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Mama L [17]
3 years ago
7

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

servoir of the nozzle, the pressure and temperature are 10 atm and 300 K, respectively. At the nozzle exit, the pressure is 1 atm. Calculate the temperature and density of the flow at the exit. Assume that the flow is isentropic and (of course) compressible.
Physics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

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a

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7 0
3 years ago
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5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw
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For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle  is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle   is mathematically given as

\theta=n(\lambda/d)

Therefore

\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)

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2 years ago
A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\
zloy xaker [14]

Answer:

  t = 1.77 s

Explanation:

The equation of a traveling wave is

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where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

      v = λ f

Where the frequency and period are related

     f = 1 / T

we substitute

      v = λ / T

let's develop the initial equation

    y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

 

         2π /λ = 1.65

          λ = 2π / 1.65

          λ = 3.81 m

         2π / T = 4.64

          T = 2π / 4.64

          T = 1.35 s

we seek the speed of the wave

           v = 3.81 / 1.35

           v = 2.82 m / s

           

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          v = d / t

           t = d / v

           t = 5 / 2.82

           t = 1.77 s

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3 years ago
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5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
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