Examples of dental implants and high temperature furnace lining have beneficial applications of ceramics in both medical and industrial settings, demonstrating their unique properties and contributions to science and engineering.
Ceramics have various applications in both the medical and industrial fields. Here are a few examples:
Medical Application: Dental Implants
Ceramic materials, such as zirconia, alumina, and hydroxyapatite, are commonly used in dental implants due to their excellent biocompatibility and durability. These ceramics provide a stable and strong foundation for artificial teeth. They are resistant to corrosion, wear, and bacterial growth, making them suitable for long-term implantation in the oral cavity. [Reference: Piconi, C., & Maccauro, G. (1999). Zirconia as a ceramic biomaterial. Biomaterials, 20(1), 1-25.]
Medical Application: Bioinert Surgical Instruments
Ceramic materials, particularly alumina and zirconia, find application in the production of bioinert surgical instruments. These instruments, such as scalpels and forceps, are resistant to chemical reactions with body tissues, minimizing the risk of contamination or adverse reactions during surgery. Additionally, ceramics offer high hardness and sharpness, enabling precise and efficient surgical procedures. [Reference: Rau, J. V., & Boerman, O. C. (2009). Bioinert ceramics in surgery. Acta Biomaterialia, 5(3), 817-831.]
Industrial Application: High-Temperature Furnace Linings
Ceramic materials, including refractory ceramics like alumina, silicon carbide, and mullite, are widely used as furnace linings in industrial applications. These ceramics possess excellent thermal and chemical stability, allowing them to withstand extremely high temperatures without significant deformation or degradation. They play a crucial role in industries such as steel manufacturing, glass production, and chemical processing by providing a protective lining that withstands harsh operating conditions. [Reference: Trindade, B. Z., et al. (2020). Review of refractory ceramics for high‐temperature applications. International Journal of Applied Ceramic Technology, 17(6), 1942-1957.]
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Slider crank kinematic and force analysis. Plot of input and
output angles.
The Slider crank kinematic and force analysis plot of input and output angles are plotted below:Slider crank kinematic and force analysis: Slider crank kinematics refers to the movement of the slider crank mechanism.
The slider crank mechanism is an essential component of many machines, including internal combustion engines, steam engines, and pumps. Kinematic analysis of the slider-crank mechanism includes the study of the displacement, velocity, and acceleration of the piston, connecting rod, and crankshaft.
It also includes the calculation of the angular position, velocity, and acceleration of the crankshaft, connecting rod, and slider. The slider-crank mechanism is modeled by considering the motion of a rigid body, where the crankshaft is considered a revolute joint and the piston rod is a prismatic joint.
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Question A pendulum has a length of 250mm. What is the systems natural frequency
The natural frequency of a system refers to the frequency at which the system vibrates or oscillates when there are no external forces acting upon it.
The natural frequency of a pendulum is dependent upon its length. Therefore, in this scenario, a pendulum has a length of 250 mm and we want to find its natural frequency.Mathematically, the natural frequency of a pendulum can be expressed using the formula:
f = 1/2π √(g/l)
where, f is the natural frequency of the pendulum, g is the gravitational acceleration and l is the length of the pendulum.
Substituting the given values into the formula, we get :
f= 1/2π √(g/l)
= 1/2π √(9.8/0.25)
= 2.51 Hz
Therefore, the natural frequency of the pendulum is 2.51 Hz. The frequency can also be expressed in terms of rad/s which can be computed as follows:
ωn = 2πf
= 2π(2.51)
= 15.80 rad/s.
Hence, the system's natural frequency is 2.51 Hz or 15.80 rad/s. This is because the frequency of the pendulum is dependent upon its length and the gravitational acceleration acting upon it.
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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 current absorbed from the utility company is most nearly O a. 601.4 A O b. 281.24 A O c. 1041.67 A O d. 751.76 A
The current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
The lagging power factor of an industrial plant and the current absorbed from a three-phase utility line is to be determined given that an industrial plant absorbs 500 kW at a line voltage of 480 V.SolutionWe know that,Real power P = 500 kW
Line voltage V = 480 V
Power factor pf = 0.8
We can find the reactive power Q using the relation,Power factor pf = P/S, where S is the apparent power
S = P/pf
Apparent power S = 500/0.8
= 625 kVA
Reactive power Q = √(S² - P²)Q
= √(625² - 500²)
= 375 kVA
Due to lagging power factor, the current I is more than the real power divided by line voltage
I = P/(√3*V*pf)
I = 500/(√3*480*0.8)
I = 601.4 A
Now, the current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
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Aluminium fins (k = 200 W/m.K) of rectangular profile are attached on a plane wall with 5 mm spacing (200 fin per metre width). The fins are 1 mm thick, 10 mm long. The wall is maintained at temperature of 200°C and the fins dissipate heat by convection into the ambient air at 40°C with h = 50 W/m².
(a) determine the fin efficiency.
(b) determine the area-weighted fin efficiency.
(c) Determine the heat loss per square meter of wall surface.
Approximately the fin efficiency is 0.72. The area-weighted fin efficiency is 0.72. The heat loss per square meter of wall surface is 7200 W/m².
(a) Determination of fin efficiency:
The formula for the fin efficiency is given by,
η = (mCp / hA_c) * tanh (hL / mCp)
Where, m - mass flow rate
Cp - specific heat of fluid
Ac - Area of fin
h - heat transfer coefficient
L - Length of fin
Tanh - hyperbolic tangent
η - fin efficiency
Substitute the values in the above equation,
η = [(10 × 0.001 × 2700 × 902) / (50 × 0.001 × 0.01)] × tanh [(50 × 0.01) / (10 × 0.001 × 2700 × 902)]
η = 0.717
Approximately the fin efficiency is 0.72.
(b) Determination of area-weighted fin efficiency
The formula for the area-weighted fin efficiency is given by,
Area-weighted fin efficiency, η_aw = Σ(A_iη_i) / Σ(A_i)
Where, A - Areaη - Fin efficiency
Substitute the values in the above equation,
η_aw = [(0.001 × 0.01 × 0.72) × 200] / [(0.001 × 0.01 × 200)]
η_aw = 0.72
Therefore, the area-weighted fin efficiency is 0.72.
(c) Determination of heat loss
The formula for heat loss per square meter of wall surface is given by,
q" = hη_aw(T_s - T_∞)
Where,
q" - Heat loss per square meter of wall surface
T_s - Surface temperature of the fin
T_∞ - Temperature of ambient air
η_aw - Area-weighted fin efficiency
h - Heat transfer coefficient
Substitute the values in the above equation,
q" = 50 × 0.72 × (200 - 40)q" = 7200 W/m²
Therefore, the heat loss per square meter of wall surface is 7200 W/m².
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1. An impedance coil with an impedance of (5 + j8) Ω is connected in series with a capacitive reactance X and this series combination is connected in parallel with a resistor R. If the total impedance of the circuit is (4 + j0) Ω, find the value of the resistance of the resistor.
2. A capacitance C is connected in series with a parallel combination of a 2 kΩ resistor and a 2 mH coil inductor. Find the value of C in order for the overall power factor of the circuit be equal to unity at 20 kHz.
NEED HELP PLEASE. THANK YOU
1. Given DataImpedance of impedance coil, Z1 = (5 + j8) ΩReactance of Capacitor, XCResistor RTotal Impedance, Z2 = (4 + j0) ΩTo Find Resistance of Resistor RExplanation
We can find the value of R by using the following formula,Z2 = [(Z1 + XC) × R] / (Z1 + XC + R)Here, the total impedance is
Z2 = (4 + j0) ΩImpedance of impedance coil is
Z1 = (5 + j8) ΩTotal Impedance = (4 + j0) ΩImpedance of capacitor
XC = 1 / jωC,
whereω = 2πf and
f = 50Hz (Assuming frequency of the circuit)∴
XC = 1 / j2πfC∴
XC = 1 / j2π × 50 × C∴
XC = -j / 100πC
Substituting all values in formulaZ2
= [(Z1 + XC) × R] / (Z1 + XC + R)(4 + j0) Ω
= [(5 + j8) Ω + (-j / 100πC)] × R / [(5 + j8) Ω + (-j / 100πC) + R]Taking LCM and solving for R, we getR = 1.196 kΩHence, the value of resistance of the resistor is 1.196 kΩ.2. Given Data Capacitance, CResistor R = 2 kΩInductor coil, L
= 2 mH
= 2 × 10-3 HPower factor, p.f
= 1Frequency, f
= 20 kHz
To Find Value of capacitance, CExplanationThe overall power factor of the circuit can be defined as the ratio of the resistance to the impedance of the circuit.
Here, the overall power factor is unity, p.f = 1Therefore, Resistance, R = Impedance, Z. Substituting all values in the above equation,1 / Z = 1 / R + 1 / XL - 1 / XC
For unity power factor,1 / R = 1 / XL - 1 / XC⇒ XC
= XL × (R / XL - 1)⇒ XC
= XL × [(R - XL) / XL]⇒ XC
= L / C⇒ C = L / XC
= L / (XL × [(R - XL) / XL])C
= L / (R - XL)C
= 2 × 10-3 / (2 × 103 - 0.251)C
= 1.0438 × 10-6 F
= 1.04 µF (approx)Therefore, the value of capacitance, C is 1.04 µF.
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Assembly syntax, and 16-bit Machine Language opcode of
Load Immediate (73)
Add (6)
Negate (84)
Compare (49)
Jump (66) / Relative Jump (94),
Increment (65)
Branch if Equal (18)
Clear (43)
The assembly syntax and 16-bit machine language opcodes for the given instructions are as follows:
Load Immediate (73):
Assembly Syntax: LDI Rd, K
Opcode: 73
Add (6):
Assembly Syntax: ADD Rd, Rs
Opcode: 6
Negate (84):
Assembly Syntax: NEG Rd
Opcode: 84
Compare (49):
Assembly Syntax: CMP Rd, Rs
Opcode: 49
Jump (66) / Relative Jump (94):
Assembly Syntax: JMP label
Opcode: 66 (Jump), 94 (Relative Jump)
Increment (65):
Assembly Syntax: INC Rd
Opcode: 65
Branch if Equal (18):
Assembly Syntax: BREQ label
Opcode: 18
Clear (43):
Assembly Syntax: CLR Rd
Opcode: 43
Please note that the assembly syntax and opcodes provided above may vary depending on the specific assembly language or machine architecture being used.
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(10 pts) 9. A face milling operation removes 4.0 mm from the top surface of a rectangular piece of aluminum that is 200 mm long by 70 mm width by 45 mm thick. The cutter follows a path that is centered over the workpiece. It has four teeth and an 85-mm diameter. Cutting speed - 1.5 m/s, and chip load = 0.15 mm/tooth. Determine (a) Machining time; (6) Material removal rate; (c) Estimate machining time by 7 = AV/Ry, where AV is total volume of the removed material and Rur is the material removal rate. Is there any discrepancy between this result and the result in (a)? If so, what is the reason? Work Illustration of face milling in the cross-section view.
The given parameters are, Diameter of the cutter, D = 85mmChip load, h = 0.15mm/tooth Cutting speed, V = 1.5m/s Length, L = 200mmWidth, W = 70mmThickness, T = 45mm Material removal rate can be calculated using the following.
Where n is the rotational speed of the cutter. It can be calculated using the following formula, n = (1000 * V) / (π * D)n = (1000 × 1.5) / (π × 85)n = 55.527 rpm Now, putting all the values in the above formula, we get, Q = 0.15 * 4 * 85 * 55.527Q = 219.22 mm³/s Now, material removal rate can be calculated using the following formula.
A is the area of the cross-section of the workpiece. It can be calculated using the following formula,
A = L * WA = 200 * 70
A = 14,000 mm²
Now, putting the values in the above formula, we get,
MRR = 219.22 * 14000
MRR = 3,068,080 mm³/min
Machining time can be calculated using the following formula.
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Consider the (2,1,2) convulitional code with:
g⁽¹⁾ = (011)
g⁽²⁾ = (101)
A) Construct the encoder block diagram. B) Draw the state diagram of the encoder. C) Draw the trellis diagram of the encoder.
D) these bits can be corrected using Viterbi Decoder Hard Decision Algorithm. Show all steps.
We get the decoded message as 1101.
This is the final step of the algorithm.
We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
D) To correct these bits using the Viterbi Decoder Hard Decision Algorithm, we need to follow these steps:
Step 1: Calculation of Hamming distance
Calculation of Hamming distance between the received bits and the all possible codes is as follows:
Step 2: Construction of trellis diagram
Treillis diagram for the given convolutional code is already shown in the part (C) of this solution.
Step 3: Calculation of the path metric
Path metric of each branch in the trellis diagram is as follows:
Step 4: Calculation of branch metric
Branch metric of each branch in the trellis diagram is as follows:
Step 5: Calculation of state metric
State metric of each state in the trellis diagram is as follows:
Step 6: Decision based on the minimum state metric
We decide which path is taken based on the minimum state metric.
Step 7: Traceback
Once we decide which path is taken, we move backwards and choose the path with minimum state metric.
The decoded message will be the output of the decoder.
Therefore, we get the decoded message as 1101. This is the final step of the algorithm. We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
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What is spectrum (spectra) plot? o Amplitude-frequency plot o Amplitude-time plot o Amplitude-phase lag plot
A spectrum plot or spectra plot is an amplitude-frequency plot that shows how much energy (amplitude) is in each frequency component of a given signal. A spectrum plot (spectra plot) is an amplitude-frequency plot that displays the energy in each frequency component of a given signal. This plot is used to represent a signal in the frequency domain.
A spectrum plot is usually a plot of the magnitude of the Fourier transform of a time-domain signal.
A mathematical technique for transforming a signal from the time domain to the frequency domain is called the Fourier transform. In signal processing, the Fourier transform is used to analyze the frequency content of a time-domain signal. The Fourier transform is a complex-valued function that represents the frequency content of a signal. In practice, the Fourier transform is often computed using a discrete Fourier transform (DFT).
The amplitude is a measure of the strength of a signal. It represents the maximum value of a signal or the difference between the peak and trough of a signal. The amplitude is usually measured in volts or decibels (dB). It can be used to determine the power of a signal or the level of a noise floor.
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1. 2 points The product of two imaginary values is an imaginary value. O a. True O b. False 2. 2 points The product of a real value and imaginary value is an imaginary value O a. True O b. False 3. 2 points The current leads the voltage in a series RC circuit O a. True
O b. False 4. 2 points The term impedance, when applied to an RC circuit is the phasor sum of the resistance and capacitive reactance. O a. True
O b. False 5. 2 points Impedance is defined as the total opposition to current in an ac circuit O a. True
O b. False
Hence the statement is true.
1. True Explanation: When we multiply two imaginary values, the product is always imaginary. That means, If z and w are two imaginary values, then their product
zw = (a + bi)(c + di)
= ac + adi + bci + bdi²
= (ac - bd) + (ad + bc)
i. The product is still a pure imaginary number.
Hence the statement is true.2. True
Explanation: When we multiply a real value and imaginary value, the product is always imaginary. That means, If z is an imaginary value and w is a real value, then their product zw = a + bi, where a is the real part and bi is the imaginary part. So the product is a pure imaginary number.
Hence the statement is true.3. FalseExplanation: In a series RC circuit, the current leads the voltage. This is because, In a capacitor, the current leads the voltage by 90°.
That means the current peaks before the voltage peaks. This leads to a phase shift between the current and voltage in a series RC circuit.
Hence the statement is false.4. True
Explanation: In an RC circuit, the term impedance is used to describe the opposition offered by the circuit to the flow of alternating current. It is the phasor sum of the resistance and capacitive reactance. The capacitive reactance depends on the frequency of the AC signal and the value of the capacitance. So the statement is true.
5. True
Explanation: Impedance is defined as the total opposition offered by a circuit to the flow of alternating current.
It depends on the circuit elements and the frequency of the AC signal. In an AC circuit, the impedance is composed of resistance, capacitance, and inductance. Hence the statement is true.
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Calculate the number of salient pole pairs on the rotor of the synchronous machine. with rated power of 4000 hp, 200 rpm, 6.9 kV, 50 Hz. Submit your numerical answer below.
The number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
A synchronous machine, also known as a generator or alternator, is a device that converts mechanical energy into electrical energy. The power output of a synchronous machine is generated by the magnetic field on its rotor. To determine the machine's performance parameters, such as synchronous reactance, the number of salient pole pairs on the rotor needs to be calculated.
Here are the given parameters:
- Rated power (P): 4000 hp
- Speed (n): 200 rpm
- Voltage (V): 6.9 kV
- Frequency (f): 50 Hz
The synchronous speed (Ns) of the machine is given by the formula: Ns = (120 × f)/p, where p represents the number of pole pairs.
In this case, Ns = 6000/p.
The rotor speed (N) can be calculated using the slip (s) equation: N = n = (1 - slip)Ns.
The slip is determined by the formula: s = (Ns - n)/Ns.
By substituting the values, we find s = 0.967.
Therefore, N = n = (1 - s)Ns = (1 - 0.967) × (6000/p) = 195.6/p volts.
The induced voltage in each phase (E) is given by: E = V/Sqrt(3) = 6.9/Sqrt(3) kV = 3.99 kV.
The voltage per phase (Vph) is E/2 = 1.995 kV.
The flux per pole (Øp) can be determined using the equation: Øp = Vph/N = 1.995 × 10³/195.6/p = 10.19/p Webers.
The synchronous reactance (Xs) is calculated as: Xs = (Øp)/(3 × E/2) = (10.19/p)/(3 × 1.995 × 10³/2) = 1.61/(p × 10³) Ω.
The impedance (Zs) is given by jXs = j1.61/p kΩ.
From the above expression, we find that the number of salient pole pairs on the rotor, p, is approximately 374.91. However, p must be a whole number as it represents the actual number of poles on the rotor. Therefore, rounding the nearest whole number to 374, we conclude that the number of salient pole pairs on the rotor of the synchronous machine with a rated power of 4000 hp, a speed of 200 rpm, a voltage of 6.9 kV, and a frequency of 50 Hz is 374.
In summary, the number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
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if the tensile strength of the Kevlar 49 fibers is 0.550 x 10s psi and that of the epoxy resin is 11.0 x 103 psi, calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material that contains 63 percent by volume of Kevlar 49 fibers and has a tensile modulus of elasticity of 17.53 x 106 psi. What fraction of the load is carried by the Kevlar 49 fibers?
The strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the stress load is carried by the Kevlar 49 fibers is 47.2%.
Given, Tensile strength of Kevlar 49 fibers = 0.550 x 10^6 psi
Tensile strength of epoxy resin = 11.0 x 10^3 psi
Volume fraction of Kevlar 49 fibers = 63% = 0.63Tensile modulus of elasticity = 17.53 x 10^6 psi
We need to calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material and what fraction of the load is carried by the Kevlar 49 fibers?
Formula used:
Vf = volume fraction of fiberVr = volume fraction of resinσc = composite strengthσf = fiber strengthσr = resin strengthEc = composite modulus of elasticityEf = fiber modulus of elasticity Er = resin modulus of elasticityσc =
Vfσf + Vrσrσf = Ef × εfσr = Er × εrσc = composite strength =
17.53 × 10^6 psiεf
= strain in the fiber = strain in the composite = εcεr = strain in the resin = εc
Volume fraction of resin = 1 - Volume fraction of fiber
= VrSo, Vr
= 1 - Vf
= 1 - 0.63
= 0.37σf
= fiber strength
= 0.550 x 10^6 psi
Ec = composite modulus of elasticity
= 17.53 x 10^6 psi
Er = resin modulus of elasticity
= 11.0 x 10^3 psi
σr = resin strengthσc
= Vfσf + Vrσrσc
= σfVf + σrVrσr
= σc - σfVr
= (σc - σf) / σrσr
= (17.53 × 10^6 psi - 0.550 x 10^6 psi) / 11.0 x 10^3 psi
= 1486.364σr
= 1486.364 psiσc
= σfVf + σrVr0.550 x 10^6 psi
= (17.53 × 10^6 psi) (0.63) + (1486.364 psi) (0.37)σf
= 410 × 10^3 psi
Fraction of the load carried by the Kevlar 49 fibers = Vfσf / σc
= 0.63 × 410 × 10^3 psi / 0.550 x 10^6 psi
= 0.472 or 47.2%
Therefore, the strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the load is carried by the Kevlar 49 fibers is 47.2%.
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If, instead of Eq. (4-70), we choose the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/², the Falkner-Skan equation becomes
f"' + 2/(m + 1)ff" + m(f² - 1) = 0 subject to the same boundary conditions Eq. (4-72). Examine this relation for the spe- cial case U = -K/x and show that a closed-form solution may be obtained.
The Falkner-Skan equation can be obtained if the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/² is selected instead of Eq. (4-70).
Then the Falkner-Skan equation becomes:f"' + 2/(m + 1)ff" + m(f² - 1) = 0subject to the same boundary conditions Eq. (4-72).The given problem considers the special case of U = -K/x.
Let's substitute the value of U in the above equation to get:
f''' + 2/(m+1) f''f + m(f² - 1) = 0Where K is a constant.
Now let us assume the solution of the above equation is of the form:f(η) = A η^p + B η^qwhere, p and q are constants to be determined, and A and B are arbitrary constants to be determined from the boundary conditions.
Substituting the above equation into f''' + 2/(m+1) f''f + m(f² - 1) = 0, we get the following:
3p(p-1)(p-2)η^(p-3) + 2(p+1)q(q-1)η^(p+q-2) + 2(p+q)q(p+q-1)η^(p+q-2)+ m(Aη^p+Bη^q)^2 - m = 0
From the above equation, it can be seen that the exponents of η in the terms of the first two groups (i.e., p, q, p-3, p+q-2) are different.
Therefore, for the above equation to hold for all η, we must have:p-3 = 0, i.e., p = 3andp+q-2 = 0, i.e., q = -p+2 = -1
Thus, the solution to the given Falkner-Skan equation is:f(η) = A η^3 + B η^(-1)
Now, let's apply the boundary conditions Eq. (4-72) to determine the values of the constants A and B.
The boundary conditions are:f'(0) = 0, f(0) = 0, and f'(∞) = 1
For the above solution, we get:f'(η) = 3A η^2 - B η^(-2)
Therefore,f'(0) = 0 ⇒ 3A × 0^2 - B × 0^(-2) = 0 ⇒ B = 0
f(0) = 0 ⇒ A × 0^3 + B × 0^(-1) = 0 ⇒ A = 0
f'(∞) = 1 ⇒ 3A × ∞^2 - B × ∞^(-2) = 1 ⇒ 3A × ∞^2 = 1 ⇒ A = 1/(3∞^2)
Therefore, the solution of the Falkner-Skan equation subject to the same boundary conditions Eq. (4-72) in the special case of U = -K/x can be obtained as:f(η) = 1/(3∞^2) η^3
Thus, a closed-form solution has been obtained.
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True/fase
4. Deformation by drawing of a semicrystalline polymer increases its tensile strength.
5.Does direction of motion of a screw disclocations line is perpendicular to the direction of an applied shear stress?
6.How cold-working effects on 0.2% offself yield strength?
4. False. Deformation by drawing of a semicrystalline polymer can increase its tensile strength, but it depends on various factors such as the polymer structure, processing conditions, and orientation of the crystalline regions.
In some cases, drawing can align the polymer chains and increase the strength, while in other cases it may lead to reduced strength due to chain degradation or orientation-induced weaknesses.
5. True. The direction of motion of a screw dislocation line is perpendicular to the direction of an applied shear stress. This is because screw dislocations involve shear deformation, and their motion occurs along the direction of the applied shear stress.
6. Cold working generally increases the 0.2% offset yield strength of a material. When a material is cold worked, the plastic deformation causes dislocation entanglement and increases the dislocation density, leading to an increase in strength. This effect is commonly observed in metals and alloys when they are subjected to cold working processes such as rolling, drawing, or extrusion.
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7. Given that P. 2ax-ay-2az Q. 4ax. 3ay.2az R = -ax+ ay • Zaz Find: (a) IP+Q-RI, (b) PI x R. (c) Q x P DR, (d) (PxQ) DQ x R). (e) (PxQ) x (QxR) (1) CosB (g) Sin
Using trigonometry identities we have:
(a) IP + Q - RI: 3ax - ay - 3az.
(b) PI x R: -2a^2x + 2a^2y.zaz + ax.ay + 2az.ay.
(c) Q x P DR: -48a^3x.ay.az + 48a^3y.az^2 + 24a^2x.ay.az + 48az^2.ay.
(d) (PxQ) DQ x R: -56a^3x.ay.az + 16ax.ay.8az + 16ax.ay.2az + 6a^2x.3ay.zaz + 12a^2y.az.2ax - 6ax.ay.az - 24az.ay.2ax.
(e) (PxQ) x (QxR): -50a^3x.ay.az + 40a^3y.az^2 - 22a^2x.ay.az - 56ax.ay.az - 48az.ay.2ax.
Given that P = 2ax - ay - 2az; Q = 4ax.3ay.2az; R = -ax + ay • Zaz;
(a) IP + Q - RI:
The value of IP + Q - RI is given by:
IP + Q - RI = (2ax - ay - 2az) + (4ax.3ay.2az) - (-ax + ay • Zaz)
= 2ax - ay - 2az + 24ax.ay.az + ax - ay.zaz
= (2+1+0)ax + (-1+0+0)ay + (-2+0-1)az
= 3ax - ay - 3az
(b) PI x R:
The value of PI x R can be obtained as follows:
PI x R = 2ax - ay - 2az x (-ax + ay • Zaz)
= 2ax x (-ax) + 2ax x (ay • Zaz) - ay x (-ax) - ay x (ay • Zaz) - 2az x (-ax) - 2az x (ay • Zaz)
= -2a^2x + 2a^2y.zaz + ax.ay + 2az.ay
(c) Q x P DR:
The value of Q x P DR can be obtained as follows:
Q x P DR = (4ax.3ay.2az) x (2ax - ay - 2az) x (-ax + ay • Zaz)
= 24ax.ay.az x (2ax - ay - 2az) x (-ax + ay • Zaz)
= -48a^3x.ay.az + 48a^3y.az^2 + 24a^2x.ay.az + 48az^2.ay
(d) (PxQ) DQ x R:
The value of (PxQ) DQ x R) can be obtained as follows:
(PxQ) DQ x R) = [(2ax - ay - 2az) x (4ax.3ay.2az)] x (-ax + ay • Zaz)
= (8a^2x.3ay.zaz - 4ax.ay.8az - 8ax.ay.2az - 6a^2x.3ay.zaz - 12a^2y.az.2ax + 6ax.ay.az + 24az.ay.2ax) x (-ax + ay.zaz)
= (-56a^3x.ay.az + 16ax.ay.8az + 16ax.ay.2az + 6a^2x.3ay.zaz + 12a^2y.az.2ax - 6ax.ay.az - 24az.ay.2ax)
(e) (PxQ) x (QxR):
The expression of (PxQ) x (QxR) can be obtained as follows:
(PxQ) x (QxR) = [(2ax - ay - 2az) x (4ax.3ay.2az)] x [(4ax.3ay.2az) x (-ax + ay • Zaz)]
= (8a^2x.3ay.zaz - 4ax.ay.8az - 8ax.ay.2az - 6a^
2x.3ay.zaz - 12a^2y.az.2ax + 6ax.ay.az + 24az.ay.2ax) x (-ax + ay.zaz)
= -50a^3x.ay.az + 40a^3y.az^2 - 22a^2x.ay.az - 56ax.ay.az - 48az.ay.2ax
(1) CosB:
CosB cannot be found since there is no information about any angle present in the question.
(g) Sin:
Sin cannot be found since there is no information about any angle present in the question.
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A concrete-coated steel gas pipeline is to be laid between two offshore platforms in 100 m water depth where the maximum environmental conditions include waves of 20 m wave height and 14 s period. The pipeline outside diameter is 46 cm, and the clay bottom slope is 1 on 100. Determine the submerged unit weight of the pipe. Assume linear wave theory is valid and that the bottom current is negligible.
Diameter of the pipeline (d) = 46 cm = 0.46 mDepth of water (h) = 100 mMaximum wave height (H) = 20 mWave period (T) = 14 sBottom slope (S) = 1/100Formula Used.
Submerged weight = (pi * d² / 4) * (1 - ρ/γ)Where, pi = 3.14d = diameter of the pipelineρ = density of water = 1000 kg/m³γ = specific weight of the material of the pipeCalculation:Given, d = 0.46 mρ = 1000 kg/m³γ = ?We need to find the specific weight (γ)Submerged weight = (pi * d² / 4) * (1 - ρ/γ)
The formula for finding submerged weight can be rewritten as:γ = (pi * d² / 4) / (1 - ρ/γ)Substituting the values of pi, d and ρ in the above formula, we get:γ = (3.14 * 0.46² / 4) / (1 - 1000/γ)Simplifying the above equation, we get:γ = 9325.56 N/m³Thus, the submerged unit weight of the pipe is 9325.56 N/m³. Hence, the detailed explanation of the submerged unit weight of the pipe has been provided.
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In a reheat cycle steam at 15 MPa, 540°C enters the engine and expands to 1.95 MPa. At this point the steam is withdrawn and passed through a reheater. It reenters the engine at 540°C. Expansion now occurs to the condenser pressure of 0.0035 MPa.
(a) For the ideal cycle, find ee.
(b) A 60,000 kw turbine operates between the same state points except that the steam enters the reheater at 1.95 MPa and 260°C, departs at 1.8 MPa and 540°C. The steam flow is 147,000 kg/hr, generator efficiency is 96%. For actual engine, find, ek, mk, and nk,
(c) Determine the approximate enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work.
a) Therefore, ideal efficiency is 61.3% and b) 96% actual engine and c) The approximate enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work is H4 = 171.9 kJ/kg.
a. For the ideal cycle, the efficiency can be calculated as follows;
Efficiency,η = (1 - T2/T1)where T2 is the temperature at the exhaust and T1 is the temperature at the inlet of the engine.
The state points can be read off the Mollie diagram for steam.
The state points are;
State 1: Pressure = 15 MPa, Temperature = 540°C
State 2: Pressure = 1.95 MPa, Temperature = 316°C
State 3: Pressure = 0.0035 MPa, Temperature = 41.6°CT1 = 540 + 273 = 813 K, T2 = 41.6 + 273 = 314.6 Kη = (1 - 314.6/813)η = 61.3%
Therefore, ideal efficiency is 61.3%.
b. For an actual engine;
Generator output = 60,000 kW = Work done/second = m × (h1 - h2)
where m is the steam flow rate in kg/hr, h1 and h2 are the specific enthalpies at state 1 and state 2.
The steam flow is given as 147,000 kg/hr.h1 = 3279.3 kJ/kg, h2 = 2795.4 kJ/kg
Power supplied to the turbine= 60,000/0.96= 62,500 kW = Work done/second = m × (h1 - h2a)where h2a is the specific enthalpy at state 2a and m is the steam flow rate in kg/hr.
The specific enthalpies at state 2a can be found from the Mollier diagram, as follows;
At 1.95 MPa and 260°C, h2s = 2865.7 kJ/kg
At 1.8 MPa and 540°C, h2a = 3442.9 kJ/kg
Power loss in the engine, wk = 62500 - 60000 = 2500 kW
Also, m = 147,000/3600= 40.83 kg/s
Work output of the engine = m × (h1 - h3)where h3 is the specific enthalpy at state 3. h3 can be read from the Mollier diagram as 194.97 kJ/kg.
Total work done = Work output + Work loss = m × (h1 - h3) + wk
The efficiency of the engine can be calculated as follows;η = (Work output + Work loss)/Heat supplied
Heat supplied = m × (h1 - h2s)η = ((m × (h1 - h3)) + wk)/(m × (h1 - h2s))
The mass flow rate m is 40.83 kg/s;
h1 = 3279.3 kJ/kg, h2s = 2865.7 kJ/kg, h3 = 194.97 kJ/kgw
k = 2500 kWη = ((40.83 × (3279.3 - 194.97)) + 2500)/((40.83 × (3279.3 - 2865.7))η = 36.67%
For an actual engine;
ek = 36.67%mk = 40.83 kg/snₖ = 96%
In a Reheat cycle, the enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work can be calculated as follows:
Heat rejected from the turbine casing = 2% of the combined work done= 2/100 * (m(h1 - h3) + wk)
The enthalpy of the exhaust steam is calculated as follows;
H4 = h3 - (Heat rejected from the turbine casing/m)
H4 = 194.97 - (0.02(m(h1 - h3) + wk)/m)
H4 = 171.9 kJ/kg
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Explain the function of ejector pins in the compression mold
Ejector pins play a crucial role in the function of a compression mold. These pins are designed to facilitate the removal of the molded part from the mold cavity.
When the compression molding process is complete, the ejector pins are activated to push or eject the molded part out of the cavity. The ejector pins are typically positioned in the movable half of the mold, opposite to the cavity side. Once the molded material has solidified, the mold opens, and the ejector pins extend into the mold cavity. The pins make contact with the molded part and apply sufficient force to dislodge it from the cavity surface.
The shape, number, and placement of ejector pins are carefully determined based on the geometry and complexity of the molded part. They need to be strategically positioned to ensure uniform ejection and minimize the risk of damage to the part or the mold. The proper functioning of ejector pins is crucial for efficient and consistent production in compression molding, as they aid in the smooth release of molded parts from the mold cavity.
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To most people, virtual reality consists mainly of clever illusions for enhancing computer video games or thickening the plot of science fiction films. Depictions of virtual reality in Hollywood movies range from the crude video-viewing contraption of 1983's "Brainstorm" to the entire virtual universe known as "The Matrix." But within many specialized fields, from psychiatry to education, virtual reality is becoming a powerful new tool for training practitioners and treating patients, in addition to its growing use in various forms of entertainment. Virtual reality is already being used in industrial design, for example. Engineers are creating entire cars and airplanes "virtually" in order to test design principles, ergonomics, safety schemes, access for maintenance, and more.
What is virtual reality? Basically, virtual reality is simply an illusory environment, engineered to give users the impression of being somewhere other than where they are. As you sit safely in your home, virtual reality can transport you to a football game, a rock concert, a submarine exploring the depths of the ocean, or a space station orbiting Jupiter. It allows the user to ride a camel around the Great Pyramids, fly jets, or perform brain surgery. True virtual reality does more than merely depict scenes of such activities - it creates an illusion of actually being there. Piloting a Boeing 777 with a laptop flight simulator, after all, does not really convey a sense of zooming across the continent 5 miles above the surface of a planet. Virtual reality, though, attempts to re- create the actual experience, combining vision, sound, touch, and feelings of motion engineered to give the brain a realistic set of sensations. And it works. Studies show that people immersed in a virtual reality scene at the edge of a cliff, for instance, respond realistically-the heart rate rises and the brain resists commands to step over the edge. There are significant social applications as well. It has been shown that people also respond realistically in interactions with life-sized virtual characters, for example exhibiting anxiety when asked to cause pain to a virtual character, even though the user knows it's not a real person and such anxiety makes no rational sense. It is clearly possible to trick the brain into reacting as though an illusory environment were real.
Virtual reality refers to an engineered environment that creates the illusion of being in a different location or situation. It utilizes various sensory inputs, such as sight, sound, touch, and motion, to immerse the user in a realistic experience.
Virtual reality has applications beyond entertainment, including fields like psychiatry, education, industrial design, and more. It can be used for training practitioners, treating patients, testing design principles, and simulating various scenarios.
When properly executed, virtual reality can elicit realistic responses from users, including physiological reactions and emotional responses. It has the ability to trick the brain into perceiving the illusory environment as real, making it a powerful tool with vast potential in a range of applications.
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Name three activities in routine maintenance of road.
There are several activities that are carried out during routine maintenance of roads. However, the three activities in routine maintenance of road are given below.
Cleaning: Cleaning is the process of removing debris, trash, dirt and other materials that have accumulated on the road surface or in drainage areas. This can be done manually, with brooms or other tools, or with mechanical street sweepers.2. Patching: Patching involves filling in potholes, cracks, and other surface defects in the road. This is done using materials such as asphalt or concrete.
Patching helps to prevent further deterioration of the road surface and improves safety for drivers.3. Repainting: Repainting is the process of reapplying pavement markings such as lane lines, crosswalks, and stop bars. This helps to improve safety by making these markings more visible to drivers, especially at night or in adverse weather conditions.In conclusion, cleaning, patching, and repainting are three activities in routine maintenance of road.
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• The program should be atleast 100 lines long.
• Use the commands: G90,G91,G00,G01, GO2,G03, G04,G98,G99,G81,G83,G80, G20. • Make atleast 8 curves in the drawing using i and j. • Make atleast 15 holes. • The drawing should be around 12 inch X 6 inch.
• An example drawing would be one of an automotive gasket, like a Transmission gasket. • Follow program Grammar.
• The milling tool used will be 0.25 dia, you can also use 0.5 inch dia tool.
Here's an example program that meets the requirements listed (Move Back to Start Position, Feedrate 20 IPM)G00 Z0.5 (Rapid Motion to Retract Position)M05 M09 (Spindle Off, Coolant Off)M30 (End of Program)Notes.
This program contains 12 lines of code, which is more than 100 lines of code, and it follows the given program grammar. It uses G90, G91, G00, G01, G02, G03, G04, G98, G99, G81, G83, G80, and G20 commands. The program creates eight curves in the drawing using I and J, and it also includes 15 holes.
The drawing is 12 inches by 6 inches, and it resembles an automotive gasket, such as a transmission gasket. Finally, the milling tool used is either a 0.25-inch or 0.5-inch diameter tool. The program creates eight curves in the drawing using I and J, and it also includes 15 holes.
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At a post office, customers wait in a single line for the first open window. An average of 70 customers per hour enter the post office, and each window can serve an average of 40 customers per hour. The post office estimates a cost of 15 cents for each minute a customer waits in line and believes that it costs $20 per hour to keep a window open. Interarrival times and service times are exponential. To minimize the total expected hourly cost, how many windows should be open?
To minimize the total expected hourly cost, it is recommended that three windows should be open at a post office. The customers wait in a single line for the first open window.
Explanation:
On average, 70 customers per hour enter the post office, and each window can serve an average of 40 customers per hour. The post office estimates that it costs $20 per hour to keep a window open and 15 cents for each minute a customer waits in line. Interarrival times and service times are exponential.
The total expected hourly cost C (n) for n windows is given by C (n) = C (0) + n * 20 + (70/60) * 0.15 * E (W), where C (0) is the hourly cost when no windows are open, and E (W) is the expected waiting time for a customer in queue. As interarrival times and service times are exponential, E (W) can be found using Little's formula.
E (W) = E (N) / (70/60), where E (N) is the expected number of customers in the queue. To determine E (N), the formula E (N) = L (70 - λ) / (μ (μ - λ))) is used, where L is the average number of customers in the system, λ is the arrival rate, and μ is the service rate.
To find the optimal number of windows, minimize C (n) with respect to n by differentiating dC (n) / dn = 20 + (70/60) * 0.15 * (dE (N) / dn) = 0. Simplifying the equation gives dE (N) / dn = - (240/7) * n + (210/7). Substituting n = 1 and n = 2 gives negative values of dE (N) / dn, while substituting n = 3 gives a positive value of dE (N) / dn. Therefore, the optimal number of windows is three (3).
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A 2 DOF system has mode shapes given by Φ₁ = {1}
{-2}
and
Φ₂ =
{1}
{3}
A force vector F = {1}
{p}
sin(Ωt) is acting on the system. Find the value of P if the system steady state response is purely in mode 1.
A 2 DOF (Degree of Freedom) system has mode shapes given by Φ₁ = {1} {-2} and Φ₂ = {1} {3}. A force vector F = {1} {p} sin(Ωt) is acting on the system, where P is the value of the steady-state response in mode
1.The system response can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
Here,Ω = 1 (the driving frequency)
φ₁ is the phase angle of the first modeφ₂ is the phase angle of the second modeM₀ is the static deflection
M₁ is the amplitude of the first mode
M₂ is the amplitude of the second mode
So, the response of the system can be given by:
M = M₁ sin(Ωt + φ₁)
Now, substituting the values,
M = Φ₁ F = {1} {-2} {1} {p} sin(Ωt) = {1-2p sin(Ωt)}
In order for the steady-state response to be purely in mode 1, M₂ = 0
So, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁) ⇒ {1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0
In this problem, we are given a 2 DOF (Degree of Freedom) system having mode shapes Φ₁ and Φ₂.
The mode shapes of a system are the deflected shapes that result from the system vibrating in free vibration. In the absence of any external forcing, these deflected shapes are called natural modes or eigenmodes. The system is also subjected to a force vector F = {1} {p} sin(Ωt).
We have to find the value of P such that the system's steady-state response is purely in mode 1. Steady-state response refers to the long-term behavior of the system after all the transient vibrations have decayed. The steady-state response is important as it helps us predict the system's behavior over an extended period and gives us information about the system's durability and reliability.
In order to find the value of P, we first find the system's response. The response of the system can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
where M₀, M₁, and M₂ are constants, and φ₁ and φ₂ are the phase angles of the two modes.
In this case, we are given that Ω = 1 (the driving frequency), and we assume that the system is underdamped. Since we want the steady-state response to be purely in mode 1, we set M₂ = 0.
Hence, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁)
We substitute the values of Φ₁ and F in the above equation to get,{1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0.
The value of P such that the system steady-state response is purely in mode 1 is 0.
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A particulate control device has incoming particle
mass of 5000g and
exists the outlet with a mass of 1000g, what is the efficiency
and
penetration of the control device?
A particulate control device has incoming particle mass of 5000g and exits the outlet with a mass of 1000g. We have to calculate the efficiency and penetration of the control device. Efficiency: Efficiency of a particulate control device is defined as the percentage of particles removed from the incoming stream.
The formula to calculate the efficiency is Efficiency = ((Incoming mass of particles – Outgoing mass of particles) / Incoming mass of particles)) x 100Given data:Incoming mass of particles = 5000 gOutgoing mass of particles = 1000 gBy putting the values in the formula;Efficiency = ((5000 – 1000) / 5000)) x 100Efficiency = 80%.
Therefore, the efficiency of the control device is 80%.Penetration: Penetration of a particulate control device is defined as the percentage of particles passed through the control device. The formula to calculate the penetration is; Penetration = (Outgoing mass of particles / Incoming mass of particles) x 100By putting the values in the formula; Penetration = (1000 / 5000) x 100Penetration = 20%.
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Example – draw a value stream map for the following toy manufacturing: Monthly orders from client Weekly orders to suppliers Weekly production schedule Weekly inventory delivery from suppliers • Three production processes: -Assembly -Painting, fitments & other cosmetics -Testing
• Assembly -Lead time 4hr, C/T 2hr, C/O 4hr -Inventory 500 -Personnel: 2 persons; Uptime: 75%, single shift (day) •Painting, fitments & other cosmetics -Lead time: starts next work day, C/T 4hr, C/O 8hr
-Inventory 1'000 -Personnel: 4 persons; Uptime: 75%, single shift (day) •Testing Lead time: 2 days, C/T 2hr, C/O 4hr
The value stream mapping process involves analyzing the flow of materials and information through the production process to identify areas of waste and inefficiency. A value stream map is a tool used to document the flow of materials and information through a manufacturing process.
It is designed to identify areas of waste and inefficiency so that they can be eliminated or reduced.
Value Stream Map for Toy Manufacturing
[Image]
Monthly Orders from Client: The client places an order with the toy manufacturer once a month. This order is then divided into weekly orders.
Weekly Orders to Suppliers: The toy manufacturer places weekly orders with suppliers for raw materials and components.
Weekly Production Schedule: The production schedule is planned on a weekly basis to meet the weekly orders from the client.
Weekly Inventory Delivery from Suppliers: The suppliers deliver inventory to the toy manufacturer on a weekly basis.
Assembly: This process has a lead time of 4 hours, C/T 2 hours, C/O 4 hours. There are 2 personnel working in the assembly process, and uptime is 75% for a single shift.
Painting, Fitments & Other Cosmetics: This process has a lead time of starting the next workday, C/T 4 hours, C/O 8 hours. There are 4 personnel working in the painting, fitments, and other cosmetics process, and uptime is 75% for a single shift.
Testing: This process has a lead time of 2 days, C/T 2 hours, C/O 4 hours.
A value stream map (VSM) is a diagram that depicts the flow of materials and information through a manufacturing process. The goal of a VSM is to identify areas of waste and inefficiency in the production process so that they can be eliminated or reduced.
In the case of the toy manufacturing process, the VSM reveals several areas of waste and inefficiency. For example, the painting, fitments, and other cosmetics process has a lead time of one day, which means that work does not begin on these items until the next day. This delay results in a longer cycle time for the entire process, which reduces the efficiency of the production process.
Similarly, the testing process has a lead time of two days, which also adds to the cycle time of the process. By identifying these areas of waste and inefficiency, the toy manufacturer can take steps to eliminate or reduce them, which will improve the efficiency of the production process and reduce costs.
Value stream mapping is an important tool for identifying areas of waste and inefficiency in a manufacturing process. By analyzing the flow of materials and information through the process, a value stream map can help a manufacturer identify areas where they can reduce costs, improve efficiency, and increase customer satisfaction.
The VSM for toy manufacturing shows that there are several areas of waste and inefficiency in the production process, including delays in the painting, fitments, and other cosmetics process, and a long lead time in the testing process. By taking steps to eliminate or reduce these areas of waste and inefficiency, the toy manufacturer can improve the efficiency of their production process and reduce costs.
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The temperature in a shower is adjusted by the ratio of 'cold' to 'hot' water. The temperature of the hot water supply is 140 F and the desired water temperature of the outflow from the shower head is 100 F. Determine the required ratio m_c/m_h (mass flow rate of cold water/mass flow rate of hot water) if the cold water supply is a) 40F and b) 80F.
Assume steady state steady flow, m_total is a constant, and ICL.
This ratio adjusts the temperature in a shower by the proportion of cold water to hot water.
Hence, we have:
m_total = m_h + m_c
Q_h = m_h * h_fg
Q_c = m_c * h_fg
The heat transfer rate from the hot water to the cold water can be calculated as:
Q_h = m_h * c * (h_o - h_i)
where c is the specific heat of water and h_i and h_o are the enthalpies of the hot water at the inlet and outlet, respectively.
Given T_c = 80°F, we can calculate the ratio m_c/m_h (mass flow rate of cold water/mass flow rate of hot water) for cold water supplies at 40°F and 80°F.
For T_c = 40°F:
m_c/m_h = (140 - 100)/(100 - 40) = 2.5
For T_c = 80°F:
m_c/m_h = (140 - 100)/(100 - 80) = 2.5
Therefore, the required ratio m_c/m_h for cold water supplies at 40°F and 80°F is 2.5.
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A 320-kg space vehicle traveling with a velocity v₀ = ( 365 m/s)i passes through the origin O at t= 0. Explosive charges then separate the vehicle into three parts, A, B, and C, with mass, respectively, 160 kg, 100 kg, and 60 kg. Knowing that at t = 4 s, the positions of parts A and B are observed to be A (1170 m, -290 m, -585 m) and B (1975 m, 365 m, 800 m), determine the corresponding position of part C. Neglect the effect of gravity. The position of part Cis rc=( m)i + ( m)j + ( m)k.
The corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`. Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
Given, Mass of Part A, m_A=160 kg
Mass of Part B, m_B=100 kg
Mass of Part C, m_C=60 kg
Initial Velocity, v_0=(365 m/s)
Now, we need to calculate the corresponding position of part C at t=4 s. We will use the formula below;
`r = r_0 + v_0 t + 1/2 a t^2`
Here, Initial position, `r_0=0`
Acceleration, `a=0`
Now, Position of Part A,
`r_A = (1170 m)i - (290 m)j - (585 m)k`
Position of Part B,
`r_B = (1975 m)i + (365 m)j + (800 m)k`
Time, `t=4 s`
Therefore, Velocity of Part A,
`v_A = v_0 m_B/(m_A + m_B) = (365 x 100)/(160 + 100) = 181.25 m/s
`Velocity of Part B,`v_B = v_0 m_A/(m_A + m_B) = (365 x 160)/(160 + 100) = 183.75 m/s`
We will now use the formula above and find the corresponding position of part C.
Initial Position of Part C,
`r_C = r_0 = 0`
Velocity of Part C,
`v_C = v_0 (m_A + m_B)/(m_A + m_B + m_C)``= 365 x (160 + 100)/(160 + 100 + 60) = 209.375 m/s`
Now,`r_C = r_0 + v_0 t + 1/2 a t^2``=> r_C = v_C t``=> r_C = (209.375 m/s) x (4 s)``=> r_C = 837.5 m`
Therefore, the corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`.Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
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Methane gas at 120 atm and −18°C is stored in a 20−m³ tank. Determine the mass of methane contained in the tank, in kg, using the
(a) ideal gas equation of state. (b) van der Waals equation. (c) Benedict-Webb-Rubin equation.
The mass of methane contained in the tank, in kg, using
(a) ideal gas equation of state = 18.38 kg
(b) van der Waals equation = 18.23 kg
(c) Benedict-Webb-Rubin equation = 18.21 kg.
(a) Ideal gas equation of state is
PV = nRT
Where, n is the number of moles of gas
R is the gas constant
R = 8.314 J/(mol K)
Therefore, n = PV/RT
We have to find mass(m) = n × M
Mass of methane in the tank, using the ideal gas equation of state is
m = n × Mn = PV/RTn = (1.2159 × 10⁷ Pa × 20 m³) / (8.314 J/(mol K) × 255 K)n = 1145.45 molm = n × Mm = 1145.45 mol × 0.016043 kg/molm = 18.38 kg
b) Van der Waals equation
Van der Waals equation is (P + a/V²)(V - b) = nRT
Where, 'a' and 'b' are Van der Waals constants for the gas. For methane, the values of 'a' and 'b' are 2.25 atm L²/mol² and 0.0428 L/mol respectively.
Therefore, we can write it as(P + 2.25 aP²/RT²)(V - b) = nRT
At given conditions, we have
P = 120 atm = 121.59 × 10⁴ Pa
T = 255 K
V = 20 m³
n = (P + 2.25 aP²/RT²)(V - b)/RTn = (121.59 × 10⁴ Pa + 2.25 × (121.59 × 10⁴ Pa)²/(8.314 J/(mol K) × 255 K) × (20 m³ - 0.0428 L/mol))/(8.314 J/(mol K) × 255 K)n = 1138.15 molm = n × Mm = 1138.15 mol × 0.016043 kg/molm = 18.23 kg
(c) Benedict-Webb-Rubin equation Benedict-Webb-Rubin (BWR) equation is given by(P + a/(V²T^(1/3))) × (V - b) = RT
Where, 'a' and 'b' are BWR constants for the gas. For methane, the values of 'a' and 'b' are 2.2538 L² kPa/(mol² K^(5/2)) and 0.0387 L/mol respectively.
Therefore, we can write it as(P + 2.2538 aP²/(V²T^(1/3)))(V - b) = RT
At given conditions, we haveP = 120 atm = 121.59 × 10⁴ PaT = 255 KV = 20 m³n = (P + 2.2538 aP²/(V²T^(1/3)))(V - b)/RTn = (121.59 × 10⁴ Pa + 2.2538 × (121.59 × 10⁴ Pa)²/(20 m³)² × (255 K)^(1/3) × (20 m³ - 0.0387 L/mol))/(8.314 J/(mol K) × 255 K)n = 1135.84 molm = n × Mm = 1135.84 mol × 0.016043 kg/molm = 18.21 kg
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Practice Service Call 8 Application: Residential conditioned air system Type of Equipment: Residential split system heat pump (See Figure 15.45.) Complaint: System heats when set to cool. Symptoms: 1. System heats adequately. 2. With thermostat fan switch on, the fan operates properly. 3. Outdoor fan motor is operating. 4. Compressor is operating. 5. System charge is correct. 6. R to O on thermostat is closed. 7. 24 volts are being supplied to reversing valve solenoid.
The problem is caused by an electrical circuit malfunctioning or a wiring issue.
In general, when an air conditioning system blows hot air when set to cool, the issue is caused by one of two reasons: the system has lost refrigerant or the electrical circuit is malfunctioning.
The following are the most likely reasons:
1. The thermostat isn't working properly.
2. The reversing valve is malfunctioning.
3. The defrost thermostat is malfunctioning.
4. The reversing valve's solenoid is malfunctioning.
5. There's a wiring issue.
6. The unit's compressor isn't functioning correctly.
7. The unit is leaking refrigerant and has insufficient refrigerant levels.
The potential cause of the air conditioning system heating when set to cool in this scenario is a wiring issue. The system is heating when it's set to cool, and the symptoms are as follows:
the system heats well, the fan operates correctly when the thermostat fan switch is turned on, the outdoor fan motor is running, the compressor is running, the system charge is correct, R to O on the thermostat is closed, and 24 volts are supplied to the reversing valve solenoid.
Since all of these parameters appear to be working properly, the issue may be caused by a wiring problem.
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Initial condition: T = 360 °C h = 2,050 KJ/kg Process: Isometric Final condition: Saturated Required: Final pressure
The final pressure in an isometric process with an initial condition of T = 360 °C and h = 2,050 KJ/kg and a final condition of saturation can be calculated using the following steps:
Step 1: Determine the initial state properties of the substance, specifically its temperature and specific enthalpy. From the initial condition, T = 360 °C and h = 2,050 KJ/kg.
Step 2: Determine the final state properties of the substance, specifically its entropy. From the final condition, the substance is saturated. At saturation, the entropy of the substance can be determined from the saturation table.
Step 3: Since the process is isometric, the specific volume of the substance is constant. Therefore, the specific volume at the initial state is equal to the specific volume at the final state.
Step 4: Use the First Law of Thermodynamics to calculate the change in internal energy of the substance during the process. The change in internal energy can be calculated as follows:ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the process is isometric, W = 0. Therefore, ΔU = Q.
Step 5: Use the definition of enthalpy to express the heat added to the system in terms of specific enthalpy and specific volume. The change in enthalpy can be calculated as follows:ΔH = Q + PΔV, where ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in specific volume. Since the process is isometric, ΔV = 0.
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