Change in Temperature due to Enthalpy Change

Liquid water at 25^oC 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

ΔH + Δu²/2 + gΔz = Q + W_s

since there is no shaft work, W_s = 0

and flow is horizontal, Δz = 0

Therefore,

ΔH + Δu²/2 = Q

and 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 = mC_pΔT and, m = 1000 g; C_p = 4.184 J/g.^oC

Therefore,

ΔT = 71.11/(1000 x 4.184) = 0.017^oC

Temperature change = 0.017^oC