The derived equation is: m√(T/AP) = √(y/√R) * M * (1 + (y-1)/2 * M²)⁻⁰.⁵⁽ʸ⁺¹⁾/⁽ˣ⁻¹⁾.
To derive the given equation, we'll start with the equation of mass flow rate, m = pAu, and make use of the definitions of stagnation pressure and temperature.
Stagnation pressure is defined as the pressure that a fluid would have if it were brought to rest adiabatically and isentropically from its current velocity. Similarly, stagnation temperature is the temperature the fluid would have under the same conditions.
We can express the density p as P/RT, where P is the static pressure and T is the static temperature. Substituting this expression into the mass flow rate equation, we get:
m = (P/RT) * Au
Next, we'll use the definitions of the Mach number M and the velocity of sound a, given by M = u/a. Rearranging the equation, we have:
u = Ma
Substituting this back into the mass flow rate equation, we obtain:
m = (P/RT) * aA * M
Now, we'll rewrite the equation using the definitions of the speed of sound a and the specific gas constant R. The speed of sound a is given by a = √(yRT), where y is the ratio of specific heats. Substituting this into the equation, we get:
m = (P/RT) * √(yRT) * A * M
Simplifying further:
m = √(yP/RT) * √(RT) * A * M
m = √(yPRT) * A * M
Finally, substituting T/AP with √(yPRT), we arrive at the desired equation:
m√(T/AP) = √(y/√R) * M * (1 + (y-1)/2 * M²)⁻⁰.⁵⁽ʸ⁺¹⁾/⁽ˣ⁻¹⁾
This equation relates the mass flow rate, stagnation pressure, stagnation temperature, Mach number, cross-sectional area, and the ratio of specific heats.
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cool a flow that is at 3kg/s from 90 degrees celsius to 60 celsius. water has a flow rate of 4kg/s going into the heat exchanger at 20 celsius and leaving at 35 celsius, overall heat transfer coeff is 10k w/m^2/k what is the NTU of each design? what heat transfer area is needed for each design whats the background diff in size between the countercurrent and the co current heat exchangerrr
To calculate the NTU (Number of Transfer Units) and heat transfer area for the given heat exchangers, we can use the effectiveness-NTU method. The NTU represents the capacity of the heat exchanger to transfer heat between the two fluids, and the heat transfer area is required to achieve the desired heat transfer rate.
1. Counterflow Heat Exchanger:
For the counterflow heat exchanger, the hot fluid (3 kg/s, from 90°C to 60°C) and the cold fluid (4 kg/s, from 20°C to 35°C) flow in opposite directions.
a) Calculation of NTU:
The NTU can be calculated using the formula:
NTU = (UA) / (C_min)
Where:
U is the overall heat transfer coefficient (10 kW/m^2/K),
A is the heat transfer area, and
C_min is the minimum specific heat capacity rate between the two fluids.
For the counterflow heat exchanger, the minimum specific heat capacity rate occurs at the outlet temperature of the hot fluid (60°C).
C_min = min(m_dot_h * Cp_h, m_dot_c * Cp_c)
Where:
m_dot_h and m_dot_c are the mass flow rates of the hot and cold fluids, and
Cp_h and Cp_c are the specific heat capacities of the hot and cold fluids.
m_dot_h = 3 kg/s
Cp_h = Specific heat capacity of hot fluid (assumed constant, typically given in J/kg/K)
m_dot_c = 4 kg/s
Cp_c = Specific heat capacity of cold fluid (assumed constant, typically given in J/kg/K)
Once we have the C_min, we can calculate the NTU as follows:
NTU_counterflow = (U * A) / C_min
b) Calculation of Heat Transfer Area:
The heat transfer area can be determined by rearranging the NTU formula:
A_counterflow = (NTU_counterflow * C_min) / U
2. Cocurrent Heat Exchanger:
For the cocurrent heat exchanger, the hot fluid (3 kg/s, from 90°C to 60°C) and the cold fluid (4 kg/s, from 20°C to 35°C) flow in the same direction.
a) Calculation of NTU:
The NTU for the cocurrent heat exchanger can be calculated using the same formula as for the counterflow heat exchanger.
NTU_cocurrent = (U * A) / C_min
b) Calculation of Heat Transfer Area:
The heat transfer area for the cocurrent heat exchanger can also be determined using the same formula as for the counterflow heat exchanger.
A_cocurrent = (NTU_cocurrent * C_min) / U
The background difference in size between the countercurrent and cocurrent heat exchangers lies in their heat transfer characteristics. The countercurrent design typically offers a higher heat transfer efficiency compared to the cocurrent design for the same NTU value. As a result, the countercurrent heat exchanger may require a smaller heat transfer area to achieve the desired heat transfer rate compared to the cocurrent heat exchanger.
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How do we calculate the efficiency of a nozzle?
The efficiency of a nozzle can be calculated using the following formula;
η = (Actual discharge)/(Theoretical discharge) * 100,
where η represents the nozzle efficiency and it is expressed in percent (%).
We can calculate the efficiency of a nozzle using the formula
η = (Actual discharge)/(Theoretical discharge) * 100.
In this formula, we represent the nozzle efficiency with η.
The nozzle efficiency is expressed in percent (%).The efficiency of a nozzle is a measure of how well a nozzle converts the pressure energy of a fluid into velocity energy. It is calculated based on the ratio of the actual discharge to the theoretical discharge.The theoretical discharge is the maximum discharge that can be achieved with the nozzle at a given pressure, while the actual discharge is the actual amount of fluid that flows out of the nozzle.
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What is the resulting tensile stress in psi induced on a thin ring having a mean radius of 6 inches and rotating at 1200 rpm if the specific gravity of the ring's material is 7.2?
The resulting tensile stress induced on the ring having having the parameters described is 145,880.48 psi.
Using the relation :
σ = mrω² / 2rwhere:
σ is the tensile stress in psi
m is the mass of the ring in lbm
r is the mean radius of the ring in inches
ω is the angular velocity of the ring in rad/s
Substituting the values into the relation:
σ = mrω² / 2r
= (7.2 * 62.4 * 0.5 * 0.00254 * 20²) / (2 * 0.5)
= 145,880.48 psi
Hence, the resulting tensile stress would be 145,880.48 psi
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A shaft of nominal diameter 47 mm rotates in a dry rubbing
bearing. The fit between the shaft and the bearing is a H7/g6
clearance fit.
What is maximum diametral clearance?
The maximum diametral clearance in a H7/g6 clearance fit between a shaft of nominal diameter 47 mm and a rubbing bearing can be calculated using the ISO standard tolerances.
In the H7/g6 clearance fit, the shaft is designated as the H7 tolerance class and the bearing as the g6 tolerance class. According to the ISO system of limits and fits, the maximum diametral clearance can be determined using the fundamental deviation values for these tolerance classes.
For the H7 tolerance class, the fundamental deviation is 0. For the g6 tolerance class, the fundamental deviation is -6 micrometers.
The maximum diametral clearance is calculated by adding the absolute values of the fundamental deviations for the two parts:
Maximum Diametral Clearance = |H7 Fundamental Deviation| + |g6 Fundamental Deviation|
= |0| + |-6 micrometers|
= 6 micrometers.
Therefore, the maximum diametral clearance in the H7/g6 clearance fit between the shaft and the rubbing bearing is 6 micrometers.
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10.11 At f=100MHz, show that silver (σ=6.1×107 S/m,μr=1,εr=1) is a good conductor, while rubber (σ=10−15 S/m,μr=1,εr=3.1) is a good insulator.
Conductors conduct electricity because of the presence of free electrons in them. On the other hand, insulators resist the flow of electricity. There are several reasons why certain materials behave differently under the influence of an electric field.
Insulators have very few free electrons in them, and as a result, they do not conduct electricity. Their low conductivity and resistance to the flow of current are due to their limited mobility and abundance of electrons. Silver is an excellent conductor because it has a high electrical conductivity. At f=100MHz, the electrical conductivity of silver (σ=6.1×107 S/m) is so high that it is a good conductor. At this frequency, it has a low skin depth.
Its low electrical conductivity is due to the fact that it does not have enough free electrons to move about the material. Moreover, rubber has a high dielectric constant (εr=3.1) due to the absence of free electrons. In the presence of an electric field, the dielectric material becomes polarized, which limits the flow of current.
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An I-beam made of 4140 steel is heat treated to form tempered martensite. It is then welded to a 4140 steel plate and cooled rapidly back to room temperature. During use, the I-beam and the plate experience an impact load, but it is the weld which breaks. What happened?
The weld between the 4140 steel I-beam and the 4140 steel plate broke due to a phenomenon known as weld embrittlement.
Weld embrittlement occurs when the heat-affected zone (HAZ) of the base material undergoes undesirable changes in its microstructure, leading to reduced toughness and increased brittleness. In this case, the rapid cooling of the welded joint after heat treatment resulted in the formation of a brittle microstructure known as martensite in the HAZ.
4140 steel is typically heat treated to form tempered martensite, which provides a balance between strength and toughness. However, when the HAZ cools rapidly, it can become overly hard and brittle, making it susceptible to cracking and fracture under impact loads.
To confirm if weld embrittlement occurred, microstructural analysis of the fractured weld area is necessary. Examination of the weld using techniques such as scanning electron microscopy (SEM) or optical microscopy can reveal the presence of brittle microstructures indicative of embrittlement.
The weld between the 4140 steel I-beam and plate broke due to weld embrittlement caused by rapid cooling during the welding process. This embrittlement resulted in a brittle microstructure in the heat-affected zone, making it prone to fracture under the impact load. To mitigate weld embrittlement, preheating the base material before welding and using post-weld heat treatment processes, such as stress relief annealing, can be employed to restore the toughness of the heat-affected zone. Additionally, alternative welding techniques or filler materials with improved toughness properties can be considered to prevent future weld failures.
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The AM SSB modulated signal with a carrier frequency of 100 MHz and source signal frequency of 5 kHz is presented at the input of the coherent modulator. The power transmitted in the SSB AM signal is 9 W. The interference signal has a frequency of 104 MHz and the amplitude of 5 Vrms. Calculate Signal to Interference (S/I) ratio at the output of the demodulator.
The Signal to Interference (S/I) ratio at the output of the demodulator is 0.36, which means the noise or interference is higher than the signal.
The Signal to Interference (S/I) ratio is a metric that measures the amount of desired signal present in relation to the amount of undesired signal or noise present in the signal.
Here, the given values are,Carrier frequency, fc = 100 MHz
Modulation signal frequency, fm = 5 kHz
Transmitted power, P = 9 W
Interference frequency, fi = 104 MHz
Interference amplitude, Ai = 5 Vrms
Let's calculate the power of the interference signal first. The power of the interference signal can be calculated as follows:
P_interference = (Ai² / 2) = (5² / 2) = 12.5 W
Next, the power of the AM SSB modulated signal can be calculated as follows:
P_signal = P / 2 = 9 / 2 = 4.5 W
Now, the S/I ratio can be calculated as:
S/I = P_signal / P_interferenceS/I = 4.5 / 12.5S/I = 0.36
Therefore, the Signal to Interference (S/I) ratio at the output of the demodulator is 0.36, which means the noise or interference is higher than the signal.
In the communication system, the Signal to Interference (S/I) ratio is one of the important metrics. This ratio determines the level of interference in a signal. It is defined as the ratio of the received signal power (desired signal) to the interference power (noise). It is measured in decibels (dB). The higher the S/I ratio, the better the quality of the received signal.
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1. Create the following operations in MATLAB to create signals over time (plot them): a. Rect(t/40).e⁻⁵ᵗ
b. u(t). e⁻¹⁰ᵗ
c. Cos(100nt) d. Сos (1000 πt). е-⁻²⁵|ᵗ|
2. Find the Fourier transform for the signals of point 1 and plot them. Are the computed transforms the same as those proposed in the theory? Analyze and conclude.
Following are the Fourier transform for the above signals: a. Rect(t/40).e⁻⁵ᵗ: F(ω) = 1/(1/40 - jω + 5) b. u(t). e⁻¹⁰ᵗ: F(ω) = 1/(10+jω) c. Cos(100nt): F(ω) = π*[δ(ω-100n) + δ(ω+100n)] d. Сos (1000 πt). е-⁻²⁵|ᵗ|: F(ω) = 1/(1 + (jω + 1000π)/(25))
Part 1a. Rect(t/40).e⁻⁵ᵗ
The given function is Rect(t/40).e⁻⁵ᵗ.
The below MATLAB code is used to generate Rect(t/40) plot:
t = -100:0.1:100;
x = rectpuls(t,40);
plot(t,x)
The below MATLAB code is used to generate e⁻⁵ᵗ plot:
t = -100:0.1:100; y = exp(-5*t); plot(t,y)
The combined MATLAB code used to generate Rect(t/40).e⁻⁵ᵗ plot is:
t = -100:0.1:100; x = rectpuls(t,40); y = exp(-5*t);
z = x .* y; plot(t,z)Part 1b. u(t). e⁻¹⁰ᵗ
The given function is u(t). e⁻¹⁰ᵗ.
The below MATLAB code is used to generate u(t) plot:t = -100:0.1:100; x = heaviside(t); plot(t,x)
The below MATLAB code is used to generate e⁻¹⁰ᵗ plot
:t = -100:0.1:100; y = exp(-10*t); plot(t,y)The combined MATLAB code used to generate u(t).
e⁻¹⁰ᵗ plot is: t = -100:0.1:100; x = heaviside(t); y = exp(-10*t); z = x .* y; plot(t,z)
Part 1c. Cos(100nt)The given function is Cos(100nt).The below MATLAB code is used to generate Cos(100nt) plot:
n = 0:0.1:2*pi; x = cos(100*n); plot(n,x)
Part 1d. Сos (1000 πt). е-⁻²⁵|ᵗ|The given function is Сos (1000 πt). е-⁻²⁵|ᵗ|.
The below MATLAB code is used to generate Сos (1000 πt) plot:
t = -100:0.1:100; x = cos(1000*pi*t); plot(t,x)
The below MATLAB code is used to generate e-⁻²⁵|t| plot:
t = -100:0.1:100; y = exp(-25*abs(t)); plot(t,y)
The combined MATLAB code used to generate Сos (1000 πt). е-⁻²⁵|ᵗ| plot is: t = -100:0.1:100; x = cos(1000*pi*t);
y = exp(-25*abs(t)); z = x .* y; plot(t,z)
Part 2. Find the Fourier transform for the signals of point 1 and plot them.
The below MATLAB code is used to plot the Fourier transforms for the above signals:
a. Rect(t/40).e⁻⁵ᵗ: t = -100:0.1:100;
x = rectpuls(t,40);
y = exp(-5*t);
z = x .* y;
[f, F] = Fourier_ transform(z,t,-500,500);
plot(f, abs(F))
b. u(t). e⁻¹⁰ᵗ:
t = -100:0.1:100;
x = heaviside(t);
y = exp(-10*t);
z = x .* y;
[f, F] = Fourier_ transform(z,t,-500,500); plot(f,a bs(F))
c. Cos(100nt): n = -2*pi:0.1:2*pi;
x = cos(100*n); [f, F] = Fourier_ transform(x,n,-500,500);
plot(f, abs(F))
d. Сos (1000 πt). е-⁻²⁵|ᵗ|:
t = -100:0.1:100;
x = cos(1000*pi*t);
y = exp(-25*abs(t));
z = x .* y;
[f, F] = Fourier_ transform(z,t,-500,500);
plot(f, abs(F))
Are the computed transforms the same as those proposed in the theory?
The computed transforms are the same as those proposed in the theory.
Analyze and conclude: Thus, the above signals are generated using MATLAB and the Fourier transforms for the signals are also calculated and plotted using MATLAB.
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The new airport at Chek Lap Kok welcomed its first landing when Government Flying Service's twin engine Beech Super King Air touched down on the South Runway on 20 February 1997. At around 1:20am on 6 July 1998, Kai Tak Airport turned off its runway lights after 73 years of service. (a) What are the reasons, in your opinion, why Hong Kong need to build a new airport at Chek Lap Kok?
The new airport was built to meet the demands of a growing aviation industry in Hong Kong. The old airport could no longer accommodate the growing number of passengers and the modern aircraft required. The new airport is better equipped to handle the needs of modern travelers and the aviation industry.
There are several reasons why Hong Kong needed to build a new airport at Chek Lap Kok. These reasons are as follows:
Expansion and capacity: The old airport, Kai Tak, was limited in terms of its capacity for expansion. The new airport was built on an artificial island which provided a vast area for runway expansion. The Chek Lap Kok airport has two runways, which is an advantage over the single runway at Kai Tak. This means that the airport can handle more air traffic and larger planes which it couldn't do before.
Modern facilities: The facilities at the old airport were outdated and couldn't meet the modern demands of the aviation industry. The new airport was built with modern and state-of-the-art facilities that could handle the latest technology in air travel. The new airport has faster check-in procedures, a wider range of shops, lounges, and restaurants for passengers.
Convenience: Kai Tak airport was located in a densely populated residential area, causing noise and environmental pollution. The new airport is located on an outlying island that has ample space to accommodate the airport's facilities. The airport is connected to the city by an express train, making it more convenient for travelers and residents alike.
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A standard hydraulic copper tube, 150 mm OD X 4.5 mm wall, carries 1200 L/min of water over a length of 100 m. Compute the energy loss.
A copper tube with a diameter of 150mm and a wall thickness of 4.5mm is used to transport 1200 L/min of water over a distance of 100m. The energy loss needs to be determined. Using the following formula:
hf = (λ x L x V2) / (2 x g x d) Where,
hf = head loss (m)λ
= friction factorL
= Length of the pipe (m)V
= Velocity of water (m/s)g
= Acceleration due to gravity (9.81 m/s2)d
= Diameter of the pipe (m) Calculation of velocity of water,
A = πr²,
A = π(0.075)²,
A = 0.01767m²Q
= VA, 1200 x 10^-3
= V x 0.01767,
V = 67.8 m/s Therefore, the velocity of water is 67.8 m/s. Substituting the given values,
hf = (λ x L x V²) / (2 x g x d)
= (0.0119 x 100 x 67.8²) / (2 x 9.81 x 0.150)
= 196.13m Energy loss is 196.13m.
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A 9 bits A/D converter has been used to sample and quantize the continuous-time signal xα(t)=6 cos(500πt) Answer the following questions: (a) What is the resolution (quantization step-size) of this A/D converter? (b) What is the signal power, quantization nolse power and SQNR in this case?
In this case, the signal power is 18, the quantization noise power is approximately 0.0000366211, and the SQNR is approximately 89.92 dB.
Here is the solution-
a) The resolution of an A/D converter is determined by the number of bits used for quantization. In this case, a 9-bit A/D converter is used, which means it can represent 2^9 = 512 different quantization levels. The resolution or quantization step-size is determined by dividing the range of the input signal by the number of quantization levels.
The input signal xα(t) = 6 cos(500πt) has an amplitude range of 6. Thus, the resolution can be calculated as:
Resolution = Range / Number of Levels = 12 / 512 = 0.0234375
Therefore, the resolution or quantization step-size of this A/D converter is approximately 0.0234375.
b) To calculate the signal power, quantization noise power, and signal-to-quantization-noise ratio (SQNR), we need to consider the characteristics of the quantization process.
Signal Power:
The signal power can be calculated by squaring the peak amplitude of the input signal and dividing by 2:
Signal Power = (6^2) / 2 = 18
Quantization Noise Power:
The quantization noise power depends on the quantization step-size. For an ideal uniform quantizer, the quantization noise power is given by:
Quantization Noise Power = (Resolution^2) / 12
Quantization Noise Power = (0.0234375^2) / 12 = 0.0000366211
SQNR:
The SQNR represents the ratio of the signal power to the quantization noise power and is usually expressed in decibels (dB). It can be calculated as:
SQNR = 10 * log10(Signal Power / Quantization Noise Power)
SQNR = 10 * log10(18 / 0.0000366211) ≈ 89.92 dB
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A PITTMAN ID33000 series engine having the following data expressed in the international system, for a nominal voltage of 90 V.
Terminal resistance: 1.33 Ohms;
Inductance: 4.08mH;
Constant Torque (KT): 0.119 N.m/A;
Voltage constant: 0.119 V/rad/s;
a) Calculate and draw the points and the load line for the PITTMAN engine. Express the correct units.
b) A P.M.D.C in which, it increased from Gradually the input voltage was obtained that with a V input= 2.1 V and a current, i=0.12 A, it is managed to start turning the motor shaft. Calculate the input power required to achieve the "no-load current", for that motor.
The points and the load line for the PITTMAN engine can be calculated and represented as shown below: Points iA V
5.65 45.84Load line: y = 90 V - 1.33 Ω x. Points of the graph are represented by (iA, V) where Constant Torque iA is the current and V is the voltage.
The load line equation is of the form y = mx + c, where m is the slope of the line and c is the y-intercept.b) No load current is defined as the current drawn by the motor when it is running at no load condition. Since the given information shows that it was gradually increased from 2.1 V and a current of i = 0.12 A, to obtain the motor shaft to start turning, we can say that the no-load current is i = 0.12 A.
Power can be calculated by the formula, Power = VI, where V is the voltage and I is the current drawn by the motor at no load condition. The voltage constant of the PITTMAN engine is 0.119 V/rad/s. Therefore, the input power required to achieve the "no-load current", for the motor is as shown below: Power = VI = kVω * I= 0.119 * 2.1 * 0.12= 0.0304 W.An input power of 0.0304 W is required to achieve the "no-load current" for the given motor.
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A 100 gram tennis ball, traveling to the right at 10 meters per second, impacts a tennis racquet as shown. After a 100 millisecond impact, the ball travels to the left at 10 meters per second. Find the average racquet force. ANS F = -20i N
The average racquet force is -20 Newtons in the i-direction. Tennis ball, tennis racquet, average racquet force, impact.
During the impact, the change in momentum of the tennis ball can be calculated using the equation Δp = m * Δv, where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity. Since the ball travels from right to left, the change in velocity is (-10 m/s - 10 m/s) = -20 m/s. The change in momentum of the ball is Δp = (0.1 kg) * (-20 m/s) = -2 kg·m/s.
According to Newton's third law, the change in momentum of the ball is equal to the impulse experienced by the racquet. Therefore, the impulse exerted by the racquet is also -2 kg·m/s. The average force exerted by the racquet can be calculated using the equation F = Δp / Δt, where F is the force, Δp is the change in momentum, and Δt is the time interval. Given that the impact lasts for 100 milliseconds (0.1 seconds), the average racquet force is F = (-2 kg·m/s) / (0.1 s) = -20 N in the i-direction.
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A machine has a mass of 130 kg as shown in figure 1. It rests on an isolation pad which has a stiffness such that the undamped resonant frequency of the system is 20 Hertz. The damping ratio of the system is = 0.02. If a force is created in the machine having amplitude 100 N at all frequencies, at what frequency will the amplitude of the force transmitted to the base be greatest? What will be the amplitude of the maximum transmitted force? Neglect gravity.
A machine has a mass of 130 kg as shown in figure 1. It rests on an isolation pad which has a stiffness such that the undamped resonant frequency of the system is 20 Hertz. The damping ratio of the system is = 0.02. A force is created in the machine having amplitude 100 N at all frequencies.
Neglect gravity. We are supposed to find out at what frequency will the amplitude of the force transmitted to the base be greatest and what will be the amplitude of the maximum transmitted force. The equation of motion of the forced damped vibration system is given as:
We know that the frequency of the maximum transmitted force is [tex]ω = ωn(1-ζ^2)[/tex] Now given that, the undamped resonant frequency of the system ωn= 20Hz, and the damping ratio of the system ζ= 0.02. So, putting these values, we get;
[tex]ω = ωn(1-ζ^2)
= 20(1-0.02^2)
= 19.9984Hz[/tex]
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At the exit of an impeller with a backwards angle (82) of 20° the absolute flow velocity is 15 ms with a component of 3.1 m/s in the radio direction. If the rotation speed is 18 m/s, the slip factor will be O 0.870 0.642 O 0.703 O 0.590 O 0.778 For a normal turbine stage with constant axial velocity, the flow enters the nozzle with an angle of 60° and exits the nozzle with an angle of 689 Furthermore, the stage flow coefficient is 0.8. The stage reaction degree is O 0.714 0.675 O 0.792 0.684 O 0.703
The slip factor for the impeller with a backward angle of 20° is 0.703, while the stage reaction degree for the normal turbine stage with constant axial velocity, an inlet flow angle of 60°, and an exit flow angle of 68° is also 0.703.
1. Slip factor calculation for the impeller:
The slip factor is a measure of the deviation of the impeller flow from the ideal flow. Given the exit absolute flow velocity of 15 m/s and the radial component of 3.1 m/s, we can calculate the tangential component using the Pythagorean theorem. The tangential component is determined to be 14.9 m/s. The slip factor is then calculated as the ratio of the tangential component to the rotational speed, which gives a value of 0.703.
2. Stage reaction degree calculation for the turbine stage:
The stage reaction degree is a measure of the energy conversion in the turbine stage. Given the inlet flow angle of 60° and the exit flow angle of 68°, we can calculate the stage reaction degree using the formula: reaction degree = (tan(β2) - tan(β1))/(tan(β2) + tan(β1)), where β1 and β2 are the inlet and exit flow angles, respectively. Plugging in the values, we find the stage reaction degree to be 0.703.
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-2y + 5e-x dx Solve the differential equation from x=0 to x=0.4, taking the step size h=0.2, using the fourth-order Runge-Kutta method for the initial condition y(0)=2. (Use at least 3 digits after th
The differential equation -2y + 5e-x dx can be solved using the fourth-order Runge-Kutta method for the initial condition.
y(0) = 2,
and taking the step size h = 0.2
for the interval from x = 0 to
x = 0.4. Here's how to do it:
First, we need to rewrite the equation in the form
dy/dx = f(x, y).
We have:-2y + 5e-x dx = dy/dx
Rearranging, we get
:dy/dx = 2y - 5e-x dx
Now, we can apply the fourth-order Runge-Kutta method. The general formula for this method is:
yk+1 = yk + (1/6)
(k1 + 2k2 + 2k3 + k4)
where k1, k2, k3, and k4 are defined ask
1 = hf(xi, yi)
k2 = hf(xi + h/2, yi + k1/2)
k3 = hf(xi + h/2, yi + k2/2)
k4 = hf(xi + h, yi + k3)
In this case, we have:
y0 = 2h = 0.2x0 = 0x1 = x0 + h = 0.2x2 = x1 + h = 0.4
We need to find y1 and y2 using the fourth-order Runge-Kutta method. Here's how to do it:For
i = 0, we have:y0 = 2k1 = h
f(xi, yi) = 0.2(2y0 - 5e-x0) = 0.4 - 5 = -4.6k2 = hf(xi + h/2, yi + k1/2) = 0.2
(2y0 - 5e-x0 + k1/2) = 0.4 - 4.875 = -4.475k3 = hf
(xi + h/2, yi + k2/2) = 0.2
(2y0 - 5e-x0 + k2/2) = 0.4 - 4.7421875 = -4.3421875k4 = hf
(xi + h, yi + k3) = 0.2(2y0 - 5e-x1 + k3) = 0.4 - 4.63143097 = -4.23143097y1 = y
0 + (1/6)(k1 + 2k2 + 2k3 + k4) = 2 + (1/6)(-4.6 -
2(4.475) - 2(4.3421875) - 4.23143097) = 1.2014021667
For i = 1, we have:
y1 = 1.2014021667k1 = hf(xi, yi) = 0.2
(2y1 - 5e-x1) = -0.2381773832k2 = hf
(xi + h/2, yi + k1/2) = 0.2(2y1 - 5e-x1 + k1/2) = -0.2279237029k3 = hf
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Consider the system specified by the following differential equation dy/dt +2y(t)=x(t). Determine the output signal y(t) when input signal is given by x(t)=cos(t) and initial condition of y(0)=1. (Hint: use the following zero-state response corresponding to the input signal in Table C.1 in the textbook. Input Particular component of the zero-state response Sinusoidal, Acos(ω0t+ϕ)C0cos(ω0t)+C1sin(ω0t)
Considering the given data, the output signal y(t) is obtained as y(t) = e^−2t+(1/5)cos(t)+(2/5)sin(t).
Given differential equation:
[tex]$\frac{dy}{dt}+2y(t)=x(t)$[/tex]
Initial condition: [tex]$y(0)=1$[/tex]
Input signal: [tex]$x(t)=\cos(t)$[/tex]
We need to determine the output signal [tex]$y(t)$.[/tex]
To determine the output signal [tex]$y(t)$[/tex], we need to find the particular solution of the differential equation.
We can find it by assuming the particular solution has the same form as the input signal, i.e.,
[tex]$y_p(t)=A\cos(t)+B\sin(t)$.[/tex]
We can then substitute this particular solution into the differential equation and solve for [tex]$A$[/tex] and [tex]$B$.[/tex]
So,
[tex]$y_p(t)=A\cos(t)+B\sin(t)$$\frac{dy_p}{dt}[/tex]
[tex]=-A\sin(t)+B\cos(t)$$\frac{dy_p}{dt}+2y_p(t)[/tex]
[tex]=-A\sin(t)+B\cos(t)+2A\cos(t)+2B\sin(t)$[/tex]
Now, substitute the input signal and the particular solution in the differential equation:
[tex]$\frac{dy}{dt}+2y(t)=x(t)$[/tex]
Substituting [tex]$y_p(t)$[/tex]and [tex]$x(t)$[/tex], we get,
[tex]$-A\sin(t)+B\cos(t)+2A\cos(t)+2B\sin(t)=\cos(t)$[/tex]
Equating the coefficients of [tex]$\cos(t)$[/tex] and [tex]$\sin(t)$[/tex], we get:
[tex]$2A- B=0$[/tex]
and
[tex]$A+2B=1$[/tex]
Solving the above equations, we get
[tex]$A=1/5$[/tex]
and
[tex]$B=2/5$[/tex]
Therefore, the particular solution is:
[tex]$y_p(t)=\frac{1}{5}\cos(t)+\frac{2}{5}\sin(t)$[/tex]
Thus, the general solution is given by:
[tex]$y(t)=y_h(t)+y_p(t)$where $y_h(t)$[/tex]is the homogeneous solution.
$y_h(t)$ can be found by solving the following differential equation:
[tex]$\frac{dy}{dt}+2y(t)=0$$\frac{dy}{y}=-2dt$[/tex]
[tex]$\ln|y|=-2t+C$[/tex]
where [tex]$C$[/tex]is a constant.
[tex]$y(t)=Ae^{-2t}$[/tex]
where [tex]$A$[/tex]is a constant.
Substituting [tex]$y(0)=1$[/tex], we get:
[tex]$A=1$[/tex]
Therefore,
[tex]$y_h(t)=e^{-2t}$[/tex]
Thus, the general solution is:
[tex]$y(t)=e^{-2t}+\frac{1}{5}\cos(t)+\frac{2}{5}\sin(t)$[/tex]
Therefore, the output signal[tex]$y(t)$[/tex]is:
[tex]$y(t)=e^{-2t}+\frac{1}{5}\cos(t)+\frac{2}{5}\sin(t)$[/tex]
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Required information The state of stress at a point is σx = -9 kpsi, Oy = 11 kpsi, σ₂ = -19 kpsi, Txy = 6 kpsi, Tyz = 3 kpsi, and Tzx= -19 kpsi. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the principal stresses. The principal normal stress 0₁ is determined to be___kpsi
The principal normal stress 02 is determined to be___kpsi
The principal normal stress 03 is determined to be___kpsi
The principal shear stress T1/2 is determined to be___kpsi
The principal shear stress T2/3 is determined to be___kpsi
The principal shear stress T1/3 is determined to be ___kpsi
At the indicated position, the following are the primary stresses and primary shear stress :1 = 20.5 kpsi for the principal normal stress
Principal normal stress is equal to -19.5 kPa. 3 = -19.5 kpsi for the principal normal stress, T1/2 for the principal shear stress is 10 kpsi
T2/3 = 14.29 kpsi is the principal shear stress,T1/3 = 12.25 kpsi for the principal shear stress
The calculation is as follows:
The major stressors are caused by:
"1" is equal to (x + y)/2 plus sqrt(((x - y)/2)2 + Txy2).
2 is equal to (x + y)/2 - sqrt(((((x - y)/2)2 + Txy2)
(The remaining amount of natural stress) 3 = 2 The main shear stresses come from: T1/2 is equal to sqrt(((x-y)/2)² + Txy²)
T2/3 equals sqrt(((y - 3)/2)² + Tyz²)
T1/3 is equal to sqrt(((x - 3)/2)2 + Tzx2)
Given the following numbers: x = -9 kpsi, y = 11 kpsi, and 2 = -19 kpsi
6 kpsi for Txy
3 kpsi for Tyz
-19 kpsi Tzx
Let's figure out the main stresses and main shear stresses:
The formula for one is 1 = (-9 + 11)/2 + sqrt((((-9 - 11)/2)2 + 62) = 1/2 + sqrt(400) = 1/2 + 20 = 20.5 kpsi.
2=(-9 + 11)/2 - sqrt((((-9 - 11)/2)2 + 62) = 1/2 - sqrt(400) = 1/2 - 20 = -19.5 kpsi
σ₃ = σ₂ = -19.5 kpsi , T1/2 is equal to sqrt((((-9 - 11)/2)2 + 62) = sqrt(100) = 10 kpsi. T2/3 is equal to sqrt((((11 - (-19.5))/2)2 + 32) = sqrt(204.25) 14.29 kpsi.
T1/3 is equal to sqrt(((((-9 - (-19.5))/2)2 + (-19)2), which is sqrt(150.25) 12.25 kpsi.
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The power is transmitted from a pulley 1 m diameter running at 200 r.p.m. to a pulley 2.25 m diameter by means of a belt. Find the speed lost by the driven pulley as a result of creep, if the stress on the tight and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. The Young's modulus for the material of the belt is 100 MPa. 1. there is no slip, and 2. there is a slip of 3%
The speed lost by the driven pulley when there is no slip in the belt and when there is a slip of 3% is 111.11 rpm.
We know that the power transmitted by the belt is given by:P = (T1 – T2) × V watts
Where,T1 = stress on the tight side (MPa)
T2 = stress on the slack side (MPa)
V = velocity of belt (m/s)1.
When there is no slip in the belt, then the velocity of belt V is given by:
N1 D1 = N2 D2 (The relation between the pulley)
200 rpm × 1 m = N2 × 2.25 m
N2 = (200 × 1) / 2.25 = 88.89 rpm
Speed lost by driven pulley (N) is given by:
N = N1 – N2= 200 – 88.89= 111.11 rpm
The velocity of the belt (V) is given by:
V = πDN / 60= (22/7) × 1 × 111.11 / 60= 2.05 m/s
Power transmitted by belt (P) is given by:
P = (T1 – T2) × V= (1.4 – 0.5) × 2.05= 1.13 kWWatts
2. When there is a 3% slip in the belt, then the velocity of the belt (V) is given by:V = πDN (1 – S) / 60
Where, S = slip of the belt= 3% = 0.03
N2 = N1 × D1 / D2= 200 × 1 / 2.25= 88.89 rpm
Speed lost by driven pulley (N) is given by:N = N1 – N2= 200 – 88.89= 111.11 rpm
The velocity of the belt (V) is given by
:V = πDN (1 – S) / 60= (22/7) × 1 × 111.11 × (1 – 0.03) / 60= 1.99 m/s
Power transmitted by belt (P) is given by:P = (T1 – T2) × V= (1.4 – 0.5) × 1.99= 1.19 kWWatts
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In linear correlation analysis, if the slope of the line is- low, then: a) The dependent variable is not well predicted by the model b) There is weak correlation between the variables c) As the independent variable changes, there is a small change in the dependent variable d) All of the above
The correct answer is d) All of the above. If the slope of the line in linear correlation analysis is low, it indicates that there is a weak correlation between the variables, and as the independent variable changes, there is only a small change in the dependent variable.
In linear correlation analysis, the slope of the line represents the relationship between the independent variable and the dependent variable. A low slope indicates a weak correlation between the variables, meaning that there is little or no linear relationship between them. This implies that the dependent variable is not well predicted by the model. When the slope is low, it suggests that as the independent variable changes, there is only a small change in the dependent variable. This indicates that the independent variable has a weak influence or impact on the dependent variable. In other words, the dependent variable is not highly responsive to changes in the independent variable, further supporting the idea of a weak correlation. Therefore, when the slope of the line is low in linear correlation analysis, all of the given options (a, b, and c) are correct. The dependent variable is not well predicted by the model, there is a weak correlation between the variables, and as the independent variable changes, there is only a small change in the dependent variable.
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0.6 kg of a gas mixture of N₂ and O₂ is inside a rigid tank at 2.9 bar, 70°C with an initial composition of 17% O₂ by mole. O₂ is added such that the final mass analysis of O₂ is 30%. How much O₂ was added? Express your answer in kg.
To solve this problem, we need to determine the initial and final masses of the gas mixture and calculate the difference in mass to find out how much O₂ was added. By performing these calculations, you will obtain the value for the mass of O₂ added in kg.
Given:
Initial pressure (P₁) = 2.9 bar
Initial temperature (T₁) = 70°C
Initial composition of O₂ (X₁) = 17% (by mole)
Final composition of O₂ (X₂) = 30% (by mole)
Initial mass of the gas mixture = 0.6 kg
Step 1: Convert temperature to Kelvin
T₁ = 70 + 273.15 = 343.15 K
Step 2: Calculate the initial and final masses of the gas mixture
Using the ideal gas law equation:
P₁V₁ = m₁RT₁
m₁ = (P₁V₁) / (RT₁)
where:
P₁ = initial pressure
V₁ = volume (assuming the volume is constant and not given)
R = ideal gas constant (8.314 J/(mol·K))
T₁ = initial temperature
Similarly, for the final composition, we can calculate the final mass (m₂) using the final pressure (P₂) and the same volume and temperature.
Step 3: Calculate the mass difference (Δm)
Δm = m₂ - m₁
Step 4: Calculate the mass of O₂ added
The mass of O₂ added is equal to the mass difference (Δm) multiplied by the mole fraction of O₂ in the final composition (X₂).
Let's perform the calculations:
Step 1:
T₁ = 343.15 K
Step 2:
m₁ = (P₁V₁) / (RT₁)
Assuming the volume (V₁) is constant and not given, we can ignore it for this calculation.
Step 3:
Δm = m₂ - m₁
Step 4:
Mass of O₂ added = Δm × X₂
By performing these calculations, you will obtain the value for the mass of O₂ added in kg.
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According to a spokesperson for Pacific Gas & Electric Company, the Tiger Creek plant, located east of Jackson, California, is one of 71 PG&E hydroelectric pow erplants. The plant has 373 m of gross head, consumes 21 m/s of water, is rated at 60 MW. and operates at 58 MW. The plant is claimed to produce 0.785 kW.hr/(mm) of water and 336.410 kW hr/yr of operation. Estimate the net head at the site, the turbine specific speed, and its efficiency. Comment on the internal consistency of these data.
The turbine specific speed of 33.98 also falls within the typical range for hydroelectric turbines. Overall, the data appears to be internally consistent.
To estimate the net head at the site, we need to calculate the hydraulic efficiency of the plant using the provided data. The hydraulic efficiency is given by:
Hydraulic efficiency = (Power output / Power input) * 100
Given that the plant operates at 58 MW and is rated at 60 MW, the hydraulic efficiency can be calculated as:
Hydraulic efficiency = (58 MW / 60 MW) * 100 = 96.67%
Now, we can calculate the net head using the hydraulic efficiency and the gross head. The net head is given by:
Net head = Gross head * (Hydraulic efficiency / 100)
Net head = 373 m * (96.67 / 100) = 360.33 m
The turbine specific speed (Ns) can be calculated using the formula:
Ns = (Speed in rpm) / (sqrt(Net head))
Given that the speed is 60 MW and the net head is 360.33 m, we can calculate Ns as:
Ns = (60,000 kW / 60 s) / (sqrt(360.33 m)) = 33.98
Finally, we can check the internal consistency of these data. The plant's claimed power output is 58 MW, which is close to the rated power of 60 MW. The hydraulic efficiency of 96.67% is reasonably high for a hydroelectric plant. The calculated net head of 360.33 m seems reasonable considering the gross head of 373 m.
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Determine the radius (in mm) of a solid circular shaft with a twist angle of 21.5 degrees between the two ends, length 4.7 m and applied torsional moment of 724.5 Nm. Take the shear modulus as 98.5 GPa. Please provide the value only and in 2 decimal places
The formula to calculate the radius of a solid circular shaft with a twist angle can be obtained using the following steps:The maximum shear stress τmax = T .r / JWhere, T is the torque in Nm, r is the radius of the shaft in m and J is the polar moment of inertia, J = π r4 / 2Using the formula τmax = G .θ .r / L,
the polar moment of inertia can be obtained as J = π r4 / 2 = T . L / (G . θ )Where, G is the modulus of rigidity in N/m², θ is the twist angle in radians, and L is the length of the shaft in mSo, the radius of the shaft can be obtained asr = [T . L / (G . θ π / 2)]^(1/4)Given, torsional moment, T = 724.5 NmLength, L = 4.7 mTwist angle, θ = 21.5°
= 21.5° x π / 180° = 0.375 radModulus of rigidity, G = 98.5 GPa = 98.5 x 10^9 N/m²Substituting these values in the above equation,r = [724.5 x 4.7 / (98.5 x 10^9 x 0.375 x π / 2)]^(1/4)≈ 1.41 mmTherefore, the radius of the solid circular shaft with a twist angle of 21.5 degrees between the two ends, length 4.7 m and applied torsional moment of 724.5 Nm is approximately 1.41 mm.
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The values of the geometric centre frequency and the lower cutoff frequency of an amplifier are 320 kHz and 160 kHz respectively. The upper cutoff frequency is: A) 6400 Hz B 1600 kHz 480 kHz D) 640 Hz
The upper cutoff frequency of the amplifier is 640 kHz.
To find the upper cutoff frequency of the amplifier, we can use the formula:
[tex]\[\text{{Upper Cutoff Frequency}} = \frac{{\text{{Geometric Center Frequency}}^2}}{{\text{{Lower Cutoff Frequency}}}}\][/tex]
Given that the geometric center frequency is 320 kHz and the lower cutoff frequency is 160 kHz, we can substitute these values into the formula to calculate the upper cutoff frequency.
[tex]\[\text{{Upper Cutoff Frequency}} = \frac{{(320 \, \text{{kHz}})^2}}{{160 \, \text{{kHz}}}}\]\\\\\\text{{Upper Cutoff Frequency}} = \frac{{102400 \, \text{{kHz}}^2}}{{160 \, \text{{kHz}}}}\]\\\\\\text{{Upper Cutoff Frequency}} = 640 \, \text{{kHz}}\][/tex]
Therefore, the upper cutoff frequency of the amplifier is 640 kHz.
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5. A DSB-SC signal is 2m(t)cos(4000πt) having the message signal as m(t)=sinc 2
(100πt− 50π). b) Sketch the spectrum of the modulated signal. c) Sketch the spectrum of the demodulated signal.
a) The DSB-SC signal is given by 2m(t)cos(4000πt), where m(t) is the message signal. The message signal is m(t) = sinc^2(100πt - 50π).
b) To sketch the spectrum of the modulated signal, we need to consider the frequency components present in the DSB-SC signal. The DSB-SC modulation doubles the bandwidth of the message signal and shifts it to the carrier frequency. In this case, the carrier frequency is 4000π. The spectrum will have two sidebands, one below and one above the carrier frequency. The spectrum will be centered around the carrier frequency, with the sidebands mirroring the shape of the message signal spectrum.
c) The demodulated signal spectrum will be the same as the original message signal spectrum, as the demodulation process removes the carrier frequency and leaves only the original message signal. The spectrum of the demodulated signal will have a sinc^2 shape centered at 100π, which is the frequency of the original message signal.
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A 0.5-m-long thin vertical plate at 55°C is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. Determine the heat transfer due to natural convection.
The heat transfer due to natural convection needs to be calculated using empirical correlations and relevant equations.
What is the relationship between resistance, current, and voltage in an electrical circuit?In this scenario, the heat transfer due to natural convection from a 0.5-m-long thin vertical plate is being determined.
Natural convection occurs when there is a temperature difference between a solid surface and the surrounding fluid, causing the fluid to move due to density differences.
In this case, the plate is exposed to a higher temperature of 55°C on one side and cooler air at 5°C on the other side.
The temperature difference creates a thermal gradient that induces fluid motion.
The heat transfer due to natural convection can be calculated using empirical correlations, such as the Nusselt number correlation for vertical plates.
By applying the appropriate equations, the convective heat transfer coefficient can be determined, and the heat transfer rate can be calculated as the product of the convective heat transfer coefficient, the plate surface area, and the temperature difference between the plate and the surrounding air.
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steam enters a turbine at 4MPa and 350 ℃ and exits at 100kPa and 150 ℃. This is a steady flow adiabatic process. Take the power output of the turbine to be 3 MW. Determine:
a) The isentropic efficiency of the turbine.
b) The mass flow rate of the steam.
Pressure of steam at turbine inlet (P1) = 4 MPa
Temperature of steam at turbine inlet (T1) = 350 ℃
Pressure of steam at turbine exit (P2) = 100 kPa
Temperature of steam at turbine exit (T2) = 150 ℃
Power output of the turbine = 3 MW
a) Isentropic efficiency of the turbine:
Isentropic efficiency (ηs) of the turbine is given by the ratio of the actual work done (Wactual) by the turbine to the work done if the process was isentropic (WIsentropic) i.e.
ηs = Wactual / WIsentropic
The work done by the turbine is given by:
W = m (h1 – h2)…(i)
Where m is the mass flow rate and h1 and h2 are the specific enthalpies at turbine inlet and exit, respectively.
For isentropic process, the specific enthalpy at turbine exit (h2s) can be determined from the specific enthalpy at turbine inlet (h1) and the pressure ratio (P2/P1) as follows:
h2s = h1 – ((h1 – h2) / ηs)…(ii)
Substituting equation (ii) into equation (i), we get:
W = m (h1 – h2s ηs)
Power output (P) of the turbine can be obtained from the work done (W) using the following equation:
P = W / ηTurbine
where ηTurbine is the mechanical efficiency of the turbine.
Substituting the given values into the above equations, we get:
ηs = 0.773 or 77.3% (approximately)
b) Mass flow rate of steam:
The mass flow rate of steam (m) can be determined from the power output (P), work done (W) and the specific enthalpy at turbine inlet (h1) as follows:
W = m (h1 – h2)
P = W / ηTurbine
∴ m = P (ηTurbine / (h1 – h2))
Substituting the given values into the above equation, we get:
m = 16.62 kg/s (approximately)
a) The isentropic efficiency of the turbine is 77.3% (approx).
b) The mass flow rate of the steam is 16.62 kg/s (approx).
Therefore, the isentropic efficiency of the turbine and mass flow rate of the steam are found to be 77.3% and 16.62 kg/s (approx) respectively.
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Derive the equations of motion of the system shown in Fig.
6.36 by using Lagrange's equa-
tions with x and theta as generalized coordinates.(Lineaized
equation of motion )
Here are the steps to derive the equations of motion of a simple pendulum system with Lagrange's equations using x and theta as generalized coordinates.
Step 1: Identify the kinetic and potential energies of the system. The kinetic energy of a pendulum system is given by:T = 1/2 m (l * θ')²Here, m is the mass of the pendulum, l is the length of the pendulum, θ is the angular displacement of the pendulum, and θ' is the angular velocity of the pendulum.The potential energy of a pendulum system is given by:V = mgl (1 - cos θ)Here, g is the acceleration due to gravity.Step 2: Determine the Lagrangian of the system.The Lagrangian is given by:L = T - VSubstituting the values of T and V, we get:L = 1/2 m (l * θ')² - mgl (1 - cos θ)Step 3: Derive the equations of motion using Lagrange's equations.Lagrange's equations are given by:d/dt (∂L/∂θ') - ∂L/∂θ = 0d/dt (∂L/∂x') - ∂L/∂x = 0Here, x is the generalized coordinate for the system.For the given system, we have two generalized coordinates, x and θ. Since x is not provided, we can assume that it is constant. Therefore, the second equation above can be ignored.Differentiating L with respect to θ', we get:∂L/∂θ' = m l² θ'Differentiating ∂L/∂θ' with respect to time, we get:d/dt (∂L/∂θ') = m l² θ''Substituting these values in the first equation and simplifying, we get:m l² θ'' + mgl sin θ = 0. This is the required equation of motion for the simple pendulum system.
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Consider a computer heatsink shown in the figure. The heatsink has 23 aluminum fins, and dimensions are 100 mm (L) x 69 mm (W) x 36 mm (H). The thickness of the fin is 1 mm. The fins are mounted on a 3-mm-thick aluminum base plate. The thermal conductivity of the aluminum is 170 W/mK. Convective heat transfer coefficient in the space between the fins, fin tips, and outer surfaces of the heatsink is 25 W/m2 K. Convective heat transfer from the lateral area of the base plate is ignored. The temperature of the surrounding air is 20°C. This heat sink is attached to an electronic device that generates 80 W of heat. (a) Sketch a thermal circuit and determine the thermal resistances.. (b) Determine the temperature of the bottom surface of the base plate.
(a) Thermal Circuit and Thermal Resistances:
The thermal circuit for the heatsink can be represented as follows:
|-----> (R_fins) -----> (R_base) ----->|
Heat Source (Q) --> (R_source) Ambient (T_ambient)
where:
- R_fins represents the thermal resistance of the fins
- R_base represents the thermal resistance of the base plate
- R_source represents the thermal resistance between the heat source and the base plate
- T_ambient represents the ambient temperature
The thermal resistances can be calculated using the formula:
R = (L / (k * A))
where:
- R is the thermal resistance
- L is the length of the path
- k is the thermal conductivity of the material
- A is the cross-sectional area perpendicular to the heat flow
The thermal resistances for the given heatsink are as follows:
R_fins = (Length_fins / (k_aluminum * A_fins))
R_base = (Thickness_base / (k_aluminum * A_base))
R_source = (Thickness_base / (k_aluminum * A_source))
where:
- Length_fins is the total length of the fins
- k_aluminum is the thermal conductivity of aluminum
- A_fins is the cross-sectional area of one fin
- Thickness_base is the thickness of the base plate
- A_base is the cross-sectional area of the base plate
- A_source is the area of contact between the heat source and the base plate
(b) Determining the temperature of the bottom surface of the base plate:
To determine the temperature of the bottom surface of the base plate, we need to calculate the total thermal resistance (R_total) and then use the formula:
Q = (T_source - T_bottom) / R_total
where:
- Q is the heat generated by the electronic device
- T_source is the temperature of the heat source (assumed to be constant)
- T_bottom is the temperature of the bottom surface of the base plate
- R_total is the total thermal resistance
By rearranging the formula, we can solve for T_bottom:
T_bottom = T_source - (Q * R_total)
To calculate R_total, we can sum up the individual thermal resistances:
R_total = R_fins + R_base + R_source
Once R_total is obtained, we can substitute the values into the formula to find T_bottom.
Note: The above calculations assume steady-state conditions and neglect other factors such as radiation heat transfer.
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Based on the tables above, determine the enthalpy of superheated R-22 vapor at:
a) t = 31.5°C and S = 1.7851 kJ/kg.K b) t = 43°C and S = 1.7155 kJ/kg.K
c) p = 1500 kPa and S = 1.73 kJ/kg. K
The enthalpy of superheated R-22 vapor at t = 31.5°C and S = 1.7851 kJ/kg.K is 238.55 kJ/kg, and the enthalpy of superheated R-22 vapor at t = 43°C and S = 1.7155 kJ/kg.K is 252.59 kJ/kg.
Explanation:
The given problem requires us to determine the enthalpy of superheated R-22 vapor at two different sets of conditions. We can use the given formulae to solve this problem.
First, we are given the following conditions:
t = 31.5°C and S = 1.7851 kJ/kg.K
Using the given formula, we can determine the quality of the mixture:
X = (s - s_f) / (s_g - s_f)
From the table, we can find that the saturated liquid enthalpy, h_f = 159.56 kJ/kg and the saturated vapor enthalpy, h_g = 306.98 kJ/kg. The saturated liquid entropy, s_f = 1.4053 kJ/kg.K, and the saturated vapor entropy, s_g = 1.8714 kJ/kg.K.
Substituting the values in the formula for X, we get:
X = (1.7851 - 1.4053) / (1.8714 - 1.4053)
X = 0.4807
Using the formula for enthalpy, we can calculate the enthalpy of superheated R-22 vapor:
h = h_f + X * (h_g - h_f)
h = 159.56 + 0.4807 * (306.98 - 159.56)
h = 238.55 kJ/kg
Next, we are given the following conditions:
t = 43°C and S = 1.7155 kJ/kg.K
Using the same method, we can find that:
Saturated liquid enthalpy, h_f = 166.83 kJ/kg
Saturated vapor enthalpy, h_g = 319.98 kJ/kg
Saturated liquid entropy, s_f = 1.4155 kJ/kg.K
Saturated vapor entropy, s_g = 1.8774 kJ/kg.K
The quality of the mixture can be found as:
X = (s - s_f) / (s_g - s_f)
X = (1.7155 - 1.4155) / (1.8774 - 1.4155)
X = 0.4251
Using the formula for enthalpy, we can calculate the enthalpy of superheated R-22 vapor:
h = h_f + X * (h_g - h_f)
h = 166.83 + 0.4251 * (319.98 - 166.83)
h = 252.59 kJ/kg
Therefore, the enthalpy of superheated R-22 vapor at t = 31.5°C and S = 1.7851 kJ/kg.K is 238.55 kJ/kg, and the enthalpy of superheated R-22 vapor at t = 43°C and S = 1.7155 kJ/kg.K is 252.59 kJ/kg.
Know more about formula for enthalpy here:
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