Temperature Change During Throttling

2. Temperature Change During Throttling

Recall from Chapter 2 that the internal energy of unit mass of any fluid can be changed by frictional process and by performing boundary deformation work (Pdv). For in- compressible fluid, dv=0 and hence boundary deformation work is zero; thus "u" can change only with a frictionless process (e.g., flow of liquids through pipes) where the mechanical part of energy “vdP” is converted into internal energy using frictional processes. Thus a combi- nation of both of the processes Pdv and vdP result in temperature change during the throttling process.

a. Incompressible Fluid Assume that an incompressible fluid is throttled from a higher to a lower pressure un- der steady state conditions. Let us follow unit mass as it enters and exits the throttling device. Since dv=0, there is no deformation work and hence u cannot change due to Pdv. However the unit mass is pushed into the throttling device with a pump work of vP i , while the same mass is

pushed out with a pump work of vP e . Thus there must have been destruction of mechanical energy from vp i to vP e which is converted into thermal energy (see Chapter 2) resulting in an increase of u and hence an increase of T during throttling. The energy increases by an amount du = –vdP ≈ –v(P e –P i ). (Alternately dh =du+ pdv+ vdp = du +0+ vdp=0.) Also, recall that µ JT

≈ – v/c P < 0. Further throttling is inherently irreversible process and hence entropy always in- creases in adiabatic throttling process (e.g., increased T causes increased s).

b. Ideal Gas Now consider a compressible ideal gas. Visualize the throttling process as a two step procedure. First the specific volume is maintained as though fluid is incompressible and the energy rises by the amount du = –vdP. This leads to gas heating. Now let the volume increase during the second stage (i.e expansion to low pressure). The volume increase cannot change the intermolecular potential energy, since the gas is ideal, but Pdv work is performed. This leads to decrease in the internal energy. The total energy change is du = – vdP – Pdv = – d(Pv), i.e., or du + d(Pv) = dh = 0. For ideal gases , dh = c p dT, and, hence, dT = 0. In this case, the

energy decrease by the Pdv deformation work equals the energy increase due to pumping (=- vdP)

c. Real Gas In a real gas the additional work required to overcome the intermolecular attraction forces (or the increase in the molecular intermolecular potential energy, “ipe”) must be ac- counted for. Then the temperature can decrease if increase in “ipe” is significant. Consider the relation

dh = du + d(Pv) = du +P dv + v dP. Since du = c v dT + (T ( ∂P/∂T) v – P) dv,

dh =c v dT + (T( ∂P/∂T) v – P)dv + Pdv + vdP. (161) The terms on the RHS of this equation, respectively, denote (1) the change in the temperature

due to change in the molecular translational, vibrational, and rotational energies, (2) the inter- molecular potential energy change, (3) the deformation work, and (4) the work required for pumping. During throttling, dh = 0, the intermolecular potential energy increases, the work deformation is positive, and dP < 0 so that vdP <0. In case if vdP ≈ 0, then it is apparent that

the net change in the molecular translational, vibrational, and rotational energies is negative, i.e., dT < 0. Recall (15), i.e.,

µ JT = – (v –T ( ∂v/∂T) P )/c p = – v/c p + {T ( ∂v/∂T) P }/c p .

[h 0 (T) -h(T,P)]/RT c A

Figure 21: Use of the enthalpy correction charts to de- termine the inversion conditions.

The first term on the RHS of this expression represents the heating effect due to flow work, while the second term accounts for the entire Pdv work and the energy required to overcome the intermolecular forces. Generally, both the terms are important for fluids.