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## Q. 4.P.11

Air is flowing at the rate of 30 kg/m²s through a smooth pipe of 50 mm diameter and 300 m long. If the upstream pressure is 800 kN/m², what will the downstream pressure be if the flow is isothermal at 273 K? Take the viscosity of air as 0.015 mN s/m² and assume that volume occupies 22.4 m³. What is the significance of the change in kinetic energy of the fluid?

## Verified Solution

$(G / A)^2 \ln \left(P_1 / P_2\right)+\left(P_2^2-P_1^2\right) / 2 P_1 v_1+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0$       (equation 4.55)

The specific volume at the upstream condition is:

$v_1=(22.4 / 29)(273 / 273)(101.3 / 800)=0.098 m ^3 / kg$

$G / A=30 kg / m ^2 s$

∴                  $\operatorname{Re}=(0.05 \times 30) /\left(0.015 \times 10^{-3}\right)=1.0 \times 10^5$

For a smooth pipe, $R / \rho u^2=0.0032$ from Fig. 3.7.
Substituting gives:

$(30)^2 \ln \left(800 / P_2\right)+\left(P_2^2-800^2\right) \times 10^6 /\left(2 \times 800 \times 10^3 \times 0.098\right)$

$+4(0.0032)(300 / 0.05)(30)^2=0$

and the downstream pressure, $P_2=\underline{\underline{793 kN / m ^2}}$,

The kinetic energy term $=(G / A)^2 \ln (800 / 793)=7.91 kg ^2 / m ^4 s ^2$
This is insignificant in comparison with $69,120 kg ^2 / m ^4 s ^2$ which is the value of the other terms in equation 4.55.