By Eq. (2.23b),

\dot{n} = uA\rho (2.23b)

u = \frac{\dot{n} }{A\rho } = \frac{\dot{n} V^{ig} }{A}

where

A = \frac{\pi }{4} D^{2} = (\frac{\pi }{4})(5 \times 10^{-2} )^{2} = 1.964 \times 10^{-3} m^{2}

The appropriate value of the gas constant for calculating the upstream molar volume is R = 83.14 \times 10^{-6}  bar·m^{3}·mol^{-1}·K^{-1} . Then

V^{ig}_{1} = \frac{RT_{1} }{P_{1} } = \frac{(83.14 \times 10^{-6} )(293.15K)}{6 bar} = 4.062 × 10^{-3} m^{3}·mol^{-1}

Then,

u_{1} = \frac{(1 mol· s^{-1})(4.062 \times  10^{-3} m^{3} ·mol^{-1})}{1.964 \times 10^{-3} m^{2}}= 2.069  m· s^{-1}

If the downstream temperature is little changed from the upstream temperature, then to a good approximation:

V^{ig}_{2} =V^{ig}_{1}       and       u_{2} = 2 u_{1} = 4.138   m· s^{−1}

The rate of change in kinetic energy is therefore:

\dot{m} \Delta  \left(\frac{1}{2}u^{2}\right) = \dot{n} ℳ \Delta  \left(\frac{1}{2}u^{2}\right) = (1 \times 29 \times  10^{-3}kg·s^{−1} )\frac{(4.138^{2} – 2.069^{2})m^{2}. s^{-2} }{2} = 0.186 kg·m^{2}. s^{-3}  = 0.186 J· s^{-1}

In the absence of heat transfer and work, the energy balance, Eq. (2.30), becomes:

\Delta \left(H + \frac{1}{2} u^{2} + zg \right)\dot{m} = \dot{\varrho } +\dot{W_{s} }     (2.30)

\Delta \left(H^{ig} + \frac{1}{2} u^{2} \right) \dot{m} = \dot{m} \Delta H^{ig} + \dot{m} \Delta \left(\frac{1}{2} u^{2} \right)= 0

\left(\dot{m} lℳ\right) C^{ig}_{P}\Delta T + \dot{m}\Delta \left(\frac{1}{2} u^{2} \right) = \dot{n} C^{ig}_{P} \Delta T + \dot{m} \Delta \left(\frac{1}{2} u^{2} \right) = 0

Then

(1)(29.100)\Delta T = -\dot{m\Delta }(\frac{1}{2}u^{2} ) = -0.186

and

ΔT = −0.0064 K

Clearly, the assumption of negligible temperature change across the valve is justified. Even for an upstream pressure of 10 bar and a downstream pressure of 1 bar and for the same flow rate, the temperature change is only −0.076 K. We conclude that, except for very unusual conditions, ΔH^{ig} = 0 is a satisfactory energy balance.