Question

republic of cameroon
peace - work - fatherland
the university of bamenda
higher technical teachers training college bamenda
directorate of studies
p.o. box 39 bambili

republique du cameroun
paix - travail - patrie
universite de bamenda
ecole normale superieure denseignement technique
direction des etudes
tel 33 05 10 69

school: h.t.t.t.c
department: ceft
lecture(s): dr. kouamou nguessia.
course code: ceft2222
course title: material and fluid mechanics
option: bpw, 200
date:
hall:
time: 1hr 30min
nature: 2^{nd} semester exam
instructions: ...answer all the questions...

exercise 1 (8marks)
pressure gage b is to measure the pressure at point a in a water flow. if the pressure at b is \\(87\text{ kpa}\\), estimate the pressure at a in \\(\text{ kpa}\\). assume all fluids are at \\(20^{\circ}\text{c}\\). see figure below.

\\(\gamma_{\text{water}} = 9790\text{ n/m}^3\\); \\(\gamma_{\text{mercury}} = 133,100\text{ n/m}^3\\);
\\(\gamma_{\text{oil}} = 8720\text{ n/m}^3\\)

exercise 2 (12marks)
the open tank of the figure above has two piezometers a and b and contains two immiscible liquids (case of water and oil). find:
a) the height of the free surface of the liquid in the piezometer a
b) the height of the liquid surface in the piezometer b
c) the total pressure at the bottom of the tank.
nb: in c the pressure is known because it is atmospheric pressure (\\(p_{\text{atm}} = 10\text{bar}\\)). take the gravity \\(g = 9.81\text{ n/kg}\\).
the density of water is \\(\
ho_{\text{w}} = 1000\text{ kg/m}^3\\)

good luck

Explanation:

🆕 New Concept Discovered: Manometry
Using hydrostatic pressure changes across fluid interfaces to find unknown pressures.

Step 1: Identify given values and convert units

Identify the specific weights (\(\gamma\)) of the fluids and the geometric heights from the diagram, converting all dimensions from centimeters to meters:

• \(\gamma_{\text{water}} = 9790\text{ N/m}^3\)
• \(\gamma_{\text{mercury}} = 133,100\text{ N/m}^3\) (noting the French notation \(133.100\) on the paper represents \(133,100\))
• \(\gamma_{\text{oil}} = 8720\text{ N/m}^3\)
• Pressure at gage B: \(p_B = 87\text{ kPa} = 87,000\text{ Pa}\)

Heights of the fluid columns:

• Height difference from point A to the water-mercury interface:

$$ h_1 = 5\text{ cm} + 4\text{ cm} = 9\text{ cm} = 0.09\text{ m} $$

• Height of the mercury column (from the bottom bend to the mercury-oil interface):

$$ h_2 = 11\text{ cm} - 4\text{ cm} = 7\text{ cm} = 0.07\text{ m} $$

• Height of the oil column (from the mercury-oil interface up to the center of gage B):

$$ h_3 = 6\text{ cm} = 0.06\text{ m} $$

Step 2: Set up the hydrostatic manometer equation

Start at point A and write the pressure balance equation by moving through the tube to gage B:

• As we go down in a fluid, pressure increases: \(+ \gamma \cdot h\)
• As we go up in a fluid, pressure decreases: \(- \gamma \cdot h\)

Starting at \(p_A\):

1. Move down through the water column of height \(h_1 = 0.09\text{ m}\) to the bottom of the U-tube:

$$ p_A + \gamma_{\text{water}} \cdot h_1 $$

1. Move up through the mercury column of height \(h_2 = 0.07\text{ m}\) to the mercury-oil interface:

$$ p_A + \gamma_{\text{water}} \cdot h_1 - \gamma_{\text{mercury}} \cdot h_2 $$

1. Move up through the oil column of height \(h_3 = 0.06\text{ m}\) to the center of gage B:

$$ p_A + \gamma_{\text{water}} \cdot h_1 - \gamma_{\text{mercury}} \cdot h_2 - \gamma_{\text{oil}} \cdot h_3 = p_B $$

Step 3: Solve for the pressure at point A (\(p_A\))

Rearrange the equation to isolate \(p_A\):

$$ p_A = p_B - \gamma_{\text{water}} \cdot h_1 + \gamma_{\text{mercury}} \cdot h_2 + \gamma_{\text{oil}} \cdot h_3 $$

Substitute the numerical values:

$$ p_A = 87,000 - (9790 \cdot 0.09) + (133,100 \cdot 0.07) + (8720 \cdot 0.06) $$

Calculate each term:

• Water term: \(9790 \cdot 0.09 = 881.1\text{ Pa}\)
• Mercury term: \(133,100 \cdot 0.07 = 9317\text{ Pa}\)
• Oil term: \(8720 \cdot 0.06 = 523.2\text{ Pa}\)

Combine the values:

$$ p_A = 87,000 - 881.1 + 9317 + 523.2 $$

$$ p_A = 95,959.1\text{ Pa} $$

Convert the final pressure back to kilopascals (kPa):

$$ p_A \approx 95.96\text{ kPa} $$

Answer:

The estimated pressure at point A is 95.96 kPa (or 96.0 kPa).