Fluid Mechanics Test 6

Concept:

Hydro Electric Power Output:

The theoretical amount of power generated when water with flow rate Q is allowed to fall through net head of H is equal to, \({{\rm{P}}_{{\rm{th}}}} = {{\rm{\gamma }}_{\rm{w}}}{\rm{QH}}\), where γ_w is unit weight of water.

But as there is losses in various components of the plants like in pumps, turbines, pipes, so actual power output will be less than the theoretical output and their ratio is known as plant efficiency.

\(\therefore {\rm{Plant\;Efficiency\;}}\left( {\rm{\eta }} \right) = \frac{{{\rm{Actual\;Power\;Output}}}}{{{\rm{Theoritical\;Power\;Output}}}} = \frac{{\rm{P}}}{{{{\rm{P}}_{{\rm{th}}}}}}\)

∴ P = η × P_th = η × γ_w × Q × H

Load Factor: It is the ratio of average load over a certain period to peak load over that period.

\(\therefore {\rm{Load\;Factor}},{\rm{\;}}{{\rm{L}}_{\rm{f}}} = \frac{{{\rm{Average\;Load\;}}\left( {{\rm{\;or\;Power}}} \right)}}{{{\rm{Peak\;Load\;}}\left( {{\rm{\;or\;Power}}} \right)}} = \frac{{\rm{P}}}{{{{\rm{P}}_{{\rm{peak}}}}}}\)

Calculation:

Given, Flow rate Q= 26 cumec =26 m³/s, Net head available, H = 20 m, Plant efficiency, η=0.6, unit weight of water, γ_w = 10 kN/m³ and load factor L_f = 0.75

∴ Power output (P) = ηγ_wQH = 0.6 × 10 × 26 × 20 = 3120 kW

\(\therefore {\rm{Load\;Factor\;}}\left( {{{\rm{L}}_{\rm{f}}}} \right) = \frac{{\rm{P}}}{{{{\rm{P}}_{{\rm{peak}}}}}}\)

\( \Rightarrow 0.75 = \frac{{3120}}{{{{\rm{P}}_{{\rm{peak}}}}}}\)

∴ P_peak = 4160 kW.

In single-acting reciprocating pump , head available at cylinder (H) is given by

H= H_atm – H_{suction head} – H_{suction acceleration} – H_{suction friction loss}

H_atm = 10 m (Given)

H_{Suction friction loss} = Neglected as not given

H_{Suction head} = 1.5 m (Given)

H_{Suction acceleration} = + 3.5 (At beginning of stroke)

HSuction acceleration = - 3.5 (At the end of stroke)

HSuction acceleration = 0 (At mid-stroke)

Case 1:

At beginning of stroke:

H = 10 - 1.5 - 3.5 = 5 m

Case 2:

At end of stroke:

H = 10 - 1.5 - (-3.5) = 8 m

Concept:

Runoff River Plant: It utilises current flow in stream and haspondage to improve load factor and no storage is required here. Most of the small hydro power plants are Runoff river plant as the power is generated only when enough water is available from the river.

When the stream flow reduces below the design flow value, the generation will reduce as the water does not flow through the intake structure into the turbine.

Reservoir Plant: It utilises annually average flow with storage and hence it requires large storage. It consists of a reservoir with gate-controlled outlets wherein surface water may be retained for a considerable period of time and released for use at a time when the normal flow of the stream is insufficient to satisfy requirements.

Pumped storage plant: This generates power during peak hours, but during the off-peak hours, water is pumped back from the tail water pool to head water pool. So these plants use reversible turbines which functions both as turbine and pump.

Concept:

Specific speed of a centrifugal pump is given by:

\({N_S} = \frac{{N\sqrt Q }}{{{H^{\frac{3}{4}}}}}\) 

Where,

H = Head of water

Q = Discharge through the pump

N = Operating speed of the pump

Calculation:

Q = 82 m³/s

N = 1500 rpm

H = 49 m

\({N_S} = \frac{{1500\sqrt {82} }}{{{{\left( {49} \right)}^{\frac{3}{4}}}}} = 733.42\) 

Hence, specific speed will be 734.

Concept:

The theoretical discharge through piston is given by:

\({Q_{th}} = \frac{{ALN}}{{60}}\)

Where,

A = Cross –sectional area

L = Stroke length

N = Speed

D = Dia. Of piston

Calculation:

Given, L = 25 cm, N = 135 rpm, D = 30 cm

\({Q_{th}} = \frac{{\frac{\pi }{4} \times \left( {{{0.3}^2}} \right) \times 0.25 \times 135}}{{60}}\)

Q_th = 0.0397 m³/s or 39.7 l/s

We know that,

\({\rm{\% \;slip}} = 1 - {\rm{\;}}\frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{th}}}}}}\)

\(0.04 = 1 - \frac{{{Q_{actual}}}}{{39.7}}\)

∴Q_actual = 38.11 l/s

Concept:

The quantities such as unit speed (N_u), unit power(P_u) and unit discharge (Q_u) are speed, power, and discharge respectively for a turbine working under the unit head. These are given by:

\({{\rm{N}}_{\rm{u}}} = {\rm{\;}}\frac{{\rm{N}}}{{\sqrt {\rm{H}} }}\)

\({{\rm{P}}_{\rm{u}}} = {\rm{\;}}\frac{{\rm{P}}}{{{{\rm{H}}^{3/2}}}}\)

\({{\rm{Q}}_{\rm{u}}} = {\rm{\;}}\frac{{\rm{Q}}}{{\sqrt {\rm{H}} }}\)

Calculation:

Given that,

N_u = 98; P_u = 12; H = 8.5

Using \({{\rm{N}}_{\rm{u}}} = {\rm{\;}}\frac{{\rm{N}}}{{\sqrt {\rm{H}} }}\)

 \(98 = {\rm{\;}}\frac{{\rm{N}}}{{\sqrt {8.5} }}\)

∴ N = 286 rpm

Similarly, \({\rm{P}} = {{\rm{P}}_{\rm{u}}}{{\rm{H}}^{\frac{3}{2}}} = 12 \times {\left( {8.5} \right)^{\frac{3}{2}}} = 297.4{\rm{\;KW}}\)

Specific speed of turbine, N_s is given by:

\({N_s} = \frac{{N\sqrt P }}{{{H^{\frac{5}{4}}}}}\)

N is in rpm; P is in kW and H is in m.

\({N_s} = \frac{{286\sqrt {297.4} }}{{{{8.5}^{\frac{5}{4}}}}} = 339.5{\rm{\;units}}\)

Gross head, H_g = 600 m

Head loss in friction,

\({h_f} = \frac{{{H_g}}}{3} = 200\;m\) 

Net head, H = H_g - h_f = 600 – 200 = 400 m

\({V_1} = {C_v}\sqrt {2gH} = 1\sqrt {2 \times 9.81 \times 400} = 88.59\;m/s\) 

u = speed ratio × √2gH

\(\begin{array}{l}u = 0.45 \times \sqrt {2 \times 9.81 \times 400} = 39.86\;m/s\\u = \frac{{\pi DN}}{{60}}\\D = \frac{{u \times 60}}{{\pi N}} = \frac{{39.86 \times 60}}{{\pi \times 1000}} = 761.27\;mm\end{array}\) 

Ratio of jet diameter to wheel diameter:

\(\begin{array}{l}\frac{d}{D} = \frac{1}{6}\\d = \frac{1}{6} \times D = \frac{1}{6} \times 761.27 = 126.88\;mm\end{array}\)