Answer:
a
[tex]P_G = 14.03 \ psig [/tex]
b
[tex]h_m = 0.148 \ m [/tex]
Explanation:
From the question we are told that
The pressure of the manometer when there is no gas flow is [tex]P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2[/tex]
The level of mercury is [tex]h = 950 \ mm = 0.950 \ m[/tex]
The drop in the mercury level at the visible arm is [tex]d = 39.0 = 0.039 \ m [/tex]
Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as
[tex]P_g = P_m = g * \delta h * \rho[/tex]
Here [tex] \rho [/tex] is the density of mercury with value [tex] \rho = 13.6 *10^{3} kg/m^3 [/tex]
and [tex]\delta h[/tex] is the difference in the level of gas in arm one and two
So
[tex]\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }[/tex]
[tex]\delta h = 0.802 \ m [/tex]
Generally the height of the mercury at the arm connected to the pipe is mathematically represented as
[tex]h_m = 0.950 - 0.802[/tex]
=> [tex]h_m = 0.148 \ m [/tex]
Generally from manometry principle we have that
[tex]P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0[/tex]
Here [tex]P_G[/tex] is the pressure of the gas
[tex]P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0[/tex]
[tex]P_G = 9.6724 04 *10^{4} \ N/m^2[/tex]
converting to psig
[tex]P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}[/tex]
[tex]P_G = 14.03 \ psig [/tex]
