Formula for temperature change of liquid through a valve 4

[mitchelg]

How do I calculate the temperature rise (or fall) of a liquid as it passes through a valve knowing the media (i.e. water), media density/specific heat, flow rate thru' valve, valve inlet temperature and valve inlet pressure plus pressure drop.

 

[pennpiper]

What makes you think that a fluid (water) changes temperature just by going thru a valve?

If this were true and each valves causes a one degree temp change then maybe if I string 50 valves in sequence along a line I would raise the water temp by fifty degrees?

Wow! a whole new energy saving concept.

 

[mitchelg]

For clarification, the pressure drop across the valve can be between 150 to 500 bar at flow rates between 150-1500 m3/hr.
We have observed/measured a temperature rise across the valve before but ask if there is a calculation available to enable us to predict the rise.

 

[hydtools]

Pressure drop is wasted energy, no work done, and creates heat. Calculate lost work, pressure x flow rate. Then some thermodynamic calculations to calculate heat input to fluid and surroundings to get to temperature rise. Been too long since I had to do thermo calcs, sorry.

Ted

 

[hacksaw]

high pressure valves often rely tortuous path designs to control cavitation and to bring the pressure down in stages.

are you using one of those valves?

 

[katmar]

The power absorbed in a pump is given by Pressure change x volumetric flowrate, or
Power = (Volume/time) x Pressure change

In the same way, when the liquid flows through a valve and drops in pressure this energy can only come out as heat. If the valve is well insulated then the heat will be in the liquid.

The energy to heat a body of liquid is Mass x Specific Heat x Temperature Change
If we divide this through by unit time we get
Power = (Energy/time) = (Mass/time) x Specific Heat x Temperature Change
or, if we want to work in volumetric units
Power = (Volume/time) x density x Specific Heat x Temperature Change

Temperature change is what we want so we have
Temperature Change = Power / {(Volume/time) x density x Specific Heat }

From the pump equation we can substitute the Power with (Volume/time) x Pressure change

now, Temperature Change =
{(Volume/time) x Pressure change} / {(Volume/time) x density x Specific Heat }

There is a (Volume/time) in the top and bottom, so we cancel and get
Temperature Change = Pressure change / (density x Specific Heat )

It's late on a Friday here, so I have no appetite for gc, BTUs and other arcane units - so using proper units we get
DegC = kPa / (kg/m³ x kJ/kg.DegC)

For water with a density of 1000, a specific heat of 4.18 and a pressure drop of 150 bar (15000 kPa) we get
Temperature rise = 15000 / (1000 x 4.18) = 3.6 Deg C

In reality there will be some losses to the atmosphere, but I expect this is the type of temperature rise you are seeing.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

You can also see it this way. A valve throttling is an isenthalpic process (well not properly isenthalpic as the trasformation is not reversible, but starting and final enthalpy can be assumed to be equal).

h1 = c*T1 + p1*v
h2 = c*T2 + p2*v

being:

h1 = enthalpy of state 1
h2 = enthalpy of state 2
c = specific heat
T1 = temperature of state 1
p1 = pressure of state 1
v = specific volume
T2 = temperature of state 2
p2 = pressure of state 2


h1 = h2 implies

c*T1 + p1*v = c*T2 + p2*v

rearranging

T2-T1 = (p2-p1)*v/c