Change in Temperature due to Enthalpy Change

Liquid water at 25oC flows in a straight horizontal pipe, in which there is no exchange of either heat or work with the surroundings. Its velocity is 12 m/s in a pipe with an i.d. of 2.5 cm until it flows into a section where the pipe diameter increases to 7.5 cm. What is the temperature change?

Calculations:

For the steady flow process, the first law is written as

DH + Du2/2 + gDz = Q + Ws

since there is no shaft work, Ws = 0

and flow is horizontal, Dz = 0

Therefore,

DH + Du2/2 = Q

and since there is no heat transfer, Q = 0

Therefore, DH + Du2/2 = 0

Applying continuity equations,

u1A1r1 = u2A2r2

for water r1 = r2

Therefore,

u2/u1 = A1/A2 = d12/d22 = 2.52/7.52 = 0.1111

u2 = 12 x 0.1111 = 1.333 m/s

(u22 - u12) / 2 = (1.3332 - 122) / 2 = -71.11

Therefore, DH = 71.11 J/kg

Enthalpy change per kg mass = 71.11 J

We know, DH = mCpDT and, m = 1000 g; Cp = 4.184 J/g.oC

Therefore,

DT = 71.11/(1000 x 4.184) = 0.017oC

Temperature change = 0.017oC