Water flows from one reservoir to another a height, 45 m below. a turbine (η=0.74) generates power from this flow. 1 m3/s passes through the turbine. if 11 m headloss occurs between the two reservoirs, determine the actual (i. e. useable) power generated by the turbine (in kw)

The power generated is 246.820 kW

Explanation:

Imagine or draw the control volume: Its limits would be from the surface of the  first reservoir, to the surface of the second one, which is above.

Consider the energy balance (its formulation can be studied in Cengel (Fluid mechanics, chapter five):

Where H is the energy head loss in squared velocity units. (i.e. . 2 is the final reservoir and 1 is the feeding reservoir. If the reservoirs are so big, it is possible to assume the velocity on both surfaces as zero (the mass flow is not enough to induce a appreciable velocity because of the decrease on the water level). Both reservoir surfaces should be at the same pressure, so the pressure difference will be zero (neglective). Additionally, the head loss may also consider the energy loss by heat (because there is not any considerable heat transfer), so the equation is finally:

Dividing by m*g, we find an equation for the turbine head:

h is the head loss in length units.

So, calculate :

   The turbine head is negative because the energy is going out of the fluid, but it is positive if we consider it as produced energy (the sign depends on what you are analyzing).

With   find the work:

This work calculated before is the energy that the fluid delivery to the turbine, but the turbine is not 100% efficient, so it is not the energy that the turbine produces. The ratio between the delivered and the produced energy is the efficiency (I am going to write it as e):

e=(produced energy)/(delivered by the fluid energy)

So, the produced energy, can be calculated as:

(Wd is the delivered energy).

Then,

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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