## PEC EPE Exam Fluid Mechanics Numerical Questions Civil Engineering

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### Other PEC EPE Exam Past Paper Questions

**A water tank has a base area of 5 mยฒ and is filled to a height of 3 m. Calculate the weight of the water in the tank. (Assume the density of water as 1000 kg/mยณ)**

Solution:

Volume of water in the tank = Base area ร Height

Volume of water = 5 mยฒ ร 3 m = 15 mยณ

Weight of water = Volume ร Density of water

Weight of water = 15 mยณ ร 1000 kg/mยณ = 15000 kg

Answer: Weight of water in the tank = 15000 kg

**The pressure at the bottom of a water tank is 40000 Pa. If the height of the water column above the bottom is 5 m and the density of water is 1000 kg/mยณ, calculate the atmospheric pressure acting on the water surface.**

Solution:

Pressure at the surface = Pressure at the bottom + (Density ร Gravity ร Height)

Pressure at the surface = 40000 Pa + (1000 kg/mยณ ร 9.81 m/sยฒ ร 5 m) = 48905 Pa

Answer: Atmospheric pressure acting on the water surface = 48905 Pa

**Water flows through a horizontal pipe with a velocity of 2 m/s. If the pipe’s diameter is 0.2 m and the density of water is 1000 kg/mยณ, calculate the flow rate of water through the pipe.**

Solution:

Flow rate = (Area of pipe) ร (Velocity of water)

Area of pipe = ฯ ร (Diameter/2)ยฒ = ฯ ร (0.2 m/2)ยฒ โ 0.0314 mยฒ

Flow rate = 0.0314 mยฒ ร 2 m/s โ 0.0628 mยณ/s

Answer: Flow rate of water through the pipe โ 0.0628 mยณ/s

**Water is flowing through a horizontal pipe with a velocity of 3 m/s. If the pipe’s diameter is 0.4 m, calculate the Reynolds number for the flow. (Assume the kinematic viscosity of water as 1.004 ร 10^-6 mยฒ/s)**

Solution:

Reynolds number (Re) = (Density ร Velocity ร Diameter) / Kinematic viscosity

Re = (1000 kg/mยณ ร 3 m/s ร 0.4 m) / (1.004 ร 10^-6 mยฒ/s) โ 1.196 ร 10^6

Answer: Reynolds number for the flow โ 1.196 ร 10^6

**A pump lifts water from a well to a storage tank located at a height of 15 m above the well. If the flow rate of water is 0.1 mยณ/s and the density of water is 1000 kg/mยณ, calculate the power required to operate the pump.**

Solution:

Power = (Density ร g ร Q ร H) / Efficiency

Power = (1000 kg/mยณ ร 9.81 m/sยฒ ร 0.1 mยณ/s ร 15 m) / Efficiency

Answer: The power required to operate the pump depends on the efficiency of the pump and needs to be provided.

**Water flows through a horizontal pipe with a diameter of 0.3 m. If the pressure difference between the two ends of the pipe is 2000 Pa, calculate the flow rate of water through the pipe. (Assume the density of water as 1000 kg/mยณ)**

Solution:

Flow rate = (Pressure difference) / (Density ร g)

Flow rate = 2000 Pa / (1000 kg/mยณ ร 9.81 m/sยฒ) โ 0.203 mยณ/s

Answer: Flow rate of water through the pipe โ 0.203 mยณ/s

**A cylindrical tank contains oil with a height of 4 m and a diameter of 2 m. If the density of oil is 850 kg/mยณ, calculate the weight of the oil in the tank.**

Solution:

Volume of oil in the tank = ฯ ร (Diameter/2)ยฒ ร Height

Volume of oil = ฯ ร (2 m/2)ยฒ ร 4 m โ 6.28 mยณ

Weight of oil = Volume ร Density of oil

Weight of oil = 6.28 mยณ ร 850 kg/mยณ โ 5338 kg

Answer: Weight of oil in the tank โ 5338 kg

**A rectangular water tank has dimensions 6 m by 4 m by 3 m. Calculate the pressure at the bottom of the tank. (Assume the density of water as 1000 kg/mยณ and acceleration due to gravity as 9.81 m/sยฒ)**

Solution:

Pressure at the bottom = (Density ร g ร Height)

Pressure at the bottom = (1000 kg/mยณ ร 9.81 m/sยฒ ร 3 m) = 29430 Pa

Answer: Pressure at the bottom of the tank = 29430 Pa

**A water pump delivers water at a flow rate of 0.2 mยณ/s to a storage tank located at a height of 20 m above the pump. Calculate the power required to operate the pump. (Assume the density of water as 1000 kg/mยณ and acceleration due to gravity as 9.81 m/sยฒ)**

Solution:

Power = (Density ร g ร Q ร H) / Efficiency

Power = (1000 kg/mยณ ร 9.81 m/sยฒ ร 0.2 mยณ/s ร 20 m) / Efficiency

Answer: The power required to operate the pump depends on the efficiency of the pump and needs to be provided.

**Water is flowing through a horizontal pipe with a velocity of 4 m/s. If the pipe’s diameter is 0.3 m, calculate the Reynolds number for the flow. (Assume the kinematic viscosity of water as 1.004 ร 10^-6 mยฒ/s)**

Solution:

Reynolds number (Re) = (Density ร Velocity ร Diameter) / Kinematic viscosity

Re = (1000 kg/mยณ ร 4 m/s ร 0.3 m) / (1.004 ร 10^-6 mยฒ/s) โ 1.196 ร 10^6

Answer: Reynolds number for the flow โ 1.196 ร 10^6

**A pump lifts water from a well to a storage tank located at a height of 15 m above the well. If the flow rate of water is 0.1 mยณ/s and the density of water is 1000 kg/mยณ, calculate the power required to operate the pump.**

Solution:

Power = (Density ร g ร Q ร H) / Efficiency

Power = (1000 kg/mยณ ร 9.81 m/sยฒ ร 0.1 mยณ/s ร 15 m) / Efficiency

Answer: The power required to operate the pump depends on the efficiency of the pump and needs to be provided.

**Water flows through a horizontal pipe with a diameter of 0.3 m. If the pressure difference between the two ends of the pipe is 2000 Pa, calculate the flow rate of water through the pipe. (Assume the density of water as 1000 kg/mยณ)**

Solution:

Flow rate = (Pressure difference) / (Density ร g)

Flow rate = 2000 Pa / (1000 kg/mยณ ร 9.81 m/sยฒ) โ 0.203 mยณ/s

Answer: Flow rate of water through the pipe โ 0.203 mยณ/s

**A cylindrical tank contains oil with a height of 4 m and a diameter of 2 m. If the density of oil is 850 kg/mยณ, calculate the weight of the oil in the tank.**

Solution:

Volume of oil in the tank = ฯ ร (Diameter/2)ยฒ ร Height

Volume of oil = ฯ ร (2 m/2)ยฒ ร 4 m โ 6.28 mยณ

Weight of oil = Volume ร Density of oil

Weight of oil = 6.28 mยณ ร 850 kg/mยณ โ 5338 kg

Answer: Weight of oil in the tank โ 5338 kg

**A rectangular water tank has dimensions 6 m by 4 m by 3 m. Calculate the pressure at the bottom of the tank. (Assume the density of water as 1000 kg/mยณ and acceleration due to gravity as 9.81 m/sยฒ)**

Solution:

Pressure at the bottom = (Density ร g ร Height)

Pressure at the bottom = (1000 kg/mยณ ร 9.81 m/sยฒ ร 3 m) = 29430 Pa

Answer: Pressure at the bottom of the tank = 29430 Pa

**A water pump delivers water at a flow rate of 0.2 mยณ/s to a storage tank located at a height of 20 m above the pump. Calculate the power required to operate the pump. (Assume the density of water as 1000 kg/mยณ and acceleration due to gravity as 9.81 m/sยฒ)**

Solution:

Power = (Density ร g ร Q ร H) / Efficiency

Power = (1000 kg/mยณ ร 9.81 m/sยฒ ร 0.2 mยณ/s ร 20 m) / Efficiency

**Water is flowing through a horizontal pipe with a velocity of 4 m/s. If the pipe’s diameter is 0.3 m, calculate the Reynolds number for the flow. (Assume the kinematic viscosity of water as 1.004 ร 10^-6 mยฒ/s)**

Solution:

Reynolds number (Re) = (Density ร Velocity ร Diameter) / Kinematic viscosity

Re = (1000 kg/mยณ ร 4 m/s ร 0.3 m) / (1.004 ร 10^-6 mยฒ/s) โ 1.196 ร 10^6

Answer: Reynolds number for the flow โ 1.196 ร 10^6

**A pump lifts water from a well to a storage tank located at a height of 15 m above the well. If the flow rate of water is 0.1 mยณ/s and the density of water is 1000 kg/mยณ, calculate the power required to operate the pump.**

Solution:

Power = (Density ร g ร Q ร H) / Efficiency

Power = (1000 kg/mยณ ร 9.81 m/sยฒ ร 0.1 mยณ/s ร 15 m) / Efficiency

**Water flows through a horizontal pipe with a diameter of 0.3 m. If the pressure difference between the two ends of the pipe is 2000 Pa, calculate the flow rate of water through the pipe. (Assume the density of water as 1000 kg/mยณ)**

Solution:

Flow rate = (Pressure difference) / (Density ร g)

Flow rate = 2000 Pa / (1000 kg/mยณ ร 9.81 m/sยฒ) โ 0.203 mยณ/s

Answer: Flow rate of water through the pipe โ 0.203 mยณ/s

**A cylindrical tank contains oil with a height of 4 m and a diameter of 2 m. If the density of oil is 850 kg/mยณ, calculate the weight of the oil in the tank.**

Solution:

Volume of oil in the tank = ฯ ร (Diameter/2)ยฒ ร Height

Volume of oil = ฯ ร (2 m/2)ยฒ ร 4 m โ 6.28 mยณ

Weight of oil = Volume ร Density of oil

Weight of oil = 6.28 mยณ ร 850 kg/mยณ โ 5338 kg

Answer: Weight of oil in the tank โ 5338 kg

**A rectangular water tank has dimensions 6 m by 4 m by 3 m. Calculate the pressure at the bottom of the tank. (Assume the density of water as 1000 kg/mยณ and acceleration due to gravity as 9.81 m/sยฒ)**

Solution:

Pressure at the bottom = (Density ร g ร Height)

Pressure at the bottom = (1000 kg/mยณ ร 9.81 m/sยฒ ร 3 m) = 29430 Pa

Answer: Pressure at the bottom of the tank = 29430 Pa