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attashe74 [19]
3 years ago
11

Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle

entrance (state 1), the temperature and pressure are 325 degree Celsius and 700 kPa. And the velocity is 30 m/sec. At the nozzle exit (state 2) the steam temperature and pressure are 250 degree Celsius and 350 kPa. Property values are: H1=3,112.5 kJ/kg, V1=388.61 cm3/gr H2=2,945.7 kJ/kg, V2=667.75 cm3/gr What is the velocity of the steam at the exit and what is the exit diameter?
Engineering
1 answer:
777dan777 [17]3 years ago
6 0

Answer:

U_2 = 578.359 m /s

d_e = 1.4924 cm

Explanation:

Given:

Length of nozzle L = 25 cm

Inlet diameter d_i = 5 cm

Nozzle Entrance : T_1 = 325 C , P_1 = 700 KPa , U_1 = 30 m / s , H_1 = 3112.5 KJ / kg , V_1 = 388.61 cm^3 / g

Nozzle Exit : T_2 = 250 C , P_2 = 350 KPa , U_2, H_2 = 2945.7 KJ / kg , V_2 = 667.75 cm^3 / g

To Find:

a. Velocity at exit U_2

b. Exit Diameter d_e

a.

Energy Equation is given as:

ΔH + ΔU^2 / 2 + gΔz = Q + W

Q = W = Δz = 0

Substitute the values:

(H_2 - H_1 ) + (U^2_2 - U^2_1 )  / 2 = 0

U_2 = sqrt((2* (H_1 - H_2 ) ) + U^2_1)

U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900)

U_2 = 578.359 m / s

b.

Mass Balance :

U_1 * A_1 / V_1 = U_2 * A_2 / V_2

A = pi*d^2 / 4

U_1 * d_i ^ 2 / V_1 = U_2 * d_e ^2 / V_2

d_e = d_i * sqrt ( (U_1 / U_2) * (V_2 / V_1))

d_e = 5 * sqrt ((30 / 578.359) * (667.75 / 388.61))

d_e = 1.4924 cm

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LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

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For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.  

Using the given mass m, molar mass M, we can get the following equation:  

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To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

 

Now we can calculate change of U:

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5 0
3 years ago
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2 years ago
In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had
kipiarov [429]

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

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5 0
3 years ago
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MAXImum [283]
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Answer:

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Integrating we get

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Putting t=2

8=8k/3-8

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k=6

5 0
3 years ago
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