Question

Liquidwaterat294.26Kflowsinastraighthorizontalpipeinwhichthere is no exchange of either heat or work with the surroundings. Its velocity is 9.144 m/s in a pipe with an internal dimeter of 2.54 cm until it flows into a section where the pipe dimeter of 2.54 cm until it flows into a section where the pipe dimeter abruptly increases. What does the enthalpy change of the water if the downstream dimeter is 3.81 cm?

Answer 1

It is a steady flow process, so,

(DH + Dv2)/2 + gDz = Q + Ws

Here,

Ws = 0 as it does not have shaft work,

Dz = 0 as there is a straight horizontal flow,

Q = 0 as no heat is being transferred

So,

(DH + Dv2)/2 + g(0) = (0) + (0)

DH + Dv2/2 = 0

Now continuity equations are:

A1v1= A2v2

Therefore,

A1/A2 = v2/v1 = (d1)^2/(d2)^2 = (2.54)^2 / (3.81)^2 = 6.4516/14.5161

=0.444444444439

v2 = 9.144 x 0.44444= 4.06399 m/s

Now,

DH = [(v2)^2 - (v1)^2)] / 2

DH = [(4.06399)^2 - (9.144)^2) / 2

DH = -33.54836 j/kg

Enthalpy change = 33.54836 per kg mass

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