The given data are 58-hp, three-phase induction motor, 220 V, 60 Hz, and single-phase system. First, calculate the equivalent circuit values of the motor, which are required to determine the additional capacitance for starting performance.
The equivalent circuit values of the motor per phase are as follows: R1 = 0.03 Ω, R2 = 0.012 Ω, X1 = 0.08 Ω, and X2 = 0.06 Ω.
The total impedance of the motor is Z = √(R² + X²) = √(0.08² + 0.06²) = 0.1 Ω
The starting torque of the motor is proportional to the square of the voltage per phase. Hence, to improve the starting performance, the capacitance should be increased.
The equation for calculating the capacitance is C2 = 3 * (Ist / Vph) * X2 * 10^6 ,where Ist is the rated current of the motor at full load, and Vph is the rated voltage per phase.
For a 58-hp, three-phase motor, Ist is approximately 110 A. In a single-phase system, the current per phase is √(2) times the current in a three-phase system.
The Ist in a single-phase system is approximately Ist(single-phase) = √(2) * Ist(three-phase) = √(2) * 110 = 155 A.
The additional capacitance required for best starting performance is C2 = 3 * (155 / 220) * 0.06 * 10^6 = 1272 µF.
The additional capacitance required for best starting performance is 1272 µF.
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The output of a thermistor is highly nonlinear with temperature, and there is often a benefit to linearizing the output through appropriate circuit, whether active or passive. In this example, we examine the output of an initially balanced bridge circuit in which one of the arms contains a thermistor. Consider a Wheatstone bridge as shown in Figure 8.8, but replace the RTD with a thermistor having a value of R = 10,000 22 with B = 3680 K. Here, we examine the output of the circuit over two temperature ranges: (a) 25–325°C and (b) 25–75°C. KNOWN A Wheatstone bridge where R2 = R3 = R4 = 10,000 22 and where R, is a thermistor. FIND The output of the bridge circuit as a function of temperature.
Step 1: Let RT be the resistance of the thermistor at temperature T°C.RT = R₀exp(B/T)where R₀ = 10,000 Ω, B = 3680 K and T is the temperature in °C.
Step 2: Calculate the equivalent resistance of the bridge.The equivalent resistance of the bridge is given by the formula: Req = R₂ + R₄/[R₁ + R₃]The value of the resistors R2 = R3 = R4 = 10,000 Ω.Thus, Req = 10,000 Ω + 10,000 Ω/[10,000 Ω + RT].
Step 3: Calculate the current through the bridge.Using the bridge balance equation, we have:R₂R₄ = R₁R₃exp(β (T - 25))where β = 3680 K, T is the temperature in °C and R1 = RT.
Rearranging the above equation, we have:RT = R₃R₂exp(β (T - 25))/R₁The current flowing through the bridge is given by:I = [Vcc × R₂R₄]/[R₂ + R₄][R₁ + R₃]Where Vcc is the voltage supply.
Step 4: Find the output voltage of the bridge circuit.The output voltage of the bridge is given by:Vout = Vcc [R₄/(R₂ + R₄)] - Vcc [R₁/(R₁ + R₃)]This can be simplified as:Vout = Vcc [R₄/(R₂ + R₄)][R₁ + R₃]/[R₁ + R₃] - Vcc R₁/[R₁ + R₃]Vout = Vcc[R₄(R₁ + R₃) - R₁(R₂ + R₄)]/[(R₁ + R₃)(R₂ + R₄)].
For the range 25°C to 325°C, we can vary the temperature T from 25°C to 325°C in steps of 1°C and repeat steps 1 to 4 to obtain the output voltage of the bridge circuit at each temperature.
Similarly, we can obtain the output voltage of the bridge circuit for the range 25°C to 75°C as well.
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a) Interpret how stability can be determined through Bode Diagram. Provide necessary sketch. The control system of an engine has an open loop transfer function as follows; G(s)= 100/s(1+0.1s)(1+0.2s)
(i) Determine the gain margin and phase margin. (ii) Plot the Bode Diagram on a semi-log paper. (iii) Evaluate the system's stability.
To determine stability using a Bode diagram, we analyze the gain margin and phase margin of the system.
(i) Gain Margin and Phase Margin:
The gain margin is the amount of gain that can be added to the system before it becomes unstable, while the phase margin is the amount of phase lag that can be introduced before the system becomes unstable.
To calculate the gain margin and phase margin, we need to plot the Bode diagram of the given open-loop transfer function.
(ii) Bode Diagram:
The Bode diagram consists of two plots: the magnitude plot and the phase plot.
For the given transfer function G(s) = 100/(s(1+0.1s)(1+0.2s)), we can rewrite it in the form G(s) = K/(s(s+a)(s+b)), where K = 100, a = 0.1, and b = 0.2.
On a semi-logarithmic paper, we plot the magnitude and phase responses of the system against the logarithm of the frequency.
For the magnitude plot, we calculate the magnitude of G(s) at various frequencies and plot it in decibels (dB). The magnitude is given by 20log₁₀(|G(jω)|), where ω is the frequency.
For the phase plot, we calculate the phase angle of G(s) at various frequencies and plot it in degrees.
(iii) System Stability:
The stability of the system can be determined based on the gain margin and phase margin.
If the gain margin is positive, the system is stable.
If the phase margin is positive, the system is stable.
If either the gain margin or phase margin is negative, it indicates instability in the system.
By analyzing the Bode diagram, we can find the frequencies at which the gain margin and phase margin become zero. These frequencies indicate potential points of instability.
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An industrial plant absorbs 500 kW at a line voltage of 480 V with a lagging power factor of 0.8 from a three-phase utility line. The apparent power absorbed is most nearly O a. 625 KVA O b. 500 KVA O c. 400 KVA O d. 480 KVA
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
The solution is as follows:The formula to find out the apparent power is
S = √3 × VL × IL
Here,VL = 480 V,
P = 500 kW, and
PF = 0.8.
For a lagging power factor, the apparent power is always greater than the real power; thus, the value of the apparent power will be greater than 500 kW.
Applying the above formula,
S = √3 × 480 × 625 A= 625 KVA.
So, the most nearly apparent power absorbed is 625 KVA.Answer: The correct option is O a. 625 KVA.
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Efficiency of home furnace can be improved by preheating combustion air using hot flue gas. The flue gas has temperature of Tg = 1000°C, specific heat of c = 1.1 kJ/kg°C and is available at the rate of 12 kg/sec. The combustion air needs to be delivered at the rate of 15 kg/sec, its specific heat is ca 1.01 kJ/kg°C and its temperature is equal to the room temperature, i.e. Tair,in = 20°C. The overall heat transfer coefficient for the heat exchanger is estimated to be U = 80 W/m2°C. (i) Determine size of the heat exchanger (heat transfer surface area A) required to heat the air to Tair,out 600°C assuming that a single pass, cross-flow, unmixed heat exchanger is used. (ii) Determine temperature of flue gases leaving heat exchanger under these conditions. (iii) Will a parallel flow heat exchanger deliver the required performance and if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A? (iv) Will use of a counterflow heat exchanger deliver the required performance and, if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A?
i) The size of the heat exchanger required is approximately 13.5 m².
ii) The temperature of the flue gases leaving the heat exchanger T_flue,out ≈ 311.36°C.
iii) To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
iv) The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
To solve this problem, we can use the energy balance equation for the heat exchanger.
The equation is given by:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
Where:
Q is the heat transfer rate (in watts or joules per second).
m_air is the mass flow rate of combustion air (in kg/s).
c_air is the specific heat of combustion air (in kJ/kg°C).
T_air,in is the inlet temperature of combustion air (in °C).
T_air,out is the desired outlet temperature of combustion air (in °C).
m_flue is the mass flow rate of flue gas (in kg/s).
c_flue is the specific heat of flue gas (in kJ/kg°C).
T_flue,in is the inlet temperature of flue gas (in °C).
T_flue,out is the outlet temperature of flue gas (in °C).
Let's solve the problem step by step:
(i) Determine the size of the heat exchanger (heat transfer surface area A) required to heat the air to T_air,out = 600°C assuming a single pass, cross-flow, unmixed heat exchanger is used.
We can rearrange the energy balance equation to solve for A:
A = Q / (U × ΔT_lm)
Where ΔT_lm is the logarithmic mean temperature difference given by:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = T_flue,in - T_air,out
ΔT2 = T_flue,out - T_air,in
Plugging in the values:
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - 20°C (unknown)
We need to solve for ΔT2 by substituting the values into the energy balance equation:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
15 kg/s × 1.01 kJ/kg°C × (600°C - 20°C) = 12 kg/s × 1.1 kJ/kg°C × (1000°C - T_flue,out)
Simplifying:
9090 kJ/s = 13200 kJ/s - 13.2 kJ/s * T_flue,out
13.2 kJ/s × T_flue,out = 4110 kJ/s
T_flue,out = 311.36°C
Now we can calculate ΔT2:
ΔT2 = T_flue,out - 20°C
ΔT2 = 311.36°C - 20°C
ΔT2 = 291.36°C
Now we can calculate ΔT_lm:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
Finally, we can calculate the required surface area A:
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C × 84.5°C)
A ≈ 13.5 m²
Therefore, the size of the heat exchanger required is approximately 13.5 m².
(ii) Determine the temperature of flue gases leaving the heat exchanger under these conditions.
We already determined the temperature of the flue gases leaving the heat exchanger in part (i): T_flue,out ≈ 311.36°C.
(iii) In a parallel flow heat exchanger, the hot and cold fluids flow in the same direction. The temperature difference between the two fluids decreases along the length of the heat exchanger. In this case, a parallel flow heat exchanger will not deliver the required performance because the outlet temperature of the flue gases is significantly higher than the desired outlet temperature of the combustion air.
To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
(iv) In a counterflow heat exchanger, the hot and cold fluids flow in opposite directions. This arrangement allows for better heat transfer and can achieve a higher temperature difference between the two fluids. A counterflow heat exchanger can deliver the required performance in this case.
To determine if the size of the heat exchanger will be reduced or increased, we need to recalculate the required surface area A using the new ΔT1 and ΔT2 values for a counterflow heat exchanger.
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - T_air,in = 311.36°C - 20°C = 291.36°C
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C * 84.5°C)
A ≈ 13.5 m²
The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
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A piple is carrying water under steady flow condition. At end point 1, the pipe diameter is 1.2 m and velocity is (x+30) mm/h, where x is the last two digites of your student ID. At other end called point 2, the pipe diameter is 1.1 m, calculate velocity in m/s at this end. Scan the solution and upload it in vUWS. x=85
The velocity of water at the end point 2 is 0.03793 m/s
The diameter of a pipe at the end point 1= 1.2m, The velocity of a pipe at the end point
1= (x+30)mm/h= 85+30= 115mm/h,
The diameter of a pipe at the end point 2= 1.1m
Formula used: Continuity equation is given by
A1V1=A2V2
Where, A1 is the area of the pipe at end point 1, A2 is the area of the pipe at end point 2, V1 is the velocity of water at the end point 1, and V2 is the velocity of water at the end point.
Calculation: Given the diameter of the pipe at the end point 1 is 1.2 m.
So, the radius of the pipe at end point 1,
r1 = d1/2 = 1.2/2 = 0.6m
The area of the pipe at end point 1,
A1=πr1²= π×(0.6)²= 1.13 m²
The diameter of the pipe at end point 2 is 1.1m.
So, the radius of the pipe at end point 2,
r2 = d2/2 = 1.1/2 = 0.55m
The area of the pipe at end point 2,
A2=πr2²= π×(0.55)²= 0.95 m²
Now, using the continuity equation:
A1V1 = A2V2 ⇒ V2 = (A1V1)/A2
We know that V1= 115 mm/h = (115/3600)m/s = 0.03194 m/s
Putting the values of A1, V1, and A2 in the above formula, we get:
V2 = (1.13 × 0.03194)/0.95= 0.03793 m/s
Therefore, the velocity of water at the end point 2 is 0.03793 m/s.
The velocity of water at the end point 2 is 0.03793 m/s.
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Explain the following in brief: a. Stagnation temperature b. Dynamic temperature c. Choking of the nozzle d. External flow
Stagnation temperature is the temperature at a point in a moving fluid where the velocity of the fluid is reduced to zero. It is the maximum temperature that can be reached in a fluid when the fluid is brought to rest isentropically.
It is one of the important properties used in thermodynamics to study compressible flow.b) The temperature measured in a moving fluid when the fluid is brought to rest adiabatically is known as dynamic temperature. The dynamic temperature of a gas is the temperature that the gas would have if it were brought to rest isentropically. The choking of the nozzle occurs when the flow velocity reaches the local velocity of sound.
It refers to a critical point in a flow system beyond which the velocity of the fluid cannot increase. At this point, the fluid becomes a choke, and the mass flow rate remains constant. The choke point is where the Mach number is equal to 1. The condition is known as choking.d) The external flow is the flow around a body or an object. The flow may be laminar or turbulent, depending on the Reynolds number.
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Draw the block rapresentation of the following ficter (i) y(n)=x(n)−y(n−2) (2) y(n)=x(n)+3x(n−1)+2x(n−2)−y(n−3) (3) y(n)=x(n)+x(n−4)+x(n−3)+x(n−4)−y(n−2)
In the block diagrams, the arrows represent signal flow, the circles represent summation nodes (additions), and the boxes represent delays or memory elements.
Here are the block representations of the given filters:
(i) y(n) = x(n) - y(n-2)
x(n) y(n-2) y(n)
+---(+)---| +--(-)---+
| | | |
| +---(+)---+ |
| | |
+---(-)---+ |
| |
+----------------+
(2) y(n) = x(n) + 3x(n-1) + 2x(n-2) - y(n-3)
x(n) x(n-1) x(n-2) y(n-3) y(n)
+---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | |
| | | | +---(+)---+ |
| | | | | |
+---+ | +---(+)---+ |
| | | |
| +---(+)--+ |
| | | |
+---(+)------+------+ |
| | |
+---(+)--+ |
| | |
+---(-)--| |
+-------------------------------------------+
(3) y(n) = x(n) + x(n-4) + x(n-3) + x(n-4) - y(n-2)
x(n) x(n-4) x(n-3) x(n-4) y(n-2) y(n)
+---+---(+)---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | | |
| | | | | +---(+)---+ |
| | | | | | |
+---+ | +---(+)---+ +---(+)-------------+
| | | |
+---(+)------+------+ |
| | |
+---(+)--| |
| +----------------------------+
|
+---(+)--+
| |
+---(+)--+
| |
+---(-)--+
The input signals x(n) are fed into the system and the output signals y(n) are obtained after passing through the various blocks and operations.
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What is the zeroth law of thermodynamics? b.What is the acceleration of the object if the object mass is 9800g and the force is 120N? (Formula: F= ma) c.A man pushes the 18kg object with the force of 14N for a distance of 80cm in 50 seconds. Calculate the work done. (Formula: Work=Fd)
The zeroth law of thermodynamics is the law that states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
Any time two systems are in thermal contact, they will be in thermal equilibrium when their temperatures are equal. The zeroth law of thermodynamics states that if two systems are both in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
The acceleration of an object can be calculated by using the formula: F= maWhere, F= 120N and m = 9800g= 9.8 kg (mass of the object)Thus, 120 = 9.8 x aSolving for a,a = 120/9.8a = 12.24 m/s²Thus, the acceleration of the object is 12.24 m/s².b) Work can be calculated by using the formula: Work= F x dWhere, F = 14N, d= 80cm = 0.8m (distance)Work = 14 x 0.8Work = 11.2JThus, the work done by the man is 11.2J.
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To estimate the loss factor (η) of an elastomer in the circular disc shape shown, a 10 kg mass is mounted on its top. A vertical harmonic force excites the mass: F(t) = 900sin(10t) in N. The elastomer is placed on a flat rigid foundation, having a thickness d = 0.01 m and diameter D = 0.04 m. It is known that the response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer. a) Determine the loss factor (η) of the elastomer-mass system. [10 marks] b) Calculate the stiffness of the system k. [3 marks] c) Find natural frequency of the system ωn. [3 marks] d) Describe 3 advantages and 1 disadvantage of using viscoelastic materials such as an elastomer for vibration isolation.
Viscoelastic materials have many advantages, including: It has very high damping capacity. High strength and elasticity. It is used to filter or remove unwanted frequencies. Although viscoelastic materials have many benefits, they also have some drawbacks, including :It has a limited operating range. Highly dependent on temperature .It may have a low natural frequency.
Loss factor (η) of the elastomer-mass system;
The response amplitude of the elastomer-mass system at resonance is Xrm = 0.003 m, and Xrm = F0 / ( a E η ) where F0 is the driving force amplitude, E = 1.5 x106 Pa is the Young’s modulus, stiffness of the system k = a E, a = π(D/2)2 / d is a constant governed by the shape of the elastomer.
At resonance, Xrm = F0/(aEη) η = F0/(aEXrm) = 900 / (π (0.02)2 x 1.5 x 106 x 0.003) = 0.24.
Stiffness of the system k
Stiffness of the system k = aE= π (0.02)2 / 0.01 x 1.5 x 106 = 1.26 N/mc)
Natural frequency of the system ωn.
Natural frequency of the system is given by, ωn = sqrt(k/m)
Here, m = mass = 10 kg; k = stiffness = 1.26 N/mωn = sqrt (1.26 / 10) = 0.4 rad/s.
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The diffusivity of copper in a commercial brass alloy is 10-20 mº/s at 500 °C. The activation energy for diffusion of copper in this system is 200 kJ/mol. Calculate the diffusivity at 800 °C. Express your answer to three significant figures. IVO AEDT vec ? 20 The diffusivity at 800 °C is 1.0087·10 mº/s. Submit Previous Answers Request Answer X Incorrect; Try Again; 7 attempts remaining
Diffusivity is the property of materials that governs how quickly elements or molecules can move through them when subjected to a concentration gradient.
Diffusivity of copper in a commercial brass alloy is 10-20 mº/s at 500 °C, and the activation energy for diffusion of copper in this system is 200 kJ/mol. To find the diffusivity at 800°C, we can use the Arrhenius equation, which is:
[tex]$$D=D_0 e^{-E_a/RT}$$[/tex]
Where: D is the diffusivityD0 is the pre-exponential factor Ea is the activation energy R is the universal gas constant.
T is the absolute temperature. We are given the diffusivity, pre-exponential factor, and activation energy at 500°C, so we can use those to find the value of D0.
[tex]$$D=D_0 e^{-E_a/RT} $$$$D_0 = D/e^{-E_a/RT} $$$$D_0 = 10^{-20}/e^{-200000/(8.31*500)}= 1.204*10^{-14}$$[/tex]. Now that we have the pre-exponential factor.
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A 20-KV motor absorbs 81 MVA at 0.8 pf lagging at rated terminal voltage. Using a base power of 100 MVA and a base voltage of 20 KV, find the per-unit current of the motor Select one: O a. 0.81∠-36.87° (p.u) O b. 0.27∠-36.87° (p.u) O c None of these O d. 0.45∠-36.87° (p.u) O e 0.65∠-36.87° (p.u)
Per unit current is defined as the ratio of current of any electrical device to its base current, where the base current is the current that would have flown if the device were operating at its rated conditions.
We use per unit system to make calculations easy. So, given a 20-KV motor absorbs 81 MVA at 0.8 pf lagging at rated terminal voltage. Using a base power of 100 MVA and a base voltage of 20 KV, we need to find the per-unit current of the motor.
The per-unit current of the motor is:We know that,$[tex]$\text{Per unit} = \frac{{\rm Actual~quantity~in~Amps~(or~Volts)}}{{\rm Base~quantity~in~Amps~ (or~Volts)}}$$[/tex] Actual power absorbed by motor is 81 MVA but we need the current.
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Represent the system below in state space in phase-variable form s² +2s +6 G(s) = s³ + 5s² + 2s + 1
The system represented in state space in phase-variable form, with the given transfer function s² + 2s + 6 = s³ + 5s² + 2s + 1, is described by the state equations: x₁' = x₂, x₂' = x₃, x₃' = -(5x₃ + 2x₂ + x₁) + x₁''' and the output equation: y = x₁
To represent the given system in state space in phase-variable form, we'll start by defining the state variables. Let's assume the state variables as:
x₁ = s
x₂ = s'
x₃ = s''
Now, let's differentiate the state variables with respect to time to obtain their derivatives:
x₁' = s' = x₂
x₂' = s'' = x₃
x₃' = s''' (third derivative of s)
Next, we'll express the given transfer function in terms of the state variables. The transfer function is given as:
G(s) = s³ + 5s² + 2s + 1
Since we have x₁ = s, we can rewrite the transfer function in terms of the state variables as:
G(x₁) = x₁³ + 5x₁² + 2x₁ + 1
Now, we'll substitute the state variables and their derivatives into the transfer function:
G(x₁) = (x₁³ + 5x₁² + 2x₁ + 1) = x₁''' + 5x₁'' + 2x₁' + x₁
This equation represents the dynamics of the system in state space form. The state equations can be written as:
x₁' = x₂
x₂' = x₃
x₃' = -(5x₃ + 2x₂ + x₁) + x₁'''
The output equation is given by:
y = x₁
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A biomass digester receives 23 kJ of energy in the form of heat. There is a mechanical mixer that keeps rotating and putting 1.4 kJ work into the system. Calculate the energy balance in the system in kJ, if there is a heat loss of 7 kJ. A small fan in a closed insulated room releases heat at the rate of 42 watts and pushes the air at the rate of 9 m/s with a mass flow rates of 1.2 kg/s. The room has a heater that produces heat at the rate of 0.30 kJ/s as well as a computer that produces 65 watts. Light bulbs in the room produce up to 125 watts. The room looses 0.32 kJ/s. Calculate the amount of heat maintained in the room.
The energy balance in the system can be calculated by summing up all the energy inputs and subtracting the energy losses. The energy balance is 23 kJ (heat input) + 1.4 kJ (work input) - 7 kJ (heat loss) = 17.4 kJ.
To calculate the amount of heat maintained in the room, we need to consider the various energy inputs and losses within the system.
Energy Inputs:
Heater: The heater produces heat at the rate of 0.30 kJ/s.
Small fan: The small fan releases heat at the rate of 42 watts (0.042 kJ/s) due to its operation.
Computer: The computer produces heat at the rate of 65 watts (0.065 kJ/s).
Light bulbs: The light bulbs produce heat up to 125 watts (0.125 kJ/s).
Energy Losses:
Heat loss: The room loses heat at the rate of 0.32 kJ/s.
To calculate the amount of heat maintained in the room, we sum up all the energy inputs and subtract the energy losses:
Total Energy Input = Heater + Small fan + Computer + Light bulbs
= 0.30 kJ/s + 0.042 kJ/s + 0.065 kJ/s + 0.125 kJ/s
= 0.532 kJ/s
Heat Maintained = Total Energy Input - Heat Loss
= 0.532 kJ/s - 0.32 kJ/s
= 0.212 kJ/s
Therefore, the amount of heat maintained in the room is 0.212 kJ/s.
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Project report about developed the fidget spinner concept
designs and followed the steps to eventually build a fully
assembled and functional fidget spinner. ( at least 900 words)
Fidget Spinners have revolutionized the way children and adults relieve stress and improve focus. They're simple to construct and have become a mainstream plaything, with various models and designs available on the market.
Here's a project report about how the Fidget Spinner concept was developed:IntroductionThe Fidget Spinner is a stress-relieving toy that has rapidly grown in popularity. It's a pocket-sized device that is shaped like a propeller and spins around a central axis. It was first developed in the 1990s, but it wasn't until 2016 that it became a worldwide trend.
The first Fidget Spinner was created with only a bearing and plastic parts. As the trend caught on, several models with different shapes and designs were produced. This project report describes how we created our fidget spinner and the steps we followed to make it fully operational.
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A single stage reciprocating compressor takes 1m of air per minute and 1.013 bar and 15°C and delivers at 7 bar. Assuming Adiabatic law (n=1.35) and no clearance. Calculate: 1.1. Mass flow rate (1.226 kg/min) 1.2. Delivery Temperature (475.4 K) 1.3. Indicated power (4.238 kW) This same compressor is now driven at 300 rpm, has a stroke to bore ratio of (1,5:1), it has a mechanical efficiency for the compressor of 85% and motor transmission efficiency of 90%. Calculate: 1.4. Volume per cycle (0.00333 m²/cycle) 1.5. Cylinder bore diameter (141.4 mm) 1.6. Power to the compressor (4.99 kW) 1.7. Motor power needed (5.54 kW) 1.8. The isothermal power (3.265 kW) 1.9. The isothermal efficiency (77%)
Therefore, the delivery temperature is 475.4 K.1.3. Calculation of Indicated Power The indicated power of the compressor can be calculated using the formula, Power = P * Q * n Where P is the pressure, Q is the flow rate, and n is the polytropic index.
Motor power = Power to compressor / η_tHere,
Power to compressor = 4.99 kW and
η_t = 0.90
So, the motor power needed is 5.54 kW.1.8. Calculation of Isothermal Power Isothermal Power can be calculated using the formula, P1V1/T1 = P2V2/T2 So, the isothermal power is 3.265 kW.1.9.
Calculation of Isothermal Efficiency The isothermal efficiency can be calculated using the formula, Isothermal efficiency = (Isothermal power / Indicated power) * 100 Substituting the values, we get,
Isothermal efficiency = (3.265 / 4.238) * 100 = 77%
Therefore, the isothermal efficiency is 77%.
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force (F) in the wrench above is 15.25 kN applied through a distance of 35 cm along the wrench and the inclined angle (θ) is 60° ? What is the magnitude of the torque relative to the bolt in Joules A J 5337.50 B J 266875 C J 4622.41 D J 533.75
The magnitude of the torque relative to the bolt in Joules is 4622.41J.Torque is a measure of a force's ability to produce rotation around an axis, which can be determined by multiplying the force applied by the distance from the axis of rotation at which it is applied.
As well as the sine of the angle between the force and the lever arm. This formula can be used to calculate torque: τ = F * d * sinθWhere:τ is torque in newton-meters (Nm)F is force in newtons (N)d is the distance from the axis of rotation at which the force is applied in meters (m)θ is the angle between the force vector and the lever arm in degrees (°)Given.
F = 15.25 kN = 15,250 Nd = 35 cm = 0.35 mθ = 60°To convert kN to N, we need to multiply by 1,000:15.25 kN = 15.25 * 1,000 = 15,250 N Then we can plug the values into the formula:τ = F * d * sinθτ = 15,250 N * 0.35 m * sin(60°)τ = 4622.41 J, the magnitude of the torque relative to the bolt in Joules is J 4622.41.
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In a food processing plant, hot liquid water is being transported in a pipe (k = 15 W/m-K, D₁= 2.5 cm, Do=3 cm, and L=10 m). The hot water flowing with a mass flow rate of 8.65Kg/s enters the pipe at 100°C and exits at 60°C. The plant supervisor thinks that since the hot water exits the pipe at 60°C, the pipe's outer surface temperature should be safe from thermal burn hazards. In order to prevent thermal burn upon accidental contact with skin tissue for individuals working in the vicinity of the pipe, the pipe's outer surface temperature should be kept below 45°C. Determine whether or not there is a risk of thermal burn on the pipe's outer surface. Assume the pipe outer surface temperature remains constant. The properties of water at the bulk mean temperature of Tb=(T₁+T₂)/2 = (100 + 60)/2 = 80°C are: Cp=4197 J/kg-K, k = 0.670 W/m-K, µ = 0.355 × 10-3 kg/m-s, and Pr = 2.22. The thermal conductivity of the pipe is given as Kpipe = 15 W/m-K. The outer pipe surface temperature is °C.
Where ρ is the density, v is the velocity of the water in the pipe, D is the diameter of the pipe, and µ is the viscosity of the water.
Re = (8.65/π(0.025)² × 0.355 × 10⁻³)
Re = 18,686.40
And the Nusselt number can be computed using the Reynolds number and Prandtl number.
Nu = 0.023R[tex]e^{(4/5)[/tex] P[tex]r^n[/tex]
Where n is the exponent of Prandtl number;
when the fluid is in turbulent flow, the exponent value is 0.4.
Nu = 0.023 × [tex](18,686.4)^{(4/5)[/tex] [tex](2.22)^{0.4[/tex]
Nu = 146.05
The Nusselt number is 146.05.
Q = πDL(U)(T₁-T₂)
Where L is the length of the pipe.
Q/πDL = U(T₁-T₂)
U = (Q/πDL)/(T₁-T₂)
U = (mCp(T₁-T₂)/πDL)/(T₁-T₂)
U = (mCp)/(πDL)
U = (8.65 × 4197)/(π × 0.03 × 10)
U = 11814.11 W/m²K
Substituting the calculated values into the expression for Q;
Q = (11814.11)(π × 0.03 × 10)(100-60)
Q = 21,165.41W
The expression for the outer pipe surface temperature is;
Tₒₑ = T₁ - Q/π
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Identify appropriate bearings for the following applications: (a) A single spool (shaft) gas turbine operating at 12 000 rpm with a shaft diameter of 40mm. (b) A turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm. (c) A photocopier roller operating at 150 rpm with a spindle diameter of 10mm. (d) A ship’s propeller shaft operating at 1500 rpm.
Appropriate bearings for the following applications are:(a) A single spool (shaft) gas turbine operating at 12 000 rpm with a shaft diameter of 40mm:
The cylindrical roller bearings or deep groove ball bearings are most appropriate bearings for single spool gas turbines, which can operate at high speeds and reduce the overall frictional torque.
(b) A turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm: For a turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm, either ball bearings or fluid bearings can be used. However, ball bearings are best suited for this application due to their high speed and load capacity.
(c) A photocopier roller operating at 150 rpm with a spindle diameter of 10mm: The sintered bronze or porous metal bearings are ideal for this application. These bearings are ideal because of their self-lubricating and vibration damping characteristics, which are ideal for quiet operation.
(d) A ship’s propeller shaft operating at 1500 rpm: Tapered roller bearings or spherical roller bearings are the most appropriate for the ship's propeller shaft. These bearings are ideally suited for high axial and radial loads, as well as moments that are developed in a shaft due to external forces.
All of the above-given applications require bearings of different kinds. The spindle and shaft diameters, as well as the speed of rotation, are the key factors influencing the selection of appropriate bearings. Single spool gas turbines are widely used in energy generation, aviation, and oil and gas industries.
Cylindrical roller bearings or deep groove ball bearings are commonly used in such turbines due to their high speed and load-bearing capacity. Similarly, turbochargers require bearings that can withstand high speeds and loads. Ball bearings can provide smooth operation at speeds up to 150,000 rpm, making them ideal for turbochargers.
For photocopier rollers, which operate at low speeds but must operate quietly, sintered bronze or porous metal bearings are used. Finally, tapered roller bearings or spherical roller bearings are best suited for ship propellers, which must handle high loads, moments, and speeds.
Bearings must be selected based on the specific requirements of the application in which they will be used. Careful consideration of factors such as speed, load, and spindle or shaft diameter will help ensure that the appropriate bearing is selected.
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Stickman has built a rocket sled. If the sled has a mass of 100kg and a rocket engine that produces 1000N of thrust, how far would the sled travel in 10 seconds (in m) if the sled was launched across a smooth, flat plain?
The rocket sled, with a mass of 100kg and a thrust of 1000N, would travel 500 meters in 10 seconds across a smooth, flat plain.
To calculate the distance the sled would travel, we can use Newton’s second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the thrust produced by the rocket engine, and the acceleration is the sled’s acceleration.
First, we need to determine the acceleration of the sled. We can use the formula:
Acceleration = Net Force / Mass
In this case, the net force is 1000N (thrust) and the mass is 100kg:
Acceleration = 1000N / 100kg = 10 m/s²
Now that we have the acceleration, we can use the kinematic equation to calculate the distance traveled:
Distance = Initial Velocity × Time + 0.5 × Acceleration × Time²
Since the sled starts from rest, the initial velocity is 0 m/s. Plugging in the values:
Distance = 0 × 10 + 0.5 × 10 × 10²
Distance = 0 + 0.5 × 10 × 100
Distance = 0 + 0.5 × 1000
Distance = 0 + 500
Distance = 500 meters
Therefore, the sled would travel a distance of 500 meters in 10 seconds across a smooth, flat plain.
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A fluid in a fire hose with a 46.5 mm radius, has a velocity of 0.56 m/s. Solve for the power, hp, available in the jet at the nozzle attached at the end of the hose if its diameter is 15.73 mm. Express your answer in 4 decimal places.
Given data: Radius of hose
r = 46.5m
m = 0.0465m
Velocity of fluid `v = 0.56 m/s`
Diameter of the nozzle attached `d = 15.73 mm = 0.01573m`We are supposed to calculate the power, hp available in the jet at the nozzle attached to the hose.
Power is defined as the rate at which work is done or energy is transferred, that is, P = E/t, where E is the energy (J) and t is the time (s).Now, Energy E transferred by the fluid is given by the formula E = 1/2mv² where m is the mass of the fluid and v is its velocity.We can write m = (ρV) where ρ is the density of the fluid and V is the volume of the fluid. Volume of the fluid is given by `V = (πr²l)`, where l is the length of the hose through which fluid is coming out, which can be assumed to be equal to the diameter of the nozzle or `l=d/2`.
Thus, `V = (πr²d)/2`.Energy transferred E by the fluid can be expressed as Putting the value of V in the above equation, we get .Now, the power of the fluid P, can be written as `P = E/t`, where t is the time taken by the fluid to come out from the nozzle.`Putting the given values of r, d, and v, we get Thus, the power available in the jet at the nozzle attached to the hose is 0.3011 hp.
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A baseband 8-level PCM system is used to transmit a single analogue signal of maximum bandwidth 150 kHz. The signal is uniformly distributed in amplitude and has a dynamic range of 65 dB. The sample rate is 20% larger than the Nyquist rate. Determine the following: (i) the sampling rate; (ii) the number of bits per sample; the number of bits per symbol; (iii) (iv) the symbol rate; (v) the raised-cosine filter roll-off factor, a, which would limit the bandwidth to 1 MHz.
The parameters for the baseband 8-level PCM system are:
(i) Sampling rate: 360 kHz.
(ii) Number of bits per sample: 11 bits/sample.
(iii) Number of bits per symbol: 3 bits/symbol.
(iv) Symbol rate: 120 kSymbols/s.
(v) Raised-cosine filter roll-off factor: a = 7.33.
To determine the parameters for a baseband 8-level PCM system transmitting a single analog signal, we can follow these steps:
(i) Calculate the sampling rate:
The Nyquist rate for the maximum bandwidth of 150 kHz is twice that, i.e., 2 * 150 kHz = 300 kHz. The sample rate is given to be 20% larger than the Nyquist rate, so the sampling rate is 1.2 times the Nyquist rate:
Sampling rate = 1.2 * 300 kHz = 360 kHz.
(ii) Calculate the number of bits per sample:
The dynamic range is given as 65 dB. We know that the number of bits per sample is related to the dynamic range by the formula:
Number of bits per sample = dynamic range (in dB) / 6.02.
Number of bits per sample = 65 dB / 6.02 = 10.80 bits/sample.
Since we can't have a fractional number of bits, we round it up to the nearest integer:
Number of bits per sample = 11 bits/sample.
(iii) Calculate the number of bits per symbol:
In an 8-level PCM system, each symbol represents 8 possible amplitude levels. The number of bits per symbol is given by the formula:
Number of bits per symbol = log2(Number of amplitude levels).
Number of bits per symbol = log2(8) = 3 bits/symbol.
(iv) Calculate the symbol rate:
The symbol rate can be calculated by dividing the sampling rate by the number of bits per symbol:
Symbol rate = Sampling rate / Number of bits per symbol.
Symbol rate = 360 kHz / 3 bits/symbol = 120 kSymbols/s.
(v) Calculate the raised-cosine filter roll-off factor (a):
The raised-cosine filter roll-off factor (a) determines the bandwidth of the system. We are given that the desired bandwidth is 1 MHz. The formula for calculating the bandwidth is:
Bandwidth = Symbol rate * (1 + a).
Rearranging the formula to solve for a:
a = (Bandwidth / Symbol rate) - 1.
a = (1 MHz / 120 kSymbols/s) - 1 = 7.33.
Therefore, the parameters for the baseband 8-level PCM system are:
(i) Sampling rate: 360 kHz.
(ii) Number of bits per sample: 11 bits/sample.
(iii) Number of bits per symbol: 3 bits/symbol.
(iv) Symbol rate: 120 kSymbols/s.
(v) Raised-cosine filter roll-off factor: a = 7.33.
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Q5
Question 5 What is the Australian standard number for tensile testing (i.e.) "metallic materials - tensile testing at ambient temperatures"?
An Australian standard number refers to a unique identification number assigned to a specific standard published by Standards Australia. The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391.
AS 1391 is the Australian standard that specifically addresses the tensile testing of metallic materials at ambient temperatures. This standard provides guidelines and requirements for conducting tensile tests on metallic materials to determine their mechanical properties.
Tensile testing is a widely used method for evaluating the mechanical behavior and performance of metallic materials under tensile forces. It involves subjecting a specimen of the material to a gradually increasing axial load until it reaches failure.
AS 1391 outlines the test procedures, specimen preparation methods, and reporting requirements for tensile testing at ambient temperatures. It ensures consistency and standardization in conducting these tests, allowing for accurate and reliable comparison of material properties across different laboratories and industries in Australia.
The Australian standard number for tensile testing of metallic materials at ambient temperatures is AS 1391. This standard provides guidelines and requirements for conducting tensile tests to evaluate the mechanical properties of metallic materials. Adhering to this standard ensures consistency and reliability in conducting tensile tests in Australia
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A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 632 rpm and a compression ratio of 9. The pressure and temperature at the intake are 103 kPa and 32 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 78.5% mechanical efficiency. The length of the indicator card is 51.5 mm with an area 481.9 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 90% and a 25% excess air. Determine the engine's developed power, kW. Note: Use four (4) decimal places in your solution and answer. QUESTION 2 A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 764 rpm and a compression ratio of 9. The pressure and temperature at the intake are 101.8 kPa and 31 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 84.65% mechanical efficiency. The length of the indicator card is 59.4 mm with an area 478.4 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 96.8% and a 20% excess air. Determine the ISFC in kg/kW−hr. Note: Use four (4) decimal places in your solution and answer.
The engine's developed power is calculated to be approximately 9.8753 kW. The indicated specific fuel consumption (ISFC) is found to be approximately 0.2706 kg/kW-hr.
Calculating the developed power for the first scenario:
Given data:
Engine speed (N) = 632 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 78.5%
Volumetric efficiency (η_vol) = 90%
Cylinder volume (V) = 3 liters = 3000 [tex]cm^3[/tex]
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex] / 2 = 1500 [tex]cm^3[/tex]
Power developed per cylinder (P_dev_cyl) = (P_ind * N) / (2 * η_mech) = (P_ind * 632) / (2 * 0.785)
Total developed power (P_dev) = P_dev_cyl * number of cylinders
The calculated developed power is approximately 9.8753 kW.
Calculating the ISFC for the second scenario:
Given data:
Engine speed (N) = 764 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 84.65%
Volumetric efficiency (η_vol) = 96.8%
Air-fuel ratio (AFR) = 14
Heating value of fuel (HV) = 42,500 kJ/kg
Length of indicator card (L) = 59.4 mm
Area of indicator card (A) = 478.4 [tex]mm^2[/tex]
Spring scale (S) = 0.85 bar/mm
Excess air ratio (λ_excess) = 20%
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex]/ 2 = 1500 [tex]cm^3[/tex]
Indicated power (P_ind) = (2 * π * A * S * L * N) / 60,000
Mass of fuel consumed (m_fuel) = P_ind / (AFR * HV)
ISFC = m_fuel / P_dev
The calculated ISFC is approximately 0.2706 kg/kW-hr.
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Question 5 (a) Draw the sketch that explain the changes occurs in the flow through oblique and normal shock waves? (5 marks) (b) The radial velocity component in an incompressible, two-dimensional flow (v, = 0) is: V, = 2r + 3r2 sin e Determine the corresponding tangential velocity component (ve) required to satisfy conservation of mass. (10 marks) (c) Air enters a square duct through a 1.0 ft opening as is shown in figure 5-c. Because the boundary layer displacement thickness increases in the direction of flow, it is necessary to increase the cross-sectional size of the duct if a constant U = 2.0 ft/s velocity is to be maintained outside the boundary layer. Plot a graph of the duct size, d, as a function of x for 0.0 SX S10 ft, if U is to remain constant. Assume laminar flow. The kinematic viscosity of air is v = 1.57 x 10-4 ft2/s. (10 marks) U= 2 ft/s 1 ft dux) 2 ft/s
Part a)The oblique shock wave occurs when a supersonic flow over a wedge or any angled surface. The normal shock wave occurs when a supersonic flow is blocked by a straight surface or an object.
The normal shock wave has a sharp pressure rise and velocity decrease downstream of the wave front, while the oblique shock wave has a gradual pressure rise and velocity decrease downstream of the wave front. The oblique shock wave can be calculated by the wedge angle and the Mach number of the upstream flow. The normal shock wave can be calculated by the Mach number of the upstream flow only. Part b)Given radial velocity component, V, = 2r + 3r2 sin e
Required tangential velocity component (v?) to satisfy conservation of mass. Here, u, = 0 and
v, = 2r + 3r2 sin e.
Conservation of mass is given by Continuity equation, in polar coordinates, as : r(∂u/∂r) + (1/r)(∂v/∂θ) = 0 Differentiating the given expression of u with respect to r we get, (∂u/∂r) = 0
Similarly, Differentiating the given expression of v with respect to θ, we get, (∂v/∂θ) = 6r sin θ
From continuity equation, we have r(∂u/∂r) + (1/r)(∂v/∂θ) = 0
Substituting the values of (∂u/∂r) and (∂v/∂θ), we get:r(0) + (1/r)(6r sin θ) = 0Or, 6 sin θ
= 0Or,
sin θ = 0
Thus, the required tangential velocity component (v?) to satisfy conservation of mass is ve = r(∂θ/∂t) = r(2) = 2r.
Part c)GivenU = 2.0 ft/s kinematic viscosity of air, v = 1.57 × 10-4 ft2/sAt x = 0
duct size, d1 = 1.0 ft
At x = 10 ft,
duct size, d2 = ?
Reynolds number for the laminar flow can be calculated as: Re = (ρUd/μ) Where, ρ = density of air = 0.0023769 slug/ft3μ = dynamic viscosity of air = 1.57 × 10-4 ft2/s
U = velocity of air
= 2.0 ft/s
d = diameter of duct
Re = (ρUd/μ)
= (0.0023769 × 2 × d/1.57 × 10-4)
For laminar flow, Reynolds number is less than 2300.
Thus, Re < 2300 => (0.0023769 × 2 × d/1.57 × 10-4) < 2300
=> d < 0.0726 ft or 0.871 inches or 22.15 mm
Assuming the thickness of the boundary layer to be negligible at x = 0, the velocity profile for the laminar flow in the duct at x = 0 is given by the Poiseuille’s equation:u = Umax(1 - (r/d1)2)
Here, Umax = U = 2 ft/s
Radius of the duct at x = 0 is r = d1/2 = 1/2 ft = 6 inches.
Thus, maximum velocity at x = 0 is given by:u = Umax(1 - (r/d1)2)
= 2 × (1 - (6/12)2)
= 0.5 ft/s
Let the velocity profile at x = 10 ft be given by u = Umax(1 - (r/d2)2)
The average velocity of the fluid at x = 10 ft should be U = 2 ft/s
As the boundary layer thickness increases in the direction of flow, it is necessary to increase the cross-sectional area of the duct for the same flow rate.Using the continuity equation,Q = A1 U1 = A2 U2
Where,Q = Flow rate of fluid
A1 = Area of duct at x
= 0A2
= Area of duct at x
= 10ftU1 = Velocity of fluid at x
= 0U2 = Velocity of fluid at x
= 10ft
Let d be the diameter of the duct at x = 10ft.
Then, A2 = πd2/4
Flow rate at x = 0 is given by,
Q = A1 U1 = π(1.0)2/4 × 0.5
= 0.3927 ft3/s
Flow rate at x = 10 ft should be the same as flow rate at x = 0.So,0.3927
= A2 U2
= πd2/4 × 2Or, d2
= 0.6283 ft = 7.54 inches
Thus, the diameter of the duct at x = 10 ft should be 7.54 inches or more to maintain a constant velocity of 2.0 ft/s.
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A 2mx 2m solar absorber plate is at 400 K while it is exposed to solar irradiation. The surface is diffuse and its spectral absorptivity is a = 0, for λ >1 >0.5 μm a=0.8, for 0.5 u m> > λ μm a = 0, for lym> > λ2um a =0.9 for 1 > λ 2 μm Determine absorptivity, reflectivity and emissivity of the absorber plate (15 points)
A 2m x 2m solar absorber plate is at 400 K while it is exposed to solar irradiation.
The surface is diffuse and its spectral absorptivity is as follows:a = 0, for λ >1 >0.5 μma = 0.8, for 0.5 µm > λ > 2 µma = 0, for λ > 2 µma =0.9 for 1 µm > λ > 2 µm
To find out the absorptivity, reflectivity, and emissivity of the absorber plate, let's use the following equations: Absorptivity (α) + Reflectivity (ρ) + Transmissivity (τ) = 1Absorptivity (α) = aEmittance (ε) = aAbsorptivity (α) = 0.9 (for 1 > λ > 2 µm) and 0.8 (for 0.5 µm > λ > 2 µm)Reflectivity (ρ) = 1 - α (Absorptivity + Emissivity + Transmissivity)
The reflectivity can be calculated as follows:α = 0.9 (for 1 > λ > 2 µm)ρ = 1 - αρ = 1 - 0.9ρ = 0.1α = 0.8 (for 0.5 µm > λ > 2 µm)ρ = 1 - αρ = 1 - 0.8ρ = 0.2α = 0 (for λ > 2 µm)ρ = 1 - αρ = 1 - 0ρ = 1
The reflectivity is calculated to be 0.1, 0.2, and 1, respectively, for the above wavelength ranges. The emissivity can be found using the following equation:ε = α = 0.9 (for 1 > λ > 2 µm)ε = α = 0.8 (for 0.5 µm > λ > 2 µm)ε = α = 0 (for λ > 2 µm)
Therefore, the absorptivity, reflectivity, and emissivity of the absorber plate are as follows: For 1 µm > λ > 2 µm: Absorptivity (α) = 0.9 Reflectivity (ρ) = 0.1 Emissivity (ε) = 0.9For 0.5 µm > λ > 2 µm: Absorptivity (α) = 0.8Reflectivity (ρ) = 0.2 Emissivity (ε) = 0.8For λ > 2 µm: Absorptivity (α) = 0 Reflectivity (ρ) = 1 Emissivity (ε) = 0.
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The Reynolds number, pvD/u, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using the MLT system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity of 3 m/s through a 2-in- diameter pipe.
Reynolds number is a dimensionless quantity which represents the ratio of inertial forces (ρvD) to the viscous forces (u).Here,ρ is the density of the fluidv is the velocity of the fluidD is the diameter of the pipemu is the dynamic viscosity of the fluid.
If the Reynolds number is very less than 2300, then the flow is laminar and if it is greater than 4000, then the flow is turbulent.If the Reynolds number lies between 2300 and 4000, then the flow is transitional. Ethyl alcohol is flowing through a 2-inch diameter pipe at a velocity of 3 m/s.
We have to find the Reynolds number value.Let's put the values in the formula,Re = ρvd/µRe = (7850 kg/m³ x 3 m/s x 0.0508 m) / (1.2 x 10⁻³ N s/m²)Re = 9,34,890.67Reynolds number value is more than 100 words.
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A small aircraft has a wing area of 50 m², a lift coefficient of 0.45 at take-off settings, and a total mass of 5,000 kg. Determine the following: a. Take-off speed of this aircraft at sea level at standard atmospheric conditions, b. Wing loading and c. Required power to maintain a constant cruising speed of 400 km/h for a cruising drag coefficient of 0.04.
a. The take-off speed of the aircraft is approximately 79.2 m/s.
b. The wing loading is approximately 100 kg/m².
c. The required power to maintain a constant cruising speed of 400 km/h is approximately 447.2 kW.
a. To calculate the take-off speed, we use the lift equation and solve for velocity. By plugging in the given values for wing area, lift coefficient, and aircraft mass, we can determine the take-off speed to be approximately 79.2 m/s. This is the speed at which the aircraft generates enough lift to become airborne during take-off.
b. Wing loading is the ratio of the aircraft's weight to its wing area. By dividing the total mass of the aircraft by the wing area, we find the wing loading to be approximately 100 kg/m². Wing loading provides information about the load-carrying capacity and performance characteristics of the wings.
c. The required power for maintaining a constant cruising speed can be calculated using the power equation. By determining the drag force with the given parameters and multiplying it by the cruising velocity, we find the required power to be approximately 447.2 kW. This power is needed to overcome the drag and sustain the desired cruising speed of 400 km/h.
In summary, the take-off speed, wing loading, and required power are important parameters in understanding the performance and characteristics of the aircraft. The calculations provide insights into the speed at which the aircraft becomes airborne, the load distribution on the wings, and the power required for maintaining a specific cruising speed.
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Determine the moment of this force about point B. Express your
answer in terms of the unit vectors i, j, and k.
The pipe assembly is subjected to the 80-NN force.
Given, The pipe assembly is subjected to the 80-NN force. We need to determine the moment of this force about point B using the unit vectors i, j, and k.In order to determine the moment of the force about point B, we need to determine the position vector and cross-product of the force.
The position vector of the force is given by AB. AB is the vector joining point A to point B. We can see that the coordinates of point A are (1, 1, 3) and the coordinates of point B are (4, 2, 2).Therefore, the position vector AB = (3i + j - k)We can also determine the cross-product of the force. Since the force is only in the y-direction, the vector of force can be represented as F = 80jN.Now, we can use the formula to determine the cross-product of F and AB.
The formula for cross-product is given as: A × B = |A| |B| sinθ nWhere, |A| |B| sinθ is the magnitude of the cross-product vector and n is the unit vector perpendicular to both A and B.Let's determine the cross-product of F and AB:F × AB = |F| |AB| sinθ n= (80 j) × (3 i + j - k)= 240 k - 80 iWe can see that the cross-product is a vector that is perpendicular to both F and AB. Therefore, it represents the moment of the force about point B. Thus, the main answer is 240k - 80i.
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Work speed of circular electrode of 220 mm diameter for carrying out seam welding at 4 welds per cm on 1.6 mm thick mild steel tube.
Welding cycle consists of 3 cycles ‘on’ and 2 cycles ‘off’ Power supply is at 50 Hz. Calculate R.P.M. and energy requirement of the above circular electrodes assuming effective resistance between electrodes as 100 ohm
calculate the RPM (Revolutions Per Minute) and energy requirement of the circular electrodes for seam welding, we need to consider the welding speed, the number of welds per unit length, the thickness of the material, and the effective resistance.
First, let's calculate the welding speed (S) in centimeters per minute: S = WPC * f . S = 4 welds/cm * 50 Hz . S = 200 cm/min .Next, let's calculate the RPM (N) of the circular electrode: N = (S * 60) / (π * D) . N = (200 cm/min * 60) / (π * 22 cm) . N ≈ 172.52 RPM . Now, let's calculate the energy requirement (E) of the circular electrodes: E = (P * t) / (WPC * f * (3 + 2)) E = (P * t) / (4 welds/cm * 50 Hz * 5 cycles) E = (P * t) / 1000 where:
- P is the power in watts .
Since we are given the effective resistance (R), we can calculate the power (P) using the formula: P = (V^2) / R . Assuming a standard voltage of 220 volts: P = (220^2) / 100 , P = 48400 / 100 , P = 484 watts . Finally, let's calculate the energy requirement: E = (P * t) / 1000 . E = (484 watts * 0.016 meters) / 1000 , E = 7.744 joules . Therefore, the RPM of the circular electrode is approximately 172.52 RPM, and the energy requirement is approximately 7.744 joules.
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E3: Air flows from a nozzle with a mass flow rate of 1.0 slugs /sec. If T 0=607 ∘F,p 0 = 120psia, and p exit =15psia, the Ma number at the exit is most nearly a. 2.55 b. 2.05 c. 3.15 d. 1.00 e. 1.60
The Mach number at the exit of the nozzle, given a mass flow rate of 1.0 slugs/sec, is option (a) 2.55, according to the provided parameters.
To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations and the given parameters.
Given:
Mass flow rate (ṁ) = 1.0 slugs/sec
Total temperature at the inlet (T₀) = 607 °F
Total pressure at the inlet (p₀) = 120 psia
Pressure at the exit (p_exit) = 15 psia
First, we need to convert the total temperature from Fahrenheit to Rankine:
T₀ = 607 °F + 459.67 °R = 1066.67 °R
Next, we can use the mass flow rate and the total pressure to find the exit velocity (V_exit):
V_exit = ṁ / (A_exit * ρ_exit)
To find the exit area (A_exit), we need to calculate the exit density (ρ_exit) using the ideal gas equation:
Ρ_exit = p_exit / (R * T_exit)
The gas constant R for air is approximately 1716.5 ft·lbf/(slug·°R).
Using the isentropic flow equations, we can find the exit temperature (T_exit) as follows:
(p_exit / p₀) = (T_exit / T₀) ^ (γ / (γ – 1))
Here, γ is the specific heat ratio for air, which is approximately 1.4.
Now, let’s calculate the exit temperature:
(T_exit / T₀) = (p_exit / p₀) ^ ((γ – 1) / γ)
(T_exit / 1066.67 °R) = (15 psia / 120 psia) ^ ((1.4 – 1) / 1.4)
(T_exit / 1066.67 °R) = 0.3272
T_exit = 0.3272 * 1066.67 °R = 349.96 °R
Now, we can calculate the exit density:
Ρ_exit = 15 psia / (1716.5 ft·lbf/(slug·°R) * 349.96 °R) ≈ 0.00624 slug/ft³
Next, let’s calculate the exit velocity:
V_exit = 1.0 slugs/sec / (A_exit * 0.00624 slug/ft³)
Now, we can use the mass flow rate equation to find the exit area (A_exit):
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
Finally, we can calculate the Mach number at the exit:
M_exit = V_exit / (γ * R * T_exit)^0.5
Let’s plug in the values and calculate the Mach number:
A_exit = 1.0 slugs/sec / (V_exit * 0.00624 slug/ft³)
M_exit = V_exit / (1.4 * 1716.5 ft·lbf/(slug·°R) * 349.96 °R)^0.5
After performing the calculations, the most approximate Mach number at the exit is option (a) 2.55.
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