CALCULATING THE AMOUNT OF AVAILABLE POWER

A simple formula for approximating electric power production at a hydroelectric plant is: P = ρhrgk, where P is Power in watts, ρ is the density of water (~1000 kg/m3), h is height in meters, r is flow rate in cubic meters per second, g is acceleration due to gravity of 9.8 m/s2, and k is a coefficient of efficiency ranging from 0 to 1. Efficiency is often higher (that is, closer to 1) with larger and more modern turbines.

Annual electric energy production depends on the available water supply. In some installations the water flow rate can vary by a factor of 10:1 over the course of a year. Hunza river offers fifty-two times more water in summers than during winters, the extremes of run-off in the Shigar river varied in 1986 by a factor of 87. These extremes apply in a similar manner for the Gilgit, Upper Indus and Shyok, all of which derive their major discharge from glacier melt.

The catchment area upriver of the Tarbela Dam is spread over 168,000 square kilometers (65,000 sq mi) of land largely supplemented by snow and glacier melt from the southern slopes of the Himalayas. There are a number of main Indus River tributaries upstream of the Tarbela Dam. These are the Shyok River, joining near Skardu, Hunza and Gilgit near Danyur, Astore near Bunji and the Siran River near Tarbela. Discharge for Arabian Sea - average 6,600 m3/s (230,000 cu ft/s) approx.

http://www.waterinfo.net.pk/pwss/vol1main.htm

While calculating the potential for Gilgit and Baltistan, the coefficients in the above equation are as follows:

H= 1500m - 610m (4971ft) - (2001ft) =880m (2970 ft)

R=6,600 m3/s (230,000 cu ft/s)

G=9.8 m/s2

K=0.8

Therefore

P=ρhrgk=1000*880*6,600*9.8*0.8=45534720000Watt=45,534.72 Mega Watt

Even if 50% of this potential is harnessed then it not only meets the energy Needs of Pakistan but would be available for export to other countries.

Similarly if the potential in Chitral/Swat/Kaghan/AJK and other river sources is added to this potential on the Indus alone, you can gauge the prospects for the entire country.

Implications: If only 10,000 MW is harnessed, it will generate daily revenue of RS 240 Billion even if it is sold at a paltry rate of Rs1.0 per unit. Remember the current energy in Pakistan is billed at Rs 22.00  to Rs 41.00 per unit. This volume of revenue will not only be sufficient to pay the loans/interest but it will also generate thousands of jobs for the coming generation, besides making the region envy of the world.

Message to the leadership in Pakistan: “Please heed to this rather than your coffers”.

Brig Hisamullah Beg SI(M)

Baltit, Hunza

I am seeking the Government in Gilgit and Baltistan to make a policy decision; meanwhile I have mailed quarries to agencies such as Kuwait Fund, ADB, World Bank and AIIB. This can have a meaning only if the Government of GB modifies present policy of allowing “Private Investment with Community involvement limited to 2-Mega Watt and below”. Furthermore In view of the inefficient and corrupt set up of the government (small contractors tell me that they have to pay Rs 500/00 for the file to move from one desk to the other; military monitoring team says it is filth whatever you check) it will only be possible to harness this potential through a strong “CIVIL SOCIETY”

Hydroelectric Potential at Ataabad Site (before dredging):

(Considering the Average Winter Flow)

H=55.0 m R=95 cum/sec (Minimum Winter Flow) G=9.8 m/sec/sec K=0.8 ρ ~1000 kg/m³

Power=ρhrgk =1000*55*95*9.8*0.8 = 40964000 Watts= 40.96 MW

NOTE: After 24-m dredging the potential according to my calculations has come down to 26-MW. Recent feasibility - April 2021 - by WAPDA team of experts talks of 54-MW.


RELATED LINK: GB GOVERNMENT, WAPDA, ADVOCACY, RESEARCH

Copy of the letter to Government of GB:
"Dear Sir,
I will feel obliged if you would kindly provide me the GB government's written policy /Law on this topic. For a background to this request you are invited to visit:

http://hisamullahbeg.blogspot.com/2010/03/calculating-amount-of-available-power.html
Alternately it can be posted on the official site from where I and other interested people can download."

Alternate   formula for hydropower:

The energy generated by a hydroelectric power plant can be calculated using a two-step process: first, determine the power output, and then use that to calculate the energy produced over a specific period.

1. Calculating Hydroelectric Power (P)

The fundamental formula to calculate the power output of a hydroelectric system (especially for dam-based plants) is:

$P=\eta \cdot \rho \cdot g\cdot h\cdot Q$

Where:

• P is the Power output, measured in Watts (W).
• η (eta) is the overall efficiency of the hydroelectric system. This accounts for losses in the turbine, generator, and other components. It's a dimensionless value, typically expressed as a decimal (e.g., 0.85 for 85% efficiency).
• ρ (rho) is the density of water, which is approximately 1000 kg/m3 (kilograms per cubic meter). This value can vary slightly with temperature and purity.
• g is the acceleration due to gravity, which is approximately 9.81 m/s2 (meters per second squared).
• h is the effective hydraulic head, measured in meters (m). This is the vertical distance the water falls from the intake to the turbine. For more precise calculations, "net head" is used, which accounts for head losses due to friction in the penstock (the pipe carrying water to the turbine).
• Q is the volumetric flow rate of the water, measured in cubic meters per second (m3/s). This is the volume of water passing through the turbines per unit of time.

Simplified Power Formula:

Sometimes, a simplified version is used where the density of water and acceleration due to gravity are combined into a constant:

$P\approx \eta \cdot 9.81\cdot h\cdot Q\text{(if Q is in }{\text{m}}^3\text{/s and P in kW, a conversion factor for density is implicitly included or units adjusted elsewhere)}$

However, the first, more explicit formula is generally preferred for clarity and accuracy.

For Run-of-River and Tidal Power Stations:

These systems primarily use the kinetic energy of flowing water. The formula differs slightly:

$P=0.5\cdot \eta \cdot \rho \cdot Q\cdot v^2$

Where:

• v is the flow velocity of the water in meters per second (m/s).
• The other variables (η, ρ, Q) are the same as above.

2. Calculating Hydroelectric Energy (E)

Once you have the power output (P), you can calculate the energy (E) generated over a specific period using the formula:

$E=P\cdot t$

Where:

• E is the Energy, typically measured in Joules (J), Watt-hours (Wh), or kilowatt-hours (kWh).
• P is the Power output (calculated using the formulas above), in Watts (W) or kilowatts (kW).
• t is the period for which the plant is operating, measured in seconds (s) or hours (h).

Conversion:

• If power is in Watts and time in seconds, energy will be in Joules ($1\text{ J}=1\text{ W}\cdot \text{s}$).
• If power is in kilowatts and time in hours, energy will be in kilowatt-hours. Kilowatt-hours are the most common unit for electricity billing.

In summary, to calculate hydroelectric energy:

1. Determine the appropriate power formula based on the type of hydroelectric system.
2. Gather the values for efficiency, water density, gravity, head (or velocity), and flow rate.
3. Calculate the power output (P).
4. Multiply the power (P) by the time (t) the plant operates to find the energy (E) generated.