Question

A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kJ/kg, respectively, and exit specific enthalpy of 297.31 kJ/kg. For negligible heat transfer with the surroundings, the exit velocity is:

a) 223 m/sb) 197 m/xc) 90 m/sd) 70 m/s

Answer 1

correct option is c) 90 m/s

Step-by-step explanation:

given data

inlet velocity v1 = 250 m/s

specific enthalpy h1 = 270.11 kJ/kg = 270110 J/kg

exit specific enthalpy h2 = 297.31 kJ/kg = 297310 J/kg

solution

we use here energy energy rate balance equation for get exit velocity that is express as

(h1 - h2) +
`(v1^2-v2^2)/(2)` = 0

so v2 will be

v2 =
`√(2(h1-h2)+v1^2)` ................1

put here value and we will get

v2 =
`√(2* (270.11-297.31)* 10^3 +250^2)`

solve it we get

v2 = 90 m/s

so correct option is c) 90 m/s