Power Engineering and Turbomacinery Test 1

Explanation:

• High head turbine: In this type of turbines, the net head varies from 150 m to 2000 m or even more, and these turbines require a small quantity of water. Example: Pelton wheel turbine.
• Medium head turbine: The net head varies from 30 m to 150 m, and also these turbines require a moderate quantity of water. Example: Francis turbine.
• Low head turbine: The net head is less than 30 m and also these turbines require a large quantity of water. Example: Kaplan turbine.

Important Point:

• Pelton turbine – Low discharge and high head
• Francis turbine – Medium discharge and medium head
• Kaplan Turbine – High discharge and low head

Concept:

In a turbine, the thermal efficiency is given by 

\(\eta=\frac{W_{net}}{Q_{supplied}}\)

The net work in a gas turbine is the difference of work done by the turbine and work consumed by the compressor 

W_net = W_T - W_C

Efficiency is 20%

∴ \(\eta=\frac{W_{net}}{Q_{in}}=\frac{W_{T}-W_C}{Q_{supplied}}\)

Calculation:

Given, WT = 600 kJ/kg, WC = 400 kJ/kg Q_supplied = 1000 kJ/kg

\(\eta=\frac{600-400}{1000}=0.2\)

Energy Released by the coal = 900 × 10⁷ kJ/hr = 2500 MW

Heat rejected from the power plant 2500-1000 = 1500 MW

Concept:

Work ratio is defined as net work output and work done by turbine

Work ratio = \(\frac{{Net\;work\;output}}{{positive\;workdone\;\left( {Turbine\;work} \right)}}\)

\(WR = \frac{{{C_p}\left[ {{T_3} - {T_4}} \right] - {C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}\)

\(WR = 1-\frac{{{}{{} {}} {}\left[ {{T_2} - {T_1}} \right]}}{{{}\left[ {{T_3} - {T_4}} \right]}}\)

Efficiency is defined as the ratio of net work output to the energy supplied to the cycle.

Heat added in system \( = {C_p}\left[ {{T_3} - {T_2}} \right]\)

Head rejected = C_p [T₄ – T₁]

For ideal Regenerator

T₂ = T₄ & T₂ = T₄

\(\eta = 1 - \frac{{{Q_{Rejected}}}}{{{Q_{added}}}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_3} - {T_2}} \right]}}{{{C_p}\left[ {{T_4} - {T_1}} \right]}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_3} - {T_4}} \right]}}{{{C_p}\left[ {{T_2} - {T_1}} \right]}}\)

Concept:

Specific speed of the turbine (NS)

NS = \(\frac{{{\rm{N}}\sqrt {\rm{P}} }}{{{{\rm{H}}^{5/4}}}}\)

P → shaft power

H → Head

N → Speed

Calculation:

NS = \(\frac{{{\rm{N}}\sqrt {\rm{P}} }}{{{{\rm{H}}^{5/4}}}}\)

NS = \(\frac{{135{\rm{\;}} \times {\rm{\;}}\sqrt {7223} }}{{{{\left( {25} \right)}^{5/4}}}}\)

NS = 205.24

Rankine cycle with superheating is shown in the figure.

• Increased heat addition and rejection
• Due to increase heat addition, mean temperature of heat addition increases
• The increase in mean temperature of heat addition increases the efficiency

​∴ In the superheat Rankine cycle the mean temperature of heat addition increases and hence the efficiency increases.

Concept:

For Brayton cycle

\({W_{max}} = {C_p}{\left[ {\sqrt {{T_{max}}} - \sqrt {{T_{min}}} } \right]^2}\)

Calculation:

\({W_{max}} = 1.005\;{\left[ {\sqrt {1600} - \sqrt {900} } \right]^2}\)

∴ W_max = 100.5 kJ/kg.K

Concept:

Force exerted = ρQV_r

V_r = relative velocity

V_r = ρQ (V + u)

Calculation:

F = ρQ(V + u)

Q = A (A + u)

∴ F = ρA(V + u)²

∴ F = 1000 × 2 × 10^-4 × (13)²

∴ F = 33.8 N

Concept:

In a two-stage compressor there is intercooling between the suction and delivery pressure at constant pressure by this intercooling some reduction in work supplied to the compressor another advantage of using multi-stage compression is that greater compression can be achieved compared to the single-stage compression.

For intercooling between the stages using one intercooler the perfect (ideal intercooler) conditions are

\({P_i} = \sqrt {{P_H}{P_L}} \)

Where PL and PH are the suction and delivery pressures respectively.

Calculation:

Given, PL = 2 bar, P_H = 8 bar

\({P_i} = \sqrt {{P_H}{P_L}} = \sqrt {{8}\times{2}}=4\; bar \)

Explanation:

Given:

H_p = 30 m, H_m = 10 m, N_m = 600 rpm

\(\frac{{{L_m}}}{{{L_p}}} = \frac{1}{3} = \frac{{{D_m}}}{{{D_p}}}\)

Since,

\({\left( {\frac{{\sqrt H }}{{DN}}} \right)_m} = {\left( {\frac{{\sqrt H }}{{DN}}} \right)_p}\)

\({N_p} = \sqrt {\frac{{{H_p}}}{{{H_m}}}} \times \frac{{{D_m}}}{{{D_p}}} \times {N_m}\)

\({N_p} = \sqrt {\frac{{30}}{{10}}} \times \frac{1}{3} \times 600\)

∴ N_p = 346.4 rpm

Concept:

Given:

(P_atm)_abs = 100 kPa, (P_Vap)_abs = 25 kPa, Net effective head = 10 m

σ­_Thoma = 0.95, γ_water = 10 kN/m³

Now,

From definition of σ_Thoma,

\({\sigma _{Thoma}} = \frac{{{H_a} - {H_{vp}} - z}}{H}\)

Where, z is height of runner outlet above tail race.

Now,

Z = H_atm – H_vap – H σ_Thoma

\(Z = \frac{{100\; \times \;{{10}^3}}}{{10\; \times \;{{10}^3}}} - \frac{{25\; \times \;{{10}^3}}}{{10\; \times \;{{10}^3}}} - 10 \times 0.95\)

Z = 10 – 2.5 – 9.5

∴ Z = -2 m

∴ Z = 2 m below tail water level

Concept:

\(\frac{{{P_m}}}{{D_m^2H_m^{\frac{3}{2}}}} = \frac{{{P_P}}}{{D_P^2H_P^{\frac{3}{2}}}}\)

\(\therefore \frac{{{D_p}}}{{{D_m}}} = \sqrt {\frac{{{P_p}}}{{{P_m}}}{{\left( {\frac{{{H_m}}}{{{H_p}}}} \right)}^{\frac{3}{2}}}} \)

\(\frac{{{D_p}}}{{{D_m}}} = \sqrt {\frac{{1.4\; \times\; {{10}^6}}}{{0.5\; \times \;{{10}^3}}}{{\left( {\frac{4}{{49}}} \right)}^{\frac{3}{2}}}} \)

\(\frac{{{D_p}}}{{{D_m}}} = 8.081\) 

∴ D_p : D_m ≈ 8 : 1

Efficiency of ideal regenerative cycle is exactly equal to that of the corresponding Carnot cycle. Hence it is maximum.

Explanation:

Power of a Pelton wheel is given by:

\(P=\dot{m}U\left( V-u \right)\left( 1+k\cos \beta \right)\)

where, U = bucket speed, V = jet speed, ṁ = mass flow rate

β = vane angle = 180 – 165° = 15°

Input power to the turbine \(=\frac{1}{2}\dot{m}{{V}^{2}}\)

∴ By definition of efficiency we have

\(\eta =\frac{\dot{m}U\left( V-u \right)\left( 1+k\cos \beta \right)}{\frac{1}{2}\dot{m}{{V}^{2}}}\)

On rearranging:

\(\eta =2\frac{u}{V}\left( 1-\frac{u}{V} \right)\left( 1+k\cos \beta \right)\)

Calculation

For max efficiency: \({{\left( \frac{u}{V} \right)}_{max}}=\frac{1}{2}\)

The ratio of the mean bucket speed to jet velocity is 60% that for the maximum efficiency i.e.

\(\frac{u}{V}=0.6\times \frac{1}{2}=0.3\)

k = 1 (∵ no losses) β = 15°

(^1 -h) = (^1 -h )(^1 -h )(^1 -h ).................................................... ........... i^{1 -}hm ) or n = n₁ + n₂ - n₁n₂ =(0.30 + 0.2 - 0.3 x 0.2)x 100% = 44%

with increase in pressure, state of steam shifts towards left and thus on expansion, quality of steam will decrease

Concept:

\({T_2} = {T_1}{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

Calculation:

Given:

Pressure ratio (r_p) = 5

Mass flow rate through Compressor

Inlet Temperature (T₁) = 15° = 288K

Inlet Pressure (P₁ ) = 1 bar

Isentropic Efficiency of Compressor η_c = 0.85

Assuming Compressor to be Isentropic we have

\({T_2} = {T_1}{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\({T_2} = 288{\left( 5 \right)^{\frac{{0.4}}{{1.4}}}}\)

∴ T₂ = 456.14 K

\(\therefore Isentropic\;Work = \dot m{C_p}\left( {{T_2} - {T_1}} \right)\)

Isentropic Work = 30 × 1.005 (456.14 - 288)

∴ Isentropic Work = 5069.422 kW

Also,

\({\eta _c} = \frac{{Isentropic\;Work}}{{Actual\;Work}}\)

\(\therefore Actual\;Work = \frac{{5069.422}}{{0.85}}\)

∴ Actual work ≈ 5964 kW

So, Option (1) is correct

Concept:

Second law efficiency,

\({\eta _\parallel } = \frac{{{\eta _{Rankine}}}}{{{\eta _{carnot}}}}\)

Calculation:

T₁ = 198.27°C = 471.27 K

T₂ = 65.9°C = 338.9 K

\({\eta _{carnot}} = 1 - \frac{{{T_2}}}{{{T_1}}} = 1 - \frac{{338.9}}{{471.27}} = 0.281\)

Isentropic expansion in turbine

s₁ = s₂ = s_f + xs_fg

\(x = \frac{{{s_1} - {s_f}}}{{{s_{fg}}}} = \frac{{6.5302 - 1.026}}{{6.6438}} = 0.83\)

h₂ = h_f + xh_fg

= 218.2 + 0.83 × 2300 = 2127.2

\({\eta _{Rankine}} = \frac{{{h_1} - {h_2}}}{{{h_1} - {h_4}}}\)  [h₄ = h₃, neglecting pump work]

\({\eta _{Rankine}} = \frac{{2800 - 2127.2}}{{2800 - 218.2}} = 0.260\)

Concept:

a) Volumetric efficiency of compressor,

\({\eta _v} = 1 + c - c{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{n}}}\)

b) For maximum pressure ratio, η_v = 0

Calculation:

Given:

V_C = 0.07 V_total, V_S = 0.93 V_total, n_e = 1.36, n_c = 1.30

\(C = \frac{{{V_C}}}{{{V_S}}} = \frac{{0.07\;{V_{total}}}}{{0.93\;{V_{total}}}}\)

∴ C = 0.0752

Now,

\({\eta _v} = 1 + C - C{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{{{n_e}}}}}\)

For maximum pressure ratio,

\(0 = 1 + \left( {0.0752} \right) - \left( {0.0752} \right){\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{{1.36}}}}\)

\({\left( {\frac{{{P_2}}}{{{P_1}}}} \right)_{max}} = 37.25\)

Mistake point:

a) V_c = 7% of total volume and not of swept volume

b) In volumetric efficiency formula, expansion process index is to be taken, and not of compression process index.