# Fundamentals of Engineering

## 11 (B): Fluid Statics

• The Pressure Field in a Static Liquid
• Manometers
• Forces on Submerged Surfaces and the Center of Pressure
• Archimedes Principle and Bouyancy

## Practice Problems

1)   The atmospheric pressure on an Earth-like planet (far far away) is $$80 \, kPa$$. If humans were to build a habitat on this planet what should the gage pressure on the inside be

1. $$17 \, kPa$$
2. $$21 \, kPa$$
3. $$26 \, kPa$$
4. $$31 \, kPa$$

Solution: Based on the local atmospheric pressure of $$80 \, kPa$$ and assuming that the desired habitat pressure is $$101 \, kPa$$ abs. the gage pressure inside the hab should be $$\text{gage pressure} = \text{abs pressure in hab} - \text{local atm pressure}$$ \begin{align} \text{gage pressure} &= 101 kPa - 80 kPa \\ &= 21 kPa \end{align} The correct answer is (B)

2)   The absolute pressure inside a container is $$25 \, kPa$$. What is the vacuum pressure. Assume the local atmospheric pressure to be $$101 \, kPa$$

1. $$55 \, kPa$$
2. $$68 \, kPa$$
3. $$76 \, kPa$$
4. $$91 \, kPa$$

Solution: Vacuum pressure is the difference between atmospheric pressure and abs pressure inside the container $$\text{vacuum pressure} = 101 \, kPa - 25 \, kPa = 76 \, kPa$$ The correct answer is (C)

3)   What pressure is equivalent to $$600 \, mm$$ of mercury (SG $$13.6$$).

1. $$43 \, kPa$$
2. $$58 \, kPa$$
3. $$72 \, kPa$$
4. $$80 \, kPa$$

Solution: Given the height of mercury in the column \begin{align} P_{gage} &= \rho g h \\ &= 13{,}600 \, kg/m^3 \times 9.81 \, m/s^2 \times 0.6 \, m \\ &= 80{,}050 \, Pa \\ &= 80 \, kPa \end{align} The correct answer is (D)

4)   A closed cylindrical tank, shown in the figure has a pressure $$60 \, kPa$$ inside the dome. Determine the pressure in pipe B. 1. $$89 \, kPa$$
2. $$103 \, kPa$$
3. $$124 \, kPa$$
4. $$135 \, kPa$$

Solution: Note that $$P_1 = P_2 = P_3$$, so \begin{align} P_B &= P_A + (\rho g h)_{oil} + (\rho g h)_{water} \\\\ P_B &= 60{,}000 Pa \\ &+ 800 \, kg/m^3 \times 9.81 \, m/s^2 \times 3 m \\ &+ 1000 \, kg/m^3 \times 9.81 \, m/s^2 \times 2 m\\\\ P_B &= 103 \, kPa \end{align} The correct answer is (B)

5)   Pipe $$A$$ and $$B$$ both contain water. The gage fluid in the manometer shown has a specific gravity of $$3.46$$. Calculate the differential pressure 1. $$15.5 \, kPa$$
2. $$17.9 \, kPa$$
3. $$21.5 \, kPa$$
4. $$30.1 \, kPa$$

Solution: Note that the pressure at points marked 2 and 3 are going to be the same. $$P_A + \gamma_w \cdot (0.6 m + 0.4 m) + \gamma_{gf} \cdot 0.4 m = P_B + \gamma_w \cdot (0.4 m + 0.4 m)$$ Rearrange and cancel terms to get \begin{align} P_A - P_B &= \gamma_w \cdot 0.2 \, m + \gamma_{gf} \cdot 0.4 \, m \\ &= 9810 \, N/m^3 \times 0.2 \, m + 3.46 \times 9810 \, N/m^3 \times 0.4 \, m \\ &= 15{,}539 \, Pa \\ &= 15.5 \, kPa \end{align} The correct answer is (A)

6)   A tank containing water has a gate hinged at the bottom as shown in the figure below. The gate is $$5 \, m$$ wide. Calculate the force on the gate due to water pressure. 1. $$1104 \, kN$$
2. $$1380 \, kN$$
3. $$1655 \, kN$$
4. $$2152 \, kN$$

Solution: The total force on the gate will be equal to the average pressure times the area. \begin{align} F &= P_{avg} \times A \\ &= (\gamma_w \cdot h_{avg}) \times A \\ &= 9810 \, N/m^3 \times 4.5 \, m \times 25 \, m^2 \\ &= 1{,}103{,}625 \, N \\ &= 1104 \, kN \\ \end{align} The correct answer is (A)

7)   A $$4 \, m$$ wide curved gate is located in the bottom of a reservoir containing water. Determine the magnitude of the resultant force of water on the gate. 1. $$281 \, kN$$
2. $$352 \, kN$$
3. $$470 \, kN$$
4. $$587 \, kN$$

Solution: For curved surfaces calculate the horizontal and vertical forces and then combine them to find the resultant. \begin{align} F_{hor} &= P_{avg} \times A_{projected} \\ &= \gamma_w \cdot {h_1 + h_2 \over 2} \times A_{projected} \\ &= 9810 \, N/m^3 \cdot {4+6 \over 2} \, m \times (2m \cdot 4m) \\ &= 392{,}400 \, N \\\\ F_{ver} &= \gamma_w \times (V_1 + V_2) \\ &= 9810 \, N/m^3 \times \left( (4m \cdot 2m \cdot 4m) + { \pi \cdot (2m)^2 \cdot 4m \over 4} \right) \\ &= 437{,}196 \, N \\\\ F &= \sqrt{392{,}400^2 + 437{,}196^2} \\ &= 587{,}468 \, N \\ &= 587 \, kN \end{align} The correct answer is (D)

8)   A log floats with one fourth of its volume protruding above the water surface. Determine the specific weight of the log.

1. $$2345.2 \, N/m^3$$
2. $$4690.4 \, N/m^3$$
3. $$6253.9 \, N/m^3$$
4. $$7357.5 \, N/m^3$$

Solution: Since the log is in equilibrium we can equate the weight of the log to the bouyant force. $$\gamma_{log} \times V_{log} = \gamma_{water} \times V_{disp}$$ \begin{aligned} \gamma_{log} &= \gamma_{water} \times { V_{disp} \over V_{log} } \\ &= 9810 \, N/m^3 \times \frac 34 \\ &= 7357.5 \, N/m^3 \end{aligned} The correct answer is (D)

9)   A cylindrical block of concrete is suspended by a cable and completely immersed in seawater (density $$1025 \, kg/m^3$$). The diameter is $$0.75 \, m$$, height $$1 \, m$$ and density of concrete is $$2300 \, kg/m^3$$. Calculate the tension in the cable

1. $$4245 \, N$$
2. $$5526 \, N$$
3. $$6631 \, N$$
4. $$9670 \, N$$

Solution: The tension in the cable can be found using $$T = W - F_B$$ Where, \begin{aligned} W &= m \cdot g = \rho_C \cdot V \cdot g \text{, and} \\ F_B &= \rho_W \cdot V \cdot g \end{aligned} (HINT: Use a FBD and sum forces in the vertical direction.) \begin{aligned} T &= (\rho_C - \rho_W) \cdot V \cdot g \\ &= (2300-1025) \, kg/m^3 \cdot (pi/4*0.75^2*1.0) \, m^3 \cdot 9.81 \,m/s^2 \\ &= 5526 \, N \end{aligned} The correct answer is (B)