I have a problem in specifying the ratio of losses for an air vessel item, i have an air vessel with an outlet nozzle diameter 400 mm installed with a 150 mm bypass to limit the maximum transient pressure in the rising main. my specific question is as follow,
the ratio of losses field description states that "for same flow magnitude. ratio of inflow head loss to outflow loss. default value is 2.5". i made calculations for the inflow and the outflow nozzle head losses and head losses coefficients ( k ). and found that if that the ratio of head loss will be around 70 while the ratio of head loss coefficient is around 2.5..
Is this field should be the ratio of head loss coefficient? or the ratio of head loss only, meaning: shall i divide the k coefficient of the inflow and outflow or shall i divide the head loss itself between the inflow and the outflow? specially that the minor loss coefficient (outflow) field is a head loss coefficient not a head loss value by my understanding.
i hope my question is clear.
It is the ratio of inflow to outflow headloss. For flows into the tank, the minor loss coefficient is multiplied by this value and the losses are computed using that. For flows out of the tank, HAMMER only uses the minor loss coefficient.
So, if you enter a minor loss coefficient of 1.5 and a ratio of losses of 2.5, the headloss coefficient used when the tank is filling would be 1.5 X 2.5 = 3.75.
This is given in detail in the below article;
Using the Ratio of Losses field for hydropneumatic tanks and surge tanks
Let me know if this helps.
Bentley Technical Support
Okay, i i understand you fully, but a more specific question is as followed:
Does HAMMER uses the same velocity for both calculations ( outflow from the tank & inflow to the tank ) , the velocity that is specified by the nozzle diameter field in the tank prpoerties?
Because there is a drastic difference in velocity for entering the tank through the small bypass and exiting the tank through the large nozzle !
As seen in the article that Yashodhan linked to, when the tank is filling, the minor loss coefficient is multiplied by the value entered in the "ratio of losses" field to determine the inflow headloss. So, it does indeed use the same velocity for both inflow and outflow.
In order to get the equivalent higher headloss due to the smaller, bypass pipe for inflow, you will need to determine the correct "ratio of losses". Meaning, you will need to select the ratio of losses value that will results in the same headloss as if the diameter was reduced for the bypass pipe. The Standard Headloss equation is used, so you can calculate the value based on that equation.
Jesse DringoliTechnical Support Manager, OpenFlows ProductsBentley Communities Site AdministratorBentley Systems, Inc.
Answer Verified By: Noor Anter
First of all thank you for your many contributions in the forum, you are a great asset that i consistently return to, so THANKS !
Second, so as i understand from your explanation, the ratio of losses shall be determined by ragarding both the additional losses from the bypass, and the usage of entering velocity instead of exiting velocity for the tank nozzle.
I shall calculate the ratio of headloss so that the total resulting head loss by using the exiting velocity with the headloss ratio will be equal to actual headloss calculated by real entering velocity and actual headloss coefficient.
Glad to help. Yes, your approach sounds reasonable.