# Is flow energy the same as work?

Studying thermodynamics I face this following text from Cengel's book.

Throttling valves are usually small devices, and the flow through them may be assumed to be adiabatic (q = 0) since there is neither sufficient time nor large enough area for any effective heat transfer to take place. Also, there is no work done (w=0), and the change in potential energy, if any, is very small (pe=0). Even though the exit velocity is often considerably higher than the inlet velocity, in many cases, the increase in kinetic energy is insignificant (ke=0). Then the conservation of energy equation for this single-stream steady-flow device reduces to **h2>h1 (kJ/kg)*

Previously, I thought that enthalpy meant internal energy + work energy (PV). So If the book says throttling valves have no work involved, how is energy equation about enthalpy (u + PV)?

• The book meant that there is no work except the Flow work. – lucas May 16 '16 at 5:20
• The Cengel-Boles book writes $h_2 \cong h_1$ not $>$ and then goes onto explain it, just read the paragraph below the one you quoted. – hyportnex May 16 '16 at 12:38

If you have studied the open system version of the first law, you are aware that, for this mathematical representation of the energy balance, work is divided into two parts: shaft work and work to push fluid into and out of the control volume. The work to push material into and out of the control volume is combined mathematically into the energy of the flowing stream, so that it is expressed in terms of enthalpy rather than internal energy. For a throttling device, the shaft work is zero, and the system is assumed to operate at steady state, so that the rate of change of internal energy within the device is zero. So, in this case, the enthalpy per unit mass entering the device is equal to the enthalpy per unit mass exiting.

The energy conservation equation can be written as below. I guess Cengel was making comments on each items.

$$\delta U = Q -W + m_1(h_1+\frac 12 v_1^2 + gz_1)-m_2(h_2+\frac 12 v_2^2 + gz_2)$$

For steady state, we know $\delta U =0$, $m_1=m_2$

So reading the book, we can translate that to,

1. q=0: $Q=0$
2. w=0: $W=0$
3. pe=0: $m_1gz_1 \approx m_2gz_2$
4. ke=0: $m_1\frac 12 v_1^2 \approx m_2\frac 12 v_2^2$

The leftover is then: $m_1h_1 \approx m_2h_2$ or $h_1 \approx h_2$