Question

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?

Answer 1

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