Nusselt Number dimension question

Click For Summary
SUMMARY

The Nusselt Number (Nu) is defined as Nu = hD/k, where h is the heat transfer coefficient and D is the diameter of the pipe. The discussion focuses on determining the dimensions of the thermal conductivity (k) in terms of mass (M), length (L), time (T), and temperature (θ). The heat transfer coefficient (h) is expressed as J/(m² hr °C), which can equivalently be represented as W/(m² K). The correct dimensional analysis reveals that k has units of J/(m hr °C), confirming the relationship between these variables in heat transfer calculations.

PREREQUISITES
  • Understanding of heat transfer principles
  • Familiarity with dimensional analysis
  • Knowledge of thermal conductivity and its units
  • Basic grasp of the Nusselt Number and its significance
NEXT STEPS
  • Study the derivation of the Nusselt Number in heat transfer applications
  • Learn about dimensional analysis techniques in fluid mechanics
  • Explore the relationship between heat transfer coefficients and thermal conductivity
  • Investigate the application of the Pi theorem in engineering problems
USEFUL FOR

Students and professionals in mechanical engineering, thermal engineering, and anyone involved in heat transfer analysis and fluid dynamics.

Jasones
Messages
2
Reaction score
0
So, in

Nu=\frac{hD}{k}

h is the heat transfer coefficient, and D is the diameter of the pipe in which heat transfer takes place...

..but what are the dimensions of k in terms of mass (M), length (L), time (T) and temperature (\theta)?

So far, I've worked out that the units for h are J/(m2 hr C), but I'm having difficulty expressing k in terms of what is given.

Am I wrong is thinking that J/(m2 hr C) can be expressed as W/(m2 K)? I'm thinking that this is relevant to the end units of k, but I am unsure if the dimensions are correct.
 
Last edited:
Engineering news on Phys.org
The Nussult number is dimensionless, so if you already know the units of h and D, then the units of k should be obvious: J/(m hr C)
 
We could do some major dimensional analysis with this to prove it, but I really don't feel like it. I'm guessing you've done the old pi theorem already though?
 
I have encountered a vertically oriented hydraulic cylinder that is designed to actuate and slice heavy cabling into sections with a blade. The cylinder is quite small (around 1.5 inches in diameter) and has an equally small stroke. The cylinder is single acting (i.e. it is pressurized from the bottom, and vented to atmosphere with a spring return, roughly 200lbs of force on the spring). The system operates at roughly 2500 psi. Interestingly, the cylinder has a pin that passes through its...

Similar threads

Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
36K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
Replies
2
Views
3K
Replies
2
Views
2K