Answer:
[tex]D=0.1160m[/tex]
Explanation:
From the question we are told that:
Output Power [tex]P=1.2kw[/tex]
Density [tex]\rho=1.29kg/m^3[/tex]
Wind speed [tex]V=17.5mph=>7.8m/s[/tex]
Efficiency [tex]\gamma=60\%=>0.60[/tex]
Let Betz Limit
[tex]C_p=\frac{16}{27}[/tex]
Generally the equation for Turbine Efficiency is mathematically given by
[tex]\gamma=\frac{P}{P'}[/tex]
Where
P'=input power
[tex]P' = In\ power[/tex]
[tex]P'=1/2*C_p*\rho u^3*A[/tex]
[tex]P'=1/2*C_p*\rho u^3*\frac{\pi}{4}*D^2[/tex]
Therefore the blade diameter for a two-blade propeller type rotor is
[tex]\gamma=\frac{P*2*27*4}{16*\rho u^3*\pi*D^2}[/tex]
[tex]D^2=\frac{P*2*27*4}{16*\rho u^3*\pi*\gamma}[/tex]
[tex]D^2=\frac{1.2*2*27*4}{16*1.29*7.91^3*\pi*0.60}[/tex]
[tex]D^2=0.0135[/tex]
[tex]D=\sqrt{0.0135}[/tex]
[tex]D=0.1160m[/tex]
Answer:
0.1160
Explanation: give them brainliest they deserve it
Steam enters a turbine with a pressure of 30 bar, a temperature of 400 oC, and a velocity of 160 m/s. Saturated vapor at 100 oC exits with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 350 K. Determine the rate at which entropy is produced within the turbine per kg of steam flo
Answer:
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
Explanation:
By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:
Principle of Mass Conservation
[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)
First Law of Thermodynamics
[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)
Second Law of Thermodynamics
[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)
By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:
[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)
[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)
Where:
[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.
[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.
[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.
[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.
[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.
[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.
[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.
[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.
[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.
By property charts for steam, we get the following information:
Inlet
[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
Outlet
[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]
If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:
[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]
[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]
[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]
[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 3.5 kW from the compressor to its surroundings.
Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Answer:
- 46.5171kW
Explanation:
FIrst, the value given:
P1 = 1.05 bar (Initial pressure)
P2 = 12 bar (final pressure)
Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)
Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)
Universal gas constant, Ru = 8.3143 Kj/Kgmolk
Specific gas constant, R = 0.28699 Kj/KgK
Initial temperature, T1 = 300 K
Final temperature, T2 = 400 K
Finding the volume:
P1V1 = RT1
V1 = RT1 ÷ P1
= (0.28699 Kj/KgK X 300k) ÷ 105
Note convert bar to Kj/Nm by multiply it by 100
V1 = 0.81997 m3/Kg
To get the mass flow rate:
m = volumetric flow rate / V1
= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg
= 0.4268Kg/s
Using tables for the enthalpy,
hT1 = 300.19 KJ/Kg
hT2 = 400.98 KJ/Kg
The enthalpy change = hT2 - hT1
= 100.79 KJ/Kg
Power, P = Q - (m X enthalpy change)
= - 3.5 - (0.4268 X 100.79)
= - 46.5171kW
A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity 0.92 and is subject to a free convection heat transfer coefficient of h 4.5 W/m2 K. The room air and wall temperatures are T 22 C and Tsur 20 C, respectively. If electricity costs C $0.18/kW h, determine the daily cost of leaving the charger plugged in when not in use.
Answer:
C = $0.0032 per day
Explanation:
We are given;
Dimension of cell phone; 50 mm × 45 mm × 20 mm
Temperature of charger; T1 = 33°C = 306K
Emissivity; ε = 0.92
convection heat transfer coefficient; h = 4.5 W/m².K
Room air temperature; T∞ = 22°C = 295K
Wall temperature; T2 = 20°C = 293 K
Cost of electricity; C = $0.18/kW.h
Chargers are usually in the form of a cuboid, and thus, surface Area is;
A = (50 × 45) + 2(50 × 20) + 2(45 × 20)
A = 6050 mm²
A = 6.05 × 10^(-3) m²
Formula for total heat transfer rate is;
E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)
Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴
Thus;
E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))
E_t = 0.7406 W = 0.7406 × 10^(-3) KW
Now, we know C = $0.18/kW.h
Thus daily cost which has 24 hours gives;
C = 0.18 × 0.7406 × 10^(-3) × 24
C = $0.0032 per day
Outline how the technological innovations of the Renaissance led to the Industrial Age.
Answer:
Printing press , lenses etc,
Explanation:
Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.
Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.
Hey answr this sajida Yusof
Answer:
that not even a question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Answer:
Hydrostatic force = 41168 N
Explanation:
Complete question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Let "x" be the side length submerged in water.
Then
w(x)/base = (4+3-x)/altitude
w(x)/5 = (4+3-x)/3
w(x) = 5* (7-x)/3
Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7
HF = integration of 40x - 4x^2/3
HF = 20x^2 - 4x^3/9 with limit 4 to 7
HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))
HF = 658.69 N *62.5 = 41168 N
How do guest room hotel smoke alarms work and differ then regular home versions?
Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room
identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
Answer:
L = Henry
C = Farad
Explanation:
The electrical parameter represented as L is the inductance whose unit is Henry(H).
The electrical parameter represented as C is the inductance whose unit is Farad
Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :
To obtain :
At resonance, Inductive reactance = capacitive reactance
Equate the inductive and capacitive reactance
Inductive reactance(Xl) = 2πFL
Capacitive Reactance(Xc) = 1/2πFC
Inductive reactance(Xl) = Capacitive Reactance(Xc)
2πFL = 1/2πFC
Multiplying both sides by F
F * 2πFL = F * 1/2πFC
2πF²L = 1/2πC
Isolating F²
F² = 1/2πC2πL
F² = 1/4π²LC
Take the square root of both sides to make F the subject
F = √1 / √4π²LC
F = 1 /2π√LC
Hence, the proof.
where can I find solved problems of advanced soil structure interaction?
The seismic response of nuclear power plant structures is often calculated using lumped parameter methods. A finite element model of the structure is coupled to the soil with a spring-dashpot system used to represent the interaction process. The parameters of the interaction model are based on analytic solutions to simple problems which are idealizations of the actual problems of interest. The objective of the work reported in this paper is to compare predicted responses using the standard lumped parameter models with experimental data. These comparisons are shown to be good for a fairly uniform soil system and for loadings that do not result in nonlinear interaction effects such as liftoff. 7 references, 7 figures.
A resistor, inductor, and capacitor are in parallel in a circuit where the frequency of operation can vary. The R, L, and C values are such that at the frequency omega subscript 0, the magnitude of all the impedances are equal to each other. If the frequency of operation approaches zero, which element will dominate in determining the equivalent impedance of this parallel combination?
a. The inductor.
b. The capacitor.
c. The resistor.
d. Insufficient information provided.
Answer:
Option A is correct
Explanation:
As we know
Inductive Susceptance = ½(pi)*f*L
Or Inductive Susceptance is inversely proportional to the frequency
Likewise conductive Susceptance = 2 (pi)*f*C
Conductive Susceptance is directly proportional to the frequency
When the frequency will reach the value zero, then the Inductive Susceptance will become infinite
Hence, inductor will dominate in determining the equivalent impedance of this parallel combination
Option A