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agasfer [191]
3 years ago
10

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
Engineering
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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A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>
  • Processes for manufacturing chemicals are created and designed by chemical engineers.
  • To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
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7 0
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
QveST [7]

Answer:

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

5 0
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

\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}

The Rayleigh number for vertical cavity

Ra_L  = \frac{g \beta (T_1-T_2)L^3}{\alpha v}

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

= 8.38*10^5

\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}

NU_L= \frac{hL}{k}= 4.878

\frac{hL}{k}= 4.878

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

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

h = 1.99 W/m².K

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

3 0
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{   } 

3 0
3 years ago
Write down the equation for the stoichiometric combustion of ethane (C2H6) with oxygen, and determine the stoichiometric air fue
Nana76 [90]

Answer:

Equation:

\text{C}_{2} \text{H}_{6} + \dfrac{7}{2}\text{O}_{2} \rightarrow 2\text{CO}_{2} + 3\text{H}_{2}\text{O}

Supose we have 1 kmol of ethane, according to the equation we should have 7/2 kmol of oxygen, witch corresponds to a number of mols of air of:

0.233\text{N}_{air} = 7/2

\text{N}_{air} = 15,0 \text{kmol} [\tex]As molar weight of ethane is [tex]\dfrac{M_{air}}{M_{ethane}} = \dfrac{N_{air}\cdot MW_{air}}{N_{ethane}\cdot MW_{etane}} = \dfrac{15kmol\cdot 28.9kg/kmol}{1kmol\cdot 30kg/kmol}=14.45 and air is 0.233*32 + 0.767*28 = 28.9kg/kmol, the mass ratio is:

7 0
3 years ago
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