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A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof?
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A closed tank containing water is partly filled with water and the air...
Given data:
- Height of building (H) = 3 m
- Friction losses (hf) = 1.5 m of water
- Discharge flow rate (Q) = 15 L/s
- Diameter of hose (d) = 5 cm
To determine:
- Air pressure in the tank required to deliver water to the roof of the building
Assumptions:
- Water is incompressible
- Friction losses are major losses
- Velocity of water at the outlet of the hose is negligible compared to its velocity inside the hose
Solution:
1. Calculate the velocity of water inside the hose:
- Convert the diameter of the hose from cm to m: d = 5/100 = 0.05 m
- Calculate the cross-sectional area of the hose: A = π/4 * d^2 = 0.0019635 m^2
- Calculate the velocity of water inside the hose: v = Q/A = 15/(0.0019635) = 7637.75 m/s
2. Calculate the total head loss in the system:
- Total head loss (hL) = hf + hroof + hpipe
- hroof = H = 3 m (height of building)
- hpipe = 0.0 m (no elevation change in horizontal pipe)
- hL = hf + hroof + hpipe = 1.5 + 3 + 0 = 4.5 m
3. Determine the pressure at the outlet of the hose:
- Convert the total head loss from m to Pa: hL = 4.5 * 9.81 = 44.145 Pa
- Calculate the pressure at the outlet of the hose: Pout = Patm - hL = 101325 - 44.145 = 101280.855 Pa
4. Determine the pressure at the inlet of the hose:
- Convert the velocity of water inside the hose from m/s to ft/s: v = 7637.75 * 3.281 = 25059.744 ft/s
- Calculate the friction factor (f) using the Moody chart for smooth pipes at Re = 4.92E+05 and ε/D = 0.0015: f = 0.023
- Calculate the Reynolds number (Re) for the flow through the hose: Re = (v * d) / ν = (25059.744 * 0.05) / 1E-06 = 1.2529872E+09
- Calculate the frictional head loss (hf) in the hose using the Darcy-Weisbach equation: hf = (f * L * v^2) / (2 * g * d) = (0.023 * 3 / 0.05 * 25059.744^2) / (2 * 9.81 * 0.05) = 104.121 m
- Convert the frictional head loss from m to Pa: hf = 104.121 * 9.81 = 1021.44 Pa
- Calculate the pressure at the inlet of the hose: Pin = Pout + hf = 101280.855 + 1021.44 = 102302.295 Pa
5. Determine the air pressure in the tank:
- The air pressure in the tank must be greater
- Height of building (H) = 3 m
- Friction losses (hf) = 1.5 m of water
- Discharge flow rate (Q) = 15 L/s
- Diameter of hose (d) = 5 cm
To determine:
- Air pressure in the tank required to deliver water to the roof of the building
Assumptions:
- Water is incompressible
- Friction losses are major losses
- Velocity of water at the outlet of the hose is negligible compared to its velocity inside the hose
Solution:
1. Calculate the velocity of water inside the hose:
- Convert the diameter of the hose from cm to m: d = 5/100 = 0.05 m
- Calculate the cross-sectional area of the hose: A = π/4 * d^2 = 0.0019635 m^2
- Calculate the velocity of water inside the hose: v = Q/A = 15/(0.0019635) = 7637.75 m/s
2. Calculate the total head loss in the system:
- Total head loss (hL) = hf + hroof + hpipe
- hroof = H = 3 m (height of building)
- hpipe = 0.0 m (no elevation change in horizontal pipe)
- hL = hf + hroof + hpipe = 1.5 + 3 + 0 = 4.5 m
3. Determine the pressure at the outlet of the hose:
- Convert the total head loss from m to Pa: hL = 4.5 * 9.81 = 44.145 Pa
- Calculate the pressure at the outlet of the hose: Pout = Patm - hL = 101325 - 44.145 = 101280.855 Pa
4. Determine the pressure at the inlet of the hose:
- Convert the velocity of water inside the hose from m/s to ft/s: v = 7637.75 * 3.281 = 25059.744 ft/s
- Calculate the friction factor (f) using the Moody chart for smooth pipes at Re = 4.92E+05 and ε/D = 0.0015: f = 0.023
- Calculate the Reynolds number (Re) for the flow through the hose: Re = (v * d) / ν = (25059.744 * 0.05) / 1E-06 = 1.2529872E+09
- Calculate the frictional head loss (hf) in the hose using the Darcy-Weisbach equation: hf = (f * L * v^2) / (2 * g * d) = (0.023 * 3 / 0.05 * 25059.744^2) / (2 * 9.81 * 0.05) = 104.121 m
- Convert the frictional head loss from m to Pa: hf = 104.121 * 9.81 = 1021.44 Pa
- Calculate the pressure at the inlet of the hose: Pin = Pout + hf = 101280.855 + 1021.44 = 102302.295 Pa
5. Determine the air pressure in the tank:
- The air pressure in the tank must be greater
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A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof?
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A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof?.
A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A closed tank containing water is partly filled with water and the air space above it is under pressure. A 5 c.m. hose connected to the tank discharges water to atmosphere on to the roof of a building 3 m above the level of water in the tank, if friction losses are 1.5 m of water, what air pressure must be maintained in the tank to deliver 15 L/s to the roof?.
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