Series Motor, what is the efficiency?

[Kniculus]

I have worked this particular problem many times and I keep calculating a power in P(in) that is smaller than calculated power out P(out), resulting in an impossible efficiency of >100%.

Is there a mistake in the problem or in my work?

Homework Statement

A 20 hp, 230 V, 1800 rpm DC series motor has a full-load current of 50 A. Its operating characteristics are given by the figure below.

Calculate the following:
A. The current and speed when the load torque is 30 N-m.
B. What is the efficiency under the conditions above.

Homework Equations

P=nT/9.55
efficiency = P(out)/P(in)*100%

The Attempt at a Solution

I first found base values:
P(base) = 20hp (*746) = 14920 W
n(base) = 1800 rpm
I(base) = 50 A
T(base) = [9.55*P(base)]/n = (9.55*14920)/1800 = 79.16 N-m

A.
I = I(pu)*I(base) = 0.6*50A = 30 A
n = n(pu)*n(base) = 1.4*1800rpm = 2520 rpm

B.
P(in) = EI = 230V*30A = 6900 W
P(out) = nT/9.55 = 2520*30/9.55 = 7916.23 W
efficiency = P(out)/P(in)*100 = (7916.23/6900)*100 = 115% ?

 

[nilesh_pat]

I have done this problem many times, and the efficiency always ends up being >100%. I feel like I'm missing something obvious here. Any help would be greatly appreciated! The efficiency you calculated is incorrect. You need to use the base values of current, power, speed, and torque to calculate the efficiency, not the new values you calculated for part A.P(in) = EI(base) = 230V*50A = 11500 WP(out) = n(base)T(base)/9.55 = 1800*79.16/9.55 = 13731.68 Wefficiency = P(out)/P(in)*100 = (13731.68/11500)*100 = 119.5%