My professor tells me I'm not using the right formula for this problem. can someone please help me. The formula is not delt p = density x gravity x height


The mercury manometer on the right and the pressure transducer on the left are both connected to a piston-driven pressure source filled with hydraulic oil (density = 880 kg/m^3). The bulbs connected to the transducer and the right side of the manometer are both evacuated (denisty = 0). Calculate pressures from the manometer reading.

hL (mm Oil) = 450
Manometer reading = 150
Transducer Output = 5.6 (mA)

To calculate the pressures from the manometer reading, we need to use a different formula. The formula you mentioned, delt p = density x gravity x height, is for calculating the pressure difference due to the height of a fluid column.

In this case, we need to consider the pressure difference between the two sides of the manometer. We can use the equation:

ΔP = ρgh

Where:
ΔP is the pressure difference
ρ is the density of the fluid
g is the acceleration due to gravity
h is the height difference

Given:
Density of hydraulic oil (ρ) = 880 kg/m^3
Height of oil column (h) = 450 mm
Height of mercury column (hm) = 150 mm

First, let's convert the heights from millimeters to meters:
Height of oil column (h) = 450 mm = 450/1000 m = 0.45 m
Height of mercury column (hm) = 150 mm = 150/1000 m = 0.15 m

Now, we can calculate the pressure difference for both sides of the manometer.

For the oil side:
ΔPoil = ρg (hoil)
= (880 kg/m^3)(9.8 m/s^2)(0.45 m)
= 3913.2 Pa

For the mercury side:
ΔPmercury = ρg (hmercury)
= (13500 kg/m^3)(9.8 m/s^2)(0.15 m)
= 19890 Pa

Therefore, the pressure difference from the manometer reading is:
ΔP = ΔPoil - ΔPmercury
= 3913.2 Pa - 19890 Pa
= -15976.8 Pa

The negative sign indicates that the pressure on the oil side is lower than the pressure on the mercury side.

So, the pressures from the manometer reading are:
Oil side pressure = Pressure on the oil side - Pressure difference = 0 + (-15976.8 Pa) = -15976.8 Pa
Mercury side pressure = Pressure on the mercury side = 0 Pa (since it is evacuated)

Note: The transducer output value of 5.6 mA is not directly used in this calculation. It might be relevant for some other measurements or calculations.