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4 years ago

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Consider the double-fluid manometer attached to an air pipe shown. If the specific gravity of one fluid is 14.514.5, determine t

he specific gravity of the other fluid if pressure of air was 81kpa81kpa. Take the atmospheric pressure to be 100 kpa.

1 answer:

kvv77 [185]4 years ago

5 0

Complete Question

The complete question is shown on the first uploaded image

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 $\rho_w = 1000 \ kg/m^3$

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

=> $S_g_2 = 3.128$

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