Background The design for a water supply system (illustrated in Figure 1) has be
ID: 1821411 • Letter: B
Question
BackgroundThe design for a water supply system (illustrated in Figure 1) has been proposed for Fawlty Towers, a ten story building in Brisbane. Water is to be supplied to all floors from an open storage tank on the roof of the building. Water is pumped into the tank via a single riser pipe from a pump connected to the local water supply on the ground floor. The owner of the building (Mr Basil) has proposed that the
tank NOT be pressurized to reduce the cost of the equipment. The specifications of the water supply system are given in Table 1. Assume the tank is kept full at all times.
Tank capacity 100 kL
Distribution pipes internal diameter 20.0 mm
Pump Rating (Input power) 500 W
Density of water 1000 kg/m3
Pump efficiency 0.750
Acceleration due to gravity 9.81 m/s2
Riser pipe internal diameter 10.0 cm
Note: not all data may be needed for the problem.
Task
Your company (Stubbs & Co) have been contracted to verify if the proposed design meets the requirements for the building, specifically:
Requirement 1: The maximum flow rate of water into the tank is at least 4 kL/hr
Requirement 2: The minimum gauge pressure to all floors of the building is 140 kPA
(a) Demonstrate, by showing your working, if the proposed system meets Requirement 1 or not.
(b) Demonstrate, by showing your working, if the proposed system meets Requirement 2 or not.
10 floors x 3 m / floor
pipe 3 m drop / floor for distribution pipes
3 m tank depth
Tank is on the top of the building at 30m high, pipe on every floor at 3m intervals
Explanation / Answer
a) The water needs to be pumped to the top of the tank, 33m up from the ground level. The change in potential energy for one cubic meter of water (which is also 1 kL):
mass*gravity*height = density*volume*gravity*height = 1000 kg * 9.81 m/s^2 * 33m = 323730 Joules per kL
The power required to pump 4 kL of water in an hour is to total energy divided by the time in seconds:
323730 J/kL * 4 kL/hr / 3600 seconds/hr = 359.7 W
The pump has an input power of 500 W and an efficiency of 0.75, so it produces 500*0.75 = 375 W of 'useful' power, which is greater than the calculated power requirements for the tank so yes, the requirements for part a) are met.
b) Hydrostatic water pressure is calculated by adding the force of the water at a particular depth to the ambient pressure at the surface of the water.
Pressure = ambient pressure + water density * gravity * depth
The ambient pressure is assumed to be at sea level, 1 atm = 101.3 kPa.
The lowest water pressure will be that of the top floor, all other floors will have a higher pressure due to increased depth. If the first floor has sufficient pressure, they all will. Since the depth of the pipes on the first floor is 6m, the water pressure there is:
101325 Pa + 1000 kg/m^3 * 9.81 m/s^2 * 6m = 160.2 kPa
This is greater than 140 kPa, so all floors will have at least 140 kPa hydrostatic pressure.
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