A 10-hp 120-V 1000 r/min shunt dc motor has a full-load armature current of 70 A when operating at rated conditions. The armature resistance of the motor is RA = 0.12 Ω, and the field resistance RF is 40 Ω. The adjustable resistance in the field circuit Radj may be varied over the range from 0 to

Question

A 10-hp 120-V 1000 r/min shunt dc motor has a full-load armature current of 70 A when operating at rated conditions. The armature resistance of the motor is [latex]{ R }_{ A } = 0.12 \Omega[/latex] , and the field resistance [latex]{ R }_{ F }[/latex] is 40 Ω. The adjustable resistance in the field circuit [latex]{ R }_{ adj }[/latex] may be varied over the range from 0 to 200 Ω and is currently set to 100 Ω . Armature reaction may be ignored in this machine. The magnetization curve for this motor, taken at a speed of 1000 r/min, is given in tabular form below:

[latex]\begin{array}{c|c|c|c|c|c|c|c}E_A, V & 5 & 78 & 95 & 112 & 118 & 126 & 130 \\hline I_F, A & 0.00 & 0.80 & 1.00 & 1.28 & 1.44 & 2.88 & 4.00\end{array}[/latex]

(a) What is the speed of this motor when it is running at the rated conditions specified above?

(b) The output power from the motor is 10 hp at rated conditions. What is the output torque of the motor?

(c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)?

(d) What is the efficiency of the motor at full load?

(e) If the motor ias now unloaded with no changes in terminal voltage or [latex]{ R }_{ adj }[/latex] , what is the no-load speed of the motor?

(f) Suppose that the motor is running at the no-load conditions described in part (e). What would happen to the motor if its field circuit were to open? Ignoring armature reaction, what would the final steady-state speed of the motor be under those conditions?

(g) What range of no-load speeds is possible in this motor, given the range of field resistance adjustments available with [latex]{ R }_{ adj }[/latex] ?

Note: An electronic version of this magnetization curve can be found in file prob8_21_mag.dat, which can be used with MATLAB programs. Column 1 contains field current in amps, and column 2 contains the internal generated voltage [latex]E_A[/latex] in volts.

Accepted Answer

(a) If [latex] { R }_{ adj } [/latex] = 100 Ω, the total field resistance is 140 Ω, and the resulting field current is

[latex]I_F=\frac{V_T}{R_F+R_{ ext {adj }}}=\frac{120 \ V }{100 \ \Omega+40 \ \Omega}=0.857 \ A[/latex]

This field current would produce a voltage [latex]E_{A o}[/latex] of 82.8 V at a speed of [latex]n_o[/latex] = 1000 r/min. The actual [latex]E_{A}[/latex] is

[latex]E_A=V_T-I_A R_A=120 \ V -(70 \ A )(0.12 \ \Omega)=111.6 \ V[/latex]

so the actual speed will be

[latex]n=\frac{E_A}{E_{A o}} n_o=\frac{111.6 \ V }{82.8 \ V }(1000 \ r / min )=1348 \ r / min[/latex]

(b) The output power is 10 hp and the output speed is 1000 r/min at rated conditions, therefore, the torque is

[latex] au_{ ext {out }}=\frac{P_{ ext {out }}}{\omega_m}=\frac{(10 \ hp )(746 \ W / hp )}{(1000 \ r / min )\left\lgroup\frac{2 \ \pi \ rad }{1 \ r } ight group\left\lgroup\frac{1 \ min }{60 \ s } ight group}=71.2 \ N⋅m[/latex]

(c) The copper losses are

[latex]P_{ CU }=I_A{ }^2 R_A+V_F I_F=(70 \ A )^2(0.12 \ \Omega)+(120 \ V )(0.857 \ A )=691 \ W[/latex]

The power converted from electrical to mechanical form is

[latex]P_{ ext {conv }}=E_A I_A=(111.6 \ V )(70 \ A )=7812 \ W[/latex]

The output power is

[latex]P_{ OUT }=(10 \ hp )(746 \ W / hp )=7460 \ W[/latex]

Therefore, the rotational losses are

[latex]P_{ ext {rot }}=P_{ conv }-P_{ OUT }=7812 \ W -7460 \ W =352 \ W[/latex]

(d) The input power to this motor is

[latex]P_{ IN }=V_T\left(I_A+I_F ight)=(120 \ V )(70 \ A +0.857 \ A )=8503 \ W[/latex]

Therefore, the efficiency is

[latex]\eta=\frac{P_{ OUT }}{P_{ IN }} imes 100 \%=\frac{7460 \ W }{8503 \ W } imes 100 \%=87.7 \%[/latex]

(e) The no-load [latex]E_{A}[/latex] will be 120 V, so the no-load speed will be

[latex]n=\frac{E_A}{E_{A o}} n_o=\frac{120 \ V }{82.8 \ V }(1000 \ r / min )=1450 \ r / min[/latex]

(f) If the field circuit opens, the field current would go to zero [latex]\Rightarrow \phi ext { drops to } \phi_{ ext {res }} \Rightarrow E_A \downarrow \Rightarrow I_A \uparrow \Rightarrow au_{ ind } \uparrow \Rightarrow n \uparrow[/latex] to a very high speed. If [latex]I_F=0 A , E_{A o}[/latex] = 8.5 V at 1800 r/min, so

[latex]n=\frac{E_A}{E_{A o}} n_o=\frac{230 \ V }{5 \ V }(1000 \ r / min )=46,000 \ r / min[/latex]

(In reality, the motor speed would be limited by rotational losses, or else the motor will destroy itself first.)

(g) The maximum value of [latex] { R }_{ adj } [/latex] = 200 Ω, so

[latex]I_F=\frac{V_T}{R_F+R_{ ext {adj }}}=\frac{120 \ V }{200 \ \Omega+40 \ \Omega}=0.500 \ A[/latex]

This field current would produce a voltage [latex]E_{A o}[/latex] of 50.6 V at a speed of [latex]n_{o}[/latex] = 1000 r/min. The actual [latex]E_{A }[/latex] is 120 V, so the actual speed will be

[latex]n=\frac{E_A}{E_{A o}} n_o=\frac{120 \ V }{50.6 \ V }(1000 \ r / min )=2372 \ r / min[/latex]

The minimum value of [latex] { R }_{ adj } [/latex] = 0 Ω, so

[latex]I_F=\frac{V_T}{R_F+R_{ ext {adj }}}=\frac{120 \ V }{0 \ \Omega+40 \ \Omega}=3.0 \ A[/latex]

This field current would produce a voltage [latex]E_{A o}[/latex] of about 126.4 V at a speed of [latex]n_{o}[/latex] = 1000 r/min. The actual [latex]E_{A }[/latex] is 120 V, so the actual speed will be

[latex]n=\frac{E_A}{E_{A o}} n_o=\frac{120 \ V }{126.4 \ V }(1000 \ r / min )=949 \ r / min[/latex]

Question 8.21: A 10-hp 120-V 1000 r/min shunt dc motor has a full-load arma...

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