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prisoha [69]
3 years ago
8

Tiny hydrogen bubbles are being used as tracers to visualize a flow. All the bubbles are generated at the origin (x = 0, y = 0).

The velocity is unsteady and obeys the equations: u = 1 m/s v = 2 m/s 0 ≤ t < 2 s u = 0 m/s v = −1 m/s 2 s ≤ t ≤ 4 s Plot the pathlines of bubbles that leave the origin at t = 0, 1, 2, 3, and 4 s. Mark the locations of these five bubbles at t = 4 s. Use a dashed line to indicate the position of a streakline at t = 4 s

Engineering
1 answer:
notka56 [123]3 years ago
3 0

Answer:

for t = 1 : ( x1 , y1 ) = (1,2)

for t = 2 : ( x2, y2 ) = (2,4)

for t = 3 : ( x3, y3 ) = ( 0,-3)

for t = 4 : ( x4, y4 ) = ( 0,-4)

Explanation:

Attached below is the detailed solution and the sketch

we have to make assumptions for the various values of t and integrate accordingly

for t = 1 : ( x1 , y1 ) = (1,2)

for t = 2 : ( x2, y2 ) = (2,4)

for t = 3 : ( x3, y3 ) = ( 0,-3)

for t = 4 : ( x4, y4 ) = ( 0,-4)

Given data :

origin ( x = 0, y = 0 )

u = 1 m/s,  v = 2 m/s,

0 \leq t < 2s  where u = 0 m/s   v = -1 m/s

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An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?
kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = \frac{NE}{t}

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = \frac{NE}{t} = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}          [t = 1 s for per second]

or

550 × 10³ = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}

or

550 = N × 4903.24 × 10⁻³⁴

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7 0
3 years ago
Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap
alexdok [17]

Answer:

heat rate   = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

     = (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

     = 5.6*10^-4/3.59*10^-7

     = 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient  using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

   =π * 0.005 * 2

    =0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) =  140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

     147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

   = 29.34* 0.0314(160-151.99)

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