All categories

3 years ago

What is meant by the thickness to chord ratio of an aerofoil?

Engineering

1 answer:

Karolina [17]3 years ago

6 0

Answer:

Chord ratio:

It is the ratio of thickness to the chord.It is also knows as thickness ratio.

Chord ratio measure the performance of wing when it is operating at the transonic speed.

When the speed cross the speed of sound wave then that wave creates the shock wave and these shock leads to produce drag force on the aerofoil profile.When Mach number is one then it means that if Mach increase then it will leads increase then drag force.

You might be interested in

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator

MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit $i_m=385mA=385\times 10^{-3}A=0.385A$

Inductance of the inductor $L=400mH=400\times 10^{-3}h=0.4H$

Capacitance $C=4.43\mu F=4.43\times 10^{-3}F$

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be $x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm$

Capacitive reactance will be equal to $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm$

Impedance of the circuit will be $Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm$

So maximum voltage will be $\Delta V_{max}=0.385\times 865.44=333.194volt$

(B) Phase difference will be given as $\Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70$

So current will be leading by an angle 54.70

5 0

3 years ago

What’s the difference between engineering stress and strain and true stress and strain

Nana76 [90]

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

4 0

3 years ago

Read 2 more answers

If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit

mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

$\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}$

$\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}$

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

$\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}$

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

$\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}$

choose R2, R1 such that it will maintain required ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0

3 years ago

He is going ___ in the hot air ballon

Vladimir [108]

no artical shoul be used here

5 0

2 years ago

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an

Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = $1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}$

Cyclone is to 80 % efficient

So particulate remaining = $0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136$

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0

3 years ago