**Swinburne Test of DC Machine**

#### Swinburne’s test is an indirect method of testing DC machines (either motor or generator), especially shunt and compound machines where the value of flux almost remains constant.

Figure shown below shows the arrangement for Swinburne’s Test on DC Machine.

SWINBURNE TEST OF DC MACHINE |

#### First of all ,machine whose efficiency is to be determined is operated at no load and Constant Loss of that machine is determined.

####

Let, V = Supply Voltage

Iₒ = No-load Current

Iₛₕ = Shunt field Current

Iₐₒ = Iₒ – Iₛₕ = No-load armature Current

Rₐ = Armature Resistance

*Note that the Shunt Field Current remains same irrespective of the load.*

**No-load Input Power = V×Iₒ ……….eq 1**

We know that when DC Machine is at no-load, then its input power is used only to supply losses and losses at no-load are:

**So, No-load Armature Copper losses = Iₐₒ²× Rₐ = (Iₒ – Iₛₕ)² × Rₐ ………eq 2**

Since, **No-Load Input Power = No-load Armature Copper losses + Constant Losses;**

**Constant Losses = No-Load Input Power – No-load Armature Copper losses**

So putting the values from eq1 & eq2 , we get

**Constant Loss, **

** P**c** = (V×Iₒ) – (Iₒ – Iₛₕ)² × Rₐ**

Since we have determined the Constant loss of the machine, now efficiency of the machine can be found at any other load.

**EFFICIENCY WHEN RUNNING AS MOTOR**

#### Let I = Load Current at which efficiency is to be found.

Input Power to Motor = V×I

Armature Current in case of motoring operation = Iₐ = Iₒ – Iₛₕ

Armature Copper loss = Iₐ²×Rₐ = (I – Iₛₕ)² × Rₐ

Total Loss in the Motor = Armature Copper loss + Constant Loss

Total Loss in the Motor = {(I – Iₛₕ)² × Rₐ} + Pc

**Motor Efficiency = {Input – loss}/Input**

**Motor Efficiency,**

** **

** {(V×I) – [(I – Iₛₕ)² × Rₐ] + P**c**} / V×I**

**EFFICIENCY WHEN RUNNING AS GENERATOR**

#### Let I = Load Current at which efficiency is to be found.

Input Power to Generator = V×I

Armature Current in case of generating operation = Iₐ = I + Iₛₕ

Armature Copper loss = Iₐ²×Rₐ = (I + Iₛₕ)² × Rₐ

Total Loss in the Generator = Armature Copper loss + Constant Loss

Total Loss in the Generator = {(I + Iₛₕ)² × Rₐ} + P_{c}

**Generator Efficiency = Output/{Output + loss}**

**Generator Efficiency, **

** (V×I )/{(V×I) + [(I + Iₛₕ)² × Rₐ] + Pc} **

**ADVANTAGES OF SWINBURNE’S TEST**

#### 1. Power Required for carrying out this test is small, therefore this method is an economical and convenient method of testing DC Machines.

2. Since the constant loss is already determined, so efficiency at any load can be found.