## Chapter 6

## Q. 6.P.6

A gas cylinder containing 30 m³ of air at 6 MN/m² pressure discharges to the atmosphere through a valve which may be taken as equivalent to a sharp edged orifice of 6 mm diameter (coefficient of discharge = 0.6). Plot the rate of discharge against the pressure in the cylinder. How long will it take for the pressure in the cylinder to fall to (a) 1 MN/m², and (b) 150 kN/m²? Assume an adiabatic expansion of the gas through the valve and that the contents of the cylinder remain at 273 K.

## Step-by-Step

## Verified Solution

Area of orifice =(\pi / 4)(0.006)^2=2.828 \times 10^{-5} m ^2 .

The critical pressure ratio w_c is:

w_c=[2 /(k+1)]^{k /(k-1)} (equation 4.43)

Taking k=\gamma=1.4 for air, w_c=0.527.

Thus sonic velocity will occur until the cylinder pressure falls to a pressure of:

P_2=(101.3 / 0.527)=192.2 kN / m ^2.

For pressures in excess of 192.2 kN/m², the rate of discharge is given by:

G=C_D A_0 \sqrt{\left(k P_1 / v_1\right)(2 /(k+1))^{(k+1) /(k-1)}} (equation 6.29)

If k=1.4, G=1.162 \times 10^{-5} \sqrt{\left(P_1 / v_1\right)}

If P_a and v_a are atmospheric pressure and the specific volume at atmospheric pressure respectively, P_a v_a=P_1 v_1 and v_1=P_a v_a / P_1

P_a=101,300 N / m ^2 and v_a=(22.4 / 29)=0.773 m ^3 / kg

∴ v_1=\left(101,300 \times 0.773 / P_1\right)=\left(78,246 / P_1\right)

and: G=1.162 \times 10^{-5} \sqrt{\left(P_1^2 / 78,246\right)}=4.15 \times 10^{-8} P_1 kg/s

If P_1 is expressed in MN/m², then: G=\underline{\underline{0.0415 P _1 kg / s }}.

For pressures lower than 192.2 kN/m²:

G^2=\left(A_0 C_D / v_2\right)^2 2 P_1 v_1(k / k-1)\left[1-\left(P_2 / P_1\right)^{(k-1) / k}\right] (equation 6.26)

v_2=v_a=0.773 m ^3 / kg

P_2=P_a=101,300 N / m ^2

v_1=P_a v_a / P_1

Substituting gives: \underline{\underline{G^2=2.64 \times 10^{-4}\left[1-\left(P_a / P_1\right)^{0.286}\right]}}

Thus a table of G as a function of pressure may be produced as follows:

FLOW AND PRESSURE MEASUREMENT

P < 192.2 kN/m² |
P > 192.2 kN/m² |
||

P (MN/m²) |
G (kg/s) |
P (MN/m²) |
G (kg/s) |

0.1013 | 0 | 0.2 | 0.0083 |

0.110 | 0.0024 | 0.5 | 0.0208 |

0.125 | 0.0039 | 1.0 | 0.0416 |

0.150 | 0.0053 | 2.0 | 0.0830 |

0.175 | 0.0062 | 6.0 | 0.249 |

These data are plotted in Fig. 6b, from which discharge rate is seen to be linear until the cylinder pressure falls to 0.125 MN/m².

If m is the mass of air in the cylinder at any pressure P_1 over the linear part of the curve, G= d m / d t=0.0415 P_1.

∴ d t= d m / 0.0415 P_1

m=(29 / 22.4)\left(P_1 / 0.1013\right) \times 30=383.4 P_1 kg

∴ d t=383.4 d m / 0.0415 m =9240 d m / m

and t=9240 \ln \left(m_1 / m_2\right)

At 6 MN/m² and 1 MN/m², the masses of air in the cylinder are 2308 and 383.4 kg respectively.

∴ The time for the pressure to fall to 1 MN/m² =9240 \ln (2308 / 3834.4)

=\underline{\underline{16,600 s} }(4.61 h )

As 0.15 MN/m² is still within the linear region, the time for the pressure to fall to this value is \underline{\underline{34,100 s }} \cdot(9.47 h )