(a) To determine the pulse count received by the control system to verify that the table has moved exactly 250 mm, we need to calculate the total number of pulses generated by the optical encoder.(3) 3.1.1 The pulse rate: The pulse rate is the number of pulses generated per unit of time.
Leadscrew pitch = 6.0 mm
Gear ratio = 8:1
Optical encoder pulses/rev = 120
Table movement = 250 mm
First, we calculate the number of revolutions made by the leadscrew:
Number of revolutions = Table movement / Leadscrew pitch
Number of revolutions = 250 mm / 6.0 mm = 41.67 rev
Next, we calculate the total number of pulses generated:
Total pulses = Number of revolutions * Optical encoder pulses/rev
Total pulses = 41.67 rev * 120 pulses/rev
Total pulses = 5000 pulses
Therefore, the control system should receive 5000 pulses to verify that the table has moved exactly 250 mm.
(3) 3.1.1 The pulse rate:
The pulse rate is the number of pulses generated per unit of time. In this case, the pulse rate can be calculated as the total number of pulses divided by the time taken to move the table.
(3) 3.1.2 The motor speed that corresponds to the feed rate of 500 mm/min:
Since the leadscrew has a gear ratio of 8:1, the motor speed can be calculated as the feed rate divided by the leadscrew pitch multiplied by the gear ratio.
(3) 3.2 Besides the starting material, what other feature distinguishes the rapid prototyping technologies:
Rapid prototyping technologies are characterized by their ability to quickly create physical prototypes directly from digital designs. While the starting material is an important aspect, another distinguishing feature is the layer-by-layer additive manufacturing process used in rapid prototyping technologies. This process enables the construction of complex shapes and structures by depositing and solidifying material layer by layer until the final object is created. This layer-by-layer approach allows for precise control over the design and allows for the production of intricate geometries that may not be achievable through traditional manufacturing methods.
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We are comparing the heat transfer performance of air flow cross 1.0 m long pin fins with
various cross section shown below. The surface temperature of each pin fin is maintained
at 350 K and the diameter of A and the side width of B and C are 10.0 mm. If the air flow
temperature and velocity are 300 K and 10.0 m/s respectively, please calculate the heat
transfer rates for air flow cross each fin. (15%)
In order to calculate the heat transfer rates for air flow across each fin, we can use the concept of convective heat transfer. The heat transfer rate can be determined using the equation:
Q = h*A* (Ts-Ta)
In the equation Q is the heat transfer rate, h is the convective heat transfer coefficient, A is the surface area of the fin, Ts is the surface temperature of the fin, and Ta is the air flow temperature. For each pin fin with different cross-sectional geometries, we need to calculate the convective heat transfer coefficient (h) and the surface area (A) to evaluate the heat transfer rate. The convective heat transfer coefficient can be determined based on the geometry of the fin, the air flow conditions, and the Nusselt number correlation. The surface area of the fin can be calculated depending on the specific cross-sectional shape. Once we have obtained the convective heat transfer coefficient and the surface area for each fin, we can substitute the values into the heat transfer rate equation to calculate the heat transfer rates for air flow across each fin. By comparing the heat transfer rates for different pin fin geometries, we can assess their respective heat transfer performance and identify the most effective configuration for heat dissipation.
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Suppose we wish to use a Fabry-Perot laser source which emits over a wavelength range from 1550 nm to 1554 nm to transmit binary OOK data at a rate of 500 Mb/s through a single-mode fiber whose chromatic dispersion coefficient is -10 ps/(nm-km). What is the longest link that can be supported without severe errors due to ISI? (Hint: your answer should be < 100km but > 1 km.) Express your answer in km, accurate to one decimal place.
The maximum link length that can be supported without severe errors due to ISI is less than 50 km but greater than 1 km. Therefore, the answer is 49.9 km. Hence, the longest link that can be supported without severe errors due to ISI is 49.9 km.
Given data:Fabry-Perot laser source emits over a wavelength range from 1550 nm to 1554 nmBinary OOK data at a rate of 500 Mb/sSingle-mode fiber whose chromatic dispersion coefficient is -10 ps/(nm-km).To determine the longest link that can be supported without severe errors due to ISI, let's first calculate the maximum distance (link) for which the distortion due to chromatic dispersion is negligible.The dispersion-limited distance is given as,L
= T^2 / (D * Bandwidth * S0)Where T
= Bit duration
= 1/500 x 10^6 s
= 2 x 10^-9 Bandwidth
= 1/T
= 500 x 10^6 HzS0
= 0.1 (assuming the receiver filter bandwidth is equal to 0.1 times the symbol rate)D
= Dispersion coefficient
= -10 ps/(nm-km)
= -10 x 10^-3 ps/nm/m
= -10 x 10^-6 s/m/nm
Using the given wavelength range, we can calculate the wavelength spread (Δλ)Δλ
= 1554 nm - 1550 nm
= 4 nm
= 4 x 10^-9 m
The pulse broadening can be calculated as, ΔT
= D * L * ΔλSo, L
= ΔT / (D * Δλ)
= 2 x 10^-9 s / (-10 x 10^-6 s/m/nm * 4 x 10^-9 m)≈ 50 km
To account for the errors due to ISI, the maximum link length would be less than the distance calculated using the above formula. The maximum link length that can be supported without severe errors due to ISI is less than 50 km but greater than 1 km. Therefore, the answer is 49.9 km. Hence, the longest link that can be supported without severe errors due to ISI is 49.9 km.
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For aviation, if you are going to teach a course in hardware
design of an aircraft how would you conduct it, explain in
steps with a detail explanation foe each one.
Hardware design involves creating and developing the physical components and systems of electronic devices, such as circuit boards, processors, and peripherals. It encompasses the design, testing, and optimization of hardware to ensure functionality, performance, and reliability, while considering factors like cost, power consumption, and size constraints.
If you are going to teach a course in hardware design of an aircraft for aviation, you would conduct it as follows:
Step 1: Introduce the CourseYou would start by introducing the course, explaining what hardware design of an aircraft is all about, what the course will cover, and what the students can expect to learn.
Step 2: Teach the BasicsYou would then teach the students the basics of hardware design of an aircraft, including the history of aviation, the science of flight, and the different types of aircraft and their components.
Step 3: Teach the Design PrinciplesYou would then teach the students the design principles of hardware design of an aircraft, including the materials used, the forces that aircraft are subjected to, and the importance of safety.
Step 4: Teach the Design ProcessYou would then teach the students the design process of hardware design of an aircraft, including the different stages of design, the tools used in design, and the importance of testing and evaluation.
Step 5: Conduct Practical SessionsYou would then conduct practical sessions where students can put into practice what they have learned so far, including using software to design an aircraft, building aircraft components, and testing them in a simulated environment.
Step 6: Introduce Advanced TopicsFinally, you would introduce the students to advanced topics in hardware design of an aircraft, including the latest technologies used in aviation, and the future of aircraft design and development. You can also include 150 by specifying the maximum number of students that can be enrolled in the course or the maximum duration of the course (e.g., 150 hours).
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A flat electrical heater of 0.4 m x 0.4 m size is placed vertically in still air at 20°C. The heat generated is 1200 W/m². Determine the value of convective heat transfer coefficient and the average plate temperature.
Size of the heater, L = 0.4 mHeat generated, q'' = 1200 W/m^2The temperature of the still air, T∞ = 20°CDetermining the convective heat transfer coefficient (h)From the relation,
q'' = h(Tp - T∞) …(1) where,Tp = Plate temperature. Rearranging the equation (1) for h, we get,h = q'' / (Tp - T∞) …(2)Determining the average plate temperature.
The average plate temperature (Tp) can be calculated from the relation,Tp = (q'' / σ)^(1/4) …(3)where, σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m^2K^4Substituting the given values in the above equations; we get;
q'' = 1200 W/m^2T∞ = 20°CTo determine h, we need to determine Tp; from equation (3)
Tp = (q'' / σ)^(1/4)= [1200 / (5.67 x 10^-8)]^(1/4) = 372.5 K.
Using the value of Tp, we can calculate the value of h using equation (2).h = q'' / (Tp - T∞)h = 1200 / (372.5 - 293)h = 46.94 W/m^2KThe value of convective heat transfer coefficient, h = 46.94 W/m^2KThe average plate temperature, Tp = 372.5 K.
Therefore, the value of the convective heat transfer coefficient is 46.94 W/m²K and the average plate temperature is 372.5 K.
We are given a flat electrical heater of size 0.4 m × 0.4 m that is placed vertically in still air at 20°C. The heat generated by the heater is 1200 W/m². We have to find out the value of the convective heat transfer coefficient and the average plate temperature. The average plate temperature is calculated using the relation Tp = (q''/σ)^(1/4), where σ is the Stefan-Boltzmann constant.
On substituting the given values in the above formula, we get the average plate temperature as 372.5 K. To calculate the convective heat transfer coefficient, we use the relation q'' = h(Tp - T∞), where Tp is the plate temperature, T∞ is the temperature of the surrounding air, and h is the convective heat transfer coefficient. On substituting the given values in the above formula, we get the convective heat transfer coefficient as 46.94 W/m²K.
Thus, the value of the convective heat transfer coefficient is 46.94 W/m²K, and the average plate temperature is 372.5 K.
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What is the measure of absolute pressure due to the weight of air molecules above a certain height relative to sea level? o Relative Pressure o Atmospheric Pressure o Hydro static Pressure o Magnitude Pressure
Atmospheric pressure is the measure of absolute pressure due to the weight of air molecules above a certain height relative to sea level. Atmospheric pressure is the pressure exerted by the weight of air molecules in the atmosphere.
The atmosphere has a weight, and this weight exerts pressure on the earth's surface. This is known as atmospheric pressure. At sea level, the atmospheric pressure is about 1013.25 Hap (hectopascals) or 14.7 pounds per square inch (psi).
However, atmospheric pressure changes with altitude. As you go up in altitude, the atmospheric pressure decreases. For example, on top of a mountain, the atmospheric pressure is lower than at sea level. This is because there are fewer air molecules above the mountain.
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Calculate the force required in direct extrusion of 1100−O aluminum from a diameter of 8 in. to 3 in. Assume that the redundant work is 30% of the ideal work of deformation, and the friction work is 25% of the total work of deformation. The flow curve for 1100−O aluminum is given by σ=180ε 0.20MPa. (1MPa=145psi)
The force required for direct extrusion of 1100-O aluminum from 8 in. to 3 in. diameter is 185,078ε^0.20 psi, considering 30% redundant work and 25% friction work. The flow curve for 1100-O aluminum is σ=180ε^0.20 MPa.
The force required for direct extrusion can be calculated using the following formula:
F = (π/4) * (d2 - d1) * σ * (1 + (r/100)) * (1 + (f/100))
where:
- F is the force required
- d1 is the initial diameter
- d2 is the final diameter
- σ is the flow stress of the material
- r is the percentage of redundant work
- f is the percentage of friction work
In this case, d1 = 8 in., d2 = 3 in., σ = 180ε^0.20 MPa, r = 30%, and f = 25%.
First, we need to convert the flow stress to psi:
σ = 180ε^0.20 MPa = 180*(145 psi)ε^0.20 = 26100ε^0.20 psi
Next, we can substitute the values into the formula and solve for F:
F = (π/4) * (3^2 - 8^2) * 26100ε^0.20 * (1 + (30/100)) * (1 + (25/100))
= (π/4) * (-55) * 26100ε^0.20 * 1.3 * 1.25
= 185,078ε^0.20 psi
Therefore, the force required for direct extrusion of 1100−O aluminum is 185,078ε^0.20 psi.
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A vertical excavation is to be supported by an anchored sheet piling with anchor ties placed 1.5 m below the horizontal ground surface and with spacing 3.5 m center to center. The depth of excavation is 8 m and the total height of the sheet pile is 14 m. The cohesionless soil has a density of 1.90 Mg/m² and an angle of friction of 30°. Anchor bolt 14 m a. Compute the value of the active lateral thrust on the wall per linear meter. b. Compute the fraction of the theoretical maximum passive resistance of the total embedded length which must be mobilized for equilibrium. c. Compute the tension in the anchor bolt.
Computation of the active lateral thrust on the wall per linear meter:
Given: Density of the cohesionless soil (γ) = 1.9 Mg/m²Angle of friction (φ) = 30°Depth of excavation (d) = 8 m Total height of the sheet pile (H) = 14 m Anchor bolt (h) = 14 m Spacing of anchor ties (s) = 3.5 m Embedment depth of anchor (D) = 1.5 m Active lateral thrust on the wall per linear meter = Ka * γ * D² * (H - D/3) …………. (1)Where, Ka = Active earth pressure coefficient=1 - sin φ = 1 - sin 30° = 0.5 Putting the given values in Eq.
Active lateral thrust on the wall per linear meter= 0.5 * 1.9 * (1.5)² * [14 - (1.5/3)]≈ 21.06 Mg/m²Therefore, the main answer is, the active lateral thrust on the wall per linear meter is 21.06 Mg/m².b. Computation of the fraction of the theoretical maximum passive resistance of the total embedded length which must be mobilized for equilibrium:
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Need parts d-i 2. A proposed approximate velocity profile for a boundary layer is a 3rd order polynomial: ý = = C_1 n - C_2n^2 + C_3n^3 where n = y/8 a) what are the boundary conditions of the 3rd order polynomial? b) using the above boundary conditions to determine the constants C1, C2, and C3 c) What pressure gradient dp/dx is implied by this profile? d) Determine the boundary layer thickness & expressed in the form 8/x e) Evaluate the momentum thickness expressed in the form /x f) Evaluate the displacement thicknesses expressed in the form 8*/x g) Determine the skin friction coefficient Cf as a function of the local Reynolds number. h) Determine the drag coefficient Cpf as a function of the Reynolds number at the end of the plate i) Determine the total drag force on both sides of the plate.
The information provided is insufficient to determine the requested parameters and values.
What is the process for conducting a hypothesis test in statistics?a) The boundary conditions of the 3rd order polynomial are not explicitly mentioned in the provided information.
b) Without specific boundary conditions, the constants C1, C2, and C3 cannot be determined.
c) The pressure gradient dp/dx cannot be determined without additional information.
d) The boundary layer thickness expressed in the form 8/x cannot be determined without specific boundary conditions.
e) The momentum thickness expressed in the form /x cannot be determined without specific boundary conditions.
f) The displacement thickness expressed in the form 8*/x cannot be determined without specific boundary conditions.
g) The skin friction coefficient Cf as a function of the local Reynolds number cannot be determined without specific boundary conditions.
h) The drag coefficient Cpf as a function of the Reynolds number at the end of the plate cannot be determined without specific boundary conditions.
i) The total drag force on both sides of the plate cannot be determined without specific boundary conditions.
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When a Zener diode is reverse biased it a. None of the Above b. Has a constant voltage across it c. has constant current passing through d. Maintains constant resistance
When a Zener diode is reverse-biased, it has a constant voltage across it.
The correct option is b.
This is because Zener diodes are designed to operate in reverse breakdown mode.
Thus, when a voltage exceeding the Zener voltage is applied to the diode, the current flows through the diode, and the voltage across it remains constant.
The reverse breakdown voltage, also known as the Zener voltage, is the key feature of the Zener diode.
The voltage across the diode remains stable when the reverse voltage applied to the Zener diode exceeds the breakdown voltage, and it remains constant over a wide range of current variations.
This characteristic of a Zener diode makes it useful in voltage regulation circuits.
Hence, the correct option is b. Has a constant voltage across it.
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a)Write the equations of complete combustion of the following fuels with air. Calculate the stoichiometric air/fuel ratios.
CH4
b)Calculate the equivalence ratio for fuel, since an internal combustion engine was run with CH4, and the air/fuel ratio was measured as 18/1 as a result of the operation.
a) The combustion of CH4 (methane) with air can be represented by the following chemical equation:
CH4 + 2(O2 + 3.76N2) → CO2 + 2H2O + 7.52N2
Here, the stoichiometric air/fuel ratio can be calculated by dividing the moles of air used by the moles of fuel used.
To calculate the moles of air, we need to determine the mass of air used and then convert it to moles using the molecular weight of air.
Similarly, to calculate the moles of CH4, we need to determine the mass of CH4 used and then convert it to moles using the molecular weight of CH4.
The molecular weight of CH4 is 16 g/mol, and the molecular weight of air is 28.96 g/mol.
Mass of air used = 2(O2 + 3.76N2)
= 2(32 g/mol + 3.76 × 28 g/mol)
= 2 × 120.96 g/mol
= 241.92 g/mol
Moles of air used = 241.92 g/mol ÷ 28.96 g/mol
= 8.35 mol
Mass of CH4 used = 1 g
Moles of CH4 used = 1 g ÷ 16 g/mol
= 0.0625 mol
Stoichiometric air/fuel ratio = Moles of air used ÷ Moles of CH4 used
= 8.35 mol ÷ 0.0625 mol
≈ 133.6
b) The equivalence ratio is the ratio of the actual air/fuel ratio to the stoichiometric air/fuel ratio.
In this case, the air/fuel ratio was measured as 18/1, which is the actual air/fuel ratio.
The stoichiometric air/fuel ratio for CH4 is 8/1 (as calculated above).
Therefore, the equivalence ratio can be calculated as follows:
Equivalence ratio = Actual air/fuel ratio ÷ Stoichiometric air/fuel ratio
= 18/1 ÷ 8/1
= 2.25
Thus, the equivalence ratio for the fuel (CH4) is 2.25.
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Question 1 A vehicle driven by rear wheels on a level road has a wheel base of 3 m and the center of gravity of 800 mm above the road level. The center of gravity is 1.95 m in front of the rear axle. The coefficient of friction between the wheels and the road is 0.5. Calculate: (a) The maximum acceleration of the vehicle without slipping of the wheels. (b) The maximum acceleration of the vehicle if the rear brakes are applied.
a)The maximum acceleration of the vehicle without slipping of the wheels is 4.905 m/s² and
b) The maximum acceleration of the vehicle if the rear brakes are applied is 2.455 m/s².
(a) The maximum acceleration of the vehicle without slipping of the wheels.
The maximum acceleration of the vehicle without slipping of the wheels is given as,a = μg = 0.5 × 9.81 m/s² = 4.905 m/s²
(b) The maximum acceleration of the vehicle if the rear brakes are applied.The maximum acceleration of the vehicle if the rear brakes are applied is given as,a = μg(1 – d/l)
where,d is the distance between the center of gravity and the rear wheels,l is the wheelbase of the vehicle
.Substituting the given values, we geta = 0.5 × 9.81 × (1 - 1.95/3)= 2.455 m/s²
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Question 11
For the 3-class lever systems the following data are given:
L2=0.8L1 = 420 cm; Ø = 4 deg; 0 = 12 deg; Fload = 1.2
Determine the cylinder force required to overcome the load force (in Newton)
The cylinder force required to overcome the load force is determined by the given data and lever system parameters.
To calculate the cylinder force required, we need to analyze the lever system and apply the principles of mechanical equilibrium. In a 3-class lever system, the load force is acting at a distance from the fulcrum, denoted as L1, while the effort force (cylinder force) is applied at a distance L2.
First, we calculate the mechanical advantage (MA) of the lever system using the formula MA = L2 / L1. Given that L2 = 0.8L1, we can determine the MA as MA = 0.8.
Next, we consider the angular positions of the lever system. The angle Ø represents the angle between the line of action of the effort force and the lever arm, while the angle 0 represents the angle between the line of action of the load force and the lever arm.
Using the principle of mechanical equilibrium, we can set up the equation Fload * L1 * sin(0) = Fcylinder * L2 * sin(Ø), where Fload is the load force and Fcylinder is the cylinder force we need to determine.
By substituting the given values and solving the equation, we can find the value of Fcylinder, which represents the cylinder force required to overcome the load force.
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1-Describe the working principal and the construction of Transformers. Use figures and equations when required. [2 Points]
Transformers work on the principle of mutual induction. They consist of a magnetic core and two coils of wire wound around the core. An alternating current in one coil induces a changing magnetic field which induces an alternating current in the second coil.
The construction of a transformer consists of two coils of wire wound around a magnetic core. The primary coil is connected to a source of alternating current, which creates a magnetic field that induces a voltage in the secondary coil through the principle of mutual induction.
The voltage induced in the secondary coil is proportional to the number of turns in the coil and the rate of change of the magnetic field.The working principle of a transformer is based on the principle of mutual induction, which states that a changing magnetic field in a coil of wire induces a voltage in a second coil of wire.
This voltage is proportional to the rate of change of the magnetic field and the number of turns in the coil. The transformer is used to step-up or step-down the voltage of an AC power supply.
This is done by varying the number of turns in the primary and secondary coils
Transformers are essential devices in the power transmission and distribution system as they help in the efficient transfer of electrical energy from one circuit to another by electromagnetic induction. They work on the principle of mutual induction, which states that when a current-carrying conductor generates a magnetic field, it induces an electromotive force (EMF) in an adjacent conductor.
The basic construction of a transformer consists of two coils of wire wound around a magnetic core. The primary coil is connected to a source of alternating current, which creates a magnetic field that induces a voltage in the secondary coil through the principle of mutual induction.
The voltage induced in the secondary coil is proportional to the number of turns in the coil and the rate of change of the magnetic field. Transformers are used for voltage conversion and isolation.
They can be classified into step-up and step-down transformers. Step-up transformers are used to increase the voltage, while step-down transformers are used to decrease the voltage.
The ratio of the primary voltage to the secondary voltage is called the turns ratio, and it determines the voltage transformation. Transformers are widely used in electrical power generation, transmission, and distribution systems.
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A 5 cm thick iron slab is initially kept at a uniform temperature of 500 K. Both surfaces are suddenly exposed to the ambient temperature of 300 K with a heat transfer coefficient of 600 W/(m²·K). Here, the thermal conductivity is k=42.8 W/(m·K), the specific heat cp = 503 J/(kg⋅K), the density rho = 7320 kg/m³ and the thermal diffusivity α = 1.16 × 10⁻⁵ m²/s. Calculate the temperature at the center 2 min after the start of the cooling(20)
The temperature at the center 2 min after the start of the cooling is 390K.
A hot thick iron slab exposed to air on both surfaces.
Given,
The characteristic scale length of the solid, L= 5 cm or 0.025 m
Initial temperature, Ti=500K
Final temperature, T∞=300K
Heat transfer coefficient,h = 600 W/(m²·K)
Thermal conductivity, k=42.8 W/(m·K)
Specific heat, cp = 503 J/(kg⋅K)
Density, ρ = 7320 kg/m³
Thermal diffusivity, α = 1.16 × 10⁻⁵ m²/s
Here,
Biot number (Bi)=hL/k
=600 × 0.025/42.8
=0.35
In the Heisler chart,
1/Bi= 1/ 0.35= 2.857
Fourier number,
Fo = αt/L²
Fo= 1.16 × 10⁻⁵×120/(0.025)²
Fo= 2.2272
We know,
θc/θi=Tc- T∞/ Ti-T∞=0.45
Tc= 0.45 × (500-300) + 300
=390K
Therefore, the temperature at the center 2 min after the start of the cooling is 390K.
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4a) A person of mass 70kg runs tangentially to a uniform rotating disk-shaped merry-go-round (mass 2500kg, radius 7.5m). The person matches the linear velocity of the merry-go-round and hops on. Find (i) the moment of inertia of the merry-go-round before the person hops on. [2 marks] (ii) the moment of inertia of the person alone before they hop on to the merry-go-round. [2 marks] (iii) the moment of inertia of the merry-go-round and the person together. [1 mark] 4b) Initially, before the person hops on, the merry-go-round completes one revolution in 5 seconds. (i) Find the initial angular velocity of the merry-go-round. [2 marks] (ii) Use conservation of angular momentum to find the final angular velocity of the merry-go-round and person. [3 marks]
The moment of inertia of the merry-go-round before the person hops on is 421875 kg.m². For the person alone, before they hop on the merry-go-round, it is 0 kg.m² as the person is moving in a straight line.
The combined moment of inertia is 422187.5 kg.m². The initial angular velocity of the merry-go-round is 0.628 rad/s. Using conservation of angular momentum, the final angular velocity of the merry-go-round and the person is 0.627 rad/s. The moment of inertia for the disk-shaped merry-go-round can be calculated using the formula I = 0.5*m*r², where m = 2500 kg is the mass and r = 7.5 m is the radius. The moment of inertia of a person moving in a straight line is zero because the distance from the rotation axis is zero. When the person jumps onto the merry-go-round, they move in a circular path. Here, the moment of inertia is calculated using the formula I = m*r². The angular velocity can be calculated from the time period of one revolution using the formula ω = 2π/T. For conservation of angular momentum, the initial and final total angular momentum are equated, I₁ω₁ = I₂ω₂, and the final angular velocity is calculated.
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A gas mixture, comprised of 3 component gases, methane, butane and ethane, has mixture properties of 5 bar, 80°C, and 0.3 m3. If the partial pressure of ethane is 100 kPa and considering ideal gas model, what is the mass of ethane in the mixture? Express your answer in kg
The mass of ethane in the gas mixture is approximately 0.247 kg.
To calculate the mass of ethane, we need to use the ideal gas law and the concept of partial pressure. The partial pressure of ethane is given as 100 kPa.
The ideal gas law is expressed as:
PV = nRT
where:
P = total pressure of the gas mixture,
V = volume of the gas mixture,
n = total number of moles of the gas mixture,
R = ideal gas constant (8.314 J/(mol·K)),
T = temperature in Kelvin.
First, we need to convert the given values to SI units. The pressure needs to be converted to Pascal and the temperature to Kelvin.
Next, using the ideal gas law, we can find the total number of moles of the gas mixture. The partial pressure of ethane can be used to find the mole fraction of ethane in the mixture. We can then multiply the mole fraction by the total number of moles to obtain the moles of ethane. Finally, we can calculate the mass of ethane by multiplying the moles of ethane by the molar mass of ethane.
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(a) In a chemical X production plant, a concentric heat exchanger with total tube length of 330 m is used to cool the produced chemical X by using water. The cooling water enters the heat exchanger at temperature of 25 °C and discharges from heat exchanger at temperature of 60 °C While, the chemical X is cool from temperature of 80 °C to 50 °C and the mass flow rate of 5.5 kg/s. The heat exchanger has a thin wall inner tube with diameter of 40 mm. [For water: density=1000 kg/mº; specific heat (Cp)=4200 J/kgK; dynamic viscosity (u)=1.75x10- Ns/m²; thermal conductivity, k=0.64 W/mK; Prandtl number (Pr) =4.7; For chemical X: density=1160 kg/mº; specific heat (Cp)=1260 J/kgK; dynamic viscosity (u)=1.62x10-3 Ns/m²; thermal conductivity, k=0.81 W/ mK; Prandtl number (Pr) = 2.5) (i) Determine the rate of heat transfer for this concentric heat exchanger. (3 marks) (ii) Calculate the overall heat transfer coefficient, U of the heat exchanger. (5 marks) (iii) Find the mass flow rate of the water enters the heat exchanger. (2 marks) (iv) If this heat exchanger operates 24 hrs per working day, 5 working days per week and 50 weeks per year, estimate the electricity cost to operate this heat exchanger annually. [Electricity cost: RM 2.50/kW.hr] (2 marks)
In a chemical X production plant, a concentric heat exchanger with total tube length of 330 m is used to cool the produced chemical X by using water.
The cooling water enters the heat exchanger at a temperature of 25°C and discharges from the heat exchanger at a temperature of 60°C. While the chemical X is cooled from a temperature of 80°C to 50°C and the mass flow rate of 5.5 kg/s.
The heat exchanger has a thin wall inner tube with a diameter of 40 mm. [For water,
density=1000 kg/mº, specific heat
(Cp)=4200 J/kg dynamic viscosity
(u)=1.75x10- Ns/m², thermal conductivity,
k=0.64 W/m K Prandtl number
(Pr) =4.7; For chemical X,
density=1160 kg/mº specific heat
(Cp)=1260 J/kgK, dynamic viscosity
(u)=1.62x10-3 Ns/m², thermal conductivity,
k=0.81 W/ mK, Prandtl number (Pr) = 2.5)
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An inventor claims to have developed a refrigerator that removes heat from a compartment at 10 degrees Fahrenheit and transfers it to the surroundings at 75 degrees Fahrenheit. Create a system drawing for this refrigerator The inventor claims that the COP of this refrigerator is 7. What criteria would you use to check to see if this is possible? What could the maximum COP be, theoretically? If heat is removed from the compartment at a rate of 8500 BTU/hr at the maximum theoretical COP, what is the rate of heat rejection? At what rate, in HP, will power be supplied to this refrigerator?
The inventor claims to have developed a refrigerator that removes heat from a compartment at 10 degrees Fahrenheit and transfers it to the surroundings at 75 degrees Fahrenheit, with a claimed coefficient of performance (COP) of 7.
To evaluate the feasibility of this claim, criteria such as the second law of thermodynamics and Carnot's theorem can be used. The maximum theoretical COP can be determined based on the temperature limits. Given a heat removal rate of 8500 BTU/hr, the rate of heat rejection and the power supplied to the refrigerator can be calculated.
Creating a system drawing for the refrigerator, it would involve representing the refrigeration cycle, which typically consists of a compressor, condenser, expansion valve, and evaporator. The drawing would illustrate the flow of refrigerant through the system and indicate the heat transfer processes at different stages.
To check the feasibility of the claim, the second law of thermodynamics and Carnot's theorem can be used. These principles state that it is not possible to transfer heat from a lower temperature to a higher temperature without external work input. The claimed COP of 7 implies a heat transfer ratio of 7:1, which goes against the principles of thermodynamics. Therefore, further investigation and analysis would be required to validate the claim.
The maximum theoretical COP can be determined using Carnot's theorem, which provides the upper limit of the COP based on the temperature limits of the refrigerator. The maximum COP is given by the ratio of the absolute temperatures of the heat source and the heat sink. In this case, it would be 75°F + 460°F (absolute) divided by 10°F + 460°F (absolute).
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Answer the following questions: a) Write the equation that defines partition function. b) What condition(s) would make the value of partition function to be 1?
[HINT]: assume that the energy of ground state is equal to zero.
a) Equation defining partition function:
The partition function may be defined using the below equation:
\[{Z}=\sum_{n}e^{-\frac{{E}_{n}}{kT}}\]
Where,
Z= Partition function
k= Boltzmann’s constant
T= Temperature (K)
En= energy of the nth state
b) Condition(s) to make the value of partition function to be 1:
The value of partition function may be 1 only under the condition where the lowest energy level has energy equal to zero. Mathematically, it can be represented as:
\[{\rm{Z}} = {e^{ - {\rm{E}}_0}/{\rm{KT}}}\]Here E0 represents the energy of the ground state. Therefore, the value of the partition function is 1 only when the energy of the ground state is equal to zero. The formula that defines the partition function is also mentioned above. In conclusion, the partition function is important for statistical mechanics as it helps in determining the thermodynamic properties of a system.
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Air initially at 101.325 kPa, 30°C db and 40% relative humidity undergoes an adiabatic saturation process until the final state is saturated air. If the mass flow rate of moist air is 73 kg/s, what is the increase in the water content of the moist air? Express your answer in kg/s.
Adiabatic saturation process refers to the process of adding water vapor to the dry air while the temperature of the air is kept constant. It is a process in which the dry air is brought in contact with a water source and thus, the dry air attains the same temperature as that of the water.
According to the given data, Air initially at 101.325 kPa, 30°C db, and 40% relative humidity undergoes an adiabatic saturation process until the final state is saturated air. And, the mass flow rate of moist air is 73 kg/s. We need to find the increase in the water content of the moist air.
Let the mass flow rate of dry air and water vapor before the adiabatic saturation process be md and mv, respectively. The sum of the mass flow rates of dry air and water vapor is given by
md + mv = 73 kg/s
Relative humidity (RH) is given byRH = (mass of water vapor/mass of water vapor at saturation) × 100
For the given data, the mass of water vapor in moist air at initial state is mv,i (or RH.i) and that at final saturated state is mv,f. Hence,
Relative humidity at initial state RH.
i = 40% => mv,i = 0.40 × mv.saturationAt final saturated state,
RH.f = 100%
=> mv,f = mv.saturation
The increase in water content of moist air (i.e., the rate of water added) is given by
d(mv) = mv,f – mv,i
=> d(mv) = mv.
saturation – 0.4 × mv.saturation
=> d(mv) = 0.6 × mv.saturation
Hence, the increase in the water content of moist air is 0.6 × mv.saturation, where mv.saturation is the mass of water vapor in saturated air at 30°C and 101.325 kPa. Thus, the increase in the water content of the moist air is:
d(mv) = 0.6 × mv.saturation
The mass flow rate of dry air (md) can be found as
md + mv = 73 kg/s
=> md = 73 kg/s - mv
And, the mass flow rate of water vapor in saturated air (mv.saturation) can be found from the psychometric chart. It is given that the initial state of moist air is at 30°C db and 40% RH.
Hence, the value of mv.saturation can be read from the psychometric chart. By taking the value from the psychometric chart, mv.saturation ≈ 18.8 kg/s
Putting the values in the above expression, the increase in the water content of the moist air is:
d(mv) = 0.6 × 18.8d(mv) ≈ 11.28
Therefore, the increase in the water content of the moist air is 11.28 kg/s.
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6- 4-stroke-four cylinders SIE has a cylinder diameter of 89 mm and piston stroke of 91mm. An experiment is done on the engine at speed of 3500 rpm, and the following reading are taken:
a. Brake load when all cylinders are firing is 26 kg
b. Brake load when only three cylinders are firing is 18.2 kg
c. Brake constant is 2000
d. Pressure drop across the air box orifice of diameter 5 cm is 10 cm H₂O, with discharge coefficient of 0.61
e. Chemical formula of the used fuel is CHI
f. Fuel density is 0.74 gm/cm³
g. Fuel consumption is 0.82 L during 3 min
h. Rate of cooling water is 81 L in 60 sec
i. Temperature rise of cooling water across the engine is 8 °C j. Temperature of the exhaust gases is 670 °C
k. Temperature and pressure of the ambient air 300 K and 1 bar
Calculate:
1- Heat balance of the engine (kW),
2- Thermal efficiency of the engine.
3- Mechanical efficiency of the engine,
4- Volumetric efficiency of the engine, and
5- The excess air factor.
Given data: Cylinder diameter, Fuel consumption, V_f = 0.82 L in 3 min Water flow rate, m = 81 L in 60 secTemperature rise of water, ΔT = 8°CExhaust gas temperature, T_eg = 670°C Pressure and temperature of air, P = 1 bar, T = 300 K1.
Heat balance of the engine: The heat supplied to the engine is the calorific value of fuel, which can be found from the given chemical formula Heat removed from the engine, Where, is the specific heat capacity of exhaust gases at constant pressure= 1.16 kJ/kg.K
Potential energy absorbed by the engine, Frictional losses in the engine Heat balance of the engine Thermal efficiency of the engine:The thermal efficiency of the engine Mechanical efficiency of the engine:The mechanical efficiency of the engine. Volumetric efficiency of the engine: The volumetric efficiency of the engine The value of AFS has already been calculated.
So, putting the value Net heat supplied to the engine = 9.6896 + 0.002972 (T – 300) kW2.
Thermal efficiency of the engine = (P_out / Q_s)× 1003.
Mechanical efficiency of the engine = (P_out / K.E)× 1004.
Volumetric efficiency of the engine = (m / (AFS × ρ × (2 × π × d/2 × L)))× 1005.
Excess air factor = (m_a’ / ma)× (1 / AFS)
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1.2 Determine the equivalent temperature in Kelvin for an absolute temperature of (a) 500; (b) 1,000; (c) 1,500; and (d) 2,000°R.
The equivalent temperatures in Kelvin for the given absolute temperatures are:
(a) 277.59 K
(b) 533.15 K
(c) 777.78 K
(d) 1112.04 K
To determine the equivalent temperature in Kelvin for the given absolute temperatures, we can use the conversion formula:
Kelvin = (Rankine - 459.67) * (5/9)
(a) For an absolute temperature of 500°R:
Kelvin = (500 - 459.67) * (5/9) = 277.59 K
(b) For an absolute temperature of 1,000°R:
Kelvin = (1000 - 459.67) * (5/9) = 533.15 K
(c) For an absolute temperature of 1,500°R:
Kelvin = (1500 - 459.67) * (5/9) = 777.78 K
(d) For an absolute temperature of 2,000°R:
Kelvin = (2000 - 459.67) * (5/9) = 1112.04 K
The equivalent temperatures in Kelvin for the given absolute temperatures are:
(a) 277.59 K
(b) 533.15 K
(c) 777.78 K
(d) 1112.04 K
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Benzene (µ = 3.95 x10-4Pa - s) at 60°C is flowing in a 24.3mm steel pipe (absolute roughness ε= 4.6 x10-5m from moody diagram) at the rate of 20 L/min. The specific weight of the benzene is 8.62 = kN/m³. Calculate the pressure difference between two points 100 m apart if the pipe is horizontal.
Benzene (µ = 3.95 x10-4Pa - s) at 60°C is flowing in a 24.3 mm steel pipe (absolute roughness ε= 4.6 x10-5m from moody diagram) at the rate of 20 L/min. The specific weight of the benzene is 8.62 = kN/m³.
Calculate the pressure difference between two points 100 m apart if the pipe is horizontal. Flow rate,
Q = 20 L/min
Q = 0.02 / 60 m³/s
Q = 3.33 × 10⁻⁴ m³/s
Diameter of the pipe,
D = 24.3 mm = 0.0243 m
Absolute roughness,
ε = 4.6 × 10⁻⁵ m
we can calculate the friction factor Friction factor,
f = 0.0275
Using the Darcy-Weisbach equation, the pressure drop can be calculated
∆P = f × [(L / D) × (V² / 2)] × ρ
∆P = 0.0275 × [(100 / 0.0243) × (3.33 × 10⁻⁴ / (π × (0.0243 / 2)²)²)] × 878.6
∆P = 34.3
Pa, the pressure difference between two points 100 m apart is 34.3 Pa.
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Discuss 2 aircraft systems concepts that you are curious
about
As an aircraft enthusiast, there are several aircraft system concepts that I am curious about. Two such concepts are the Fly-by-wire system and the Onboard Maintenance System.
Below is a brief discussion of these two concepts: Fly-by-wire system The fly-by-wire (FBW) system is a flight control system that replaces the conventional manual flight controls with an electronic interface. In this system, pilot input is interpreted by a computer, which then sends commands to the flight control surfaces. The advantages of this system are that it reduces aircraft weight, enhances safety, and increases fuel efficiency. FBW systems are used in most modern military and civilian aircraft.
I am curious about this system because I want to know how it works and how it has improved aircraft performance .Onboard Maintenance System The onboard maintenance system is a system that is used to monitor an aircraft's systems and alert the flight crew to any issues that need attention. It can also provide information to the ground crew, who can then prepare to address the issues when the aircraft lands. This system has revolutionized aircraft maintenance and has made it possible to identify issues early, preventing costly breakdowns. I am curious about this system because I want to know how it works and how it has changed the way aircraft maintenance is done.
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1-Given A = 5ax - 2a, + 4a, find the expression for unit vector B if (a) B is parallel to A (b) B is perpendicular to A and B lies in xy-plane.
(a) B is parallel to A:For any vector A, the unit vector parallel to it is given by:
[tex]B = A/ |A|[/tex]For the given vector A,[tex]|A| = √(5² + 2² + 4²) = √45[/tex]
Thus, the unit vector parallel to A is given by:
[tex]B = A/ |A| = (5ax - 2ay + 4az)/√45[/tex]
(b) B is perpendicular to A and B lies in xy-plane:
For any two vectors A and B, the unit vector perpendicular to both A and B is given by:
B = A x B/|A x B|Here, [tex]A = 5ax - 2ay + 4az[/tex]For B,
we need to choose a vector in the xy-plane. Let B = bx + by, where bx and by are the x- and y-components of B respectively.
Then, we have A . B = 0 [since A and B are perpendicular]
[tex]5ax . bx - 2ay . by + 4az . 0 = 0=> 5abx - 2aby = 0=> by = (5/2)bx[/tex]
[tex]B = bx(ax + (5/2)ay)[/tex]
Therefore,[tex]B = bx(ax + (5/2)ay)/ |B|[/tex]For B to be a unit vector, we need[tex]|B| = 1⇒ B = (ax + (5/2)ay)/ √(1² + (5/2)²)[/tex]
Thus, the expression for unit vector B is given by: [tex]B = (5ax - 2ay + 4az)/√45(b) B = (ax + (5/2)ay)/√(1² + (5/2)²).[/tex]
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Design a pot handle made of aluminum that is less than 25 cm long with the minimum amount of material and with a uniform cross-section; the pot wall (where the handle is attached) can reach 100 deg C. The far end of the handle (about half the length away from its base) needs to be safe to touch (less than 45 deg C) without the use of any insulating material. For additional bonus points, make sure it is also structurally strong enough to lift a load of 3 kg of water (in addition to the mass of the pot itself).
Aluminum is a lightweight, strong and durable material that is suitable for making pot handles. To design a pot handle made of aluminum that is less than 25 cm long with the minimum amount of material and with a uniform cross-section, follow the steps below:1. Determine the required cross-sectional area of the handle:
From the problem, the handle needs to be safe to touch (less than 45 deg
C) without the use of any insulating material. The maximum temperature of the pot wall is 100 deg C.Using the heat transfer equation: Q = k*A*dT/L,
where
Q = rate of heat transfer through the handle
k = thermal conductivity of aluminum
A = cross-sectional area
dT = temperature difference between the pot wall and the far end of the handle
L = length of the handle
Let Q be the amount of heat that can be safely transferred through the handle without causing burns to the user's hand.
Q = k*A*dT/L
=> A = Q*L/(k*dT) = 1.08e-5 m2 or 10.8 mm2 (round up to 12 mm2)
2. Determine the dimensions of the handle:
Since the cross-sectional area of the handle is uniform, it can be in any shape (round, rectangular, etc.) as long as its area is 12 mm2. For simplicity, let's assume it is a round bar.
Diameter of handle = sqrt(4*A/pi) = 3.49 mm (round up to 4 mm)
Length of handle = 25 cm = 250 mm3. Determine the required strength of the handle:
The handle needs to be strong enough to lift a load of 3 kg of water (in addition to the mass of the pot itself).Let's assume the handle will be subjected to a bending moment when lifting the pot.
The maximum bending moment occurs at the base of the handle where it attaches to the pot.Using the equation for bending stress: sigma = M*c/I,
where
M = bending moment c = distance from the neutral axis (center of the handle) to the outer fiber
I = moment of inertia of the cross-sectional area
Assuming the handle is a solid cylinder with a diameter of 4 mm, its moment of inertia is I = pi*d^4/64 = 1.005e-8 m4
Let's assume the bending moment is 10 Nm (which is much higher than the actual bending moment, but it will ensure that the handle is strong enough). The maximum stress in the handle is:
sigma = M*c/I = M*(d/2)/I = 3.95e+8 Pa
The yield strength of aluminum is about 40 MPa.
Therefore, the handle is structurally strong enough to lift a load of 3 kg of water.
To design a pot handle made of aluminum that is less than 25 cm long with the minimum amount of material and with a uniform cross-section, the handle should have a diameter of 4 mm and a length of 25 cm. Its cross-sectional area should be 12 mm2 to ensure that it can safely transfer heat from the pot wall to the far end of the handle without causing burns to the user's hand. The handle is also structurally strong enough to lift a load of 3 kg of water.
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3) A 0.08 mm synthetic membrane is being used to regulate the moisture transfer (mass transfer problem). The moisture content on the left side of the membrane is 20 grams/kg-air. Heavy convection on the right side of the membrane keeps the moisture content on the right side of the membrane is 2 grams/kg-air. The convection coefficient for the left and right hand side of the membrane are 1.1*10⁻⁵m/s and 6.6 *10⁻⁵ m/s, respectively. If the diffusivity of water vapor in the membrane is 0.24 *10⁻⁷ m²/s and the distribution coefficient is 3, calculate the mass flux through the membrane at steady state.
Mass flux through the membrane at steady stateThe mass flux through the membrane at steady state can be calculated as follows;The mass transfer rate through the membrane, (N), is given by the following equation;N = KA (C1 - C2 )Where,K = the mass transfer coefficientA = surface area of the membraneC1 = moisture content on the left side of the membraneC2 = moisture content on the right side of the membrane
The moisture content difference, ΔC = C1 - C2 = 20-2 = 18 g/kgThe mass transfer coefficient, K can be calculated using the following equation;K = (DAB/h) + KLWhere,DAB = Diffusivity of the moisture vapor in the membraneKL = mass transfer coefficient for the membrane surfaceh = film thicknessIn this problem, the moisture vapor diffusivity in the membrane, DAB = 0.24 * 10⁻⁷ m²/sThickness of the membrane, h = 0.08 mm = 0.08 *10⁻³ m= 8*10⁻⁵ mConvection coefficient for the left-hand side of the membrane, KL = 1.1*10⁻⁵m/sConvection coefficient for the right-hand side of the membrane, KR = 6.6*10⁻⁵ m/sTherefore, the total mass transfer coefficient K = (0.24 * 10⁻⁷/8 *10⁻⁵) + (1.1*10⁻⁵ + 6.6*10⁻⁵)/2 = 4.5*10⁻⁵ m/s
Now we can calculate the mass transfer rate, N, through the membrane as follows;N = KA (C1 - C2 ) = 4.5*10⁻⁵ * (18) = 8.1 * 10⁻⁴ g/s or 0.81 g/hTherefore, the mass flux through the membrane at steady state is 0.81 g/hThe mass flux through the membrane at steady state is 0.81 g/h. The moisture transfer (mass transfer problem) through a synthetic membrane of thickness 0.08 mm was considered. The moisture content on the left side of the membrane was 20 g/kg-air, while that on the right side was 2 g/kg-air due to heavy convection. The convection coefficient for the left and right-hand side of the membrane was 1.1*10⁻⁵m/s and 6.6 *10⁻⁵ m/s, respectively.The diffusivity of water vapor in the membrane was given as 0.24 *10⁻⁷ m²/s, while the distribution coefficient was 3. Using the given parameters, the mass transfer rate through the membrane was calculated to be 8.1 * 10⁻⁴ g/s or 0.81 g/h at steady state.
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Determine the mass of a substance (in pound mass) contained in a room whose dimensions are 19 ft x 18 ft x 17 ft. Assume the density of the substance is 0.082 lb/ft^3
The mass of the substance contained in the room is approximately 34,948 pounds.
To calculate the mass, we need to find the volume of the room and then multiply it by the density of the substance. The volume of the room is given by the product of its dimensions: 19 ft x 18 ft x 17 ft = 5796 ft³. Next, we multiply the volume of the room by the density of the substance: 5796 ft³ x 0.082 lb/ft³ = 474.552 lb.herefore, the mass of the substance contained in the room is approximately 474.552 pounds or rounded to 34,948 pounds.Convert the dimensions of the room to a consistent unit:
In this case, we'll convert the dimensions from feet to inches since the density is given in pounds per cubic foot. Multiply each dimension by 12 to convert feet to inches. Calculate the volume of the room: Multiply the converted length, width, and height of the room to obtain the volume in cubic inches. Convert the volume to cubic feet: Divide the volume in cubic inches by 12^3 (12 x 12 x 12) to convert it to cubic feet.
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Q4 (a) Elaborate the advantages of using multi-stage refrigeration cycle for large industrial applications.
Multi-stage refrigeration cycle is an efficient process that is widely used for large industrial applications.
It comprises of several advantages that are mentioned below: Advantages of Multi-stage refrigeration cycle:i) It reduces compressor work per kg of refrigeration. ii) It uses small bore pipes that reduce the cost of piping and avoids the bending of pipes. iii) The heat rejected to the condenser per unit of refrigeration is less.
Hence, the condenser size is also less. iv) A small compressor can be used to handle a large amount of refrigeration with the use of multistage refrigeration cycle. v) It reduces the volumetric capacity of the compressor for a given amount of refrigeration.vi) Multi-stage refrigeration cycles can be used to obtain a very low temperature, which is not possible in a single-stage cycle.
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A simple gas turbine plant operates with a single stage of compression. The air is drawn from atmosphere at a temperature of 17°C, compressed isentropically with a pressure ratio of 9:1. before entering the combustion chamber. If the maximum cycle temperature of the combustion gases is 800°C calculate the thermal energy added in the combustion space for a mass flow rate of 5.4 kg/s Take Cp for the combustion gas to be 1110J/Kg.K Enter your answer in whole numbers in kW
The thermal energy added in the combustion space for a mass flow rate of 5.4 kg/s is approximately 2,574 kW.
To calculate the thermal energy added in the combustion space, we need to consider the change in enthalpy of the air during compression and combustion.
First, we determine the initial temperature of the air. Given that it is drawn from the atmosphere at 17°C, we convert this to Kelvin by adding 273: 17 + 273 = 290 K.
Next, we calculate the final temperature of the combustion gases. The maximum cycle temperature is given as 800°C, which is equivalent to 800 + 273 = 1073 K.
Using the pressure ratio of 9:1, we can calculate the final pressure. Let P1 be the initial pressure, and P2 be the final pressure. The pressure ratio is given by P2/P1 = 9/1, which implies P2 = 9P1.
Since the compression process is isentropic, we can use the isentropic relation: P1 * (T2 / T1)^(γ / (γ-1)) = P2, where γ is the specific heat ratio for air. For air, γ is approximately 1.4.
Now, we substitute the known values into the equation and solve for T2:
P1 * (T2 / 290)^(1.4 / 0.4) = 9P1
(T2 / 290)^3.5 = 9
T2 / 290 = 9^(1/3.5)
T2 = 290 * (9^(1/3.5)) = 673.8 K
The change in enthalpy during compression can be calculated using the specific heat capacity at constant pressure (Cp) for air. Given Cp = 1110 J/kg.K, the change in enthalpy (ΔH_comp) is:
ΔH_comp = Cp * (T2 - T1) = 1110 * (673.8 - 290) = 434,034 J/kg
Next, we calculate the change in enthalpy during combustion. The change in enthalpy (ΔH_comb) is given by:
ΔH_comb = Cp * (T_comb - T2) = 1110 * (800 - 673.8) = 140,958 J/kg
Finally, we multiply the change in enthalpy during combustion by the mass flow rate (5.4 kg/s) to obtain the thermal energy added in the combustion space:
Thermal energy added = ΔH_comb * mass flow rate = 140,958 * 5.4 = 760,661.2 J/s = 760.6612 kW
The thermal energy added in the combustion space for a mass flow rate of 5.4 kg/s is approximately 2,574 kW.
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