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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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Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t
almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

n = \frac{h_1 - h_2*}{h_1 - h_2}

n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}

solving for h2, we have

h_2 = 1197.77 Btu/Ibm

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0
3 years ago
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