Why does boundary work in throttling process is zero when change in volume is not? For a steady-state steady flow process, total volume remain constant and therefore $\newcommand{\dd}[1]{\text{d}#1}\dd{V}=0$.
Now assume a throttling process where:
$$\dd{H}=\dd{Q}+\dd{W}_{\text{s}}+\dd{W}_{\text{boundary}}=0$$
$$\dd{H}=\dd{U}+\dd{(pV})=0$$
Then we write
$$\dd{U}=-p\dd{V}-V\dd{p}$$
If the working fluid is a real gas the $\dd{U}=-\dd{(pV)}$ as I see in many textbooks. Could you please help where I am going wrong?
 A: Boundary work is normally associated with the expansion and contraction of the boundaries of a system. The familiar example is a gas in a cylinder fitted with a movable piston.  The system is the gas. The boundaries are the cylinder walls and the face of the piston. Since the piston is movable, it permits the gas to expand or contract depending on whether the gas is doing work on the surroundings or the surroundings is doing work on the gas, respectively.
Throttling is basically an irreversible process in which a given quantity of working fluid flows from a region of high pressure, typically through a nozzle or valve, to a region of low pressure allowing the fluid to expand and increase its volume so that $P_{f}V_{f}=P_{i}V_{i}$ where $i$ and $f$ indicate initial and final states of the process, respectively. The nozzle or valve though which the fluid passes does not expand or contract. 
Regarding your first equation for $dH$ that you say is for a throttling process:
$dW$(Boundary) is normally expressed as $pdV$.  As already indicated,  boundary work is not relevant to nozzles or valves involved in a throttling process.
Presumably by $dW_s$ you mean shaft work. Shaft work is normally the work output and/or input during a steady flow open system process. It is typically the work done on the working fluid by compressors and pumps (work input) or work done by the working fluid by turbines (work output).  Shaft work is not relevant to a throttling process.
Regarding your equation for $du$, $vdP$ is flow work and is normally associated with the work needed to move a working fluid into or out of a control volume where pressure differentials, $dP$, exist across entrance and exit of the control volume. Flow work is not relevant to the throttling process.
Hope this helps.
A: In an open system throttling process, if you adopt a Lagrangian frame of reference and focus on a specific parcel of mass, its volume is not constant.  So even through the size of the control volume remains constant, the size of each gas parcel passing through the control volume does not.  You are aware that the total amount of work that is done on the surroundings with an open system is comprised of the shaft work plus the work needed to push mass into and out of the control volume, right?
