Natural gas (methane) flows from a 3 -in.- -diameter gas main, through a 1 -in.- -diameter pipe, and into the burner of a furnace at a rate of $100 \mathrm{ft}^{3} /$ hour. Determine the pressure in the gas main if the pressure in the 1 -in. pipe is to be 6 in. of water greater than atmosphere pressure. Neglect viscous effects.

Identify and note down known values from the problem. Flow rate of the methane gas is \(100 \, \mathrm{ft}^{3}/\mathrm{hour}\), diameter of the gas main is \(3 \, \mathrm{in.}\), diameter of the pipe is \(1 \, \mathrm{in.}\), and the pressure difference between the pipe and the atmosphere is \(6 \, \mathrm{in.} \, \text{of water}\).

Calculate the velocities in the gas main and in the pipe using the equation \(v = Q/A\), where \(v\) is the velocity, \(Q\) is the volume flow rate, and \(A\) is the cross-sectional area. The cross-sectional area \(A\) can be calculated using the equation \(A = \pi (D/2)^2\), where \(D\) is the diameter. Be sure to convert units appropriately, when necessary.

Bernoulli's equation is \(P_1 + 0.5*\rho*v_1^2 + \rho*g*h_1 = P_2 + 0.5*\rho*v_2^2 + \rho*g*h_2\). Here, \(P\) is the pressure, \(\rho\) is the fluid density, \(v\) is the velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height. Since the vertical height difference and the speed of the fluid at the two points will not affect the pressure significantly in this problem (neglect viscous effects), we can assume \(h_1 = h_2\) and the equation simplifies to \(P_1 + 0.5*\rho*v_1^2 = P_2 + 0.5*\rho*v_2^2\). \(P_1\) is the pressure in the gas main and \(P_2\) is the pressure in the pipe which is equal to [atmospheric pressure + \(6 \, \mathrm{in.} \, \text{of water}\)]. Calculate \(P_1\) by rearranging the simplified Bernoulli's equation for \(P_1\).

Upon rearranging, the equation becomes \(P_1 = P_2 + 0.5*\rho*v_2^2 - 0.5*\rho*v_1^2\). Use the known and calculated values from previous steps to solve for \(P_1\). Remember to use proper units.