Question

Consider the double-fluid manometer attached to an air pipe shown. If the specific gravity of one fluid is 14.514.5, determine the specific gravity of the other fluid if pressure of air was 81kpa81kpa. Take the atmospheric pressure to be 100 kpa.

Answer 1

Complete Question

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Answer:

The  value  is  $S_g_2 = 3.128$

Explanation:

From the question we are told that  

  The specific gravity of the first fluid  is  $S_g_1 = 14.5$

  The  pressure of air is  $P_a = 81kPa = 81*10^{3} \ Pa$

  The  atmospheric pressure is  $P_A = 100 kPa = 100 *10^{3} Pa$

Generally the specific gravity of the second  fluid is mathematically represented as

            $S_g_2 = S_g_1 * \frac{h_1}{ h_2 } + \frac{P_a - P_A}{\rho_w * g * h_2 }$

Here  $\rho_w$ is the density of water with value $\rho_w = 1000 \ kg/m^3$

=>   $S_g_2 = [14.5 * \frac{0.22}{0.4} ] + [\frac{81*10^{3}-100*10^{3}}{ 1000 * 9.8 * 0.4} ]$

=> $S_g_2 = 3.128$