Homework 1 Assignment Print View

1. Award: 50 out of 50.00 points The pressure in a natural gas pipeline is measured by the manometer shown in the figure with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure at the bottom of the natural gas pipeline. Take the height of the mercury column h to be 5 in. Take the density of water to be ρw = 62.4 lbm/ft3, and the specific gravity of mercury to be 13.6 and its density is ρHg = 848.6 lbm/ft3. The absolute pressure at the bottom of the natural gas pipeline is 18.1 psia. References Worksheet Difficulty: Medium The pressure in a natural gas pipeline is measured by the manometer shown in the figure with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure at the bottom of the natural gas pipeline. Take the height of the mercury column h to be 5 in. Take the density of water to be ρw = 62.4 lbm/ft3, and the specific gravity of mercury to be 13.6 and its density is ρHg = 848.6 lbm/ft3. The absolute pressure at the bottom of the natural gas pipeline is 17.5 ± 10% psia. Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible. 3. The pressure throughout the natural gas (including the tube) is uniform since its density is low. Starting with the pressure at point 1 in the natural gas pipeline, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give P1 - ρHg ghHg - ρwater ghwater = Patm Solving for P1, P1 = Patm + ρHg gHg hHg + ρwater gh1 5 P = 14.2 psia + (32.2 ft/s2 ) ((848.6 lbm/ft 3 ) ( 12 ft) + (62.4 lbm/ft3 ) ( 24 ft)) ( 12 P = 17.5 psia 1 lbf 2 32.2 lbm⋅ft/s )( 1 ft2 144 in 2 ) 2. Award: 50 out of 50.00 points Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in the figure below, such that hw = 71 cm, hHg = 10 cm, hair = 70 cm, and hsea = 30 cm. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρsea = 1035 kg/m3, the density of mercury to be ρHg = 13600 kg/m3, and the density of water to be ρwater = 1000 kg/m3. The pressure difference between the two pipelines is 3.33 kPa. References Worksheet Difficulty: Medium Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in the figure below, such that hw = 71 cm, hHg = 10 cm, hair = 70 cm, and hsea = 30 cm. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρsea = 1035 kg/m3, the density of mercury to be ρHg = 13600 kg/m3, and the density of water to be ρwater = 1000 kg/m3. The pressure difference between the two pipelines is Explanation: The following assumptions have been made: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible. 3.33 ± 10% kPa. Starting with the pressure in the fresh waterpipe (point 1), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 give P2 = P1 + ρwghw - ρHg ghHg − ρair ghair + ρsea ghsea Rearranging and neglecting the effect of the air column on pressure, P1 - P2 = − ρwghw + ρHg ghHg − ρsea ghsea P1 - P2 = g(ρHg hHg - ρwhw - ρsea hsea ) kN P1 - P2 = (9.81 m/s2 )[(13600 kg/m3 )(0.1 m) - (1000 kg/m3 )(0.71 m) - (1035 kg/m3 )(0.3 m)] ( 10001kg⋅m/s ) 2 P1 - P2 = 3.33 kN/m2 = 3.33 kPa