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Chapter 6

Q. 6.P.14

A gas cylinder containing air discharges to atmosphere through a valve whose characteristics may be considered similar to those of a sharp-edged orifice. If the pressure in the cylinder is initially 350 kN/m², by how much will the pressure have fallen when the flowrate has decreased to one-quarter of its initial value? The flow through the valve may be taken as isentropic and the expansion in the cylinder as isothermal. The ratio of the specific heats at constant pressure and constant volume is 1.4.

Step-by-Step

Verified Solution

From equation 4.43:
the critical pressure ratio, w_c=[2 /(k+1)]^{k /(k-1)}=(2 / 2.4)^{1.4 / 0.4}=0.528
If the cylinder is discharging to atmospheric pressure, sonic velocity will occur until the cylinder pressure has fallen to (101.3 / 0.528)=192  kN / m ^2
The maximum discharge when the cylinder pressure exceeds 192 kN/m² is given by:

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

If P_a and v_a are the pressure and specific volume at atmospheric pressure, then:

1 / v_1=P_1 / P_a v_a

and:           G_{\max }=C_D A_0 \sqrt{\frac{k P_1^2}{P_a v_a}\left(\frac{2}{k+1}\right)^{(k+1) /(k-1)}}

=C_D A_0 P_1 \sqrt{\left[\left(k / P_a v_a\right)(2 / k+1)\right]^{(k+1) /(k-1)}}

If G_{350} and G_{192} are the rates of discharge at 350 and 192 kN/m² respectively, then:

G_{350} / G_{192}=(350 / 192)=1.82

or:                                G_{192}=0.55 G_{350}

For pressures below 192 kN/m²:

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

Substituting for 1 / v_1=P_1 / P_a v_a and v_2=v_a gives:

G=\frac{C_D A_0}{v_a} \sqrt{2 P_a v_a\left(\frac{k}{k-1}\right)\left[1-\left(\frac{P_2}{P_1}\right)^{(k-1) / k}\right]}

and:             G^2=\left(C_D A_0 / v_a\right)^2 2 P_a v_a[k /(k-1)]\left[1-\left(P_2 / P_1\right)^{(k-1) / k}\right]

=\left(C_D A_0 / v_a\right)^2 2 P_a v_a \times 3.5\left[1-\left(P_2 / P_1\right)^{0.286}\right]

When P_1=192 kN / m ^2,  G_{192}=0.55 G_{350},  P_2, atmospheric pressure, 101.3 kN/m² and:

\left(0.55 G_{350}\right)^2=\left(C_D A_0 / v_a\right)^2 2 P_a v_a \times 3.5\left[1-(101.3 / 192)^{0.286}\right]

When the final pressure P_1 is reached, the flowrate is 0.25G_{350}.

∴              \left(0.25 G_{350}\right)^2=\left(C_D A_0 / v_a\right)^2 2 P_a v_a \times 3.5\left(1-\left(101.3 / P_1\right)^{0.286}\right)

Dividing these two equations gives:

\left(\frac{0.55}{0.25}\right)^2=\frac{1-(101.3 / 192)^{0.286}}{1-\left(101.3 / P_1\right)^{0.286}}

and:                                         P_1=\underline{\underline{102.3  kN / m ^2}}