Applied Challenge

Unit 1 Engineering Project

A building's water supply: one realistic problem that uses Lessons 1, 2, 4, 6, 10, 11, 12 and 13.

Put the whole unit to work on one real problem. This project counts as the Applied Challenge: it opens once all 13 lessons are completed and takes about 30 minutes. Use the Project preset of the Lesson 11 tank simulation and the Lesson 13 Pump Power calculator.

The Situation

A five-storey apartment block has an underground sump and a 10,000 L rooftop tank. A pump at ground level lifts water 18 m from the sump's water surface to the tank inlet at 1.5 L/s. Pump efficiency 55%, motor efficiency 80%. The tank's water surface is 16 m above the ground-floor taps. Residents use 10,000 L a day.

You can work through the project now. Log in and complete all 13 lessons to save it and earn the Unit 1 Engineer badge.

  1. Identify The Working Fluid Lessons 1 and 4

    Is water a fluid here, and can we treat it as incompressible at these pressures?
    Choose one
    Show the model answer
    Water is a fluid, because it cannot hold a shear stress at rest. Treat it as incompressible, because its bulk modulus (2.2 GPa) is huge compared with the pressures here.

  2. Draw The System Boundary For The Pump Lesson 10

    Draw a boundary around the pump casing. What kind of system is it?
    Choose one
    Show the model answer
    Boundary around the pump casing: water crosses in and out and shaft work enters, so it is an open system. Surroundings: sump, pipes, motor, air.

  3. Scenario A: Fill The Empty Tank While Residents Draw Water Lesson 11

    Set up a control volume around the rooftop tank. With the pump on (1.5 L/s) and residents drawing 0.5 L/s, how many seconds does it take to fill the empty 10,000 L tank?
    s
    Show the model answer
    dm/dt = 1.5 − 0.5 = 1.0 kg/s, so the tank gains 1 kg every second and fills in 10,000 s, about 2.8 h. With no draw it would take 10,000 ÷ 1.5 = 6,667 s, about 1.85 h.

  4. Pressure At A Ground-Floor Tap Lessons 2 and 6

    What is the gauge pressure at a ground-floor tap when nothing flows? Give your answer in kPa.
    kPa
    Show the model answer
    p = ρgh = 1000 × 9.81 × 16 = 157 kPa gauge, about 1.6 bar. When taps open, the pressure is lower because of flow speed and pipe friction.

  5. List The Quantities And Their SI Units Lesson 12

    Which unit is the SI unit of mass flow rate?
    Choose one
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    Flow rate Q (m³/s), mass flow ṁ (kg/s), height H (m), pressure p (Pa), power P (W), energy E (J; also kWh), time t (s).

  6. Scenario B: Pump Power And Daily Energy Lesson 13

    The pump must deliver all 10,000 L the residents use in a day. What is the electrical input power of the motor, in watts? Then work out the energy per day.
    W
    Show the model answer
    Hydraulic power = 1000 × 9.81 × 0.0015 × 18 = 265 W. Shaft power = 265 ÷ 0.55 = 482 W. Electrical input = 482 ÷ 0.80 = 602 W. The pump runs 10,000 ÷ 1.5 = 6,667 s (1.85 h) a day, using 0.602 × 1.85 = 1.11 kWh per day.

  7. Daily Energy Lesson 13

    Using Scenario B, how many kWh does the motor use per day?
    kWh
    Show the model answer
    0.602 kW × 1.85 h = 1.11 kWh per day.

  8. Optional: Effect Of Efficiency Lesson 13

    Replace the pump with a 70% efficient one (motor still 80%). At ₹8 per kWh, and if the better pump costs ₹3,000 more, after how many years does it pay for itself?
    years
    Show the model answer
    Shaft 378 W, electrical 473 W, 0.88 kWh per day. Saving: about 0.24 kWh per day, about 87 kWh per year. At ₹8 per kWh that is about ₹700 a year, so ₹3,000 ÷ ₹700 per year gives a payback of about 4.3 years.

Assumptions

H is the static lift only (real pipe friction adds several metres of head); efficiencies are constant; water at 20 °C with ρ = 1000 kg/m³.