Question 9.1: Determine the pressure difference between the benzene at A a...
Determine the pressure difference between the benzene at A and the air at B in Figure 9.12.
Given: Manometer geometry and gage fluids; heights of fluid columns.
Find: Pressure difference between A and B.
Assume: No relative motion of fluid elements (hydrostatics); fluids have constant, uniform density, and it is appropriate to evaluate densities at 20˚C.
We look up the properties of the fluids used in our manometer at 20˚C and find the following:
| Fluid | Density ρ (kg/m³) |
| Water | 998.0 |
| Mercury | 13,550.0 |
| Air | 1.2 |
| Benzene | 881.0 |
| Kerosene | 804.0 |
We know that in a fluid at rest, the pressure depends only on the elevation
in the fluid. Thus, in any continuous length of the same fluid, two points at the same elevation must be at the same pressure. Manometers are based on this principle. We find the requested pressure difference by starting at point A and working our way through the manometer, noting that the pressure increases when the fluid level drops and that pressure decreases when the fluid level rises:
• Point 1: P_1 = P_A + ρ_{_B}gh_1
• Point 2: P_2 = P_1 – ρ_{_M}gh_2
• Point 3: P_3 = P_2 – ρ_{_K}gh_3
• Point 4: P_4 = P_3 + ρ_{_W}gh_4
• At B: P_B = P_4 – ρ_{_A}gh_5
So,
P_B = P_A + ρ_{_B}gh_1 – ρ_{_M}gh_2 – ρ_{_K}gh_3 + ρ_{_W}gh_4 + ρ_{_A}gh_5 .
P_B-P_A = (881 kg/m³)(9.81 m/s²)(0.2 m) – (13,550 kg/m³)(9.81 m/s²)(0.08 m)
–(804 kg/m³)(9.81 m/s²)(0.32 m) + (998 kg/m³)(9.81 m/s²)(0.26 m)
–(1.2 kg/m³)(9.81 m/s²)(0.09 m)
= –8885 Pa
or
P_A-P_B = 8.9 kPa.