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g100num [7]
3 years ago
15

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

lled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 15.5 psig. The level of mercury in the open arm is h=950.0 mm above the lowest part of the manometer. When the gas is not flowing, the pressure is the same everywhere in the pipe.
Required:
a. What is the gas pressure psig at this moment?
b. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe? (in mm) When gas is flowing, the mercury level in the visible arm drops by 39.0 mm.
Physics
1 answer:
Svetllana [295]3 years ago
6 0

Answer:

a

P_G  = 14.03 \  psig  

b

h_m =   0.148 \  m

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

Here \rho is the density of mercury with value \rho = 13.6 *10^{3} kg/m^3

and \delta h is the difference in the level of gas in arm one and two

So

\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

\delta h  = 0.802 \  m

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

h_m =   0.950 -  0.802

=> h_m =   0.148 \  m

Generally from manometry principle we have that

P_G + \rho * g  * d   -  \rho *  g  * [h - (h_m + d)] = 0

Here P_G is the pressure of the gas

P_G +13.6 *10^{3} * 9.8  * 0.039    -  13.6 *10^{3}  *  9.8  * [0.950 - (0.148 + 0.039)] = 0

P_G  =  9.6724 04 *10^{4} \  N/m^2

converting to  psig

P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

P_G  = 14.03 \  psig

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