Solved Problems on Flow Processes

SOLVED PROBLEMS ON FLOW PROCESSES

1. Solved Problem on Boiler

Problem 4.24

A boiler generates steam at 3 bar and 0.85 dry from water at 45°C; 540 kJ/s heat is added during the evaporation. Calculate the amount of steam generated per hour.

Given data:

p = 3 bar = 300 kPa

x = 0.85

Tw = 45°C

Heat added = 540 kJ/s

To find:

Steam generated per hour

Solution:

From saturated water table of pressure scale, corresponding to 3 bar,

2. Solved Problems on Turbine

Problem 4.25

Steam enters an adiabatic turbine at 10 MPa and 500°C at the rate of 3 kg/s and leaves at 50 kPa. If the power output of turbine is 2 MW, determine the velocity of steam at exit of turbine.

Given data:

p1 = 10 MPa = 100 bar

T1 = 500°C

m = 3 kg/s

p2 = 50 kPa = 0.5 bar

W = 2 MW = 2000 kW

To find:

C2

Solution:

From superheated enthalpy and superheated entropy tables, corresponding to 100 bar and 500°C,

h1 = 3374.6 kJ/kg

s1 = 6.599 kJ/kg K

In adiabatic process, entropy remains constant,

s1 = s2

⸫ s2 = 6.599 kJ/kg K

From saturated water table of pressure scale, corresponding to 0.5 bar,

Problem 4.26

150 kg/s of steam at 25 bar and 300°C expands isentropically in a steam turbine to 0.33 bar. Determine the power output of the turbine.

Given data:

m = 150 kg/s

p1 = 25 bar = 2500 kPa

T1 = 300°C

p2 = 0.33 bar = 33 kPa

To find:

W

Solution:

From superheated enthalpy and superheated entropy tables, corresponding to 25 bar and 300°C,

h1 = 3010.4 kJ/kg 

s1 = 6.648 kJ/kgK

For isentropic process,

s1 = s2

⸫ s2 = 6.648 kJ/kg K

From saturated water table of pressure scale, corresponding to 0.33 bar,

hf2 = 298.6 kJ/kg,

sf2 = 0.971 kJ/kg K,

hfg2 = 2330.6 kJ/kg,

s fg2 = 6.766 kJ/kg K

s2 = sf2 +x2 sfg2

From saturated water table of pressure scale, corresponding to 0.33 bar,

6.648 = 0.971 + x2 × 6.766

x2 = 0.839

h2 = hf2 + x2 hfg2

= 298.6 + 0.839 × 2330.6 = 2253.97 kJ/kg

Turbine work, W = m (h1 - h2)

= 150 (3010.4 - 2253.97)

= 113464.5 kW = 113.47 MW  Ans.

3. Solved Problems on Nozzle

Problem 4.27

A nozzle is supplied with steam of 1 MPa at 200°C with a velocity of 100 m/s. The expansion takes place to a pressure of 300 kPa. Assuming isentropic efficiency of nozzle to be 90%, find the final velocity.

Given data:

p1 = 1 MPa = 10 bar = 1000 kPa

T1 = 200°C

C1 = 100 m/s

p2 = 300 kPa = 3 bar

ηisentropic = 90%

To find:

C2

Solution:

From saturated water table of pressure scale, corresponding to 10 bar, Tsat = 179.9°C 

Since T1 > Tsat, the state would be in the superheated region.

From superheated enthalpy and superheated entropy tables, at 10 bar and 200°C,

h1 = 2826.8 kJ/kg 

s2 = 6.692 kJ/kgK,

From saturated water table of pressure scale, corresponding to 3 bar,

hf2 = 561.5 kJ/kg,

hfg2 = 2163.2 kJ/kg

sf2 = 1.672 kJ/kgK,

sfg2 = 5.319 kJ/kgK

s2s = sf2 + x2 sfg2

s1= s2s = 6.692 kJ/kgK

⸫ 6.692 = 1.672 + x2s × 5.319

x2s = 0.944

⸫ h2s = hf2 + x2s hfg2 = 561.5 + 0.944 × 2163.2 = 2603.56 kJ/kg

Problem 4.28

In a steam generator, the steam generating tubes receive heat from hot gases passing over the oxide surface evaporating water inside the tubes. Flue gas flow rate is 20 kg/s with an average specific heat of 1.04 kJ/kgK. The gas temperature decreases from 650°C to 400°C while generating steam at 300°C water enters the tube as a saturated liquid and leaves with a quality of 90%. Assume environment temperature as To = 27°C. Determine the water flow rate, availability of hot fluid and cold fluid, irreversibility and second law efficiency.

Given data:

m1 = 20 kg/s

C1 = 1.04 kJ/kg K

T1 = 650°C = 650 + 273 = 923 K

T2 = 400°C

400 + 273 = 673 K

T3 = 300°C (Saturated liquid)

x4 = 0.9

To = 27°C = 27 + 273 = 300 K

To find:

m2, ψ1, ψ2, I and ηII

Solution:

From saturated steam table, corresponding to 300°C,

Problem 4.29

Steam flows in a pipeline at 1.5 MPa. After expanding to 0.1 MPa in a throttling calorimeter, the temperature is found to be 120°C. Find the quality of steam in the pipeline and also calculate availabilities, irreversibility and second law efficiency. Assume To = 25°C.

Given data:

p1 =1.5 MPa = 15 bar

p2 = 0.1 MPa = 1 bar

T2 = 120°C = 120 + 273 = 393 K

To = 25°C = 25 + 273 = 298 K

To find:

x1, ψ1, ψ2, I and ηII

Solution:

From the superheated steam table, corresponding to p2 = 1 bar and T2 = 120°C,

h2 = 2716.2 kJ/kg 

s2 = 7.4606 kJ/kgK

For throttling process,

h1 = h2 = 2716.2 kJ/kg

But at p1 = 15 bar,

h1 = 844.89 kJ/kg; 

sf = 2.3315 kJ/kgK;

hfg = 1947.3 kJ/kg

sfg = 6.4448 kJ/kgK

h1 = hf + x1 hfg

2716.2 = 844.89 + x1 × 1947.3