Turbomachinery Test 1

Concept:

In semicircular vane turbines, theoretical efficiency is given as:

\(\eta = \frac{{1 + \cos \phi }}{2}\)

Calculation:

Maximum value of cos ϕ = 1

\(\therefore {\eta _{max}} = \frac{{1 + \cos \phi }}{2}\)

\({\eta _{max}} = \frac{{1 + 1}}{2}\)

∴ η_max = 100 %

Concept:

For a centrifugal pump, impeller diameter always refers to the outlet diameter even if not specified. The power at the pump impeller is greater than the power given to water at outlet of pump. This ratio of power given to water at outlet to the available power at impeller is called manometric efficiency and is given as:

\({\eta _m} = \frac{{g{H_m}}}{{{V_{w2}} . {U_2}}}\)

Where, H_m is the net head (manometric head) V_w2 is the whirl velocity at outlet, V₂ is impeller speed at outlet.

Also, \({U_2} = \frac{{\pi {D_2}N}}{{60}}\)

Calculation:

H_m = 14.5 m, N= 1000 rpm, D₂ = 300 mm

η_{man θ} = 0.95

\({U_2} = \frac{{\pi \times 0.3 \times 1000}}{{60}} = 5.7\;m/s\)

Explanation:

Given:

H_v = 0.35 m, σ_c = 0.12, H = 50 m, N = 350 rpm, P_atm = 1.03 kgf/cm²

\({H_{atm}} = \frac{{{P_{atm}}}}{{\rho g}} = \frac{{1.03 \times 9.81 \times {{10}^4}}}{{{{10}^3} \times 9.81}} = 10.3\;m\)

Now,

Critical cavitation factor (σ_c) is given by:

\({\sigma _c} = \frac{{{H_{atm}} - {H_s} - {H_v}}}{H}\)

\(0.12 = \frac{{10.3 - {H_s} - 0.35}}{{50}}\)

10.3 - H_s – 0.35 = 50 × 0.12 = 6

Concept:

\(1)\;\frac{{N\sqrt Q }}{{{H^{\frac{3}{4}}}}},\;\;2)\frac{{gH}}{{{N^2}{D^2}}},\;\;3)\;\frac{Q}{{N{D^3}}},\;\;4)\;\frac{P}{{\rho {N^3}{D^5}}}\)

The above four dimensions terms are used as similarity parameters for a pump. For a geometrically similar pump, these parameters must be same for model and prototype

Depending on the requirement of question, one or more such relationships may be used.

The given questions demands the use of only (i) i.e.

\(\frac{{{N_1}\sqrt {{Q_1}} }}{{H_1^{0.75}}} = \frac{{{N_2}\sqrt {{Q_2}} }}{{H_2^{0.75}}}\) 

Calculation:

\({N_1} = {N_2},{Q_2} = \frac{{{Q_1}}}{2} = \frac{{20}}{2} = 10\;L/s\)

H₁ = 15 m

H₂ = ?

\(\therefore \frac{{\sqrt {20} }}{{{{15}^{0.75}}}} = \frac{{\sqrt {10} }}{{H_2^{0.75}}}\) 

Explanation:

Given:

P = 3000 kW, n_o = 0.85, H = 300 m, ϕ = u/V = 0.44 C_V = 0.97, N_s = 14

Now, 

\({{N}_{s}}=\frac{N\sqrt{P}}{{{H}^{\frac{5}{4}}}}\) [Note: ‘P’ is in kW]

\(∴ N=\frac{{{N}_{s}}.{{H}^{\frac{5}{4}}}}{\sqrt{P}}=\frac{14\times {{300}^{\frac{5}{4}}}}{\sqrt{3000}}\)

∴ N = 319.3 rpm

Now, 

Velocity of nozzle outlet \(V={{C}_{V}}\sqrt{2gH}\) and bucket speed \(U=\phi \sqrt{2gH}\)

\(∴ U=0.44\times \sqrt{2\times 9.8\times 300}\)

∴ U = 33.74 m/s

Now, 

\(U=\frac{\pi DN}{60}\Rightarrow D=\frac{60\times U}{\pi N}\)

\(\text{D}=\frac{60\times 33.74}{\pi \times 319.3}\)

∴ D = 2.01 m

Note: To calculate the jet diameter, we have the relation

Calculation:

Given:

Shaft power (P) = 25 MW = 25 × 10⁶ W, H = 40 m, speed ratio = 2, Flow ratio = 0.5, D_b = 0.35 D₀,

η₀ = 0.88

Now,

Speed of turbine,

\({u_1} = \frac{{\pi {D_0}N}}{{60}}\; - - - \left( 1 \right)\)

Now,

we have to calculate u₁ & D₀.

Since,

\({\rm{Speed\;ratio}} = 2 = \frac{{{u_1}}}{{\sqrt {2gH} }}\)

\({u_1} = 2 \times \sqrt {2 \times 9.81 \times 40} \)

∴ u₁ = 56.028 m/s

Now,

\({\rm{Flow\;ratio}} = 0.5 = \frac{{{V_{{f_1}}}}}{{\sqrt {2gH} }}\)

\({V_{{f_1}}} = 0.5 \times \sqrt {2 \times 9.81 \times 40} \)

\(\therefore {V_{{f_1}}} = 14.007\;m/s\) 

Since,

\({\rm{overall\;efficiency\;}}\left( {{\eta _0}} \right) = \frac{{S.P}}{{W.P}}\)

\(0.88 = \frac{{25 \times {{10}^6}}}{{\rho QgH}}\)

\(Q = \frac{{25 \times {{10}^6}}}{{0.88 \times {{10}^3} \times 9.81 \times 40}} = 72.398\;{m^3}/s\)

Also,

\(Q = \frac{\pi }{4}\left( {D_0^2 - D_b^2} \right) \times {V_{{f_1}}}\)

\(72.398 = \frac{\pi }{4}\left[ {D_0^2 - {{\left( {0.35{D_0}} \right)}^2}} \right] \times 14.007\)

∴ D₀ = 2.738 m

Now,

From equation (1)

\(56.028 = \frac{{\pi \times 2.738 \times N}}{{60}}\)

∴ N = 390.81 rpm