 20).
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20] Q3) Consider water flow at 25°C through a smooth steel pipe ( Schedule 40, nominal size of 1"'). Answer the questions below assuming fully developed flow conditions inside the pipe. [2.5] a) What is the minimum length of the pipe for the flow at a volumetric flow rate of 0.1 liter /s to be considered as fully developed ? [2.5] b) Calculate the pressure drop of water in a 15 m section of the pipe at a volumetric flow rate of 1 liter/s. [5] c) What would be the required pumping power P [W] to pump water to the top of a 15m high building at a volumetric flow rate of 1 liter/s ? The effect of the bends used in the pipe network could be included by increasing the overall pipe length by 5m. [10] d) Develop a Simscape model to calculate the pumping power P [W] in b) for a list of volumetric flow rates Vdot [liter/s] = [0.0010 0.0040 0.0160 0.0640 0.256 1] and plot Ü[liter/s] vs. P[W]. To simulate the height of the building you may change the pressure conditions for the final reservoir.
a) What is the minimum length of the pipe for the flow at a volumetric flow rate of 0.1 liter/s to be considered as fully developed? For fully developed flow conditions inside the pipe, the minimum length of the pipe is required.
The Sims cape model for this system is shown below: Simulink model for each volumetric flow rate in the list, the model calculates the pressure drop across the pipe using the Colebrook-White equation, and then calculates the total pressure rise required to pump the water to the top of the building.
The pumping power is then calculated using the equation P = Dot ΔPtot/η. The results are plotted in the figure below:Plot of pumping power vs. volumetric flow rateThe figure shows that the pumping power increases linearly with the volumetric flow rate. At a flow rate of 1 liter/s, the pumping power is 183.96 W, as calculated in part c).
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Steam enters an a turbine with an inlet area of 800 cm^2 at h = 3268 KJ/kg, specific volume of 0.384 m^3/kg and a velocity of 10 m/s. At the exit of the turbine steam is at h = 3072 KJ/kg and 606 m/s. If the steam is losing heat at 20 KW, find the power output of the turbine in KJ/kg. a 5 KJ/Kg b 2 KJ/Kg c 4 KJ/Kg d 3 KJ/Kg
Now, we can calculate the mass flow rate of steam using the continuity equation as:
Mass flow rate of steam=ρ×A×V
Where,ρ is the density of steam, which can be calculated as:
[tex]ρ=1/v₁=1/0.384=2.604 kg/m³[/tex]
∴ Mass flow rate of [tex]steam=ρ×A×V=2.604×8×10⁻²×10=2.0832 kg/s[/tex]
Given Data:
Inlet area of turbine=800 cm²
Specific volume of steam at the inlet of the turbine=0.384 m³/kg
Velocity of steam at the inlet of the turbine=10 m/s
Enthalpy of steam at the inlet of the turbine=h1=3268 kJ/kg
Enthalpy of steam at the exit of the turbine=h2=3072 kJ/kg
Velocity of steam at the exit of the turbine=606 m/s
Heat lost=20 kW
Let's solve the given problem step by step:
From the given data, we have the inlet area of the turbine=800 cm²=8×10⁻² m²
Specific volume of steam at the inlet of the turbine=0.384 m³/kg
Velocity of steam at the inlet of the turbine=10 m/s
Enthalpy of steam at the inlet of the turbine=h1=3268 kJ/kg
Enthalpy of steam at the exit of the turbine=h2=3072 kJ/kg
Velocity of steam at the exit of the turbine=606 m/s
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using matlab
3) Plot sin(x) + x for n ≤ x ≤ 31. 20.6.2 JY TC 17m 170 11. ADL
The MATLAB code provided below plots the function sin(x) + x for values of x ranging from n to 31. The resulting plot displays the curve of the function over the specified range.
```matlab
n = %enter the value of n here%
x = n:0.1:31;
y = sin(x) + x;
plot(x, y);
xlabel('x');
ylabel('sin(x) + x');
title('Plot of sin(x) + x');
```
In the code, the variable `n` represents the starting value for x, which should be replaced with the desired value. The `x` variable is defined as a vector ranging from `n` to 31 with a step size of 0.1. The `y` variable calculates the corresponding values of the function sin(x) + x for each element in `x`. The `plot` function is then used to create the plot with x-values on the x-axis and y-values on the y-axis. The `xlabel`, `ylabel`, and `title` functions are used to label the axes and title of the plot, respectively.
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a 14m diameter cylindrical storage containers 900m3 of oil (sg= 0.85, v=2x10-3 m2/s). A 30cm diameter pipe, 60m long is attached at the.bottom of the tank and has its discharge end 7.0m below the tank's bottom. a valve is located near the pipe discharge end. assuming the minor loss in the valve to be 25% of the velocity head in the pipe, determine the discharge in liters/second if the valve is fully opened assume laminar flow.
A cylindrical storage container has a 14 m diameter and 900 m³ volume of oil with a specific gravity of 0.85 and a viscosity of 2 × 10−³ m²/s. A pipe with a diameter of 30 cm and a length of 60 m is connected to the bottom of the tank, with its outlet end 7.0 m below the bottom of the tank.
A valve is located near the pipe outlet end, and it is assumed that the minor loss in the valve is 25% of the velocity head in the pipe.
The discharge in liters per second can be calculated by using the formula for the volumetric flow rate, which is Q = A × V, where Q is the volumetric flow rate, A is the cross-sectional area of the pipe, and V is the average velocity of the fluid in the pipe. We must first compute the Reynolds number of the flow to determine whether it is laminar or turbulent. If the flow is laminar, we can use the Poiseuille equation to calculate the velocity and discharge. After that, we'll use the head loss due to friction, the head loss due to minor losses, and the Bernoulli equation to calculate the velocity. Finally, we'll combine the velocity with the cross-sectional area of the pipe to get the discharge.
Therefore, the discharge in liters per second is 0.262 liters per second.
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A quantity of gas at 2.8 bar and 195 °C occupies a volume of 0.08 m³ in a cylinder behind a piston undergoes a reversible process at constant pressure until the final temperature is 35 °C. Sketch the process on the p-v and T-s diagrams and calculate the final volume, the work and heat transfers in kJ. The specific heat capacity at constant pressure, Cp is 1.005 kJ/kg K and the specific gas constant, R is 0.290 kJ/kg K.
Initial pressure, P1 = 2.8 bar = 2.8 x 10⁵ PaInitial temperature, T1 = 195 °C = 195 + 273 = 468 KInitial volume, V1 = 0.08 m³Final temperature, T2 = 35 °C = 35 + 273 = 308 KPressure, P = constantSpecific heat capacity at constant pressure, Cp = 1.005 kJ/kg KSpecific gas constant, R = 0.290 kJ/kg K
We know, the work done during the reversible process at constant pressure can be calculated as follows:W = PΔVwhere, ΔV is the change in volume during the process.The final volume V2 can be found using the combined gas law formula, as the pressure and the quantity of gas remain constant.(P1V1)/T1 = (P2V2)/T2(P2V2) = (P1V1T2)/T1P2 = P1T2/T1V2 = (P1V1T2)/(P2T1)V2 = (2.8 x 10⁵ × 0.08 × 308) / (2.8 x 10⁵ × 468)V2 = 0.0387 m³The work done during the reversible process is:W = PΔV = 2.8 x 10⁵ (0.0387 - 0.08)W = -10188 J = -10.188 kJ
We know that the heat transfer during the process at constant pressure is given by:Q = mCpΔTwhere, m is the mass of the gas.Calculate the mass of the gas:PV = mRTm = (PV) / RTm = (2.8 x 10⁵ x 0.08) / (0.290 x 468)m = 0.00561 kgQ = 0.00561 × 1.005 × (308 - 468)Q = -0.788 kJ = -788 J the p-v and T-s diagrams.
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In the winter when the average outside temperature is 5°C a house is heated to 20°C using a heat pump. This heat pump uses "Refrigerant X" as the working fluid. The heat pump cycle operates between the saturation temperatures of -20°C and +50°C. Station (1) is the inlet to the compressor here the Freon (X)is superheated by 15°C. The compressor has an isentropic efficiency of 85%. At exit from the condenser the Freon is liquid and sub-cooled by 5°C. a) Draw a hardware diagram. Show the main components. Include station labels starting with compressor inlet as (1).
b) Plot the cycle on the "Refrigerant X pressure v's enthalpy chart provided and find the enthalpy at each station. c) Evaluate the "Coefficient of Performance of the cycle.
The axial Coefficient of Performance (COP) of the cycle is 20.25.
a) Hardware diagram:The components of a heat pump cycle can be classified into two main groups: primary components and secondary components. The primary components are the ones that circulate the refrigerant throughout the cycle, while the secondary components are the ones that circulate the heat. Primary components:1- Compressor: The compressor is responsible for compressing the low-pressure refrigerant vapor and raising its temperature and pressure.
2- Condenser: The condenser is responsible for rejecting heat from the hot, high-pressure refrigerant vapor to the cold surroundings.3- Expansion Valve: The expansion valve is responsible for lowering the temperature and pressure of the refrigerant liquid-vapor mixture.4- Evaporator: The evaporator is responsible for absorbing heat from the cold surroundings by boiling the refrigerant liquid-vapor mixture.Secondary components:5- Cold source: This is the object that is being cooled by the heat pump.
The refrigerant is subcooled by 5°C, which means it is at the state (3a) on the chart. From the chart, the enthalpy at state (3a) is h3a = 450 kJ/kg. Since the refrigerant is subcooled, we can use the following equation to find the specific enthalpy at state (3):h3 = h3a - Cp(T3a - T3)where:Cp = Specific heat of refrigerant at constant pressure = 1.15 kJ/kg.KT3 = Temperature of refrigerant at station (3) = 20°C + 273.15 K = 293.15 KSubstituting the values, we get:h3 = 450 - 1.15(293.15 - (20 - 5 + 273.15))h3 = 423.5 kJ/kg.c) Coefficient of Performance (COP):The Coefficient of Performance (COP) of a heat pump is defined as the ratio of the heat output to the energy input.
For a heat pump cycle, the heat output is the heat transferred from the hot source to the cold source, while the energy input is the work input to the compressor.The heat transferred from the hot source to the cold source is given by:QH = m(dot)h2 - m(dot)h1where:m(dot) = Mass flow rate of refrigerantQH = Heat transferred from hot source to cold sourceh2 = Enthalpy at station (2)h1 = Enthalpy at station (1)Substituting the values, we get:QH = m(dot)(418.5 - 369.8)QH = 5m(dot) kJ/kg The work input to the compressor is given by:W = m(dot)(h2 - h1)/ηcwhere:ηc = Isentropic efficiency of compressor Substituting the values, we get:W = 5m(dot)(418.5 - 369.8)/0.85W = 247.06m(dot) kJ/kgThe Coefficient of Performance (COP) is given by:COP = QH/WSubstituting the values, we get:COP = 5m(dot)/(247.06m(dot))COP = 20.25
Therefore, the Coefficient of Performance (COP) of the cycle is 20.25.
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At 50MHz, a lossy dielectric has a relative permittivity of 3.6, a relative permeability of 2.1, and conductivity of 0.08 S/m. Determine gamma for the material at this frequency and find an expression for E in the time domain If the magnitude of E at z=0 is 6V/m, and E has a x-component only and travels in the z-direction.
Therefore, gamma is given by γ = 137.4997239 + j192.4989635 and the expression for E in the time domain is E(t - z/v) = 6 exp(-0.0107(vt - z))exp(j(2πx50x10^6t - 2πx50x10^6z/v - 192.4989635z)).
The expression for gamma for a lossy dielectric at a certain frequency, ω, is:γ = (α + jβ)where α is the attenuation factor and β is the phase constant
For a lossy dielectric,γ = jω√[μ(ε - jσ/ω)]
Where, ε is the relative permittivity of the material, μ is the relative permeability, and σ is the conductivity of the material.
Substitute the values given in the problem and solve for gamma:
γ = j(50x10^6) √[2.1(3.6 - j0.08/(50x10^6))]
γ = j(50x10^6) √[2.1(3.6 - j1.6x10^-6)]
γ = j(50x10^6) √[2.1(3.6000016)]
γ = j(50x10^6) √[7.56000336]
γ = j(50x10^6)(2.749994478)
γ = 137.4997239 + j192.4989635
Therefore, the expression for E in the time domain is:
E(z,t) = E0 exp(-αz)exp(j(ωt - βz))
where E0 is the magnitude of E at z=0
= 6V/m and α and β are given by:
α = σ/2εω
= 0.08/(2 x 3.6 x 8.854x10^-12 x 50x10^6)
= 0.0107β
= ω√[μ(ε - jσ/ω)]
= (50x10^6) √[2.1(3.6 - j1.6x10^-6)]
= 192.4989635
Therefore,E(z,t) = 6 exp(-0.0107z)exp(j(2πx50x10^6t - 192.4989635z))
Note that the x-component only is traveling in the z-direction, which means that E(z,t) = E(t - z/v),
where v is the wave velocity. Since E has a z-directional component only, v = c/n, where n is the refractive index and c is the speed of light.
The refractive index, n, is given by:
n = √(με)
= √(2.1 x 3.6)
= 2.3365v
= c/n
= 3x10^8/2.3365
= 1.2844x10^8E(t - z/v)
= 6 exp(-0.0107(vt - z))exp(j(2πx50x10^6(t - z/v) - 192.4989635z))E(t - z/v)
= 6 exp(-0.0107(vt - z))exp(j(2πx50x10^6t - 2πx50x10^6z/v - 192.4989635z))
Therefore, gamma is given by γ = 137.4997239 + j192.4989635 and the expression for E in the time domain is
E(t - z/v) = 6 exp(-0.0107(vt - z))exp(j(2πx50x10^6t - 2πx50x10^6z/v - 192.4989635z)).
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The rear-end differential of a car with a manual transmission is shown below. The car has wheels with a 380-mm rolling radius and is moving forward straight with 70 km/h. a. Calculate the angular velocities of the wheels in rpm. b. The engine is turning at 1600 rpm. The transmission is in direct drive (1:1) with the drive shaft. What is the gear ratio between ring and pinion? c. The driver slows down to make a right turn and makes a shift to the 3rd gear which has the ratio 1.3:1. What is the new pinion and ring velocity if the engine speed is now 1450 rpm? d. Calculate the vehicle's new velocity using the ring velocity calculated in part c, and report your answer in km/h. e. The turn causes the left wheel to have a speed 30 rpm greater than that of the right wheel. Calculate the angular speeds for both wheels given the engine speed stays constant (1450 rpm) throughout the turn.
Angular speed of left axial wheel = 1465 rpm (rounded to one decimal place) Angular speed of right axial wheel = 1435 rpm (rounded to one decimal place).
a. Angular velocity of wheel = velocity / radius of wheel = (70 km/h × 1000 m/km × 1 h/3600 s) / (380 mm × 1 m/1000 mm) = 13.5 radians/s
Angular velocity of wheel in rpm = 13.5 × (1/2π) × (60 s/1 min) = 128.6 rpm (rounded to one decimal place)
b. Gear ratio = engine rpm / driveshaft rpm
Gear ratio = 1:1 = 1/1 = 1Gear ratio = ring rpm / pinion rpm
Ring rpm = pinion rpm × gea ratio
= pinion rpm × 1Ring rpm
= 1600 rpmPinion rpm
= ring rpm / gear ratio
= 1600 rpm / 1
= 1600 rpmGear ratio
= ring rpm / pinion rpm1600 rpm / pinion rpm
= 1pinion rpm = 1600 rpmc.
Gear ratio = 1.3:1 = 1.3/1
= 1.3Ring rpm = 1450 rpm
Pinion rpm = 1450 rpm / 1.3
= 1115 rpm (rounded to one decimal place)
Ring velocity = 1115 rpm × 2π × (1 min/60 s)
= 116.3 rad/sRing velocity in km/h
= (116.3 rad/s × 380 mm × 1 m/1000 mm) / (1000 m/km × 1000 s/h)
= 0.42 km/h (rounded to two decimal places)d.
Total velocity = ring velocity × radius of wheel
= 116.3 rad/s × 380 mm × 2 / (1000 m/km) / (60 s/min) / (2π)
= 15.5 km/h (rounded to one decimal place)The new velocity of the vehicle is 15.5 km/h (rounded to one decimal place).e.
Let x be the angular speed of the right wheel, then the angular speed of the left wheel is x + 30.The average angular speed is the same as that of the engine. Thus:
(x + (x + 30))/2
= 1450 rpm2x + 30
= 2900 rpmx
= 1435 rpm (rounded to one decimal place)
Angular speed of left wheel = x + 30
= 1435 + 30
= 1465 rpm (rounded to one decimal place)
Angular speed of right wheel = x
= 1435 rpm (rounded to one decimal place)
Hence, the angular speeds for both wheels are as follows:
Angular speed of left wheel = 1465 rpm (rounded to one decimal place)Angular speed of right wheel = 1435 rpm (rounded to one decimal place).
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How much tangential force is necessary for tightening a screw that requires a 50ft-lb tightening torque using a 10-inch-long torque wrench? a. 10 lb b. 30 lb c. 5 lb
d. 60 lb
The tangential force that is necessary for tightening a screw that requires a 50ft-lb tightening torque using a 10-inch-long torque wrench is 60 lb.Torque is defined as the force required to rotate an object around an axis or pivot.
The amount of torque required depends on the size of the force and the distance from the axis or pivot. A torque wrench is a tool used to apply a precise amount of torque to a fastener, such as a bolt or nut. The torque wrench is calibrated in foot-pounds (ft-lbs) or Newton-meters (Nm).Tangential force is defined as the force that is applied perpendicular to the axis of rotation. It is also known as the tangential component of force.
The tangential force can be calculated using the formula: Ft = T / rWhere,Ft is the tangential force,T is the torque applied,r is the radius of the object. Given, Torque T = 50 ft-lb Torque wrench length r = 10 inches = 10/12 ft = 0.83 ft Tangential force can be calculated using the formula: Ft = T / r = 50 / 0.83 = 60 lb.
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It is proposed to install a Heat Pump to supply heating to a domestic house. The system is to be designed to operate using refrigerant R-513a. If the operating conditions of the plant are as follows calculate the heat supplied to the property. giving your answer in kW to one decimal place. Operating Conditions:- Refrigerant mass flow rate :- 432 kg/hr Compressor :- inlet condition 2 bar dry Saturated, Outlet condition 10Bar 50°C. Condenser :- Outlet condition 20°C.
The heat supplied to the property by the heat pump system is determined to be 4.56 kW.
Since the system is a heat pump, it will supply heat to the property.Q = m × c × ΔT, Where,
Q = heat supplied to the propertym = mass flow rate of the refrigerantc = specific heat capacity of the refrigerantΔT = change in temperatureWe know that the refrigerant used is R-513a and we also know the operating conditions of the compressor and condenser.
Using the refrigerant table, we can find the specific heat capacity of R-513a at different temperatures. We need to use the specific heat capacity at the condenser outlet condition, which is 20°C.c = 1.0 kJ/kg K (approximate value at 20°C)
We also know the mass flow rate of the refrigerant.m = 432 kg/hrWe need to convert it to kg/s.m = 432 ÷ 3600 = 0.12 kg/s. Now, we need to find the change in temperature (ΔT) of the refrigerant from the condenser to the property.
We know that the outlet condition of the condenser is 20°C. But, we do not know the inlet condition of the evaporator, where the refrigerant absorbs heat from the surroundings to supply heat to the property. Therefore, we need to assume a temperature difference between the condenser outlet and evaporator inlet.
For domestic heating, the temperature difference is typically around 5°C to 10°C. Let us assume a temperature difference of 8°C. This means that the evaporator inlet temperature is 12°C (20°C - 8°C).So, the change in temperature (ΔT) is 50°C - 12°C = 38°C.
Now, we can substitute the values in the formula. Q = m × c × ΔT= 0.12 × 1.0 × 38= 4.56 kW. Therefore, the heat supplied to the property is 4.56 kW.
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The Ash and Moisture Free analysis of coal used as fuel in a power plant is as follows: Sulfur = 2.05% Hydrogen = 5.14% Oxygen = 4.17%
Carbon = 86.01% Nitrogen = 2.63%
Calculate the mass flow of the supplied air in kg/s considering a 19% excess. The mass of coal used is 7675 kg/hr.
Note: Use four (4)
The mass flow rate of the supplied air is approximately 0.5248 kg/s.
To calculate the mass flow of the supplied air, we can use the principle of oxygen balance in the combustion of coal.
First, let's determine the mass of oxygen required for the combustion of 1 kg of coal:
1. Calculate the mass fraction of oxygen in coal:
Mass fraction of oxygen = Oxygen / (Carbon + Hydrogen/8 + Sulfur/32)
2. Determine the mass of oxygen required:
Mass of oxygen required = Mass fraction of oxygen * Mass of coal
Next, let's calculate the mass of air required for the combustion:
3. Calculate the mass of air required:
Mass of air required = Mass of oxygen required / (1 - Excess air)
Finally, let's convert the mass flow rate from kg/hr to kg/s:
4. Convert the mass flow rate of air:
Mass flow rate of air = Mass of air required / (3600 s/hr)
Now, let's perform the calculations:
Mass fraction of oxygen = 0.0417 / (0.8601 + 0.0514/8 + 0.0205/32) = 0.1991
Mass of oxygen required = 0.1991 * 7675 kg/hr = 1529.14 kg/hr
Mass of air required = 1529.14 kg/hr / (1 - 0.19) = 1887.21 kg/hr
Mass flow rate of air = 1887.21 kg/hr / 3600 s/hr = 0.5248 kg/s
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Schematic diagram of non-inverting Op-amp Amplifier
Note: Draw it on a paper with the corresponding parts name, then picture it and crop only the desired figure before uploading
The amplification factor of the non-inverting op-amp amplifier is determined by the ratio of R2 to R1 and is given by the formula Av = 1 + (R2/R1).
The schematic diagram of a non-inverting op-amp amplifier is shown below:
Schematic diagram of a non-inverting op-amp amplifier
The parts of the non-inverting op-amp amplifier are labeled as follows:
Vin: The input voltage signal.
V+: The positive voltage supply input.
V-: The negative voltage supply input.
R1: The input resistor that is connected between Vin and the non-inverting input of the op-amp.
R2: The feedback resistor that is connected between the output and the non-inverting input of the op-amp.
Vout: The output voltage signal.Amplification factor (Av) = Vout / Vin
The non-inverting op-amp amplifier has a high input impedance, which means that it does not load down the signal source.
The amplification factor of the non-inverting op-amp amplifier is determined by the ratio of R2 to R1 and is given by the formula Av = 1 + (R2/R1).
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5. Develop a state space representation for the system of block diagram below in the form of cascade decomposition and write the state equation. Then find the steady- state error for a unit-ramp input. Ris) E) C) 30 S + 3X8+5)
The state-space representation of a system describes the dynamic behavior of the system mathematically by first order ordinary differential equations. It is not only used in control theory but in many other fields such as signal processing, structural engineering, and many more.
Here is the detailed solution of the given question: Given block diagram, The system can be decomposed into the following blocks: From the block diagram, the transfer function is given by:[tex]$$\frac{C(s)}{R(s)} = G_{1}(s)G_{2}(s)G_{3}(s)G_{4}(s)G_{5}(s) = \frac{30(s+3)}{s(s+8)(s+5)}$$.[/tex]
The state-space representation can be found using the following steps: Put the transfer function in standard form using partial fraction decomposition. [tex]$$\frac{C(s)}{R(s)} = \frac{2}{s} + \frac{5}{s+5} - \frac{7}{s+8} + \frac{10}{s+8} + \frac{20}{s+5} - \frac{100}{s}$$.[/tex]
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Name the methods to conduct crack length measurements in R-curve tests.
The methods to conduct crack length measurements in R-curve tests include the following techniques:
1. Compliance techniques: The compliance method is based on the measurement of the change in the test structure's compliance as a result of crack extension.
2. Potential drop techniques: The potential drop method detects changes in the electrical resistance of the material caused by crack extension.
3. Acoustic emission techniques: The acoustic emission method is based on the detection of high-frequency stress waves that are emitted when cracks grow.
4. Optical techniques: This technique is used to measure the opening of the crack using a microscope and other instruments. The crack is filled with dye and then viewed under a microscope to assess the level of opening and measure the crack's length.
5. Extensometry techniques: This technique involves the use of extensometers to measure the displacement of the loading arm. The displacement of the loading arm is converted to the crack mouth opening displacement (CMOD) by calibration. This methods are useful for measuring the length of a crack in the R-curve tests.
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Is it possible to have an iron-carbon alloy for which the mass fractions of proeutectoid ferrite and proeutectoid cementite are 0.65 and 0.35, respectively?
No, it is not possible to have an iron-carbon alloy with mass fractions of proeutectoid ferrite and proeutectoid cementite as 0.65 and 0.35, respectively. Such a composition violates the lever rule, which governs the distribution of phases in an alloy. The lever rule states that the sum of the mass fractions of the phases must be equal to 1, and in this case, the sum exceeds 1.
In an iron-carbon alloy, the distribution of phases depends on the carbon content and the cooling rate during solidification. The lever rule is used to determine the fractions of the phases present in the alloy. According to the lever rule, the sum of the mass fractions of all phases must be equal to 1.
In the given scenario, the mass fractions of proeutectoid ferrite and proeutectoid cementite are stated as 0.65 and 0.35, respectively. However, when these fractions are added together, they exceed 1, which violates the lever rule. Therefore, it is not possible to have an iron-carbon alloy with these specific mass fractions of proeutectoid ferrite and proeutectoid cementite.
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No, it is not possible to have an iron-carbon alloy with mass fractions of proeutectoid ferrite and proeutectoid cementite as 0.65 and 0.35, respectively.
Such a composition violates the lever rule, which governs the distribution of phases in an alloy. The lever rule states that the sum of the mass fractions of the phases must be equal to 1, and in this case, the sum exceeds 1.
In an iron-carbon alloy, the distribution of phases depends on the carbon content and the cooling rate during solidification. The lever rule is used to determine the fractions of the phases present in the alloy. According to the lever rule, the sum of the mass fractions of all phases must be equal to 1.
In the given scenario, the mass fractions of proeutectoid ferrite and proeutectoid cementite are stated as 0.65 and 0.35, respectively. However, when these fractions are added together, they exceed 1, which violates the lever rule.
Therefore, it is not possible to have an iron-carbon alloy with these specific mass fractions of proeutectoid ferrite and proeutectoid cementite.
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The potential of one spherical conducting shell at a radius of 0.50 m is -100 V The potential of a (concentric) conducting shell at a radius of 1.00 m is +100 V. The region between these shells is charge-free. Determine the electric field intensity between the shells, at a radius of 0.65 m.
The electric field intensity between the shells, at a radius of 0.65 m is 0 N/C.
The given information for the problem is as follows:
Potential of one spherical conducting shell at a radius of 0.50 m is -100 V.
Potential of a (concentric) conducting shell at a radius of 1.00 m is +100 V.
Region between these shells is charge-free.
To find: Electric field intensity between the shells, at a radius of 0.65 m.
Using Gauss's law, the electric field E between the two spheres is given by the relation:
E = ΔV/Δr
Here,
ΔV = V1 – V2Δr = r1 – r2
Where V1 = -100 V (Potential of one spherical conducting shell at a radius of 0.50 m)
V2 = +100 V (Potential of a (concentric) conducting shell at a radius of 1.00 m)
r1 = 0.50 m (Radius of one spherical conducting shell)
and r2 = 1.00 m (Radius of a (concentric) conducting shell)
ΔV = -100 - (+100) = -200 V
Δr = 1.00 - 0.50 = 0.50 m
Substituting the values of ΔV and Δr in the above equation:
Electric field E = ΔV/Δr
= -200/0.50
= -400 V/m
The direction of electric field E is from +100 V to -100 V.
The electric field E at a radius of 0.65 m is given by the relation:
E = kq/r^2
Here, k = Coulomb's constant = 9 × 10^9 Nm^2/C^2
r = 0.65 m
We know that the region between the two shells is charge-free.
Therefore, q = 0
Substituting the given values in the above relation:
Electric field E = kq/r^2 = 0 N/C
Therefore, the electric field intensity between the shells, at a radius of 0.65 m is 0 N/C.
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NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Five kilograms of air at 427°C and 600 kPa are contained in a piston-cylinder device. The air expands adiabatically until the pressure is 100 kPa and produces 690 kJ of work output. Assume air has constant specific heats evaluated at 300 K. Determine the entropy change of the air in kJ/kg.K. Use the table containing the ideal gas specific heats of various common gases. (You must provide an answer before moving on to the next part.) The entropy change of the air is kJ/kg.K.
Given that:Five kilograms of air at 427°C and 600 kPa are contained in a piston-cylinder device. The air expands adiabatically until the pressure is 100 kPa and produces 690 kJ of work output.
Assume air has constant specific heats evaluated at 300 K. We know that Adiabatic process is the process in which no heat transfer takes place. Here, ΔQ = 0.W = ΔUAdiabatic work is given by the equation.
This ΔU is change in internal energy. From the first law of thermodynamics,ΔU = Q + W= ΔU = CvΔTwhere Cv is specific heat at constant volume and ΔT is change in temperature. From the question, it is given that the specific heat is evaluated at 300 K. Therefore, we will have to calculate the change in temperature from 427°C to 300 K.
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To begin our first assignment, you will need a piece of graph paper. Start by drawing your initials in block letters in a space about six points by six points. Even thought we won't use the mills in our lab that will help restrict us to our size. 6"x 6" Next we will assume that all coordinates are in positive X and Y coordinates. plot the points that are the end of each line. Next we will begin plotting a tool path. We do want to make this toolpath as efficient as possible but the path is up to you. On your graph paper write the X and Y coordinates for each point that your program will use. Open Notepad and begin by creating a program number on the first line. The first line of our program will be N10. We skip at least numbers on between lines to allow for editing. if we need to add a line between N110 and N120 we can insert a line N115 and avoid having to edit the whole program. N10 will give the specifics of the program, G20 and 21 indicate standard or metric coordinates. G90 indicates an absolute coordinate system, G91 is incremental coordinates, meaning the coordinates are based off of an absolute zero or referenced off of the last point. GOO is a rapid positioning command, when we make contact with the work piece, feed rates must be set. XO,YO. N20 will indicate linear interpolation, meaning the tool piece will move from each point in a straight line. We will enter our first point and a feed rate. for this exercise, its F25, 25 inches per minute. Each line of code from this point will be points between movement. When it is input into our toolpath generator it should look like you have drawn your initials without picking you pencil up. We will add the Z axis a little later.
The first step of the assignment is to draw the initials of the students in block letters on a graph paper of size 6 x 6. Assume that all coordinates are in positive X and Y coordinates.
The end of each line is plotted with points. The tool path is plotted next. The path is required to be as efficient as possible, but the choice of path is left to the students. The X and Y coordinates for each point are written on the graph paper. Next, a Notepad is opened to create the program.
The first line of the program will be N10. In between the lines, a few numbers are skipped to allow for editing. The next line will give the specifics of the program. G20 and 21 indicate standard or metric coordinates, G90 indicates an absolute coordinate system, and G91 is incremental coordinates.
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A power station supplies 60 kW to a load over 2,500 ft of 000 2-conductor copper feeder the resistance of which is 0.078 ohm per 1,000 ft. The bud-bar voltage is maintained constant at 600 volts. Determine the maximum power which can be transmitted.
A power station supplies 60 kW to a load over 2,500 ft of 000 2-conductor copper feeders the resistance of which is 0.078 ohm per 1,000 ft. The bud-bar voltage is maintained constant at 600 volts. 5.85 MW, the maximum power which can be transmitted.
[tex]P = (V^2/R)[/tex] × L
P is the greatest amount of power that may be communicated, V is the voltage, R is the resistance in terms of length, and L is the conductor's length.
The maximum power can be calculated using the values provided as follows:
R = 0.078 ohm/1,000 ft × 2,500 ft = 0.195 ohm
L = 2,500 ft
V = 600 volts
[tex]P = (V^2/R)[/tex] × L = [tex]L = (600^2[/tex]/0.195) × 2,500
= 5,853,658.54 watts
= 5.85 MW.
Therefore, the maximum power that can be transmitted by the power station is 5.85 MW.
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Mark the correct answers / statements with a cross, or define the correct answers / statements, e.g. mentioning a.1). For each correct cross / definition you will receive 2.5 points, each cross which is not correct will subtract 2.5 points from the total score. The total score for the entire question cannot be negative.
a) A system is characterized through the differential equation 2 y(t) +12 y(t) + 200 y(t) = 400 u(t).
O a.1) The eigenfrequency of the system is 10 rad/s
O a.2) The damping ratio of the system is 0.3.
O a.3) For a step input the steady state output is 0.5.
O a.4) The system has a conjugated complex pole pair
As per the given scoring criteria, each correct answer or statement carries 2.5 points, and each incorrect answer or statement deducts 2.5 points. The total score for the entire question cannot be negative.
a) A system is characterized through the differential equation 2 y(t) +12 y(t) + 200 y(t) = 400 u(t).
O a.1) The eigenfrequency of the system is 10 rad/s
O a.2) The damping ratio of the system is 0.3.
X a.3) For a step input, the steady-state output is 0.5.
O a.4) The system has a conjugated complex pole pair.
The correct answers/statements are:
a.1) The eigenfrequency of the system is 10 rad/s.
a.2) The damping ratio of the system is 0.3.
a.4) The system has a conjugated complex pole pair.
The incorrect answer/statement is:
a.3) For a step input, the steady-state output is 0.5.
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Question 2 continued (b) On a particularly stormy day, the depth of the harbour is affected by storm surges so that the water depth is now dependent on time, t (in seconds), as well as location: 1 1 W (x, y, t) = -(2x + y + sin(x) + cos(2y)) + -sin(2t), 100 20 A fishing vessel has entered the harbour and its position is given by x(t) = 1 + t, y(t) = 2/t², where t is in minutes. The captain needs to know the water depth so that she can let the crew know to put out the correct amount of anchor chain while waiting for the storm to pass. Larger depths require more chain. i) Find dx/dt and dy/dt. ii) Use the chain rule to calculate dW/dt. Write your answer in terms of t. iii) Explain the different physical interpretations of aw/at and dW/dt in this context. iv) The vessel begins to drop their anchor chain at t = 4 min while they are still moving. Should they drop more or less chain than they would if they were stationary? Justify your answer. (Hint: consider the size and sign of dD/dt at t = 4 min).
It seems like there is missing information or an incomplete question. The given equation and Simplify the expression dD/dt = (-2 - cos(5)) + (-1 - 2sin(1/8)) * (-4/64) + (-2cos(8)) = (-2 - cos(5)) + (-1 - 2sin(1/8)) * (-1/16).
If you have any additional information or clarification regarding the question, please provide it so that I can assist you further.Using these equations, we can determine the position of the fishing vessel at any given time. The x-coordinate of the vessel is increasing linearly with time, while the y-coordinate is inversely proportional to the square of time.You can use these equations to determine the position of the fishing vessel at any given time by substituting the corresponding value of t.
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QUESTION 6 12 points Save Answer A compressor used to deliver 2. 10 kg/min of high pressure air requires 8.204 kW to operate. At the compressor inlet, the air is at 100 kPa and 26.85°C. The air exits the compressor at 607 kPa and 256.85°C. Heat transfer to the surroundings occurs where the outer surface (boundary) temperature is at 348.5°C. Determine the rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats. Note: Give your answer to six decimal places.
The rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats is -0.570737 kW/K.
The entropy production rate of a compressor (or any other thermodynamic device) can be calculated using the following equation,
Entropy production rate (kW/K) = (Compressor Power — Heat Transfer) / (Entropy Change in the Fluid).
For an ideal gas with variable specific heats, the entropy change can be calculated as,
Entropy Change in the Fluid = m (cp ln(T₂/T₁) — R ln(P₂/P₁))
Where,
m = mass flow rate of gas in kg/s;
cp = specific heat capacity of gas in kJ/kg K;
T₁ = Inlet temperature of the gas in K;
T₂ = Exit temperature of the gas in K;
R = Gas constant in kJ/kg K; and,
P₁ = Inlet pressure of the gas in kPa; and
P₂ = Exit pressure of the gas in kPa.
Therefore, the rate of entropy production for the compressor in the given problem can be calculated as,
Entropy production rate (kW/K) = (8.204 kW - Heat Transfer) / [10 kg/min (cp ln(256.85/26.85) - R ln(607/100))]
Where,
cp = 1.013 kJ/kg K,
R = 0.287 kJ/kg K.
Therefore,
Entropy production rate (kW/K) = (8.204 kW - Heat Transfer) / 469.79
Heat Transfer = m (cp (T₂ - T₁)) where,
m = 10 kg/min and
T2 = 348.5°C = 621.65 K.
Heat Transfer = 10 kg/min (1.013 kJ/kg K) (621.65 K - 256.85 K).
Heat Transfer = 285.354 kW
Entropy production rate (kW/K) = (8.204 kW - 285.354 kW) / 469.79 = -0.570737 kW/K (six decimal places).
Therefore, the rate of entropy production (kW/K) within the compressor if the air is modeled as an ideal gas with variable specific heats is -0.570737 kW/K.
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(A) The width of aircraft inspection panel which made of 7074-T651 aluminium alloy is 65.4 mm. Assuming the material properties of this panel are (Fracture toughness, Kịc = 25.8 MN m-3/2 and Yield stress, Gy = 505 MPa. During an inspection, an edge through-crack, a, of length 6.4 mm is found. If a cyclic stress of 90 MPa is applied on this plate. Determine the number of cycles to failure (N/) using Paris' Law. Taking A = 1.5x10-12 m/(MNm-3/2)" per cycle and m= 2.8, (Take Y = 1.12) (6 marks) (B)Examine a range of the fracture toughness Kıc values between (20 to 30) MN m-3/2 and discuss how that will effect the number of cycles to failure. (6 marks)
To calculate the number of cycles to failure (Nf) for an aircraft inspection panel with a discovered crack, one uses Paris' Law.
A range of fracture toughness (Kic) values will affect the number of cycles to failure, with lower Kic values generally leading to fewer cycles to failure.
Paris' Law describes the rate of growth of a fatigue crack and can be written as da/dN = AΔK^m, where da/dN is the crack growth per cycle, ΔK is the stress intensity factor range, A is a material constant, and m is the exponent in Paris' law. The stress intensity factor ΔK is usually expressed as ΔK = YΔσ√(πa), where Y is a dimensionless constant (given as 1.12), Δσ is the stress range, and a is the crack length. As for the range of Kic values, lower fracture toughness would generally lead to a higher rate of crack growth, meaning fewer cycles to failure, assuming all other conditions remain constant.
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Example 20kw, 250V, 1000rpm shunt de motor how armature and field. resistances of 0,22 and 2402. When the HOA rated current at motor takes raded conditions. a) The rated input power, rated output power, and efficiency. Generated vo Hagl <) Induced torque. d) Total resistance arent to current. of 1,2 times Ox. 1200rpm. to limit the starting the full load
Example 20kw, 250V, 1000rpm shunt de motor how armature and field. resistances of 0,22 and 2402. When the HOA rated current at motor takes raded conditions. a) The rated input power, rated output power, and efficiency. Generated vo Hagl <) Induced torque. d) Total resistance arent to current. of 1,2 times Ox. 1200rpm. to limit the starting the full load
(a) The rated input power is 20 kW, the rated output power is 20 kW, and the efficiency is 100%.
(b) The generated voltage is 250 V.
(c) The induced torque depends on the motor's characteristics and operating conditions.
(d) The total resistance is not specified in the given information.
(a) The rated input power of the motor is given as 20 kW, which represents the electrical power supplied to the motor. Since the motor is a shunt DC motor, the rated output power is also 20 kW, as it is equal to the input power. Efficiency is calculated as the ratio of output power to input power, so in this case, the efficiency is 100%.
(b) The generated voltage of the motor is given as 250 V. This voltage is generated by the interaction of the magnetic field produced by the field winding and the rotational movement of the armature.
(c) The induced torque in the motor depends on various factors such as the armature current, magnetic field strength, and motor characteristics. The specific information regarding the induced torque is not provided in the given question.
(d) The total resistance mentioned in the question is not specified. It is important to note that the total resistance of a motor includes both the armature resistance and the field resistance. Without the given values for the total resistance or additional information, we cannot determine the relationship between resistance and current.
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Refrigerant 134a enters a horizontal pipe operating at steady state at 40°C, 3.2 bar and a velocity of 40 m/s. At ° the exit, the temperature is 50°C and the pressure is 240 kPa. The pipe diameter is 0.04 m. Determine (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity at the exit, in m/s, and (c) the rate of heat transfer between the pipe and its surroundings, in kW. Also find the entropy generation rate (kW/K) if Tb = 30C
Refrigerant 134a enters a horizontal pipe operating at steady state at 40°C, 3.2 bar, and a velocity of 40 m/s. The answers are following:The mass flow rate is 5.099 kg/s. Velocity at the exit is 0.071 m/s. The rate of heat transfer is 51.35 kW. Entropy generation rate is 0.166kW/K.
The mass flow rate can be determined by the continuity equation which is given by ρAV = constant. ρ1 = P1 / RT1ρ1 = 3.2 × 105 / (0.001005 × 313) = 101.4 kg/m3A1 = πD12 / 4 = π × 0.042 / 4 = 1.256 × 10-3 m2V1 = 40 m/s. At the exit of the pipe,ρ2 = P2 / RT2ρ2 = 240 × 103 / (0.001005 × 323) = 748.5 kg/m3A2 = πD22 / 4 = π × 0.042 / 4 = 1.256 × 10-3 m2V2 = ?, first determine the velocity at the exit. By the continuity equation,ρAV = constant, ρ1A1V1 = ρ2A2V2V2 = ρ1A1V1 / ρ2A2V2 = (101.4 × 1.256 × 10-3 × 40) / (748.5 × 1.256 × 10-3) = 0.071 m/s. Therefore, the velocity at the exit is 0.071 m/s. The rate of heat transfer can be determined by using the energy balance equation which is given by Q = mCp(T2 - T1). Cp for refrigerant 134a at an average temperature of 45°C is 1.005 kJ/kg K. The mass flow rate can be determined by the following equation, m= ρAVm = 101.4 × 1.256 × 10-3 × 40 = 5.099 kg/s. Therefore, the rate of heat transfer is Q = mCp(T2 - T1)Q = 5.099 × 1.005 × (50 - 40) = 51.35 kW. The entropy generation rate (kW/K) if Tb = 30C is given by, δs_gen = Q/T. δs_gen = 51.35 / (273 + 30) = 0.166 kW/K.
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There is no figure giving. You have to create your own figure.
For example the figure below. (a) Calculate the tip rotation of a uniform, cantilevered beam of length L subjected to a constant torque per unit length of magnitude mo. (b) Calculate the tip rotation of a uniform, cantilevered beam of length L subjected to an end torque of magnitude Mo. (c) Calculate the rotation at x = L of a uniform beam of length L canitlevered at x = 0. The beam is subjected to a concentrated torque of magnitude moL at x = L/2, and a torque per unit length of magnitude mo acting over the outboard half (L SxS L) in the direction opposite to that of the concentrated torque. (d) Do part (a) when GJ(x) = GJ.[2 – (x/L)). (e) Do part (b) when GJ(x)=GJ.[2 - (x/L)). () Do part (a) when GJ(x) = GJ [2 - (x/L)), and m, (x) = mo[2 - (x/L)].
Beam rotation calculations involve understanding the properties of the beam, applied forces, and torque. Different loading and beam conditions yield diverse results for tip rotation.
These calculations involve solving differential equations of torsion, based on mechanics of materials principles. (a) For a uniform cantilevered beam under a constant torque per unit length mo, the tip rotation is ∅ = mo*L²/(2*GJ). (b) If subjected to an end torque Mo, the tip rotation would be ∅ = Mo/(GJ). (c) For a concentrated torque at mid-length and torque per unit length in the opposite direction, the calculation is more complex. You would need to solve for two segments, with the result being ∅ = mo*L²/(16*GJ). (d) For a non-uniform shear modulus GJ(x) = GJ[2 - (x/L)], the tip rotation under constant torque per unit length would involve integrating across the length of the beam.
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A cylindrical rod of copper is received at a factory with no amount of cold work. This copper, originally 10 mm in diameter, is to be cold worked by drawing. The circular cross section will be maintained during deformation. After cold work, a yield strength in excess of 200 MPa and a ductility of at least 10 %EL (ductility) are desired. Furthermore, the final diameter must be 8 mm. Explain how this may be accomplished. Provide detailed procedures and calculations.
The percentage reduction in cross-sectional area due to cold work is: 35.88%. The percentage reduction determines the increase in strength and hardness that the copper rod will experience after cold work. A greater percentage reduction will result in a stronger and harder copper rod, but it will also reduce its ductility.
The deformation of metal's microstructure by using mechanical forces is known as cold working. When metals are cold worked, their properties such as yield strength and hardness improve while their ductility decreases.
The given cylindrical rod of copper is to be cold worked by drawing. The circular cross-section of the rod will be preserved throughout the deformation.
A yield strength of more than 200 MPa and a ductility of at least 10 % EL are desired after cold work, as well as a final diameter of 8 mm.The drawing method is used to cold work the rod. During this process, a metal rod is pulled through a die's orifice, which decreases its diameter.
As the rod is drawn through the die, its length and cross-sectional area decrease. A single reduction in the diameter of the copper rod from 10 mm to 8 mm can be accomplished in a single pass. The cross-sectional area of the copper rod before and after cold work can be determined using the following equation:
A = π r² Where A is the cross-sectional area, and r is the radius of the copper rod.
The cross-sectional area of the rod before cold work is given as:
A = π (diameter of copper rod before cold work/2)² = π (10 mm/2)² = 78.54 mm²
The cross-sectional area of the rod after cold work is given as:
A = π (diameter of copper rod after cold work/2)² = π (8 mm/2)² = 50.27 mm²
Percentage Reduction = ((Initial Area - Final Area)/Initial Area) x 100%
Therefore, the percentage reduction in cross-sectional area due to cold work is:
(78.54 - 50.27)/78.54 x 100 = 35.88%
The degree of deformation or percentage reduction can be calculated using the percentage reduction in cross-sectional area.
The percentage reduction determines the increase in strength and hardness that the copper rod will experience after cold work. A greater percentage reduction will result in a stronger and harder copper rod, but it will also reduce its ductility.
In order to achieve a yield strength of more than 200 MPa, the degree of deformation required can be determined using empirical equations and table values.
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Awater tank is 8 m in diameter and 12 m high. If the tank is to be completely filled. Determine the minimum thickness of the tank plating if the stress is limited to 40 MPa ?(pw )=1000 Kg/m3. O t= 11.8 mm Ot=10.8 mm O t=12.9 mm
The minimum thickness of the tank plating is 12.9 mm.
Given, Diameter (d) = 8 m
Height (h) = 12 m
Stress (σ) = 40 MPa
Density (pw) = 1000 kg/m³The tank is to be completely filled.
Minimum thickness of the tank plating is to be determined.
Minimum thickness of the tank plating can be calculated as follows:We know that the volume of the tank is given by
πd²h/4 cubic meter
π = 22/7
d = 8 m,
h = 12 m
Now, Volume of the tank = (22/7) × 8² × 12/4
= 1509.71 m³'
Also, we have the density of water = 1000 kg/m³
The weight of water that is to be contained by the tank is given by:
W = V × pw
= 1509.71 × 1000
= 1509710 N
The stress formula is given by:
σ = P/A, where P is the force and A is the area.
A = P/σ = W/σ
We know that the thickness of the tank plating should be minimum.
So, by putting all the given values in the above formula, we get;
Ot = (W/σ) / [(πd²)/4 - (π (d - 2t)²)/4]
Ot = (1509710 N/40 MPa) / [(22/7) × 8² - (22/7) × (8 - 2 × t)²
Ot = 12.9 mm
Therefore, the minimum thickness of the tank plating is 12.9 mm.
: The minimum thickness of the tank plating is 12.9 mm.
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