DC motor question

  • Thread starter theone
  • Start date
  • #1
81
0

Homework Statement


A machine, with an armature resistance of 0.16 ohms and K##\phi## is to drive a load that requires a torque that is proportional to its speed. One point on the mechanical torque-speed relation is 16 Nm at 400 r/min. If the armature terminal voltage is 50 V, at what steady-state speed will the load be driven?

Homework Equations


[/B]
##T=K\phi i_a##, for a constant machine speed

The torque speed relation is
##w = \frac{V_t}{K\phi} - \frac{R_a}{(K\phi)^2}T##

The Attempt at a Solution


the only thing missing for me to use the torque speed relation is the load torque, but I don't understand what they did here to get it. I don't know what A is.
 

Attachments

Answers and Replies

  • #2
andrevdh
Homework Helper
2,128
116
The problem states that the torque, T, is proportional to the speed, ω.
A is the proportional constant which they calculate as 0.382 Ns.
This is similar to a y=mx relation and they are calculating m from one point on the graph:
m = y/x or in this case A=T
All that needs to be done is to convert the 400 rev/min to rad/s.
 

Related Threads on DC motor question

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
782
  • Last Post
Replies
13
Views
961
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
7K
Top