The tailrace of a turbine installation is 100 m below the water level of the dam

Question

The tailrace of a turbine installation is 100 m below the water level of the dam. as shown in the figure. The total length of the pipe with a diameter of 0.5 m is 300 m and its absolute roughness is 0.05 mm. The head loss coefficient at the entrance of the piping is 0.5. If the volumetric flow rate of the water is 0.8m3/s. determine the power developed by a turbine having an efficiency of 80 percent. The density and kinematic viscosity of water are 1000 kg/m3 and 1 Times 10 6 m2/s. respectively. (Ans.

Explanation / Answer

Let the surface of the reservoir be section 1, and the exit be section 2. Obviously Z1 - Z2 = 100 m,

and p1 = p2 = patm, V1 0 and V2 = V the velocity in the pipe.

The fluid velocity in the pipe is

V = Q/A = 0.8/(*0.52/4) = 4.074 m/s

Re = Vd/ = 4.074*0.5/(1x10-6) = 2037000

e/d = 0.05 mm/0.5 m = 0.0005 m/0.5 m = 0.0001

From Moody Chart, f 0.0129

hf = f(L/d)[V2/(2g)] = 0.0129*(300/0.5)*[4.07442/(2*9.81)] = 6.55 m

Minor loss      hm = V2/(2g) K= 4.0742 /(2*9.81)*(0.5) = 0.423 m

The energy equation for this problem is

p1/ + V12/(2g)+ z1 = p2/ + V22/(2g)+ z2 +hf   + hm +ht           =>

patm/ + 02/(2g)+ 100 = patm/ + 4.07442/(2g)+6.55+0.423+ht      =>

ht = 92.18 m

The power is ( is the efficiency)

P= (gQht ) = 80%*1000*9.81*0.8*92.23 = 578749 W = 578.75 kW

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.