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## Q. 3.7

If in Ex. 3.6 the flow rate of air is $1 mol.s^{–1}$ and if both upstream and downstream pipes have an inner diameter of 5 cm, what is the kinetic-energy change of the air and what is its temperature change? For air, $C^{ig}_{p} = 29.100 J·mol^{−1}$ and the molar mass is $ℳ = 29 g·mol^{–1}$.

## Verified Solution

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.