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
Δ
H + Δu2/2 + gΔz = Q + Wssince there is no shaft work, Ws = 0
and flow is horizontal, Δz = 0
Therefore,
Δ
H + Δu2/2 = Qand since there is no heat transfer, Q = 0
Therefore, ΔH + Δu2/2 = 0
Applying continuity equations,
u1A1ρ1 = u2A2ρ2
for water ρ1 = ρ2
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, ΔH = 71.11 J/kg
Enthalpy change per kg mass = 71.11 J
We know, ΔH = mCpΔT and, m = 1000 g; Cp = 4.184 J/g.oC
Therefore,
Δ
T = 71.11/(1000 x 4.184) = 0.017oCTemperature change = 0.017oC
Last Modified on: 04-Feb-2022
Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in