Natural convection over surfaces: A 0.5-m-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. The plate surface has an emissivity of 0.73, and its midpoint temperature is 55°C.
The length of the plate = 0.5 m The heat flux on one side of the plate is uniform.T he other side is exposed to cool air at 5°C.The plate surface has an emissivity of 0.73.The midpoint temperature of the plate = 55°C.
[tex]Ra = (gβΔT L3)/ν2[/tex]
[tex]Ra = (9.81 × 0.0034 × 50 × 0.53)/(1.568 × 10-5)Ra = 3.329 × 107Nu = 0.59[/tex]
[tex]Nu - CRa1/4 = 0.59 - 0.14 (3.329 × 107)1/4[/tex]
[tex]Nu - CRa1/4 = 0.59 - 573.7[/tex]
[tex]Nu - CRa1/4 = - 573.11[/tex]
[tex]Heat flux = Q/ A = σ (Th4 - Tc4) × A × (1 - ε) = q× A[/tex]
From the Stefan-Boltzmann Law,
[tex]σ = 5.67 × 10-8 W/m2K4σ (Th4 - Tc4) × A × (1 - ε) = q × A[/tex]
Therefore,
[tex]q = 5.67 × 10-8 × 1.049 × 10-9 × (Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12(Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12 [(Th/2)4 - (5)4] × 0.5 × (1 - 0.73)q = 29.6 W/m2[/tex]
Hence, the heat flux subjected to the plate surface is 29.6 W/m2.
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A fluid in a fire hose with a 46.5 mm radius, has a velocity of 0.56 m/s. Solve for the power, hp, available in the jet at the nozzle attached at the end of the hose if its diameter is 15.73 mm. Express your answer in 4 decimal places.
Given data: Radius of hose
r = 46.5m
m = 0.0465m
Velocity of fluid `v = 0.56 m/s`
Diameter of the nozzle attached `d = 15.73 mm = 0.01573m`We are supposed to calculate the power, hp available in the jet at the nozzle attached to the hose.
Power is defined as the rate at which work is done or energy is transferred, that is, P = E/t, where E is the energy (J) and t is the time (s).Now, Energy E transferred by the fluid is given by the formula E = 1/2mv² where m is the mass of the fluid and v is its velocity.We can write m = (ρV) where ρ is the density of the fluid and V is the volume of the fluid. Volume of the fluid is given by `V = (πr²l)`, where l is the length of the hose through which fluid is coming out, which can be assumed to be equal to the diameter of the nozzle or `l=d/2`.
Thus, `V = (πr²d)/2`.Energy transferred E by the fluid can be expressed as Putting the value of V in the above equation, we get .Now, the power of the fluid P, can be written as `P = E/t`, where t is the time taken by the fluid to come out from the nozzle.`Putting the given values of r, d, and v, we get Thus, the power available in the jet at the nozzle attached to the hose is 0.3011 hp.
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Design a circuit which counts seconds, minutes and hours and displays them on the 7-segement display in 24 hour format. The clock frequency available is 36 KHz. Assume that Binary to BCD converter and BCD to 7-Segement display is already available for the design.
The 24-hour clock has two digits for hours, two digits for minutes, and two digits for seconds. Binary Coded Decimal (BCD) is a technique for representing decimal numbers using four digits in which each decimal digit is represented by a 4-bit binary number.
A 7-segment display is used to display the digits from 0 to 9.
Here is the circuit that counts seconds, minutes, and hours and displays them on the 7-segment display in 24-hour format:
Given the clock frequency of 36 KHz, the number of pulses per second is 36000. The seconds counter requires 6 digits, or 24 bits, to count up to 59. The minutes counter requires 6 digits, or 24 bits, to count up to 59. The hours counter requires 5 digits, or 20 bits, to count up to 23.The clock signal is fed into a frequency divider that produces a 1 Hz signal. The 1 Hz signal is then fed into a seconds counter, minutes counter, and hours counter. The counters are reset to zero when they reach their maximum value.
When the seconds counter reaches 59, it generates a carry signal that increments the minutes counter. Similarly, when the minutes counter reaches 59, it generates a carry signal that increments the hours counter.
The outputs of the seconds, minutes, and hours counters are then converted to BCD format using a binary to BCD converter. Finally, the BCD digits are fed into a BCD to 7-segment display decoder to produce the display on the 7-segment display.Here's a block diagram of the circuit: Block diagram of the circuit
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A vapor-compression refrigeration system utilizes a water-cooled intercooler with ammonia as the refrigerant. The evaporator and condenser temperatures are -10 and 40°C, respectively. The mass flow rate of the intercooler water is 0.35 kg/s with a change in enthalpy of 42 kJ/kg. The low-pressure compressor discharges the refrigerant at 700 kPa. Assume compression to be isentropic. Sketch the schematic and Ph diagrams of the system and determine: (a) the mass flow rate of the ammonia refrigerant, (b) the capacity in TOR, (c) the total compressor work, and (d) the COP.
In a vapor-compression refrigeration system with an ammonia refrigerant and a water-cooled intercooler, the goal is to determine the mass flow rate of the refrigerant, the capacity in TOR (ton of refrigeration), the total compressor work, and the coefficient of performance (COP).
To determine the mass flow rate of the ammonia refrigerant, we need to apply mass and energy balance equations to the system. The mass flow rate of the intercooler water and its change in enthalpy can be used to calculate the heat transfer in the intercooler and the heat absorbed in the evaporator. The capacity in TOR can be calculated by converting the heat absorbed in the evaporator to refrigeration capacity. TOR is a unit of refrigeration capacity where 1 TOR is equivalent to 12,000 BTU/hr or 3.517 kW.
The total compressor work can be calculated by considering the isentropic compression process and the pressure ratio across the compressor. The work done by the compressor is equal to the change in enthalpy of the refrigerant during compression. The COP of the refrigeration system can be determined by dividing the refrigeration capacity by the total compressor work. COP represents the efficiency of the system in providing cooling for a given amount of work input. Schematic and Ph diagrams can be sketched to visualize the system and understand the thermodynamic processes involved. These diagrams aid in determining the properties and states of the refrigerant at different stages of the cycle.
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Design a excel file of an hydropower turgo turbine in Sizing and Material selection.
Excel file must calculate the velocity of the nozel, diameter of the nozel jet, nozzle angle, the runner size of the turgo turbine, turbine blade size, hub size, fastner, angular velocity,efficiency,generator selection,frequnecy,flowrate, head and etc.
(Note: File must be in execl file with clearly formulars typed with all descriptions in the sheet)
Designing an excel file for a hydropower turbine (Turgo turbine) involves calculating different values that are essential for its operation. These values include the velocity of the nozzle, diameter of the nozzle jet, nozzle angle, runner size of the turbine, turbine blade size, hub size, fastener, angular velocity, efficiency, generator selection, frequency, flow rate, head, etc.
To create an excel file for a hydropower turbine, follow these steps:Step 1: Open Microsoft Excel and create a new workbook.Step 2: Add different sheets to the workbook. One sheet can be used for calculations, while the others can be used for data input, output, and charts.Step 3: On the calculation sheet, enter the formulas for calculating different values. For instance, the formula for calculating the velocity of the nozzle can be given as:V = (2 * g * H) / (√(1 - sin²(θ / 2)))Where V is the velocity of the nozzle, g is the acceleration due to gravity, H is the head, θ is the nozzle angle.Step 4: After entering the formula, label each column and row accordingly. For example, the velocity of the nozzle formula can be labeled under column A and given a name, such as "Nozzle Velocity Formula".Step 5: Add a description for each formula entered in the sheet.
The explanation should be clear, concise, and easy to understand. For example, a description for the nozzle velocity formula can be given as: "This formula is used to calculate the velocity of the nozzle in a hydropower turbine. It takes into account the head, nozzle angle, and acceleration due to gravity."Step 6: Repeat the same process for other values that need to be calculated. For example, the formula for calculating the diameter of the nozzle jet can be given as:d = (Q / V) * 4 / πWhere d is the diameter of the nozzle jet, Q is the flow rate, and V is the velocity of the nozzle. The formula should be labeled, given a name, and described accordingly.Step 7: Once all the formulas have been entered, use the data input sheet to enter the required data for calculation. For example, the data input sheet can contain fields for flow rate, head, nozzle angle, etc.Step 8: Finally, use the data output sheet to display the calculated values. You can also use charts to display the data graphically. For instance, you can use a pie chart to display the percentage efficiency of the turbine. All the sheets should be linked correctly to ensure that the data input reflects on the calculation sheet and output sheet.
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Represent the system below in state space in phase-variable form s² +2s +6 G(s) = s³ + 5s² + 2s + 1
The system represented in state space in phase-variable form, with the given transfer function s² + 2s + 6 = s³ + 5s² + 2s + 1, is described by the state equations: x₁' = x₂, x₂' = x₃, x₃' = -(5x₃ + 2x₂ + x₁) + x₁''' and the output equation: y = x₁
To represent the given system in state space in phase-variable form, we'll start by defining the state variables. Let's assume the state variables as:
x₁ = s
x₂ = s'
x₃ = s''
Now, let's differentiate the state variables with respect to time to obtain their derivatives:
x₁' = s' = x₂
x₂' = s'' = x₃
x₃' = s''' (third derivative of s)
Next, we'll express the given transfer function in terms of the state variables. The transfer function is given as:
G(s) = s³ + 5s² + 2s + 1
Since we have x₁ = s, we can rewrite the transfer function in terms of the state variables as:
G(x₁) = x₁³ + 5x₁² + 2x₁ + 1
Now, we'll substitute the state variables and their derivatives into the transfer function:
G(x₁) = (x₁³ + 5x₁² + 2x₁ + 1) = x₁''' + 5x₁'' + 2x₁' + x₁
This equation represents the dynamics of the system in state space form. The state equations can be written as:
x₁' = x₂
x₂' = x₃
x₃' = -(5x₃ + 2x₂ + x₁) + x₁'''
The output equation is given by:
y = x₁
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1. (a) Let A and B be two events. Suppose that the probability that neither event occurs is 3/8. What is the probability that at least one of the events occurs? (b) Let C and D be two events. Suppose P(C)=0.5,P(C∩D)=0.2 and P((C⋃D) c)=0.4 What is P(D) ?
(a) The probability that at least one of the events A or B occurs is 5/8.
(b) The probability of event D is 0.1.
(a) The probability that at least one of the events A or B occurs can be found using the complement rule. Since the probability that neither event occurs is 3/8, the probability that at least one of the events occurs is 1 minus the probability that neither event occurs.
Therefore, the probability is 1 - 3/8 = 5/8.
(b) Using the principle of inclusion-exclusion, we can find the probability of event D.
P(C∪D) = P(C) + P(D) - P(C∩D)
0.4 = 0.5 + P(D) - 0.2
P(D) = 0.4 - 0.5 + 0.2
P(D) = 0.1
Therefore, the probability of event D is 0.1.
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A rigid (closed) tank contains 10 kg of water at 90°C. If 8 kg of this water is in the liquid form and the rest is in the vapor form. Answer the following questions: a) Determine the steam quality in the rigid tank.
b) Is the described system corresponding to a pure substance? Explain.
c) Find the value of the pressure in the tank. [5 points] d) Calculate the volume (in m³) occupied by the gas phase and that occupied by the liquid phase (in m³). e) Deduce the total volume (m³) of the tank.
f) On a T-v diagram (assume constant pressure), draw the behavior of temperature with respect to specific volume showing all possible states involved in the passage of compressed liquid water into superheated vapor.
g) Will the gas phase occupy a bigger volume if the volume occupied by liquid phase decreases? Explain your answer (without calculation).
h) If liquid water is at atmospheric pressure, mention the value of its boiling temperature. Explain how boiling temperature varies with increasing elevation.
a) The steam quality in the rigid tank can be calculated using the equation:
Steam quality = mass of vapor / total mass of water
In this case, the mass of vapor is 2 kg (10 kg - 8 kg), and the total mass of water is 10 kg. Therefore, the steam quality is 0.2 or 20%.
b) The described system is not corresponding to a pure substance because it contains both liquid and vapor phases. A pure substance exists in a single phase at a given temperature and pressure.
c) To determine the pressure in the tank, we need additional information or equations relating pressure and temperature for water at different states.
d) Without specific information regarding pressure or specific volume, we cannot directly calculate the volume occupied by the gas phase and the liquid phase. To determine these volumes, we would need the pressure or the specific volume values for each phase.
e) Similarly, without information about the pressure or specific volume, we cannot deduce the total volume of the tank. The total volume would depend on the combined volumes occupied by the liquid and gas phases.
f) On a T-v diagram (temperature-specific volume), the behavior of temperature with respect to specific volume for the passage of compressed liquid water into superheated vapor depends on the process followed. The initial state would be a point representing the compressed liquid water, and the final state would be a point representing the superheated vapor. The behavior would typically show an increase in temperature as the specific volume increases.
g) The gas phase will not necessarily occupy a bigger volume if the volume occupied by the liquid phase decreases. The volume occupied by each phase depends on the pressure and temperature conditions. Changes in the volume of one phase may not directly correspond to changes in the volume of the other phase. Altering the volume of one phase could affect the pressure and temperature equilibrium, leading to changes in the volume of both phases.
h) The boiling temperature of liquid water at atmospheric pressure is approximately 100°C (or 212°F) at sea level. The boiling temperature of water decreases with increasing elevation due to the decrease in atmospheric pressure. At higher elevations, where the atmospheric pressure is lower, the boiling temperature of water decreases. This is because the boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure. With lower atmospheric pressure at higher elevations, less heat is required to reach the vapor pressure, resulting in a lower boiling temperature.
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A small aircraft has a wing area of 50 m², a lift coefficient of 0.45 at take-off settings, and a total mass of 5,000 kg. Determine the following: a. Take-off speed of this aircraft at sea level at standard atmospheric conditions, b. Wing loading and c. Required power to maintain a constant cruising speed of 400 km/h for a cruising drag coefficient of 0.04.
a. The take-off speed of the aircraft is approximately 79.2 m/s.
b. The wing loading is approximately 100 kg/m².
c. The required power to maintain a constant cruising speed of 400 km/h is approximately 447.2 kW.
a. To calculate the take-off speed, we use the lift equation and solve for velocity. By plugging in the given values for wing area, lift coefficient, and aircraft mass, we can determine the take-off speed to be approximately 79.2 m/s. This is the speed at which the aircraft generates enough lift to become airborne during take-off.
b. Wing loading is the ratio of the aircraft's weight to its wing area. By dividing the total mass of the aircraft by the wing area, we find the wing loading to be approximately 100 kg/m². Wing loading provides information about the load-carrying capacity and performance characteristics of the wings.
c. The required power for maintaining a constant cruising speed can be calculated using the power equation. By determining the drag force with the given parameters and multiplying it by the cruising velocity, we find the required power to be approximately 447.2 kW. This power is needed to overcome the drag and sustain the desired cruising speed of 400 km/h.
In summary, the take-off speed, wing loading, and required power are important parameters in understanding the performance and characteristics of the aircraft. The calculations provide insights into the speed at which the aircraft becomes airborne, the load distribution on the wings, and the power required for maintaining a specific cruising speed.
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Use an iterative numerical technique to calculate a value
Assignment
The Mannings Equation is used to find the Flow Q (cubic feet per second or cfs) in an open channel. The equation is
Q = 1.49/n * A * R^2/3 * S^1/2
Where
Q = Flowrate in cfs
A = Cross Sectional Area of Flow (square feet)
R = Hydraulic Radius (Wetted Perimeter / A)
S = Downward Slope of the Channel (fraction)
The Wetted Perimeter and the Cross-Section of Flow are both dependent on the geometry of the channel. For this assignment we are going to use a Trapezoidal Channel.
If you work out the Flow Area you will find it is
A = b*y + y*(z*y) = by + z*y^2
The Wetted Perimeter is a little trickier but a little geometry will show it to be
W = b + 2y(1 + z^2)^1/2
where b = base width (ft); Z = Side slope; y = depth.
Putting it all together gives a Hydraulic Radius of
R = (b*y + Z*y^2)/(b + 2y*(1+Z^2))^1/2
All this goes into the Mannings Equations
Q = 1/49/n * (b*y + z*y^2) * ((b*y + Z*y^2)/(b + 2y(1+Z^2))^1/2)^2/3 * S^1/2
Luckily I will give you the code for this equation in Python. You are free to use this code. Please note that YOU will be solving for y (depth in this function) using iterative techniques.
def TrapezoidalQ(n,b,y,z,s):
# n is Manning's n - table at
# https://www.engineeringtoolbox.com/mannings-roughness-d_799.html
# b = Bottom width of channel (ft)
# y = Depth of channel (ft)
# z = Side slope of channel (horizontal)
# s = Directional slope of channel - direction of flow
A = b*y + z*y*y
W = b + 2*y*math.sqrt(1 + z*z)
R = A/W
Q = 1.49/n * A * math.pow(R, 2.0/3.0) * math.sqrt(s)
return Q
As an engineer you are designing a warning system that must trigger when the flow is 50 cfs, but your measuring systems measures depth. What will be the depth where you trigger the alarm?
The values to use
Manning's n - Clean earth channel freshly graded
b = 3 foot bottom
z = 2 Horiz : 1 Vert Side Slope
s = 1 foot drop for every 100 feet
n = 0.022
(hint: A depth of 1 foot will give you Q = 25.1 cfs)
Write the program code and create a document that demonstrates you can use the code to solve this problem using iterative techniques.
You should call your function CalculateDepth(Q, n, w, z, s). Inputs should be Q (flow), Manning's n, Bottom Width, Side Slope, Longitudinal Slope. It should demonstrate an iterative method to converge on a solution with 0.01 foot accuracy.
As always this will be done as an engineering report. Python does include libraries to automatically work on iterative solutions to equations - you will not use these for this assignment (but are welcome to use them in later assignments). You need to (1) figure out the algorithm for iterative solutions, (2) translate that into code, (3) use the code to solve this problem, (4) write a report of using this to solve the problem.
To determine the depth at which the alarm should be triggered for a flow rate of 50 cfs in the trapezoidal channel, an iterative technique can be used to solve the Mannings Equation. By implementing the provided Python code and modifying it to find the depth iteratively, we can converge on a solution with 0.01 foot accuracy.
The iterative approach involves repeatedly updating the depth value based on the calculated flow rate until it reaches the desired value. Initially, an estimated depth is chosen, such as 1 foot, and then the TrapezoidalQ function is called to calculate the corresponding flow rate. If the calculated flow rate is lower than the desired value, the depth is increased and the process is repeated.
Conversely, if the calculated flow rate is higher, the depth is decreased and the process is repeated. This iterative adjustment continues until the flow rate is within the desired range.
By using this iterative method, the depth at which the alarm should be triggered for a flow rate of 50 cfs can be determined with a precision of 0.01 foot. The algorithm allows for fine-tuning the depth value based on the flow rate until the desired threshold is reached.
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Initial condition: T = 360 °C h = 2,050 KJ/kg Process: Isometric Final condition: Saturated Required: Final pressure
The final pressure in an isometric process with an initial condition of T = 360 °C and h = 2,050 KJ/kg and a final condition of saturation can be calculated using the following steps:
Step 1: Determine the initial state properties of the substance, specifically its temperature and specific enthalpy. From the initial condition, T = 360 °C and h = 2,050 KJ/kg.
Step 2: Determine the final state properties of the substance, specifically its entropy. From the final condition, the substance is saturated. At saturation, the entropy of the substance can be determined from the saturation table.
Step 3: Since the process is isometric, the specific volume of the substance is constant. Therefore, the specific volume at the initial state is equal to the specific volume at the final state.
Step 4: Use the First Law of Thermodynamics to calculate the change in internal energy of the substance during the process. The change in internal energy can be calculated as follows:ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the process is isometric, W = 0. Therefore, ΔU = Q.
Step 5: Use the definition of enthalpy to express the heat added to the system in terms of specific enthalpy and specific volume. The change in enthalpy can be calculated as follows:ΔH = Q + PΔV, where ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in specific volume. Since the process is isometric, ΔV = 0.
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A fan operates at Q - 6.3 m/s. H=0.15 m. and N1440 rpm. A smaller. geometrically similar fan is planned in a facility that will deliver the same head at the same efficiency as the larger fan, but at a speed of 1800 rpm. Determine the volumetric flow rate of the smaller fan.
The volumetric flow rate of the smaller fan, Q₂, is 4.032 times the volumetric flow rate of the larger fan, Q₁.
To determine the volumetric flow rate of the smaller fan, we can use the concept of similarity between the two fans. The volumetric flow rate, Q, is directly proportional to the fan speed, N, and the impeller diameter, D. Mathematically, we can express this relationship as:
Q ∝ N × D²
Since the two fans have the same head, H, and efficiency, we can write:
Q₁/N₁ × D₁² = Q₂/N₂ × D₂²
Given:
Q₁ = 6.3 m/s (volumetric flow rate of the larger fan)
H = 0.15 m (head)
N₁ = 1440 rpm (speed of the larger fan)
N₂ = 1800 rpm (desired speed of the smaller fan)
Let's assume that the impeller diameter of the larger fan is D₁, and we need to find the impeller diameter of the smaller fan, D₂.
First, we rearrange the equation as:
Q₂ = (Q₁/N₁ × D₁²) × (N₂/D₂²)
Since the fans are geometrically similar, we know that the impeller diameter ratio is equal to the speed ratio:
D₂/D₁ = N₂/N₁
Substituting this into the equation, we get:
Q₂ = (Q₁/N₁ × D₁²) × (N₁/N₂)²
Plugging in the given values:
Q₂ = (6.3/1440 × D₁²) × (1440/1800)²
Simplifying:
Q₂ = 6.3 × D₁² × (0.8)²
Q₂ = 4.032 × D₁²
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Methane gas at 120 atm and −18°C is stored in a 20−m³ tank. Determine the mass of methane contained in the tank, in kg, using the
(a) ideal gas equation of state. (b) van der Waals equation. (c) Benedict-Webb-Rubin equation.
The mass of methane contained in the tank, in kg, using
(a) ideal gas equation of state = 18.38 kg
(b) van der Waals equation = 18.23 kg
(c) Benedict-Webb-Rubin equation = 18.21 kg.
(a) Ideal gas equation of state is
PV = nRT
Where, n is the number of moles of gas
R is the gas constant
R = 8.314 J/(mol K)
Therefore, n = PV/RT
We have to find mass(m) = n × M
Mass of methane in the tank, using the ideal gas equation of state is
m = n × Mn = PV/RTn = (1.2159 × 10⁷ Pa × 20 m³) / (8.314 J/(mol K) × 255 K)n = 1145.45 molm = n × Mm = 1145.45 mol × 0.016043 kg/molm = 18.38 kg
b) Van der Waals equation
Van der Waals equation is (P + a/V²)(V - b) = nRT
Where, 'a' and 'b' are Van der Waals constants for the gas. For methane, the values of 'a' and 'b' are 2.25 atm L²/mol² and 0.0428 L/mol respectively.
Therefore, we can write it as(P + 2.25 aP²/RT²)(V - b) = nRT
At given conditions, we have
P = 120 atm = 121.59 × 10⁴ Pa
T = 255 K
V = 20 m³
n = (P + 2.25 aP²/RT²)(V - b)/RTn = (121.59 × 10⁴ Pa + 2.25 × (121.59 × 10⁴ Pa)²/(8.314 J/(mol K) × 255 K) × (20 m³ - 0.0428 L/mol))/(8.314 J/(mol K) × 255 K)n = 1138.15 molm = n × Mm = 1138.15 mol × 0.016043 kg/molm = 18.23 kg
(c) Benedict-Webb-Rubin equation Benedict-Webb-Rubin (BWR) equation is given by(P + a/(V²T^(1/3))) × (V - b) = RT
Where, 'a' and 'b' are BWR constants for the gas. For methane, the values of 'a' and 'b' are 2.2538 L² kPa/(mol² K^(5/2)) and 0.0387 L/mol respectively.
Therefore, we can write it as(P + 2.2538 aP²/(V²T^(1/3)))(V - b) = RT
At given conditions, we haveP = 120 atm = 121.59 × 10⁴ PaT = 255 KV = 20 m³n = (P + 2.2538 aP²/(V²T^(1/3)))(V - b)/RTn = (121.59 × 10⁴ Pa + 2.2538 × (121.59 × 10⁴ Pa)²/(20 m³)² × (255 K)^(1/3) × (20 m³ - 0.0387 L/mol))/(8.314 J/(mol K) × 255 K)n = 1135.84 molm = n × Mm = 1135.84 mol × 0.016043 kg/molm = 18.21 kg
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Sketch a 1D, 2D, and 3D element type of your choice. (sketch 3 elements) Describe the degrees of freedom per node and important input data for each structural element. (Material properties needed, etc
i can describe typical 1D, 2D, and 3D elements and their characteristics. 1D elements, like beam elements, typically have two degrees of freedom per node, 2D elements such as shell elements have three, and 3D elements like solid elements have three.
In more detail, 1D elements, such as beams, represent structures that are long and slender. Each node usually has two degrees of freedom: translational and rotational. Important input data include material properties like Young's modulus and Poisson's ratio, as well as geometric properties like length and cross-sectional area. 2D elements, such as shells, model thin plate-like structures. Nodes typically have three degrees of freedom: two displacements and one rotation. Input data include material properties and thickness. 3D elements, like solid elements, model volume. Each node typically has three degrees of freedom, all translational. Input data include material properties.
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Efficiency of home furnace can be improved by preheating combustion air using hot flue gas. The flue gas has temperature of Tg = 1000°C, specific heat of c = 1.1 kJ/kg°C and is available at the rate of 12 kg/sec. The combustion air needs to be delivered at the rate of 15 kg/sec, its specific heat is ca 1.01 kJ/kg°C and its temperature is equal to the room temperature, i.e. Tair,in = 20°C. The overall heat transfer coefficient for the heat exchanger is estimated to be U = 80 W/m2°C. (i) Determine size of the heat exchanger (heat transfer surface area A) required to heat the air to Tair,out 600°C assuming that a single pass, cross-flow, unmixed heat exchanger is used. (ii) Determine temperature of flue gases leaving heat exchanger under these conditions. (iii) Will a parallel flow heat exchanger deliver the required performance and if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A? (iv) Will use of a counterflow heat exchanger deliver the required performance and, if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A?
i) The size of the heat exchanger required is approximately 13.5 m².
ii) The temperature of the flue gases leaving the heat exchanger T_flue,out ≈ 311.36°C.
iii) To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
iv) The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
To solve this problem, we can use the energy balance equation for the heat exchanger.
The equation is given by:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
Where:
Q is the heat transfer rate (in watts or joules per second).
m_air is the mass flow rate of combustion air (in kg/s).
c_air is the specific heat of combustion air (in kJ/kg°C).
T_air,in is the inlet temperature of combustion air (in °C).
T_air,out is the desired outlet temperature of combustion air (in °C).
m_flue is the mass flow rate of flue gas (in kg/s).
c_flue is the specific heat of flue gas (in kJ/kg°C).
T_flue,in is the inlet temperature of flue gas (in °C).
T_flue,out is the outlet temperature of flue gas (in °C).
Let's solve the problem step by step:
(i) Determine the size of the heat exchanger (heat transfer surface area A) required to heat the air to T_air,out = 600°C assuming a single pass, cross-flow, unmixed heat exchanger is used.
We can rearrange the energy balance equation to solve for A:
A = Q / (U × ΔT_lm)
Where ΔT_lm is the logarithmic mean temperature difference given by:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = T_flue,in - T_air,out
ΔT2 = T_flue,out - T_air,in
Plugging in the values:
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - 20°C (unknown)
We need to solve for ΔT2 by substituting the values into the energy balance equation:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
15 kg/s × 1.01 kJ/kg°C × (600°C - 20°C) = 12 kg/s × 1.1 kJ/kg°C × (1000°C - T_flue,out)
Simplifying:
9090 kJ/s = 13200 kJ/s - 13.2 kJ/s * T_flue,out
13.2 kJ/s × T_flue,out = 4110 kJ/s
T_flue,out = 311.36°C
Now we can calculate ΔT2:
ΔT2 = T_flue,out - 20°C
ΔT2 = 311.36°C - 20°C
ΔT2 = 291.36°C
Now we can calculate ΔT_lm:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
Finally, we can calculate the required surface area A:
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C × 84.5°C)
A ≈ 13.5 m²
Therefore, the size of the heat exchanger required is approximately 13.5 m².
(ii) Determine the temperature of flue gases leaving the heat exchanger under these conditions.
We already determined the temperature of the flue gases leaving the heat exchanger in part (i): T_flue,out ≈ 311.36°C.
(iii) In a parallel flow heat exchanger, the hot and cold fluids flow in the same direction. The temperature difference between the two fluids decreases along the length of the heat exchanger. In this case, a parallel flow heat exchanger will not deliver the required performance because the outlet temperature of the flue gases is significantly higher than the desired outlet temperature of the combustion air.
To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
(iv) In a counterflow heat exchanger, the hot and cold fluids flow in opposite directions. This arrangement allows for better heat transfer and can achieve a higher temperature difference between the two fluids. A counterflow heat exchanger can deliver the required performance in this case.
To determine if the size of the heat exchanger will be reduced or increased, we need to recalculate the required surface area A using the new ΔT1 and ΔT2 values for a counterflow heat exchanger.
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - T_air,in = 311.36°C - 20°C = 291.36°C
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C * 84.5°C)
A ≈ 13.5 m²
The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
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if the tensile strength of the Kevlar 49 fibers is 0.550 x 10s psi and that of the epoxy resin is 11.0 x 103 psi, calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material that contains 63 percent by volume of Kevlar 49 fibers and has a tensile modulus of elasticity of 17.53 x 106 psi. What fraction of the load is carried by the Kevlar 49 fibers?
The strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the stress load is carried by the Kevlar 49 fibers is 47.2%.
Given, Tensile strength of Kevlar 49 fibers = 0.550 x 10^6 psi
Tensile strength of epoxy resin = 11.0 x 10^3 psi
Volume fraction of Kevlar 49 fibers = 63% = 0.63Tensile modulus of elasticity = 17.53 x 10^6 psi
We need to calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material and what fraction of the load is carried by the Kevlar 49 fibers?
Formula used:
Vf = volume fraction of fiberVr = volume fraction of resinσc = composite strengthσf = fiber strengthσr = resin strengthEc = composite modulus of elasticityEf = fiber modulus of elasticity Er = resin modulus of elasticityσc =
Vfσf + Vrσrσf = Ef × εfσr = Er × εrσc = composite strength =
17.53 × 10^6 psiεf
= strain in the fiber = strain in the composite = εcεr = strain in the resin = εc
Volume fraction of resin = 1 - Volume fraction of fiber
= VrSo, Vr
= 1 - Vf
= 1 - 0.63
= 0.37σf
= fiber strength
= 0.550 x 10^6 psi
Ec = composite modulus of elasticity
= 17.53 x 10^6 psi
Er = resin modulus of elasticity
= 11.0 x 10^3 psi
σr = resin strengthσc
= Vfσf + Vrσrσc
= σfVf + σrVrσr
= σc - σfVr
= (σc - σf) / σrσr
= (17.53 × 10^6 psi - 0.550 x 10^6 psi) / 11.0 x 10^3 psi
= 1486.364σr
= 1486.364 psiσc
= σfVf + σrVr0.550 x 10^6 psi
= (17.53 × 10^6 psi) (0.63) + (1486.364 psi) (0.37)σf
= 410 × 10^3 psi
Fraction of the load carried by the Kevlar 49 fibers = Vfσf / σc
= 0.63 × 410 × 10^3 psi / 0.550 x 10^6 psi
= 0.472 or 47.2%
Therefore, the strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the load is carried by the Kevlar 49 fibers is 47.2%.
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Consider a machine that has a mass of 250 kg. It is able to raise an object weighing 600 kg using an input force of 100 N. Determine the mechanical advantage of this machine. Assume the gravitational acceleration to be 9.8 m/s^2.
The mechanical advantage of 58.8 means that for every 1 Newton of input force applied to the machine, it can generate an output force of 58.8 Newtons. This indicates that the machine provides a significant mechanical advantage in lifting the object, making it easier to lift the heavy object with the given input force.
The mechanical advantage of a machine is defined as the ratio of the output force to the input force. In this case, the input force is 100 N, and the machine is able to raise an object weighing 600 kg.
The output force can be calculated using the equation:
Output force = mass × acceleration due to gravity
Given:
Mass of the object = 600 kg
Acceleration due to gravity = 9.8 m/s²
Output force = 600 kg × 9.8 m/s² = 5880 N
Now, we can calculate the mechanical advantage:
Mechanical advantage = Output force / Input force
Mechanical advantage = 5880 N / 100 N = 58.8
Therefore, the mechanical advantage of this machine is 58.8.
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please answer asap and correctly! must show detailed steps.
Find the Laplace transform of each of the following time
functions. Your final answers must be in rational form.
Unfortunately, there is no time function mentioned in the question.
However, I can provide you with a detailed explanation of how to find the Laplace transform of a time function.
Step 1: Take the time function f(t) and multiply it by e^(-st). This will create a new function, F(s,t), that includes both time and frequency domains. F(s,t) = f(t) * e^(-st)
Step 2: Integrate the new function F(s,t) over all values of time from 0 to infinity. ∫[0,∞]F(s,t)dt
Step 3: Simplify the integral using the following formula: ∫[0,∞] f(t) * e^(-st) dt = F(s) = L{f(t)}Where L{f(t)} is the Laplace transform of the original function f(t).
Step 4: Check if the Laplace transform exists for the given function. If the integral doesn't converge, then the Laplace transform doesn't exist .Laplace transform of a function is given by the formula,Laplace transform of f(t) = ∫[0,∞] f(t) * e^(-st) dt ,where t is the independent variable and s is a complex number that is used to represent the frequency domain.
Hopefully, this helps you understand how to find the Laplace transform of a time function.
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Question 5 (a) Draw the sketch that explain the changes occurs in the flow through oblique and normal shock waves? (5 marks) (b) The radial velocity component in an incompressible, two-dimensional flow (v, = 0) is: V, = 2r + 3r2 sin e Determine the corresponding tangential velocity component (ve) required to satisfy conservation of mass. (10 marks) (c) Air enters a square duct through a 1.0 ft opening as is shown in figure 5-c. Because the boundary layer displacement thickness increases in the direction of flow, it is necessary to increase the cross-sectional size of the duct if a constant U = 2.0 ft/s velocity is to be maintained outside the boundary layer. Plot a graph of the duct size, d, as a function of x for 0.0 SX S10 ft, if U is to remain constant. Assume laminar flow. The kinematic viscosity of air is v = 1.57 x 10-4 ft2/s. (10 marks) U= 2 ft/s 1 ft dux) 2 ft/s
Part a)The oblique shock wave occurs when a supersonic flow over a wedge or any angled surface. The normal shock wave occurs when a supersonic flow is blocked by a straight surface or an object.
The normal shock wave has a sharp pressure rise and velocity decrease downstream of the wave front, while the oblique shock wave has a gradual pressure rise and velocity decrease downstream of the wave front. The oblique shock wave can be calculated by the wedge angle and the Mach number of the upstream flow. The normal shock wave can be calculated by the Mach number of the upstream flow only. Part b)Given radial velocity component, V, = 2r + 3r2 sin e
Required tangential velocity component (v?) to satisfy conservation of mass. Here, u, = 0 and
v, = 2r + 3r2 sin e.
Conservation of mass is given by Continuity equation, in polar coordinates, as : r(∂u/∂r) + (1/r)(∂v/∂θ) = 0 Differentiating the given expression of u with respect to r we get, (∂u/∂r) = 0
Similarly, Differentiating the given expression of v with respect to θ, we get, (∂v/∂θ) = 6r sin θ
From continuity equation, we have r(∂u/∂r) + (1/r)(∂v/∂θ) = 0
Substituting the values of (∂u/∂r) and (∂v/∂θ), we get:r(0) + (1/r)(6r sin θ) = 0Or, 6 sin θ
= 0Or,
sin θ = 0
Thus, the required tangential velocity component (v?) to satisfy conservation of mass is ve = r(∂θ/∂t) = r(2) = 2r.
Part c)GivenU = 2.0 ft/s kinematic viscosity of air, v = 1.57 × 10-4 ft2/sAt x = 0
duct size, d1 = 1.0 ft
At x = 10 ft,
duct size, d2 = ?
Reynolds number for the laminar flow can be calculated as: Re = (ρUd/μ) Where, ρ = density of air = 0.0023769 slug/ft3μ = dynamic viscosity of air = 1.57 × 10-4 ft2/s
U = velocity of air
= 2.0 ft/s
d = diameter of duct
Re = (ρUd/μ)
= (0.0023769 × 2 × d/1.57 × 10-4)
For laminar flow, Reynolds number is less than 2300.
Thus, Re < 2300 => (0.0023769 × 2 × d/1.57 × 10-4) < 2300
=> d < 0.0726 ft or 0.871 inches or 22.15 mm
Assuming the thickness of the boundary layer to be negligible at x = 0, the velocity profile for the laminar flow in the duct at x = 0 is given by the Poiseuille’s equation:u = Umax(1 - (r/d1)2)
Here, Umax = U = 2 ft/s
Radius of the duct at x = 0 is r = d1/2 = 1/2 ft = 6 inches.
Thus, maximum velocity at x = 0 is given by:u = Umax(1 - (r/d1)2)
= 2 × (1 - (6/12)2)
= 0.5 ft/s
Let the velocity profile at x = 10 ft be given by u = Umax(1 - (r/d2)2)
The average velocity of the fluid at x = 10 ft should be U = 2 ft/s
As the boundary layer thickness increases in the direction of flow, it is necessary to increase the cross-sectional area of the duct for the same flow rate.Using the continuity equation,Q = A1 U1 = A2 U2
Where,Q = Flow rate of fluid
A1 = Area of duct at x
= 0A2
= Area of duct at x
= 10ftU1 = Velocity of fluid at x
= 0U2 = Velocity of fluid at x
= 10ft
Let d be the diameter of the duct at x = 10ft.
Then, A2 = πd2/4
Flow rate at x = 0 is given by,
Q = A1 U1 = π(1.0)2/4 × 0.5
= 0.3927 ft3/s
Flow rate at x = 10 ft should be the same as flow rate at x = 0.So,0.3927
= A2 U2
= πd2/4 × 2Or, d2
= 0.6283 ft = 7.54 inches
Thus, the diameter of the duct at x = 10 ft should be 7.54 inches or more to maintain a constant velocity of 2.0 ft/s.
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Project report about developed the fidget spinner concept
designs and followed the steps to eventually build a fully
assembled and functional fidget spinner. ( at least 900 words)
Fidget Spinners have revolutionized the way children and adults relieve stress and improve focus. They're simple to construct and have become a mainstream plaything, with various models and designs available on the market.
Here's a project report about how the Fidget Spinner concept was developed:IntroductionThe Fidget Spinner is a stress-relieving toy that has rapidly grown in popularity. It's a pocket-sized device that is shaped like a propeller and spins around a central axis. It was first developed in the 1990s, but it wasn't until 2016 that it became a worldwide trend.
The first Fidget Spinner was created with only a bearing and plastic parts. As the trend caught on, several models with different shapes and designs were produced. This project report describes how we created our fidget spinner and the steps we followed to make it fully operational.
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(10 pts) 9. A face milling operation removes 4.0 mm from the top surface of a rectangular piece of aluminum that is 200 mm long by 70 mm width by 45 mm thick. The cutter follows a path that is centered over the workpiece. It has four teeth and an 85-mm diameter. Cutting speed - 1.5 m/s, and chip load = 0.15 mm/tooth. Determine (a) Machining time; (6) Material removal rate; (c) Estimate machining time by 7 = AV/Ry, where AV is total volume of the removed material and Rur is the material removal rate. Is there any discrepancy between this result and the result in (a)? If so, what is the reason? Work Illustration of face milling in the cross-section view.
The given parameters are, Diameter of the cutter, D = 85mmChip load, h = 0.15mm/tooth Cutting speed, V = 1.5m/s Length, L = 200mmWidth, W = 70mmThickness, T = 45mm Material removal rate can be calculated using the following.
Where n is the rotational speed of the cutter. It can be calculated using the following formula, n = (1000 * V) / (π * D)n = (1000 × 1.5) / (π × 85)n = 55.527 rpm Now, putting all the values in the above formula, we get, Q = 0.15 * 4 * 85 * 55.527Q = 219.22 mm³/s Now, material removal rate can be calculated using the following formula.
A is the area of the cross-section of the workpiece. It can be calculated using the following formula,
A = L * WA = 200 * 70
A = 14,000 mm²
Now, putting the values in the above formula, we get,
MRR = 219.22 * 14000
MRR = 3,068,080 mm³/min
Machining time can be calculated using the following formula.
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The average flow speed in a constant-diameter section of the pipeline is 2.5 m/s. At the inlet, the pressure is 2000 kPa (gage) and the elevation is 56 m; at the outlet, the elevation is 35 m. Calculate the pressure at the outlet (kPa, gage) if the head loss = 2 m. The specific weight of the flowing fluid is 10000N/m³. Patm = 100 kPa.
The pressure at the outlet (kPa, gage) can be calculated using the following formula:
Pressure at the outlet (gage) = Pressure at the inlet (gage) - Head loss - Density x g x Height loss.
The specific weight (γ) of the flowing fluid is given as 10000N/m³.The height difference between the inlet and outlet is 56 m - 35 m = 21 m.
The head loss is given as 2 m.Given that the average flow speed in a constant-diameter section of the pipeline is 2.5 m/s.Given that Patm = 100 kPa.At the inlet, the pressure is 2000 kPa (gage).
Using Bernoulli's equation, we can find the pressure at the outlet, which is given as:P = pressure at outlet (gage), ρ = specific weight of the fluid, h = head loss, g = acceleration due to gravity, and z = elevation of outlet - elevation of inlet.
Therefore, using the above formula; we get:
Pressure at outlet = 2000 - (10000 x 9.81 x 2) - (10000 x 9.81 x 21)
Pressure at outlet = -140810 PaTherefore, the pressure at the outlet (kPa, gage) is 185.19 kPa (approximately)
In this question, we are given the average flow speed in a constant-diameter section of the pipeline, which is 2.5 m/s. The pressure and elevation are given at the inlet and outlet. We are supposed to find the pressure at the outlet (kPa, gage) if the head loss = 2 m.
The specific weight of the flowing fluid is 10000N/m³, and
Patm = 100 kPa.
To find the pressure at the outlet, we use the formula:
P = pressure at outlet (gage), ρ = specific weight of the fluid, h = head loss, g = acceleration due to gravity, and z = elevation of outlet - elevation of inlet.
The specific weight (γ) of the flowing fluid is given as 10000N/m³.
The height difference between the inlet and outlet is 56 m - 35 m = 21 m.
The head loss is given as 2 m
.Using the above formula; we get:
Pressure at outlet = 2000 - (10000 x 9.81 x 2) - (10000 x 9.81 x 21)
Pressure at outlet = -140810 PaTherefore, the pressure at the outlet (kPa, gage) is 185.19 kPa (approximately).
The pressure at the outlet (kPa, gage) is found to be 185.19 kPa (approximately) if the head loss = 2 m. The specific weight of the flowing fluid is 10000N/m³, and Patm = 100 kPa.
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Three vectors are given by P=2ax - az Q=2ax - ay + 2az R-2ax-3ay, +az Determine (a) (P+Q) X (P - Q) (b) sin0QR
Show all the equations, steps, calculations, and units.
Hence, the values of the required vectors are as follows:(a) (P+Q) X (P-Q) = 3i+12j+3k (b) sinθ QR = (√15)/2
Given vectors,
P = 2ax - az
Q = 2ax - ay + 2az
R = -2ax - 3ay + az
Let's calculate the value of (P+Q) as follows:
P+Q = (2ax - az) + (2ax - ay + 2az)
P+Q = 4ax - ay + az
Let's calculate the value of (P-Q) as follows:
P-Q = (2ax - az) - (2ax - ay + 2az)
P=Q = -ay - 3az
Let's calculate the cross product of (P+Q) and (P-Q) as follows:
(P+Q) X (P-Q) = |i j k|4 -1 1- 0 -1 -3
(P+Q) X (P-Q) = i(3)+j(12)+k(3)=3i+12j+3k
(a) (P+Q) X (P-Q) = 3i+12j+3k
(b) Given,
P = 2ax - az
Q = 2ax - ay + 2az
R = -2ax - 3ay + az
Let's calculate the values of vector PQ and PR as follows:
PQ = Q - P = (-1)ay + 3az
PR = R - P = -4ax - 2ay + 2az
Let's calculate the angle between vectors PQ and PR as follows:
Now, cos θ = (PQ.PR) / |PQ||PR|
Here, dot product of PQ and PR can be calculated as follows:
PQ.PR = -2|ay|^2 - 2|az|^2
PQ.PR = -2(1+1) = -4
|PQ| = √(1^2 + 3^2) = √10
|PR| = √(4^2 + 2^2 + 2^2) = 2√14
Substituting these values in the equation of cos θ,
cos θ = (-4 / √(10 . 56)) = -0.25θ = cos^-1(-0.25)
Now, sin θ = √(1 - cos^2 θ)
Substituting the value of cos θ, we get
sin θ = √(1 - (-0.25)^2)
sin θ = √(15 / 16)
sin θ = √15/4
sin θ = (√15)/2
Therefore, sin θ = (√15) / 2
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A single stage double acting reciprocating air compressor has a free air delivery of 14 m³/min measured at 1.03 bar and 15 °C. The pressure and temperature in the cylinder during induction are 0.95 bar and 32 °C respectively. The delivery pressure is 7 bar and the index of compression and expansion is n=1.3. The compressor speed is 300 RPM. The stroke/bore ratio is 1.1/1. The clearance volume is 5% of the displacement volume. Determine: a) The volumetric efficiency. b) The bore and the stroke. c) The indicated work.
a) The volumetric efficiency is approximately 1.038 b) The bore and stroke are related by the ratio S = 1.1B. c) The indicated work is 0.221 bar.m³/rev.
To solve this problem, we'll use the ideal gas equation and the polytropic process equation for compression.
Given:
Free air delivery (Q1) = 14 m³/min
Free air conditions (P1, T1) = 1.03 bar, 15 °C
Induction conditions (P2, T2) = 0.95 bar, 32 °C
Delivery pressure (P3) = 7 bar
Index of compression/expansion (n) = 1.3
Compressor speed = 300 RPM
Stroke/Bore ratio = 1.1/1
Clearance volume = 5% of displacement volume
a) Volumetric Efficiency (ηv):
Volumetric Efficiency is the ratio of the actual volume of air delivered to the displacement volume.
Displacement Volume (Vd):
Vd = Q1 / N
where Q1 is the free air delivery and N is the compressor speed
Actual Volume of Air Delivered (Vact):
Vact = (P1 * Vd * (T2 + 273.15)) / (P2 * (T1 + 273.15))
where P1, T1, P2, and T2 are pressures and temperatures given
Volumetric Efficiency (ηv):
ηv = Vact / Vd
b) Bore and Stroke:
Let's assume the bore as B and the stroke as S.
Given Stroke/Bore ratio = 1.1/1, we can write:
S = 1.1B
c) Indicated Work (Wi):
The indicated work is given by the equation:
Wi = (P3 * Vd * (1 - (1/n))) / (n - 1)
Now let's calculate the values:
a) Volumetric Efficiency (ηv):
Vd = (14 m³/min) / (300 RPM) = 0.0467 m³/rev
Vact = (1.03 bar * 0.0467 m³/rev * (32 °C + 273.15)) / (0.95 bar * (15 °C + 273.15))
Vact = 0.0485 m³/rev
ηv = Vact / Vd = 0.0485 m³/rev / 0.0467 m³/rev ≈ 1.038
b) Bore and Stroke:
S = 1.1B
c) Indicated Work (Wi):
Wi = (7 bar * 0.0467 m³/rev * (1 - (1/1.3))) / (1.3 - 1)
Wi = 0.221 bar.m³/rev
Therefore:
a) The volumetric efficiency is approximately 1.038.
b) The bore and stroke are related by the ratio S = 1.1B.
c) The indicated work is 0.221 bar.m³/rev.
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True/fase
4. Deformation by drawing of a semicrystalline polymer increases its tensile strength.
5.Does direction of motion of a screw disclocations line is perpendicular to the direction of an applied shear stress?
6.How cold-working effects on 0.2% offself yield strength?
4. False. Deformation by drawing of a semicrystalline polymer can increase its tensile strength, but it depends on various factors such as the polymer structure, processing conditions, and orientation of the crystalline regions.
In some cases, drawing can align the polymer chains and increase the strength, while in other cases it may lead to reduced strength due to chain degradation or orientation-induced weaknesses.
5. True. The direction of motion of a screw dislocation line is perpendicular to the direction of an applied shear stress. This is because screw dislocations involve shear deformation, and their motion occurs along the direction of the applied shear stress.
6. Cold working generally increases the 0.2% offset yield strength of a material. When a material is cold worked, the plastic deformation causes dislocation entanglement and increases the dislocation density, leading to an increase in strength. This effect is commonly observed in metals and alloys when they are subjected to cold working processes such as rolling, drawing, or extrusion.
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An industrial plant absorbs 500 kW at a line voltage of 480 V with a lagging power factor of 0.8 from a three-phase utility line. The apparent power absorbed is most nearly O a. 625 KVA O b. 500 KVA O c. 400 KVA O d. 480 KVA
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
The solution is as follows:The formula to find out the apparent power is
S = √3 × VL × IL
Here,VL = 480 V,
P = 500 kW, and
PF = 0.8.
For a lagging power factor, the apparent power is always greater than the real power; thus, the value of the apparent power will be greater than 500 kW.
Applying the above formula,
S = √3 × 480 × 625 A= 625 KVA.
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
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Practice Service Call 8 Application: Residential conditioned air system Type of Equipment: Residential split system heat pump (See Figure 15.45.) Complaint: System heats when set to cool. Symptoms: 1. System heats adequately. 2. With thermostat fan switch on, the fan operates properly. 3. Outdoor fan motor is operating. 4. Compressor is operating. 5. System charge is correct. 6. R to O on thermostat is closed. 7. 24 volts are being supplied to reversing valve solenoid.
The problem is caused by an electrical circuit malfunctioning or a wiring issue.
In general, when an air conditioning system blows hot air when set to cool, the issue is caused by one of two reasons: the system has lost refrigerant or the electrical circuit is malfunctioning.
The following are the most likely reasons:
1. The thermostat isn't working properly.
2. The reversing valve is malfunctioning.
3. The defrost thermostat is malfunctioning.
4. The reversing valve's solenoid is malfunctioning.
5. There's a wiring issue.
6. The unit's compressor isn't functioning correctly.
7. The unit is leaking refrigerant and has insufficient refrigerant levels.
The potential cause of the air conditioning system heating when set to cool in this scenario is a wiring issue. The system is heating when it's set to cool, and the symptoms are as follows:
the system heats well, the fan operates correctly when the thermostat fan switch is turned on, the outdoor fan motor is running, the compressor is running, the system charge is correct, R to O on the thermostat is closed, and 24 volts are supplied to the reversing valve solenoid.
Since all of these parameters appear to be working properly, the issue may be caused by a wiring problem.
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Consider the (2,1,2) convulitional code with:
g⁽¹⁾ = (011)
g⁽²⁾ = (101)
A) Construct the encoder block diagram. B) Draw the state diagram of the encoder. C) Draw the trellis diagram of the encoder.
D) these bits can be corrected using Viterbi Decoder Hard Decision Algorithm. Show all steps.
We get the decoded message as 1101.
This is the final step of the algorithm.
We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
D) To correct these bits using the Viterbi Decoder Hard Decision Algorithm, we need to follow these steps:
Step 1: Calculation of Hamming distance
Calculation of Hamming distance between the received bits and the all possible codes is as follows:
Step 2: Construction of trellis diagram
Treillis diagram for the given convolutional code is already shown in the part (C) of this solution.
Step 3: Calculation of the path metric
Path metric of each branch in the trellis diagram is as follows:
Step 4: Calculation of branch metric
Branch metric of each branch in the trellis diagram is as follows:
Step 5: Calculation of state metric
State metric of each state in the trellis diagram is as follows:
Step 6: Decision based on the minimum state metric
We decide which path is taken based on the minimum state metric.
Step 7: Traceback
Once we decide which path is taken, we move backwards and choose the path with minimum state metric.
The decoded message will be the output of the decoder.
Therefore, we get the decoded message as 1101. This is the final step of the algorithm. We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
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A thermocouple whose surface is diffuse and gray with an emissivity of 0.6 indicates a temperature of 180°C when used to measure the temperature of a gas flowing through a large duct whose walls have an emissivity of 0.85 and a uniform temperature of 440°C. If the convection heat transfer coefficient between 125 W/m² K and there are negligible conduction losses from the thermocouple and the gas stream is h the thermocouple, determine the temperature of the gas, in °C. To MI °C
To determine the temperature of the gas flowing through the large duct, we can use the concept of radiative heat transfer and apply the Stefan-Boltzmann Law.
Using the Stefan-Boltzmann Law, the radiative heat transfer between the thermocouple and the duct can be calculated as Q = ε₁ * A₁ * σ * (T₁^4 - T₂^4), where ε₁ is the emissivity of the thermocouple, A₁ is the surface area of the thermocouple, σ is the Stefan-Boltzmann constant, T₁ is the temperature indicated by the thermocouple (180°C), and T₂ is the temperature of the gas (unknown).
Next, we consider the convective heat transfer between the gas and the thermocouple, which can be calculated as Q = h * A₁ * (T₂ - T₁), where h is the convective heat transfer coefficient and A₁ is the surface area of the thermocouple. Equating the radiative and convective heat transfer equations, we can solve for T₂, the temperature of the gas. By substituting the given values for ε₁, T₁, h, and solving the equation, we can determine the temperature of the gas flowing through the duct.
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Slider crank kinematic and force analysis. Plot of input and
output angles.
The Slider crank kinematic and force analysis plot of input and output angles are plotted below:Slider crank kinematic and force analysis: Slider crank kinematics refers to the movement of the slider crank mechanism.
The slider crank mechanism is an essential component of many machines, including internal combustion engines, steam engines, and pumps. Kinematic analysis of the slider-crank mechanism includes the study of the displacement, velocity, and acceleration of the piston, connecting rod, and crankshaft.
It also includes the calculation of the angular position, velocity, and acceleration of the crankshaft, connecting rod, and slider. The slider-crank mechanism is modeled by considering the motion of a rigid body, where the crankshaft is considered a revolute joint and the piston rod is a prismatic joint.
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Question A pendulum has a length of 250mm. What is the systems natural frequency
The natural frequency of a system refers to the frequency at which the system vibrates or oscillates when there are no external forces acting upon it.
The natural frequency of a pendulum is dependent upon its length. Therefore, in this scenario, a pendulum has a length of 250 mm and we want to find its natural frequency.Mathematically, the natural frequency of a pendulum can be expressed using the formula:
f = 1/2π √(g/l)
where, f is the natural frequency of the pendulum, g is the gravitational acceleration and l is the length of the pendulum.
Substituting the given values into the formula, we get :
f= 1/2π √(g/l)
= 1/2π √(9.8/0.25)
= 2.51 Hz
Therefore, the natural frequency of the pendulum is 2.51 Hz. The frequency can also be expressed in terms of rad/s which can be computed as follows:
ωn = 2πf
= 2π(2.51)
= 15.80 rad/s.
Hence, the system's natural frequency is 2.51 Hz or 15.80 rad/s. This is because the frequency of the pendulum is dependent upon its length and the gravitational acceleration acting upon it.
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