Question 9.4.6: A Kaplan turbine working under a head of 10 m and at a desig......
A Kaplan turbine working under a head of 10 m and at a design speed of 250 rpm has a flow rate of 24 m³/s. The diameter of the runner and hub are 2 m and 1 m, respectively. The inlet and outlet diameter of the draft tube are 2 m and 3 m, respectively. The pressure recorded at inlet of the draft tube is 3 m vacuum. The vapour and barometric pressure are 1.6 m and 10 m, respectively. Efficiency of draft tube is 80%. The Thoma’s cavitation factor for the turbine is given by the relation, σ = η_d K^2_f + λ K^2_u, in which η_d is the draft tube efficiency, K_f is the flow ratio, KK_u is the speed ratio and λ is a dimensionless number defined by λ =(p_2/w -p_{\min}/w)/(u^2/2g) in which u is the tangential velocity, (p_{2} /w) is the pressure head at inlet to draft tube, and (p_{\min} /w) is the minimum pressure head at a point on the blade. If overall efficiency is 90%, determine the minimum pressure on the blade.
Given:
Q = 24 m^3/s, N = 250 \ \mathrm{rpm}, D_o = 2 m, D_i = 1 m
→
A=\pi/4(D_0^2-D^2_1)=3/4\pi
Draft tube:
d_1 = 2 m, d_2 = 3 m, p_a/w = 10 m, p_ν/w = 1.6 m,
p_2/w = 10 – 3 = 7 m, η_d = 0.8
\sigma =\eta _dK_f^2+\lambda K_u^2, \lambda =(p_2/w-p_{\min}/w)(u^2/2g),\eta_o=0.9 \text{ and } H=10 m
To determine σ, we need K_f, K_u, λ, η_d \text{ and } u. K_f \text{ and } K_u can be obtained from the relations (9.1.27), where K_u, K_f and u are: (substitute the values in following steps and verify the results obtained).
Q=A\times K_f\times\sqrt{2gH} \rightarrow K_f=0.727
u=\pi D_oN/60 =K_u\times \sqrt{2gH} \rightarrow K_u=1.87 \text{ and } u=26.18 m/s
We are now in a position to express Thoma factor in terms of λ. Substituting the known values, we get
\sigma =0.423 +3.497 \lambda (A)
We shall determine it by carrying out the analysis of draft tube. Applying modified Bernoulli equation between points 1 and 2 (refer Fig. 9.4.6) to give
(p_1-p_a)/w=(C^2_2-C^2_1)/2g-H_d+h_f
(it is same as Eq. 9.4.9 except that head loss h_f has been included). Substituting for h_f in terms of draft tube efficiency from Eq. 9.4.10, we get
(p_2-p_a)/w=(C^2_2 -C_1^2)/2g-H_d (9.4.9)
\eta _d=\left\{\left\lgroup C_1^2-C_2^2\right\rgroup /2g-h_f\right\}/(C_1^2/2g), (9.4.10)
p_1/w=p_a/w-H_d-\eta _d(C_1^2/2g)
C_1 can be determined as diameter of draft tube at it inlet and Q are given.
This gives:C_1=Q/(\pi/4 d^2_1)=7.64 m/s
Substituting for η_d, C_1, p_1/w, p_a/w, we get:
H_d = height of the draft tube inlet from the tail race = 0.62 m. Cavitation factor is defined as
\sigma =(p_a/w-p\upsilon /w-H_d)/H_d=(10-1.6 -0.62)/10=0.778 (B)
Equating (A) and (B), we get, λ = (p_2/w – p_{\min}/w)/(u^2/2g) = 0.1015
→ p_{\min}/w = – 6.545 m or 6.545 (vacuum)
→ or p_{\min} = 9.81 × 6.545 = 64.2 kN/m^2 (vacuum) .