To solve this problem, we will calculate the pressure at the center of the pipe using the concept of hydrostatic pressure and manometer principles.

The key steps involve:

1. Understand the fluids and their densities:

   • Mercury has a specific gravity ($s$) of $13.55$. Its density ($\rho$) can be calculated as: $\rho_{\text{mercury}} = s \times \rho_{\text{water}} = 13.55 \times 1000 \, \text{kg/m}^3$
   • Oil has a specific gravity ($s$) of $0.80$, so: $\rho_{\text{oil}} = 0.80 \times 1000 \, \text{kg/m}^3$

2. Analyze the manometer:

   • The height difference in each fluid segment will determine the pressure contributions.
   • Assume the atmospheric pressure at the open surface is $P_0$, and calculate the pressures step by step, going from the open end (oil surface) to the center of the pipe.

3. Hydrostatic pressure relation:

   • The pressure difference due to a fluid column is: $\Delta P = \rho g h$ where:
   • $\rho$ = density of the fluid,
   • $g$ = gravitational acceleration ($\sim 9.81 \, \text{m/s}^2$),
   • $h$ = height of the fluid column.

Calculations

1. Start from the oil's open surface and add the contributions from each fluid column, keeping track of height differences and densities.
2. At the center of the pipe, sum up all pressures (positive or negative contributions).

Let me perform the detailed calculations for you.# Define given values g = 9.81 # gravitational acceleration in m/s^2 rho_water = 1000 # density of water in kg/m^3

Specific gravities and their densities

s_mercury = 13.55 # specific gravity of mercury s_oil = 0.80 # specific gravity of oil

rho_mercury = s_mercury * rho_water # density of mercury in kg/m^3 rho_oil = s_oil * rho_water # density of oil in kg/m^3

Heights of fluid columns

h_oil = 3.5 # height of the oil column in meters h_mercury = 2.0 # height of the mercury column in meters

Calculate pressure contributions

P_oil = rho_oil * g * h_oil # pressure due to oil column P_mercury = rho_mercury * g * h_mercury # pressure due to mercury column

Resulting pressure difference in Pascals

P_total = P_mercury - P_oil # net pressure difference

Convert to kPa for final answer

P_total_kPa = P_total / 1000 # in kPa

P_total_kPa238.383238.383successThe pressure at the center of the pipe, based on the given manometer readings, is approximately 238.38 kPa.

Would you like a detailed explanation of the steps or any clarifications? Here are some follow-up questions to expand your understanding:

1. How does the specific gravity of a fluid affect the pressure in a manometer?
2. Why do we subtract the pressure due to oil from the pressure due to mercury in this case?
3. What would change if the pipe were inclined instead of vertical?
4. How would the result differ if the manometer used a different fluid with a known specific gravity?
5. Can this principle be applied to other pressure measurement devices, like barometers?

Tip: Always pay attention to the direction of pressure changes when working with different fluid densities in a manometer.