The sampling period (T) plays a crucial role in the control quality of a system, particularly in systems that employ discrete-time control algorithms.
The influence of the sampling period can be summarized as follows:
1. System Responsiveness: A shorter sampling period allows for faster system response as the control algorithm can make more frequent adjustments. However, extremely short sampling periods can introduce additional computational burden and may lead to control instability or performance degradation.
2. Control Accuracy: The choice of sampling period affects the accuracy of the control system. A shorter sampling period allows for more precise tracking of the desired setpoints or reference signals. On the other hand, a longer sampling period may result in larger errors and reduced control accuracy.
3. System Stability: The sampling period influences the stability of the control system. In some cases, certain control algorithms require specific sampling periods to ensure stability. Deviating from the recommended sampling period can lead to instability or oscillations in the controlled system.
To choose an appropriate sampling period for a system, it is essential to consider the following factors:
1. System Dynamics: The inherent dynamics of the controlled system, including its response time, time constants, and natural frequencies, should be taken into account. Faster systems may require shorter sampling periods to adequately capture their dynamics.
2. Control Objectives: The desired control performance, such as tracking accuracy, disturbance rejection, or stability, should be considered. Different control objectives may have varying requirements for the sampling period.
3. Computational Resources: The available computational resources, including the processing power of the control hardware and the execution time of the control algorithm, should be considered. Very short sampling periods may strain the computational resources and lead to performance limitations.
4. Nyquist-Shannon Sampling Theorem: The sampling period should satisfy the Nyquist-Shannon sampling theorem, which states that the sampling rate should be at least twice the highest frequency component present in the system's dynamics.
Choosing an appropriate sampling period often involves a trade-off between system responsiveness, accuracy, stability, and computational resources. It is common to analyze the system's dynamics, conduct simulations or experiments, and iteratively adjust the sampling period until a satisfactory balance is achieved for the specific control application.
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This is an iron in which most of the carbon is chemically combined with the iron. What is this iron commonly called? A) White iron B) Gray iron C) Malleable iron D) Cast iron
The iron in which most of the carbon is chemically combined with the iron is commonly called Cast iron. Cast iron is an alloy of iron, carbon, and silicon that is brittle and difficult to operate.
The correct option is- D
It is used in a variety of applications, including pipes, machine tools, and automotive components, due to its excellent casting qualities. Cast iron is used to make everything from pans to pipes because it can be easily cast into a range of intricate shapes and is both durable and inexpensive.
Cast iron is mainly used to make valves, pumps, engine blocks, gearboxes, cylinder heads, and other automotive and mechanical parts. It is also used to make pipes, stoves, and cooking utensils for domestic purposes.Therefore, the correct option is D) Cast iron. Cast iron is used to make everything from pans to pipes because it can be easily cast into a range of intricate shapes and is both durable and inexpensive.
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You know that widely used plastic bottles of carbonated drinks are made from PET plastic. Considering that CO2 is permeable for PET plastic, what is the diffusion flux at 298 K? Note that permeability coefficient for carbon dioxide is 0.5⋅10−13[cm2 s⋅Pa], pressure inside and outside the bottle are 1 MPa and 0.1 kPa, respectively. Let’s consider that bottle is a hollow cylinder with diameter of 30 mm, height of 100 mm and thickness of 0.05 cm. What is expiration date of this bottle if it considers decarbonized when 500 cm3 of CO2 will diffuse? Assuming a condition of steady state diffusion, calculate diffusion flux at 350 K if activation energy is 50 kJ/mol. How fast will the bottle lose fizz in this case? Compare two cases.
The diffusion flux of carbon dioxide through the PET plastic bottle at 298 K is approximately 7.95 x 10^(-11) mol/(cm^2·s).
To calculate the diffusion flux of carbon dioxide through the PET plastic bottle at 298 K, we can use Fick's law of diffusion:
J = -D * (ΔC/Δx)
Where:
J is the diffusion flux (mol/(cm^2·s))
D is the diffusion coefficient (cm^2/s)
ΔC is the concentration difference (mol/cm^3)
Δx is the thickness of the bottle (cm)
The diffusion coefficient for carbon dioxide in PET plastic is given as 0.5 x 10^(-13) cm^2·s·Pa. However, the pressure units need to be converted to match the diffusion coefficient. Thus, 1 MPa is equal to 10^7 Pa and 0.1 kPa is equal to 100 Pa.
Using the given values, we can calculate the concentration difference (ΔC) as follows:
ΔC = (P_inside - P_outside) / (RT)
Where:
P_inside is the pressure inside the bottle (1 MPa)
P_outside is the pressure outside the bottle (0.1 kPa)
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K)
ΔC = ((1 MPa) - (0.1 kPa)) / ((8.314 J/(mol·K)) * (298 K))
= 9.9 x 10^6 Pa / 2472.972 J/mol
≈ 4.004 x 10^3 mol/cm^3
Now, we can calculate the diffusion flux (J) using the given formula:
J = -D * (ΔC/Δx)
= -(0.5 x 10^(-13) cm^2·s·Pa) * (4.004 x 10^3 mol/cm^3) / (0.05 cm)
≈ -7.95 x 10^(-11) mol/(cm^2·s)
The negative sign indicates that the diffusion flux is from inside the bottle to the outside.
The diffusion flux of carbon dioxide through the PET plastic bottle at 298 K is approximately 7.95 x 10^(-11) mol/(cm^2·s). This means that a certain amount of CO2 will continuously diffuse out of the bottle, leading to a gradual loss of fizz over time.
To determine the expiration date of the bottle when 500 cm^3 of CO2 will diffuse, we need additional information such as the initial volume of CO2 inside the bottle and the rate of diffusion.
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Q5
Question 5 What is the Australian standard number for tensile testing (i.e.) "metallic materials - tensile testing at ambient temperatures"?
An Australian standard number refers to a unique identification number assigned to a specific standard published by Standards Australia. The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391.
AS 1391 is the Australian standard that specifically addresses the tensile testing of metallic materials at ambient temperatures. This standard provides guidelines and requirements for conducting tensile tests on metallic materials to determine their mechanical properties.
Tensile testing is a widely used method for evaluating the mechanical behavior and performance of metallic materials under tensile forces. It involves subjecting a specimen of the material to a gradually increasing axial load until it reaches failure.
AS 1391 outlines the test procedures, specimen preparation methods, and reporting requirements for tensile testing at ambient temperatures. It ensures consistency and standardization in conducting these tests, allowing for accurate and reliable comparison of material properties across different laboratories and industries in Australia.
The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391. This standard provides guidelines and requirements for conducting tensile tests to evaluate the mechanical properties of metallic materials. Adhering to this standard ensures consistency and reliability in conducting tensile tests in Australia
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If the same turbine runs at two different rotation speeds but under dynamically similar conditions, which of the following is NOT true? The flow rate will be different. The work output will be different. The pressure ratio will be different The efficiency will be the same. 6 The flow coefficient will be the same.
Under dynamically similar conditions for a turbine running at two different rotation speeds, the statement that "the efficiency will be the same" is not true. Turbine efficiency is not solely dependent on dynamical similarity.
When a turbine operates under dynamically similar conditions at two different rotational speeds, most parameters like flow rate, work output, pressure ratio, and flow coefficient will differ. However, the statement "the efficiency will be the same" is not necessarily true. Turbine efficiency is influenced by several factors, including design, fluid properties, and operating conditions. While dynamical similarity tries to ensure a degree of correspondence between scenarios, the efficiency can still change with rotational speed. This variation results from influences like alterations in the Reynolds number, which could shift flow characteristics. Consequently, despite maintaining dynamical similarity.
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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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Hazard Management Evaluation Sheet Paints Safety Policy: Comments: Identify your hazard: Comments: Assess your ha7nrd: Comments: Control and communicate: Comments: Review Total Points: Overall Comments: Hazard Management Hazard identification, assessment and control are key elements of an effective environmental, health and safety system. Not only are they key parts of a health and safety program; they are also required by law under the Alberta Occupational Health and Safety Code. The intent of our Hazard Managenent Program is to allow U of A leaders and employees to proactively and collaboratively identify people, equipment, material, environmental and process factors that could potentially cause harm to individuals, our institution or our community. Armed with knowledge about our hazards we can then raise awareness of the impact of our operations and make informed decisions to effectively eliminate or reduce our risks. Foresight and planning are vital tools in the successful implementation and management of our learning and research activities here at the University. Hazard A condition or behaviouz that has the potential bo cause injugy or lass. The Hazard Management Program applies to all U of A operations and we all have a role to play in identifying, assessing and controlling hazards. Please take the time to review and implement the IIazard Management Procedure in your workplace. A template Hazard Assessment Form is also available for download to assist in recording the results of your assessment. Useful Downloads: Hazard Management Procedure Hazard Assersment Blank Form In the next few months, EHS will be developing and implementing an e-Class learning module related to Hazard Management. Please continue to watch our site for more information. In the interim you may find this Hazard Assessment and Control e-Learning module from the Alberta Government of interest. DEFINITIONS "Hazard" means an activity, arrangement, circumstance, event, occurrence, phenomezon, process, situation, or sulastance (whether arising or caused within or outside a place of work) that is an actual or potential cause or source of barm. In effect a bazard can be interpreted as anything that can cause barm in terms of human irijury or ill bealita, damage to properly, damage to the environment or a combination of all these. "Hazard Identification" is the process of recognizing that a hazard exists and defining its characteristics. "ITazard Assessment" is the overall process of detemning whether a bazard is significant. "Significant hazard" means a hazard that is an actual or potential cause or source of 8. Serious harm; or b. Harm (that is rncre than trivial) the serverity of whose elfects on any parson depend an the extent of frequency of the gersen's mexosure to the hazard; nr c. Harm that does not usually occur, or is not easily detectable, until a slgnificant time after exposure to tho harard. "Harm" means "ilness, injury or both". The term is only used in the context of harm that is more than trivial. "Serious IIann" is essentially a work-related injury, illness or condition that will result ir adaission to hospital for 48 hours or more or being off work for more than one week. Hazard Management; Robotics Identify your hazard Malfunction during normal operation tinsafe approach to the rohot work envelope Assess your hazard Bodily impact Injure a human by pinching (grippcr or joints) Pining a person against some structure Control and communicate Safeguarding robotic workstation Biartiers Operator safety Range of motion Robot operation Safety of the programer Aware of their surrounding Never be in reach of the robot Safcty of the repair preson Aware of their sumbunding Never be in reach of the robot Review and revise Review and modify the above nules
The Hazard Management Evaluation Sheet aids in the identification, assessment, control, and communication of hazards. For robotics, hazards such as malfunctions, unsafe approaches, and bodily impacts can be identified.
Hazard Management Evaluation Sheet helps in identifying hazards, assessing them, and control and communicate about them. Given below are the comments for each of these points along with the identification, assessment, and control of hazards.
Identify your hazard: In robotics, the hazards that can be identified include malfunction during normal operation, unsafe approach to the robot work envelope, and interference in operation by humans. Bodily impact and injury due to pinching or pinning a person against some structure are some other hazards that can be identified.
Assess your hazard: One needs to determine whether the identified hazard is significant or not. For robotics, the significant hazards include bodily harm, serious harm, or harm that doesn’t usually occur, or is not easily detectable until a significant time after exposure to the hazard.
Control and communicate: It is necessary to take measures to control the hazard and communicate about it. In robotics, the robotic workstation should be safeguarded, and barriers should be put in place to ensure operator safety. The range of motion should be controlled, and the programmer should be aware of their surroundings and never be in reach of the robot. The same applies to the repair person.
Review: It is essential to review and revise the above rules as per the changing circumstances. The points mentioned above should be modified as per the new conditions, if any. The Hazard Management Evaluation Sheet helps in identifying hazards and assessing them as per the potential harm they may cause.
The focus should be on control and communication, which are essential to ensure that hazards are minimized or eliminated. Reviewing and modifying the rules as per the changing situation is also an important step in effective hazard management.
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The mass of water steam in the cylinder, in kg, is A.1.082 B 0.526 C(1.003 D 0.972 0.972
The mass of water steam in the cylinder, in kg, is 0.972 (option D).The mass of water steam in the cylinder is 0.972 kg. To understand how this value was obtained.
we need to look at the problem in more detail. However, since no additional information is provided, let's make some assumptions that will help us answer the question. Let's assume that the water steam is in a cylinder under certain conditions.
We know that steam is produced when water is heated, so we can assume that the water in the cylinder is heated, causing it to evaporate and become steam.
We also know that the mass of water is usually measured in kilograms (kg), so we can assume that we are looking for the mass of steam in the cylinder in kg.Now, let's look at the options again. Option A is 1.082 kg, option B is 0.526 kg, option C is 1.003 kg, and option D is 0.972 kg.
We can immediately eliminate options A and C because they are greater than 1 kg, which is unlikely for the mass of steam in a cylinder. Option B is also too small, as it is less than 1 kg, and steam usually has a greater mass than water due to its gaseous state.
This leaves us with option D, which is 0.972 kg. This value is within the range of what we might expect for the mass of steam in a cylinder.
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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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In a food processing plant, hot liquid water is being transported in a pipe (k = 15 W/m-K, D₁= 2.5 cm, Do=3 cm, and L=10 m). The hot water flowing with a mass flow rate of 8.65Kg/s enters the pipe at 100°C and exits at 60°C. The plant supervisor thinks that since the hot water exits the pipe at 60°C, the pipe's outer surface temperature should be safe from thermal burn hazards. In order to prevent thermal burn upon accidental contact with skin tissue for individuals working in the vicinity of the pipe, the pipe's outer surface temperature should be kept below 45°C. Determine whether or not there is a risk of thermal burn on the pipe's outer surface. Assume the pipe outer surface temperature remains constant. The properties of water at the bulk mean temperature of Tb=(T₁+T₂)/2 = (100 + 60)/2 = 80°C are: Cp=4197 J/kg-K, k = 0.670 W/m-K, µ = 0.355 × 10-3 kg/m-s, and Pr = 2.22. The thermal conductivity of the pipe is given as Kpipe = 15 W/m-K. The outer pipe surface temperature is °C.
Where ρ is the density, v is the velocity of the water in the pipe, D is the diameter of the pipe, and µ is the viscosity of the water.
Re = (8.65/π(0.025)² × 0.355 × 10⁻³)
Re = 18,686.40
And the Nusselt number can be computed using the Reynolds number and Prandtl number.
Nu = 0.023R[tex]e^{(4/5)[/tex] P[tex]r^n[/tex]
Where n is the exponent of Prandtl number;
when the fluid is in turbulent flow, the exponent value is 0.4.
Nu = 0.023 × [tex](18,686.4)^{(4/5)[/tex] [tex](2.22)^{0.4[/tex]
Nu = 146.05
The Nusselt number is 146.05.
Q = πDL(U)(T₁-T₂)
Where L is the length of the pipe.
Q/πDL = U(T₁-T₂)
U = (Q/πDL)/(T₁-T₂)
U = (mCp(T₁-T₂)/πDL)/(T₁-T₂)
U = (mCp)/(πDL)
U = (8.65 × 4197)/(π × 0.03 × 10)
U = 11814.11 W/m²K
Substituting the calculated values into the expression for Q;
Q = (11814.11)(π × 0.03 × 10)(100-60)
Q = 21,165.41W
The expression for the outer pipe surface temperature is;
Tₒₑ = T₁ - Q/π
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A 2mx 2m solar absorber plate is at 400 K while it is exposed to solar irradiation. The surface is diffuse and its spectral absorptivity is a = 0, for λ >1 >0.5 μm a=0.8, for 0.5 u m> > λ μm a = 0, for lym> > λ2um a =0.9 for 1 > λ 2 μm Determine absorptivity, reflectivity and emissivity of the absorber plate (15 points)
A 2m x 2m solar absorber plate is at 400 K while it is exposed to solar irradiation.
The surface is diffuse and its spectral absorptivity is as follows:a = 0, for λ >1 >0.5 μma = 0.8, for 0.5 µm > λ > 2 µma = 0, for λ > 2 µma =0.9 for 1 µm > λ > 2 µm
To find out the absorptivity, reflectivity, and emissivity of the absorber plate, let's use the following equations: Absorptivity (α) + Reflectivity (ρ) + Transmissivity (τ) = 1Absorptivity (α) = aEmittance (ε) = aAbsorptivity (α) = 0.9 (for 1 > λ > 2 µm) and 0.8 (for 0.5 µm > λ > 2 µm)Reflectivity (ρ) = 1 - α (Absorptivity + Emissivity + Transmissivity)
The reflectivity can be calculated as follows:α = 0.9 (for 1 > λ > 2 µm)ρ = 1 - αρ = 1 - 0.9ρ = 0.1α = 0.8 (for 0.5 µm > λ > 2 µm)ρ = 1 - αρ = 1 - 0.8ρ = 0.2α = 0 (for λ > 2 µm)ρ = 1 - αρ = 1 - 0ρ = 1
The reflectivity is calculated to be 0.1, 0.2, and 1, respectively, for the above wavelength ranges. The emissivity can be found using the following equation:ε = α = 0.9 (for 1 > λ > 2 µm)ε = α = 0.8 (for 0.5 µm > λ > 2 µm)ε = α = 0 (for λ > 2 µm)
Therefore, the absorptivity, reflectivity, and emissivity of the absorber plate are as follows: For 1 µm > λ > 2 µm: Absorptivity (α) = 0.9 Reflectivity (ρ) = 0.1 Emissivity (ε) = 0.9For 0.5 µm > λ > 2 µm: Absorptivity (α) = 0.8Reflectivity (ρ) = 0.2 Emissivity (ε) = 0.8For λ > 2 µm: Absorptivity (α) = 0 Reflectivity (ρ) = 1 Emissivity (ε) = 0.
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Q5 a. Describe the meaning of Tolerance Reference Hole System. b. Explain the meaning of the symbol "m 32 x 1.5-6g"
The Tolerance Reference Hole System refers to a standardized method for specifying the dimensions and tolerances of holes in engineering drawings.The symbol "m 32 x 1.5-6g" indicates the specifications of a metric thread.
The Tolerance Reference Hole System refers to a standardized method for specifying the dimensions and tolerances of holes in engineering drawings. It establishes a set of guidelines that ensure consistency and compatibility in hole size and fit. It includes reference values for the diameter, depth, and tolerance of the hole.
The symbol "m 32 x 1.5-6g" indicates the specifications of a metric thread. "M" refers to the metric system, "32" represents the major diameter of the thread in millimeters, "1.5" signifies the pitch (the distance between corresponding points on adjacent threads) in millimeters, and "-6g" denotes the tolerance class. In this case, the tolerance class is 6g, which means that the thread has a medium tolerance range.
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Find the current flowing through a load with impedance Z
= -110 +j40 to which a voltage V = 0
+j-40 is applied. Show steps
A load with impedance Z = -110 + j40 to which a voltage V = 0 + j-40 is applied is given. Find the current flowing through it.Impedance is the complex resistance that is offered to the flow of current by the circuit. It is measured in Ohms (Ω).
It is a complex quantity having both magnitude and phase angle. It can be represented in rectangular form as well as in polar form.The rectangular form of impedance is given by:
Z = R + jX
where R is the resistance and X is the reactance. The polar form of impedance is given by: Z = |Z| ∠θwhere |Z| is the magnitude of the impedance and θ is the phase angle.Current is the flow of electric charge through a conductor. It is measured in Amperes (A).
It can be calculated using Ohm's law as:I = V/Zwhere I is the current, V is the voltage, and Z is the impedance.Substituting the given values, we get:
I = V/Z= (0 + j(-40))/(-110 + j40)
= (j(-40))/(-110 + j40)
= [(-40)j/(-110 + j40)]*[(110 - j40)/(110 - j40)]
= [-4400 + j1600]/(12100)
= (-4400/12100) + j(1600/12100)
= -0.3636 + j0.1322
Therefore, the current flowing through the load is given by -0.3636 + j0.1322 Amperes.
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E3: Air flows from a nozzle with a mass flow rate of 1.0 slugs /sec. If T 0=607 ∘F,p 0 = 120psia, and p exit =15psia, the Ma number at the exit is most nearly a. 2.55 b. 2.05 c. 3.15 d. 1.00 e. 1.60
The Mach number at the exit of the nozzle, given a mass flow rate of 1.0 slugs/sec, is option (a) 2.55, according to the provided parameters.
To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations and the given parameters.
Given:
Mass flow rate (ṁ) = 1.0 slugs/sec
Total temperature at the inlet (T₀) = 607 °F
Total pressure at the inlet (p₀) = 120 psia
Pressure at the exit (p_exit) = 15 psia
First, we need to convert the total temperature from Fahrenheit to Rankine:
T₀ = 607 °F + 459.67 °R = 1066.67 °R
Next, we can use the mass flow rate and the total pressure to find the exit velocity (V_exit):
V_exit = ṁ / (A_exit * ρ_exit)
To find the exit area (A_exit), we need to calculate the exit density (ρ_exit) using the ideal gas equation:
Ρ_exit = p_exit / (R * T_exit)
The gas constant R for air is approximately 1716.5 ft·lbf/(slug·°R).
Using the isentropic flow equations, we can find the exit temperature (T_exit) as follows:
(p_exit / p₀) = (T_exit / T₀) ^ (γ / (γ – 1))
Here, γ is the specific heat ratio for air, which is approximately 1.4.
Now, let’s calculate the exit temperature:
(T_exit / T₀) = (p_exit / p₀) ^ ((γ – 1) / γ)
(T_exit / 1066.67 °R) = (15 psia / 120 psia) ^ ((1.4 – 1) / 1.4)
(T_exit / 1066.67 °R) = 0.3272
T_exit = 0.3272 * 1066.67 °R = 349.96 °R
Now, we can calculate the exit density:
Ρ_exit = 15 psia / (1716.5 ft·lbf/(slug·°R) * 349.96 °R) ≈ 0.00624 slug/ft³
Next, let’s calculate the exit velocity:
V_exit = 1.0 slugs/sec / (A_exit * 0.00624 slug/ft³)
Now, we can use the mass flow rate equation to find the exit area (A_exit):
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
Finally, we can calculate the Mach number at the exit:
M_exit = V_exit / (γ * R * T_exit)^0.5
Let’s plug in the values and calculate the Mach number:
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
M_exit = V_exit / (1.4 * 1716.5 ft·lbf/(slug·°R) * 349.96 °R)^0.5
After performing the calculations, the most approximate Mach number at the exit is option (a) 2.55.
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Neglecting friction along the bucket, the power delivered to a Pelton wheel by liquid jet is given by P=rhoQu(V1−u){1−cos(β2)] where Q is volumetric flowrate, u is bucket speed, V1 is jet velocity, and β2 is bucket angle. Using the above relation, derive the following relationships corresponding to maximum power; (i) the bucket speed is half of jet velocity, u=2V1 (ii) the speed factor, ϕ is half of velocity coefficient, Cvo ϕ=2Cv (iii) the maximum turbine efficiency (neglecting friction in the nozzle) is ηT=21−cos(β2)
Using the above relation, derive the following relationships corresponding to maximum power, the maximum turbine efficiency (neglecting friction in the nozzle) is ηT=21−cos(β2). So the correct answer is (iii).
P = ρQu(V1 − u)
[1 − cos (β2)] where Q is volumetric flowrate bucket speedV1 is jet velocityβ2 is bucket angle Now, we have to derive the following relationships corresponding to maximum power;
(i) The bucket speed is half of the jet velocity, u=2V1
(ii) the speed factor, ϕ is half of the velocity coefficient, Cvo ϕ=2Cv
(iii) the maximum turbine efficiency (neglecting friction in the nozzle) is ηT=21−cos(β2)
Derivation:
(i) Given, u=2V1So, P = ρQ2V1(1 − cos (β2))Now, power is maximum when cos β2 = -1∴ Pmax = ρQ2V1
(ii) We know, ϕ = (V1 − u)/V1and, Cv = (V1 − u)/(V1 + u)Now, putting u = V1/2, we getϕ = 1/2Cv = 1/3∴ ϕ = 2Cv
(iii) We know,ηT = P/ρQV1Now, putting u = V1/2 and cos β2 = -1, we getηT = 1/2[1 + cos(β2)] = 21−cos(β2).
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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 biomass digester receives 23 kJ of energy in the form of heat. There is a mechanical mixer that keeps rotating and putting 1.4 kJ work into the system. Calculate the energy balance in the system in kJ, if there is a heat loss of 7 kJ. A small fan in a closed insulated room releases heat at the rate of 42 watts and pushes the air at the rate of 9 m/s with a mass flow rates of 1.2 kg/s. The room has a heater that produces heat at the rate of 0.30 kJ/s as well as a computer that produces 65 watts. Light bulbs in the room produce up to 125 watts. The room looses 0.32 kJ/s. Calculate the amount of heat maintained in the room.
The energy balance in the system can be calculated by summing up all the energy inputs and subtracting the energy losses. The energy balance is 23 kJ (heat input) + 1.4 kJ (work input) - 7 kJ (heat loss) = 17.4 kJ.
To calculate the amount of heat maintained in the room, we need to consider the various energy inputs and losses within the system.
Energy Inputs:
Heater: The heater produces heat at the rate of 0.30 kJ/s.
Small fan: The small fan releases heat at the rate of 42 watts (0.042 kJ/s) due to its operation.
Computer: The computer produces heat at the rate of 65 watts (0.065 kJ/s).
Light bulbs: The light bulbs produce heat up to 125 watts (0.125 kJ/s).
Energy Losses:
Heat loss: The room loses heat at the rate of 0.32 kJ/s.
To calculate the amount of heat maintained in the room, we sum up all the energy inputs and subtract the energy losses:
Total Energy Input = Heater + Small fan + Computer + Light bulbs
= 0.30 kJ/s + 0.042 kJ/s + 0.065 kJ/s + 0.125 kJ/s
= 0.532 kJ/s
Heat Maintained = Total Energy Input - Heat Loss
= 0.532 kJ/s - 0.32 kJ/s
= 0.212 kJ/s
Therefore, the amount of heat maintained in the room is 0.212 kJ/s.
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True or False: The convective heat transfer coefficient in
laminar flow over a flat plate INCREASES with distance
Explain your reasoning
The convective heat transfer coefficient in a fluid is directly proportional to the heat transfer surface area. This statement is False.
Convective heat transfer is the transfer of heat from one point to another in a fluid through the mixing of fluid particles. The convective heat transfer coefficient in a fluid is directly proportional to the fluid velocity, the fluid density, and the thermal conductivity of the fluid. The convective heat transfer coefficient is also indirectly proportional to the viscosity of the fluid. The heat transfer surface area only affects the total heat transfer rate. Therefore, the statement is false.
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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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The Reynolds number, pvD/u, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using the MLT system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity of 3 m/s through a 2-in- diameter pipe.
Reynolds number is a dimensionless quantity which represents the ratio of inertial forces (ρvD) to the viscous forces (u).Here,ρ is the density of the fluidv is the velocity of the fluidD is the diameter of the pipemu is the dynamic viscosity of the fluid.
If the Reynolds number is very less than 2300, then the flow is laminar and if it is greater than 4000, then the flow is turbulent.If the Reynolds number lies between 2300 and 4000, then the flow is transitional. Ethyl alcohol is flowing through a 2-inch diameter pipe at a velocity of 3 m/s.
We have to find the Reynolds number value.Let's put the values in the formula,Re = ρvd/µRe = (7850 kg/m³ x 3 m/s x 0.0508 m) / (1.2 x 10⁻³ N s/m²)Re = 9,34,890.67Reynolds number value is more than 100 words.
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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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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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To design a simply supported RCC slab for a roof of a hall 4000x9000 mm inside dimension, with 250 mm wall all around, consider the following data: d= 150 mm, design load intensity=15 kN/m², M25, Fe415. a. Find the effective span and load per unit width of the slab. b. Find the ultimate moment per unit width of the slab. c. Find the maximum shear force per unit width of the slab. d. Find the effective depth required from ultimate moment capacity consideration and comment on the safety. e. Is it necessary to provide stirrups for such a section?
Stir rups are not necessary in this slab design.
How to solve the problemsa. The effective span of the slab is the longer dimension of the hall: 9000 mm or 9 m.
The load per unit width (w) is equal to the design load intensity: 15 kN/m.
b. The ultimate moment (Mu) per unit width of the slab can be found using the formula for a simply supported slab under uniformly distributed load: Mu = w*L²/8.
Mu = 15 kN/m * (9 m)² / 8
= 151.88 kNm/m.
c. The maximum shear force (Vu) per unit width of the slab can also be found using a formula for a simply supported slab under uniformly distributed load: Vu = w*L/2.
Vu = 15 kN/m * 9 m / 2
= 67.5 kN/m.
d. Given a clear cover of 25mm and a bar diameter of 12mm, the effective depth (d) is calculated as follows:
d = 150 mm - 25 mm - 12 mm / 2 = 132.5 mm.
The ultimate moment of resistance (Mr) provided by the slab can be given by Mr = 0.138 * f * (d)²,
where fc is 25 N/mm² for M25 concrete.
Mr = 0.138 * 25 N/mm² * (132.5 mm)² = 482.25 kNm/m.
e. Since Mr > Mu (482.25 kNm/m > 151.88 kNm/m), the slab is safe for the bending moment. Therefore, stir rups are not necessary in this slab design.
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Work speed of circular electrode of 220 mm diameter for carrying out seam welding at 4 welds per cm on 1.6 mm thick mild steel tube.
Welding cycle consists of 3 cycles ‘on’ and 2 cycles ‘off’ Power supply is at 50 Hz. Calculate R.P.M. and energy requirement of the above circular electrodes assuming effective resistance between electrodes as 100 ohm
calculate the RPM (Revolutions Per Minute) and energy requirement of the circular electrodes for seam welding, we need to consider the welding speed, the number of welds per unit length, the thickness of the material, and the effective resistance.
First, let's calculate the welding speed (S) in centimeters per minute: S = WPC * f . S = 4 welds/cm * 50 Hz . S = 200 cm/min .Next, let's calculate the RPM (N) of the circular electrode: N = (S * 60) / (π * D) . N = (200 cm/min * 60) / (π * 22 cm) . N ≈ 172.52 RPM . Now, let's calculate the energy requirement (E) of the circular electrodes: E = (P * t) / (WPC * f * (3 + 2)) E = (P * t) / (4 welds/cm * 50 Hz * 5 cycles) E = (P * t) / 1000 where:
- P is the power in watts .
Since we are given the effective resistance (R), we can calculate the power (P) using the formula: P = (V^2) / R . Assuming a standard voltage of 220 volts: P = (220^2) / 100 , P = 48400 / 100 , P = 484 watts . Finally, let's calculate the energy requirement: E = (P * t) / 1000 . E = (484 watts * 0.016 meters) / 1000 , E = 7.744 joules . Therefore, the RPM of the circular electrode is approximately 172.52 RPM, and the energy requirement is approximately 7.744 joules.
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Paper Clip
For each component of the materials used in the paper clip, explore the following:
Emphasis on material selection, process selection, and cost of manufacturing.
Examine the advantages, disadvantages, and costs of the materials used?
Study the bending method for manufacturing the paper clip.
Determine at what stage of manufacture and by which method should the surface treatment be performed.
Determine the final coating of the paper clip.
The main answer:Materials used in the paper clip There are different types of materials used in the manufacturing of the paper clip. Some of the most commonly used materials include stainless steel, zinc-coated steel, plastic, and aluminum.The material selection is crucial in the manufacturing of the paper clip.
The material must be strong enough to hold papers together. Additionally, it must be flexible and malleable to allow the bending of the paper clip.Process selection is also an essential aspect of paper clip manufacturing. The production process involves wire drawing, heat treatment, wire forming, surface treatment, and finishing.Cost of manufacturing is another essential aspect of the paper clip. The manufacturing cost should be kept low to allow for a low-cost product. Advantages, disadvantages, and costs of materialsStainless steel is the most commonly used material for paper clip manufacturing. Its advantages include high durability, corrosion resistance, and high strength.
However, its main disadvantage is that it's expensive to manufacture.Zinc-coated steel is also another material used for paper clip manufacturing. Its advantages include low cost and rust resistance. However, its main disadvantage is that it's not as strong as stainless steel.Plastic is another material used for paper clip manufacturing. Its advantages include low cost and versatility. However, its main disadvantage is that it's not strong enough for heavy-duty use.Aluminum is another material used for paper clip manufacturing. Its advantages include high strength and lightweight. However, its main disadvantage is that it's expensive to manufacture.Bending method for manufacturing the paper clipThe bending method involves the use of a wire bender to shape the wire into a paper clip. The wire is first cut into a specific length and then fed into the bender, which shapes it into a paper clip.The bending method is fast and efficient and can produce paper clips in large quantities.
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Prove that the transfer function is as follows using the block diagram reduction. 160K₁/s²+(32+160K₁K₂)s + 160K₁
The given transfer function is:
[tex]$$\frac{160K_1}{s^2+(32+160K_1K_2)s + 160K_1}$$[/tex]
We can represent the above transfer function in a block diagram form as follows: Block diagram The first block represents the gain of 160K₁, the second block represents the summing point, the third block represents the gain of K₂ and the fourth block represents the plant with a transfer function of
[tex]$$\frac{1}{s^2+32s+160K_1}$$[/tex]
Now, we will perform the block diagram reduction process. First, we will merge the second and third blocks using the summing point reduction technique. Summing point reduction technique Applying summing point reduction technique, we get a new transfer function:
[tex]$$\frac{160K_1}{s^2+(32+160K_1K_2)s + 160K_1} = \frac{K_2}{1+sT_1}\left(\frac{160K_1}{s^2+2\xi\omegas+\omega_n^2}\right)$$[/tex]
where[tex]$$\omega_n = 4\sqrt{10K_1}$$$$\xi = \frac{1}{8\sqrt{10K_1}}$$$$T_1 = \frac{1}{8\sqrt{10K_1}K_2}$$[/tex]
Hence, we have proved that the given transfer function can be represented using the block diagram reduction and we got a new transfer function.
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Problem 2 (35 points) A Pitot tube, located on the undercarriage of an airship, 0.1 m aft of its leading edge, is to be used to monitor airspeed which varies from 32 to 130 km/hr. The undercarriage is approximately flat, making the pressure gradient negligible. Air temperature is 4 °C and the pressure is 84 kPa. To be outside the boundary layer, at what distance should the Pitot tube be located from the undercarriage? Assume air is an ideal gas.
A Pitot tube, located on the undercarriage of an airship, 0.1 m aft of its leading edge, is to be used to monitor airspeed which varies from 32 to 130 km/hr.
The undercarriage is approximately flat, making the pressure gradient negligible. Air temperature is 4 °C and the pressure is 84 kPa.
Assume air is an ideal gas.
Using Bernoulli’s equation, the pressure in a fluid decreases with an increase in the fluid velocity, assuming the fluid’s potential and kinetic energies are conserved.
Bernoulli's equation can be applied to air if we assume it is incompressible (i.e. the density is constant) and frictionless.
Therefore, we can write Bernoulli's equation as follows:
[tex]$$\frac{1}{2} \rho v_{1}^{2}+p_{1}=\frac{1}{2} \rho v_{2}^{2}+p_{2}$$[/tex]
where v1 is the velocity of the airship, p1 is the pressure at the bottom of the airship, v2 is the velocity at the location of the Pitot tube, and p2 is the pressure at the location of the Pitot tube.
We can solve for v2 by rearranging the equation as follows:
[tex]$$v_{2}=\sqrt{\frac{2}{\rho}\left(p_{1}-p_{2}\right)+v_{1}^{2}}$$[/tex]
The Pitot tube should be located approximately 0.369 m away from the bottom of the airship in order to be outside the boundary layer.
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To estimate the loss factor (η) of an elastomer in the circular disc shape shown, a 10 kg mass is mounted on its top. A vertical harmonic force excites the mass: F(t) = 900sin(10t) in N. The elastomer is placed on a flat rigid foundation, having a thickness d = 0.01 m and diameter D = 0.04 m. It is known that the response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer. a) Determine the loss factor (η) of the elastomer-mass system. [10 marks] b) Calculate the stiffness of the system k. [3 marks] c) Find natural frequency of the system ωn. [3 marks] d) Describe 3 advantages and 1 disadvantage of using viscoelastic materials such as an elastomer for vibration isolation.
Viscoelastic materials have many advantages, including: It has very high damping capacity. High strength and elasticity. It is used to filter or remove unwanted frequencies. Although viscoelastic materials have many benefits, they also have some drawbacks, including :It has a limited operating range. Highly dependent on temperature .It may have a low natural frequency.
Loss factor (η) of the elastomer-mass system;
The response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer.
At resonance, Xrm = F0/(aEη) η = F0/(aEXrm) = 900 / (π (0.02)2 x 1.5 x 106 x 0.003) = 0.24.
Stiffness of the system k
Stiffness of the system k = aE= π (0.02)2 / 0.01 x 1.5 x 106 = 1.26 N/mc)
Natural frequency of the system ωn.
Natural frequency of the system is given by, ωn = sqrt(k/m)
Here, m = mass = 10 kg; k = stiffness = 1.26 N/mωn = sqrt (1.26 / 10) = 0.4 rad/s.
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A2. A piston-cylinder device contains 0.8 kg of superheated water vapor at 300°C and 1 MPa. The water vapor is cooled at constant pressure until one-third of the mass condenses. The water mass then becomes a saturated liquid-vapor mixture. (a) Show the process on a sketch of T-v diagram with respect to saturation lines. Indicate on the T-v diagram the initial state and the final state of the system. (2 marks) (b) Determine the final temperature. (5 marks) (c) Determine the volume change. (5 marks) (d) Determine the additional amount of energy needed to be withdrawn from the saturated liquid-vapor mixture to complete the condensation. (5 marks)
a) Sketch the process on a T-v diagram: The process can be shown as a line connecting the initial state (superheated vapor) to the final state (saturated liquid-vapor mixture).
b) The final temperature is 151.81°C.
c) The volume change is Δv = -0.24497 m³/kg.
d) The additional energy needed to complete condensation is Q = 357.14 kJ.
Mass of superheated water vapor (m) = 0.8 kg
Initial temperature (T1) = 300°C
Pressure (P) = 1 MPa
One-third of the mass condenses. So, the remaining mass (m') is 0.8 - (1/3) × 0.8 = 0.533 kg
(a) Sketch the process on a T-v diagram:
- Initial state: At state 1, the water vapor is superheated and its temperature is 300°C. The specific volume at state 1 is v1 = 0.2465 m³/kg.
- Final state: At state 2, with one-third of the mass condensed, the remaining mass (m') is 0.533 kg. The water becomes a saturated liquid-vapor mixture. The specific volume at state 2 is v2 = 0.00253 m³/kg.
(b) The final temperature is 151.81°C.
(c) The volume change can be calculated as:
Δv = v2 - v1 = 0.00253 - 0.2465 = -0.24497 m³/kg (Negative sign indicates a decrease in volume.)
(d) To complete the condensation, the saturated liquid-vapor mixture needs to be cooled to the saturation temperature at 1 MPa, which is 151.81°C. The specific enthalpy of the saturated liquid at 151.81°C is hf = 670.68 kJ/kg. The additional energy needed to be withdrawn from the mixture can be calculated as Q = m' × hf = 0.533 × 670.68 = 357.14 kJ.
Thus, the additional amount of energy needed to be withdrawn from the saturated liquid-vapor mixture to complete the condensation is 357.14 kJ.
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A fan with 40% static efficiency has a capacity of 60,000 ft3/ hr at 60°F and barometer of 30 in Hg and gives a static pressure of 2 in WG on full delivery. What size electric motor should be used to drive this fan?
O 0.8 HP O 1.2 HP O 1.6 HP O 2 HP
To determine the size of the electric motor required to drive the fan, we need to consider the fan's capacity, static efficiency, and static pressure. The correct answer is: O 0.8 HP.
The capacity of the fan is given as 60,000 ft3/hr, which indicates the volume of air the fan can deliver in one hour. The static pressure is stated as 2 in WG (inches of water gauge), which represents the pressure the fan can generate against resistance.
The static efficiency of the fan is mentioned as 40%, which indicates the ratio of the actual work performed by the fan to the input power.
To calculate the power required to drive the fan, we can use the following formula:
Power (HP) = (Flow Rate * Static Pressure) / (Fan Efficiency * 6356)
Converting the given capacity of 60,000 ft3/hr to cubic feet per minute (CFM), we get:
Flow Rate (CFM) = 60,000 ft3/hr / 60 min/hr = 1000 ft3/min
Substituting the given values into the formula, we have:
Power (HP) = (1000 * 2) / (0.4 * 6356) ≈ 0.793 HP
Rounding off to the nearest option, the size of the electric motor required to drive this fan is 0.8 HP.
Therefore, the correct answer is: O 0.8 HP.
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Show solution properly. Moist air undergoes an adiabatic saturation process from an initial state at 101.325 kPa, 40°C db and 30% relative humidity to a final state that is saturated air. If the mass flow rate of the moist air is 75.1 kg/s, what is the increase in the water content of the moist air? Express your answer in kg/s.
Adiabatic saturation process Adiabatic saturation process involves adding water vapor to the moist air to raise its humidity ratio until it is fully saturated.
The state change between the initial and final state involves two stages: Sensible heating from T1 to T2 (where T2 is the saturation temperature at the final pressure) followed by adding water vapor adiabatically until the moist air becomes saturated.
During the adiabatic saturation process, the moist air is saturated by adding water vapor to it. This process is adiabatic since there is no exchange of heat between the moist air and surroundings.
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