Question

Throttling of fluids.Throttling is the process of converting a high pressure fluid to low pressure typically done through a valve.

Water is throttled from 20 bar , 60°C to a temperature where it is a vapor/liquid mixture with a moisture content (xL) of 0.9318. What is the temperature at the exit?

Answer 1

Assume 1 Kg of water is taken

we know throttling process is isenthalpic process (Enthalpy remain constant)

Enthalpy of water at 20 bar, 60 oC = 253 Kj/Kg . heat content = 253 Kj

after throttling it is liquid vapor mixure, amount of liquid = 0.9318 Kg as X_L = 0.9318

amount of vapor = 1-0.9318 = 0.0682

apply energy balance

heat of liquid + latent heat of evaporation + heat of vapor = heat of liquid throttled

0.9318*4.18 (60-T) + 0.0682*2260 + 0.0682 * 2.108 * (60-T) = 253 ........... (1)

as Cp heat capacity water = 4.18 Kj/Kg K latent heat of evaporation = 2260 Kj/Kg , Cp of vapor = 2.108

energy of liquid throttled = 253 Kj temperature diffrence = 60- T as they are in equilibrium

solve equation (1) T = 35.5 oC

Note: as in the above equation there is a temperature difference so there will be no impact of unit either we take in the form of oC or K