Final answer:
To determine the pressure in the gas main, we can apply the continuity equation and Bernoulli's equation. The pressure in the gas main is approximately 215199.13 N/m².
Step-by-step explanation:
To determine the pressure in the gas main, we can apply the continuity equation and Bernoulli's equation. The continuity equation states that the mass flow rate in a pipe is constant, which is given as:
(1) A1V1 = A2V2
where A is the cross-sectional area of the pipe and V is the velocity of the gas. Since the diameters are given, we can calculate the areas as:
A1 = π(3/12/2)2 = 0.1963 ft2 and A2 = π(1/12/2)2 = 0.00654 ft2
Given that the flow rate is 100 ft³/hour, we can calculate the velocity of the gas as:
V1 = 100 ft³/hour / 0.1963 ft2 = 509.02 ft/hour
Using equation (1), we can find V2, which is the velocity of the gas in the 1-in. pipe:
509.02 ft/hour = 0.00654 ft2 × V2
V2 ≈ 77819. .144 ft/hour
Now, we can use Bernoulli's equation to relate the pressure difference between the two points to the velocity change:
(2) P1 + 0.5ρV12 + ρgh1 = P2 + 0.5ρV22 + ρgh2
where P is the pressure, ρ is the density of the gas, V is the velocity, g is the acceleration due to gravity, and h is the height of the pipe above the reference point. In this case, we can assume h1 = h2 = 0, since the pipe is horizontal. We are given that the pressure in the 1-in. pipe is to be 6 in. of water greater than atmospheric pressure. Converting this to pascals, we find:
6 in. of water × 249.0889 N/m³/in. of water = 1494.5334 N/m²
Using equation (2), we can solve for P1:
1494.5334 N/m² + 0.5 × 0.681 kg/m³ × (509.02 ft/hour × 0.3048 m/ft / 3600 s/hour)2 = P1 + 0.5 × 0.681 kg/m³ × (77819.144 ft/hour × 0.3048 m/ft / 3600 s/hour)2
P1 ≈ 215199.13 N/m²
Therefore, the pressure in the gas main is approximately 215199.13 N/m².