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Question 7.57: A 10 kW, 250 V, dc shunt motor with an armature resistance o...

A 10 kW, 250 V, dc shunt motor with an armature resistance of 0.8 \Omega and a field resistance of 275 \Omega takes 3.91 A, when running light at rated voltage and rated speed.

(a) What conclusions can you draw from the above data regarding machine losses?

(b) Calculate the machine efficiency as a generator when delivering an output of 10 kW at rated voltage and speed and as a motor drawing an input of 10 kW. What assumption if any do you have to make in this computation?

(c) Determine the maximum efficiencies of the machine when generating and when motoring.

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(a) Shunt field loss,              P_{s h}=\frac{(250)^{2}}{275}=227.3 W

Rotational loss =P_{r o t}=250 \times 3.91-(3.91)^{2} \times 0.8=965 W

(b) Generator

I_{L} =\frac{10 \times 10^{3}}{250}=40 A ; I_{f}=\frac{250}{275}=0.91 A

 

I_{a} =40+0.91=40.91 A

 

P_{L} =965+227.3+(40.91)^{2} \times 0.8=2.53 kW

 

\eta_{G} =1-\frac{2.53}{10+2.53}=79.8 \%

 

Motor

I_{a} =I_{L}-I_{f}=40-0.91=39.1 A

 

P_{L} =965+227.3+(39)^{2} \times 0.8=415 kW

 

\eta_{M} =1-\frac{2.415}{10}=75.85 \%

Assumption: Stray-load loss has been neglected.

(c) The condition for maximum efficiency is

I_{a}^{2} R_{a} =P_{r o t}+P_{s h}

 

0.8 I_{a} =965+227.3=1192.3 W

or        I_{a} =38.6 A

Total loss    P_{L} =2 \times 1192.3=2384.6 W

Generator

I_{L} =I_{a}-I_{f}=38.6-0.91=37.69 A

 

P_{\text {out }} =250 \times 37.69=9422.5 W

 

\eta_{G}(\max ) =I-\frac{2384.6}{9422.5+2384.6}=79.8 \%

Motor

I_{L} =I_{a}+I_{f}=38.6+0.91=39.51 A

 

P_{\text {in }} =250 \times 39.51=9877.5 W

 

\eta_{M}(\max ) =1-\frac{2384.6}{9877.5}=75.85 \%

The MATLAB program for the Example 7.57 is shown below.

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clc

clear

Pop=l0*l000; Vt=250; Ra=0.8; Rf=275; Ia=3.91;

%% part (a)

Psh=Vt^

2/Rf;

Prot=Vt*Ia–Ia^

2*Ra;

%% part (b)

%% generator

I1=Pop /Vt;

If=Vt/Rf;

Ia=I1 + If;

Ploss=Prot+Psh+Ia^

2*Ra;

Eff_gen=(l–Ploss/(Ploss+Pop))*100

%% motor

Ia=I1–If;

Ploss=Prot+Psh+Ia^

2*Ra;

Eff_rnotor=(1–Ploss/(Pop))*100

%% part (c)

Ia=sqrt((Prot+Psh)/Ra);

Ploss_tot=2*(Prot+Psh)

%% generator

I1=Ia–If;

Pout=Vt*I1;

Eff_gen_rnax=(1–Ploss_tot/(Ploss_tot+Pout))*100

% motor

I1=Ia+If;

Pin = Vt*I1;

Eff_motor_max=(1–Ploss_tot /Pin)*100

Answer:

Eff_gen = 79.7996

Eff_ motor = 75.8498

EfC_gen_max = 79.8048

Eff_motor_max = 75.8585

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