Question 16.10: A 250 V shunt DC machine has the armature winding resistance......
A 250 V shunt DC machine has the armature winding resistance of 1 Ω and the field winding resistance of 100 Ω. If the machine is supplied by a 250 V power source and draws the line current of 25 A, calculate the ratio of speed of the machine in generating mode to the motoring one.
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Calculation amount ○ Small ○ Normal ● Large
1) 1.12
2) 1.02
3) 1.32
4) 0.97
Based on the information given in the problem, we have:
V_{t,g}=V_{t,m}=250\ V (1)
R_a=1\ \Omega (2)
R_f=100\ \Omega (3)
I_{t,m}=I_{t,g}=25\ A (4)
Figure 16.10.a and Fig. 16.10.b show the electrical circuit of shunt DC machine in generating and motoring operations, respectively.
Applying Ohm’s law for the branch of field winding in Fig. 16.10.a:
I_{f,g}=I_{f,m}=\frac{V_{t,g}}{R_{f,c} + R_{fw}}=\frac{V_{t,g}}{R_f}=\frac{250}{100}=2.5\ A (5)
Applying KCL at the top node of the circuit of Fig. 16.10.a:
I_{a,g}=I_{t,g}+I_{f,g}=25+2.5=27.2 (6)
Applying KVL in the right-hand side mesh of the circuit of Fig. 16.10.a:
E_{a,g}=V_{t,g}+R_{a}I_{a,g}=250+0.1\times 27.5=252.75\ V (7)
Applying KCL at the top node of the circuit of Fig. 16.10.b:
I_{a,m}=I_{t,m}-I_{f,m}=25-2.5=22.5 (8)
Applying KVL in the right-hand side mesh of the circuit of Fig. 16.10.b:
E_{a,m}=V_{t,m}-R_{a}I_{a,m}=250-0.1 \times 22.5=247.75\ V (9)
As we know, the induced armature voltage of an electric machine can be determined by using E_a = k_aφω. Therefore, we can write:
\frac{E_{a,g}}{E_{a,m}}=\frac{φ_{g}ω_{g}}{φ_{m}ω_{m}}=\frac{φ_{g}n_{g}}{φ_{m}n_{m}} (10)
Since I_{f,m}=I_{f,g}, we have:
φ_m=φ_g (11)
Solving (10) and (11):
\Rightarrow \frac{252.75}{247.75}=\frac{n_g}{n_m}\Rightarrow \pmb{\frac{n_g}{n_m}=1.02}
Choice (2) is the answer.
