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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 )

6b