Liquid water at 25^{o}C 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 + Du^{2}/2 + gDz = Q + W_{s}

since there is no shaft work, W_{s} = 0

and flow is horizontal, Dz = 0

Therefore,

DH + Du^{2}/2 = Q

and since there is no heat transfer, Q = 0

Therefore, DH + Du^{2}/2 = 0

Applying continuity equations,

u_{1}A_{1}r_{1} = u_{2}A_{2}r_{2}

for water r_{1} = r_{2}

Therefore,

u_{2}/u_{1} = A_{1}/A_{2} = d_{1}^{2}/d_{2}^{2} = 2.5^{2}/7.5^{2} = 0.1111

u_{2} = 12 x 0.1111 = 1.333 m/s

(u_{2}^{2} - u_{1}^{2}) / 2 = (1.333^{2} - 12^{2}) / 2 = -71.11

Therefore, DH = 71.11 J/kg

Enthalpy change per kg mass = 71.11 J

We know, DH = mC_{p}DT and, m = 1000 g; C_{p} = 4.184 J/g.^{o}C

Therefore,

DT = 71.11/(1000 x 4.184) = 0.017^{o}C

Temperature change = 0.017^{o}C