## Chapter 4

## Q. 4.P.7

Air, at a pressure of 10 MN/m² and a temperature of 290 K, flows from a reservoir through a mild steel pipe of 10 mm diameter and 30 m long into a second reservoir at a pressure P_2. Plot the mass rate of flow of the air as a function of the pressure P_2. Neglect any effects attributable to differences in level and assume an adiabatic expansion of the air. μ = 0.018 mN s/m², γ = 1.36.

## Step-by-Step

## Verified Solution

G/A is required as a function of P_2 \cdot v_2 cannot be found directly since the downstream temperature T_2 is unknown and varies as a function of the flowrate. For adiabatic flow, v_2 may be calculated from equation 4.77 using specified values of G/A and substituted in equation 4.72 to obtain the value of P_2. In this way the required data may be calculated.

8\left(R / \rho u^2\right)(l / d)=\left[\frac{\gamma-1}{2 \gamma}+\frac{P_1}{v_1}\left(\frac{A}{G}\right)^2\right]\left[1-\left(\frac{v_1}{v_2}\right)^2\right]-\frac{\gamma+1}{\gamma} \ln \left(\frac{v_2}{v_1}\right)(equation 4.77)

0.5(G / A)^2 v_1^2+[\gamma /(\gamma-1)] P_1 v_1=0.5(G / A)^2 v_2^2+[\gamma /(\gamma-1)] P_2 v_2(equation 4.72)

or: \frac{0.5(G / A)^2\left(v_1^2-v_2^2\right)+[\gamma /(\gamma-1)] P_1 v_1}{[\gamma /(\gamma-1)] v_2}=P_2

When P_2=P_1=10 MN / m ^2, G / A=0.

If G/A is 2000 kg/m²s, then:

R e=\left(0.01 \times 2000 / 0.018 \times 10^{-3}\right)=1.11 \times 10^6

When e / d=0.0002, R / \rho u^2=0.0028 from Fig. 3.7 and:

v_1=(22.4 / 29)(290 / 273)(0.1013 / 10)=0.0083 m ^3 / kg

Substituting in equation 4.77:

8(0.0028)(30 / 0.01)=\left[\frac{0.36}{2 \times 1.36}+\frac{10 \times 10^6}{0.0083}\left(\frac{1}{2000}\right)^2\right]

\times\left[1-\left(\frac{0.0083}{v_2}\right)^2\right]-\frac{2.36}{1.36} \ln \left(\frac{v_2}{0.0083}\right)

and: v_2=0.00942 m ^3 / kg.

Substituting for v_2 in equation 4.72 gives:

\times 0.00942

and: P_2=8.75 MN / m ^2.

In a similar way the following table may be produced.

G / A\left( kg / m ^2 s \right) | v_2\left( m ^3 / kg \right) | P_2\left( MN / m ^2\right) |

0 | 0.0083 | 10.0 |

2000 | 0.00942 | 8.75 |

3000 | 0.012 | 6.76 |

3500 | 0.0165 | 5.01 |

4000 | 0.025 | 3.37 |

4238 | 0.039 | 2.04 |

These data are plotted in Fig. 4a. It is shown in Section 4.5.4, Volume 1, that the maximum velocity which can occur in a pipe under adiabatic flow conditions is the sonic velocity which is equal to \sqrt{\gamma P_2 v_2}.

From the above table \sqrt{\gamma P_2 v_2} at maximum flow is:

\sqrt{1.36 \times 2.04 \times 10^6 \times 0.039}=329 m / s

The temperature at this condition is given by P_2 v_2= R T / M, and:

T_2=\left(29 \times 0.039 \times 2.04 \times 10^6 / 8314\right)=227 K

The velocity of sound in air at 227 K = 334 m/s, which serves as a check on the calculated data.