Thermal Relief Valve Sizing for Electric Heater

[leedan]

Hi all. I have frequented this site for tips and suggestions for questions I have and it has been very helpful. This is my first time posting a question. I have searched the forums to see if there is anything related to this, but I haven't been able to find something that would guide me in the right direction.

I am trying to verify that the relief valve that the vendor supplied will be sufficient as a thermal relief valve. The current pipe configuration is centered around an electric circulation heater with an upstream and downstream isolation/bypass valves. The vendor has supplied a relief valve for the heater. This is for a natural gas operation.

A coworker has specified there are two cases we need to determine if the relief valve that was supplied is sufficiently sized – 1) Fire Case, 2) If the upstream and downstream valve was closed and the heater was left on and safeguards to stop the heater from overheating failed and heater is constantly supplying heat.

Case 1) was easily evaluated by following API 521 fire case, but case 2) I am having some troubles. Specifically finding the mass flow rate.

I thought I would use the following equations:
1) A = m / (C*Kd*P1*Kb*Kc) * (T*Z/M)^(0.5) from API 520
a. In this case m = mass flow rate
b. I can determine the required effective discharge area and compare it to the manufacturer’s discharge area to confirm if it is sized properly.

2) Q = m * Cp * (T2 – T1) where
a. Q = Heat Duty of the heater [kW] (Which I have from the manufacturer’s data sheet.)
b. m = mass flow rate (I would be solving for this mass flow rate to input into equation 1.)
c. Cp = specific heat of the gas
d. T2 = relieving temperature (in this case the temperature where pressure of gas reaches Set Pressure + Overpressure)
e. T1 = 60 Degree F (I’m assuming this should be atmospheric temperature)

I have a gut feeling that I’m not approaching this with the right equation or the right assumptions to determine the flow rate, but I’m not sure how else I should proceed. It seems like I should be looking at the gas expansion rate based due to heat input, but I’m not sure what equations to use.

Any suggestions on how to approach sizing this thermal relief valve would be greatly appreciated, thanks!

 

[Latexman]

1) You should be able to estimate an external area that is wetted on the inside by the NG. See API 521. Use that to calculate a heat input. The heat input will increase the temperature of the NG, i.e. dT/dt. Then, you need an equation of state (ideal gas law?) and the mass of NG in the "system" (in the heater or trapped between the inlet and outlet valves) to calculate a flow rate, i.e. dV/dt.

2) You know the heat duty [kW]. Use that and do the same thing as number 1.

Better yet, whichever has the highest heat input, number 1 or 2, use that to size the relief. No need to do the lower heat duty.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[georgeverghese]

The sledge hammer approach suggested by @Latexman would give you a quick selection (is Qheater>Qfire) on which is the controlling case. If you want to run a calc on the required mass rate through the RV at the time of relief for heater blocked in internal heating, see if this helps ( there is no API 521 expression for this):

a)First find the conditions existing at the time of relief.
Obtain the highest gas density (kgmole/m3)at the time of the block in - say this is rho_o.

No = kgmoles of gas trapped initially = Vh.rho_o, where Vh= volume of blocked in section, m3.

Just before relief, since No is fixed, we will have Trelief = Prelief/(rho_o.R), assuming no change in gas compressibility. Note R is 8.314 kJ/kgmole/degK, P=kPa, T=degK in SI metric units.

b) Relief rate will be highest at the inception of relief, at which time Vrelief is known to be 1/rho_o.

Writing the familiar PV=RT equation as a differential,
Prelief.d(Vrelief)/dt = R.dT/dt. Since this is a constant pressure process, and we know dH/dt=CpdT/dt=Qheater

d(Vrelief)/dt = (R/P).(Qheater/Cp), where Qheater is in kW, Cp=kJ/kgmole.
In general, V=(Vh/N), where Vh is a constant. Substituting this into above eqn, we get Vh.(d/dt)(1/N) = R.Qh/(P.Cp)
or d/dt(1/N) = R.Qh/(P.Vh.Cp)
The LHS of this expression is equivalent to d/dx(1/y) where y=f(x). If we differentiate this, we get d/dx(1/y)=(-1/y^2).(dy/dx).
Similarly, d/dt(1/N) = (-1/N^2).dN/dt.
So, substituting and re arrangement gives dN/dt= -(N^2).R.Qh/(P.Vh.Cp) in kgmoles/sec, and dm/dt = dN/dt . Gas MW; in kg/sec

You can plug this initial mass rate and Trelief into this equation (1), given that N=No at inception of relief in your post to get the required RV orifice size. The negative sign for dN/dt indicates N decreases with time, which is to be expected.

Caution : Since this is a non fire case relief scenario, Trelief must be found to be less than the upper design temperature of the heater body, else you will have to raise the mechanical design temp, else provide 2oo3 voting on TSH trip. Trelief is permitted to exceed mechanical design temp only for firecase relief scenarios.

Alternatively, on Pro-II Simsci, it should be possible to model this using this scenario with the Depressure utility routine, using a constant heat input and depressurisation at constant pressure. This routine would yield lower relief mass rates than calculated above if you were to include vessel metal mass.

 

[Latexman]

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[georgeverghese]

Error in my last post at (b)
Since this is a open system, dH/dt = d/dt[NCp(T-Trelief)]=Cp.[(T-Trelief)dN/dt + NdT/dt]=Qh
Sustituting this and rearranging as before, I finally get
dN/dt = -R.N.Qh/[Cp.[(P.Vh)-(T-Trelief).R.N]]
So at start of relief, we use N=No, T=Trelief, and dN/dt reduces to
dN/dt = -Qh/(Cp.Trelief)
Also note that units for Cp is kJ/kgmole/degK, Qh is in kW

 

[chicopee]

I am assuming that you are referring to a hot water storage, then if so I don't understand the reason for going thru all that math. You have the heat input from which you can figure out the Btu/hr, you have the temperature of the water which should be 210dF, so match a relief valve to those values. Relief valves have attached tags giving you their settings that must be coincident with your determined values.