Question

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak is estimated to be 0.1 in. At the time, the upstream pressure in the pipeline is 50 psig and the downstream pressure is 40 psig. Estimate the mass flow rate of benzene through the leak. The specific gravity of benzene is 0.8794.

Answer 1

Step-by-step explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

`Q_m=AC_o√(2\rho g_cP_g)`

Here,
`Q_m` is the mass flow rate through the leak of the pipeline.
`A` is the area of the hole,
`C_o` is the discharge rate,
`\rho` is the fluid density,
`g_c` is the gravitational constant and
`P_g` is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

`d=(0.1 in)((1ft)/(12in))\\=8.33* 10^(-3)ft`

Calculate the area (A) of the hole. The area of the hole is.

`A=(\pi d^2)/(4)`

Substitute 3.14 for
`\pi` and
`8.33* 10^(-3)ft` for d and calculate A.

`A=(\pi d^2)/(4)\\\\((3.14)(8.33* 10^(-3))^2)/(4)\\\\5.45* 10^(-5)ft^2`

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is
`62.4(Ib_m)/(ft^3)`. Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

`=(0.8794)(62.4(Ib_m)/(ft^3))\\\\54.9(Ib_m)/(ft^3)`

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

`P_g`=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate
`P_g`.

Upstream x distance from upstream pressure end

`P_g`=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

`=((50psig* 43ft)+(40psig * 57ft))/(100ft)\\\\=44.3psig`

Convert the pressure from psig to
`Ib_f/ft^2`

`P_g=(44.3psig)((1(Ib_f)/(ft^2))/(1psig))(144(in^2)/(ft^2))\\\\=6,379.2(Ib_f)/(ft^2)`

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute
`5.45* 10^(-5)ft^2` for A. 0.61 for
`C_o`,
`54.9(Ib_m)/(ft^3)` for
`\rho`,
`32.17(ft.Ib_m)/(Ib_f.s^2)` for
`g_c`, and
`6,379.2(Ib_f)/(ft^2)` for
`P_g` and calculate
`Q_m`

`Q_m=AC_o√(2\rho g_cP_g)\\\\=(5.45* 10^(-5)ft^2)(0.61)\sqrt{2(54.9(Ib_m)/(ft^3))(32.17(ft.Ib_m)/(Ib_f.s^2))(6,379.2(Ib_f)/(ft^2))}\\\\(3.3245* 10^(-5)ft^2)\sqrt{22,533,031.21(Ib^2_m)/(ft^4.s^2)}\\\\=0.158(Ib_m)/(s)`

The mass flow rate of benzene through the leak in the pipeline is
`0.158(Ib_m)/(s)`