**Calculation of pressure drop in steam and water lines**

** **The pressure drop in a water & steam lines refers to the decrease in pressure that occurs as water/steam flows through a pipe or conduit due to factors such as friction and flow resistance. Several factors influence the magnitude of pressure drop in a water line:

**Pipe Characteristics:** The diameter, length, and roughness of the pipe impact the resistance to flow and consequently the pressure drop. Smaller diameter pipes and longer pipe lengths tend to result in higher pressure drops. Additionally, rougher pipe surfaces create more friction and increase pressure drop compared to smoother surfaces.

**Flow Rate: **The rate at which water/steam flows through the pipe affects the pressure drop. Higher flow rates generally result in higher pressure drops due to increased frictional resistance.

**Fluid Properties: **The physical properties of the water/steam being transported, such as viscosity and density, can influence the pressure drop. However, for water at typical temperatures and pressures, these effects are usually negligible.

Pipe Fittings and Valves: The presence of fittings, such as elbows, bends, valves, and other obstructions in the water line, can contribute to pressure drop. These components disrupt the flow and introduce additional resistance.

It's important to note that pressure drop calculations for steam lines can be complex and require a comprehensive understanding of steam properties and fluid dynamics.

**Pressure drop in water line:**

Head loss in water line for turbulent flow is given as

Head loss in meter = 4fLV^{2} / (2gD)

Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

**Example:**

**A Boiler feed pump is delivering feed water flow 50 TPH to the boiler at a distance of 70 meter.The steam drum height is 38 meter from pump suction.Calculate the pressure drop in water line, assume pipe line size is 80 NB, water density 980 kg/m3 & neglect the other losses from pipe line fittings.**

Feed water flow in m3/sec = 50 000 kg/hr / 980 kg/m3 = 51.02 m3/hr =0.014 m3/sec

Area in side the pipe line = 3.142 X 0.08^{2}/4 = 0.05 M^{2}

Feed water velocity,V = Flow / Area = 0.014 / 0.005 =2.78 m/sec

Then, head loss, H = 4 X 0.005 X 38 X 2.78^{2} / (2 X 9.81 X 0.08)

Head loss, H = 3.75 meter

Minimum head required to lift the water up to steam drum, considering pressure drop in feed water control valve is 8 kg/cm2

H = 3.75+80+38 =121.75 meter

**Pressure drop in steam line:**

Head loss in meter = 2fLdV^{2} / (500gD)

(Density of water is 500 times more than steam at atmospheric pressure)

Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

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**Example: Turbine inlet steam flow is 100 TPH & the distance between Boiler MSSV & Turbine MSSV is 82 meter.The seam pressure & temperature are 65 kg/cm2 and 490 deg C respectively.Calculate the pressure drop in steam line.**

Density of steam at above parameters = 17 kg/m3

Steam flow in m3/sec = 100 X 1000 kg/hr / 17 kg/m3=5882.35 m3/hr = 1.63 m3/sec

Assume main steam velocity being 45 m/sec

Pipe inside area A = Flow / Velocity = 1.63/45 =0.0362 m2

Now, calculate pipe diameter , A = 3.142 X D^{2}/4

D = SQRT (0.036 X 4/3.142) = 0.214 meter = 214 mm

Now, Pressure drop H =(2 X 0.005 X 82/0.224) X (17/500) X (45^{2}/9.81) =25.7 m = 2.57 kg/cm2

**For read>>>> Powerplant and calculations**

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