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3 years ago

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Which of the following statements about pitot-static systems is FALSE? a). A pitot probe measures the Total Pressure of the free

stream. b). A static port measures the Pressure of the freestream air. c). A pitot-static system measures the Dynamic Pressure using the formula Dynamic Pressure = Total minus Static. d). Dynamic Pressure is equal to the Static Pressure at a stagnation point

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

C

Explanation:

Pitot tube:

Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.

As we know that stagnation pressure is the summation of dynamic and static pressure.

Stagnation pressure = Static pressure + Dynamic pressure

So

Dynamic pressure = Stagnation pressure - Static pressure

We know that dynamic pressure

$P_{dynamic}=\dfrac{\rho V^2}{2}$

On the other hand Pitot tube measure the dynamic pressure.

So option C is correct.

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2 years ago

1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con

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Answer:

Explanation: Here it is: 67 Hope that helps! :)

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3 years ago

A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh

sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

$T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C$

to kelvin = (5 + 273)K = 278 K

From the " thermophysical properties of gases at atmospheric pressure" table; At $T_f$ = 278 K ; by interpolation; we have the following

$\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}$ → v 13.93 (10⁻⁶) m²/s

$\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}$ → k = 0.0245 W/m.K

$\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}$ → ∝ = 19.6(10⁻⁶)m²/s

$\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}$ → Pr = 0.713

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= $\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}$

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$h = \frac{4.878*0.0245}{0.06}$

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Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

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3 years ago

Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its

kirill115 [55]

<h2>Answer:</h2>

<h3>Required Answer is as follows :-</h3>

$\sf \dfrac{P _{i}}{ P_{f}} = \dfrac{3}{4}$

3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²

¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²

¾ = m₁ × (V)²/m₂ × (V²/4)

¾ = m₁/m₂ × 4

m₂ = 4 × 4/3 = 16/3m₁

$\bold{ }$

<h3>Now,</h3>

∆m = m₂ - m₁
∆m = 16m₂/3 - m₁ = 13m₁/3
Ratio = (13m₁/3)/ m₁
Ratio = 13/3

Ratio = 13:3

$\bold{ }$

<h3>Know More :-</h3>

Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg. $\bold{ }$

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3 years ago

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Nana76 [90]

Answer:

Equation:

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3 years ago