Pipe to kW

Pipe to kW

How to calculate how many kW can fit through a certain size pipe

Application: If you are undertaking a retrofit project and need to calculate if the existing pipes are suitable or not.

Note this method calculates the theoretical maximum amount of kW and does not account for internal friction losses or the length of pipe.

Pipe to kW

Internal cross-sectional area

1)

m3/s

m2

2)

m/s

Calculating how many kW can be delivered by a certain size pipe can be broken down into 4 stages:

1) mm to m2 (known pipe size to internal C.S.A) 2) m2 to m3/s 3) m3/s to l/s 4) l/s to kW

3)

Units Conversion

kW

l/s

(DTxSHC)

4)

Pipe to kW

Worked example: How many kW can be delivered through a 22mm pipe with a wall thickness of 1mm, 5K DT, 4.2 Specific Heat Capacity and a target velocity of 0.9m/s?

The first stage requires calculating internal cross-sectional area for our 22mm copper pipe:

Stage 1) mm to m2

Internal cross-sectional area

1) Calculate internal radius 2) Convert from mm to m 3) Square the radius (r2) 4) Multiply r2 by 3.14(Pi)

Here's what it looks like with the numbers kept to a maximum of 6 decimal places:

1) 22 - 2 = 20. 20 / 2 = 10 mm 2) 10 / 1000 = 0.01 m 3) 0.01 x 0.01 = 0.0001 m2 4) 0.0001 x 3.14 = 0.000314 m2

Pipe to kW

Worked example: How many kW can be delivered through a 22mm pipe with a wall thickness of 1mm, 5K DT, 4.2 Specific Heat Capacity and a target velocity of 0.9m/s?

Stage 2) m2 to m3/s

Using the Mass Flow Rate Triangle we need to do the following equation: m3/s = m2 x m/s

m3/s

m2

m/s

0.000314 x 0.9 = 0.0002826 m3/s

In the interest of keeping numbers manageable, we will round the answer the 6 decimal places: 0.0002826 becomes 0.000283 m3/s

Pipe to kW

Worked example: How many kW can be delivered through a 22mm pipe with a wall thickness of 1mm, 5K DT, 4.2 Specific Heat Capacity and a target velocity of 0.9m/s?

Stage 3) m3/s to l/s

There are 1000 litres in 1 cubic metre

To convert from m3/s to l/s we need to multiply by 1000

Units Conversion

0.000283 x 1000 = 0.283 l/s

Pipe to kW

Worked example: How many kW can be delivered through a 22mm pipe with a wall thickness of 1mm, 5K DT, 4.2 Specific Heat Capacity and a target velocity of 0.9m/s?

Stage 4) l/s to kW

Using the Heat Transfer Triangle we need to do the following equation: 0.283 x (5 x 4.2)

kW

l/s

(DTxSHC)

Always start with the part of the equation that is in brackets: 5 x 4.2 = 21 J/kg

Now we can abbreviate the equation to: 0.283 x 21 = 5.943 kW

kW are typically expressed to a maximum of 2 decimal places: 5.943 becomes 5.94 kW

Pipe to kW

Note this method calculates the theoretical maximum amount of kW and does not account for internal friction losses or the length of pipe.

There are numerous software packages available for calculating pipe sizes that can calculate these factors as well, plus other aspects of system design.

Mitsubishi Electric can provide indemnified system designs, this MUST be requested via your account manager.