Steam enters a nozzle operating at steady state at 20 bar, 2808C, with a velocity of 80 m/s. The exit pressure and temperature are 7 bar and 1808C, respectively. The mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energy, determine (a) the exit velocity, in m/s. (b) the inlet and exit flow areas, in cm2

A) 600.8 m/s

B) (i) 22.5 cm^2   (ii) 7.11 cm^2

Explanation:

Given data :

P1 = 20 bar ,  T1 = 2808°C

P2 = 7 bar ,  T2 = 1808°C

mass flow rate = 1.5 kg/s

Using the superheated vapor region in Table A-4

h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg

h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg  

A) calculate exit velocity  ( m/s )

given that we are to neglect heat transfer and potential energy

V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )

                             = [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )

                             = 600.8 m/s

B) calculate the inlet and exit flow areas ( cm^2 )

i) Inlet flow area

A1 = ( m * v1 ) / V1

    = ( 1.5 * 0.1200 )/ 80  = 22.5 cm^2

ii) exit flow area

A2 = ( m * v2 ) / V2

    = ( 1.5 * 0.2847 ) / 600.8  = 7.11 cm^2

answer:

clock

explanation:

always running (on) sing each hour (chime) and the clock face (front)