Water vapor enters an insulated nozzle operating at steady state with a velocity of 100 m/s and specific enthalpy of 3445.3 kJ/kg, and exits with specific enthalpy of 3051.1 kJ/kg. The velocity at the exit is most closely

(a) 104 m/s
(b) 636 m/s
(c) 888 m/s
(d) 894 m/s

(a) 104 m/s

Explanation:

enters with specific enthalpy  'h1' = 3445.3 kJ/kg

exits with specific enthalpy 'h2' 3051.1 kJ/kg

velocity 'V1'= 100 m/s

In order to find velocity at the exit 'V2', we use this relation

0 = Qcv - Wcv + m[(h1-h2) +  + g(z1-z2)]

0= [Qcv + m[(h1-h2) + ]

0=( h1-h2) + ( )

0= (3445.3-3051.1)+

0= 394.2 +

-394.2=

-788.4 = 10000-V2²

V2²= 10000+ 788.4

V2²= 10784.44

Taking square root on both sides

V2=103.84

V2≈ 104m/s

Therefore, The velocity at the exit is most closely  104m/s

a

explanation:

because they are cartillidge

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it