I Is the Equation h = u + (0.185)Pv Accurate for Converting Energy Units?

  • I
  • Thread starter Thread starter wrenchtime
  • Start date Start date
AI Thread Summary
The equation h = u + (0.185)Pv is proposed for converting energy units, where h and u are in Btu/lbm, P in psia, and v in ft³/lbm. It represents the relationship between enthalpy and internal energy, with (0.185)Pv signifying the area between their curves on a temperature and specific entropy graph. Testing indicates the equation is effective across various mediums, with a conversion factor of 1 foot pound equating to 0.00129 BTU. The metric equivalent of the equation is h = u + (1000)Pv, using kg/kJ for h and u, MPa for P, and m³/kg for v. The discussion confirms the equation's accuracy for energy unit conversion.
wrenchtime
Messages
9
Reaction score
0
Should h = u + Pv

be h = u + (0.185)Pv (Btu/lbm)

where units for h and u are (Btu/lbm)

P (psia) and v (ft*3/lbm)

On a Temperature and specific entropy graph

h... represents the area under the curve at a constant specific pressure

u... represents the area under the curve at a constant specific volume

(0.185)Pv... represents the area between the h and u curves

I have tested the equation on several medium and it seems to work?
 
Science news on Phys.org
Here’s a hint: 1 foot pound = 0.00129 BTU
 
To convert (psia) x (ft3 / lbm) to Btu / lbm

1 Btu / lbm = (1 lbf/in2) x (1 ft3/lbm) x (1 Btu/778 ft-lbf) x (144 in2/ 1 ft2) = 0.185 Btu/lbm

The equation in metric units is

h = u + (1000) P v

With units
h = kg /kj
u = kg /kj
P = Mpa
v = m3/kg
 
I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top