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If power P available from a hydro-scheme is given by the formula P = 9.81QH, where Q is the flow rate through the turbine in l/s and H is the head in meters, then P will be in units of

$W$

$HP$

$kJ/s$

$kWh$

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalPower System

Option A

$W$

Quick Summary:

The power available from a hydro-scheme is given by the product of the flow rate and head, modified by the gravitational constant and fluid density. The given formula $P = 9.81QH$ assumes $Q$ is in liters/second ( $l/s$ ) and $H$ is in meters ( $m$ ), which simplifies the standard mechanical power expression $P = \rho g Q H$ to units of Watts ( $W$ ).

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$P = \rho g Q H$ — Theoretical power in Watts

$P = 9.81 Q H$ — Simplified power formula for $Q$ in $l/s$

⚙️ Working Principle

Mechanical power is defined as the rate of doing work, calculated as $P = \rho \cdot g \cdot Q \cdot H$ . When $Q$ is expressed in $m^3/s$ and $\rho$ is $1000 kg/m^3$ , $P = 1000 \cdot 9.81 \cdot Q \cdot H$ . Since $1 m^3/s = 1000 l/s$ , substituting $Q_{m^3/s} = \frac{Q_{l/s}}{1000}$ results in $P = 1000 \cdot 9.81 \cdot \frac{Q_{l/s}}{1000} \cdot H$ , which simplifies exactly to $P = 9.81 \cdot Q_{l/s} \cdot H$ Watts.

📌 Key Points

▸
The constant 9.81 represents the acceleration due to gravity ( $g \approx 9.81 m/s^2$ ).
▸
Conversion factors are critical; $1 m^3 = 1000$ liters, which cancels the density factor in this simplified formula.
▸
Watts ( $W$ ) is the SI unit for power, equivalent to Joules per second ( $J/s$ ).

✅ Advantages

▸
Allows rapid calculation of hydro-potential using common engineering units.
▸
Eliminates complex conversions of volumetric flow rate during estimation.

❌ Disadvantages / Limitations

▸
Only applicable when density is assumed to be that of water ( $1000 kg/m^3$ ).
▸
Does not account for turbine efficiency ( $eta$ ).

🛠️ Applications / Uses

▸
Micro and Pico hydro project estimation.
▸
Quick site potential assessment in power systems engineering.

📄 Additional Information

▸
If efficiency $\eta$ is included, the formula becomes $P = 9.81 \eta Q H$ Watts.
▸
Option B (HP) is incorrect because 1 HP = 746 Watts.
▸
Option D (kWh) is a unit of energy, not power.

📊 Diagram / Illustration

✅

A is correct — The resulting power unit for the given formula $P = 9.81QH$ where $Q$ is in $l/s$ and $H$ is in meters is Watts ( $W$ ).

Core Concepts Used

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Hydro-power generation Unit Dimensional Analysis Gravitational potential energy

💡 EXAM TIP

Always verify units before applying constants in engineering formulas; hydro-power problems frequently use the simplified $9.81QH$ formula to save time in competitive exams.

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