I am using ASME Y14.5:2018 standard.
I have seen examples for CF as below.
Is it possible to use CF as below?
As per ASME Y14.5:2018, para 3.10 and 3.11 CF defined as "two or more interrupted features". Till date I thought of interrupted by material removal.
My bores are different in size...
Can you provide any reference standards for this?
Thank you for the suggestion.
In whats app messaging application software, if we mentioned @ symbol the list of contacts will be displayed and the concerned person will be notified separately. Here I would like to mention that the reply is to...
@Burunduk
Thank you
@CheckerHater
Small hole will be positioned wrt to datum A.
@ 3DDave
Small hole will be positioned wrt to datum A.
@ jassco
Yes. It is a solenoid body. Can you explain the A1 and A2 concept.
@ Belanger
Do you mean to provide datum B for opposite (to dia 28) bore and use...
Hi
I need to control the outer surface of the rectangle wrt bore as shown in image.
Requirement: Outer surfaces to be equally located from bore center.
Does the drawing representation right?
Thank...
Hi
Refer the below images:
Green part name: Manual override
Function of the manual override:
Rotate around its axis and the rotation stopped with the flat face at one end.
When applying torque at screw driver slot, the manual override starts rotate.
Rotation stops when the flat faces...
Thank for the input.
Below are my observations.
Density is function of pressure and temperature.
Standard temperature and pressure: 15 degree Celsius and 101325 pa.
Normal temperature and pressure: 0 degree Celsius and 101325 pa.
https://www.controlandinstrumentation.com/flow/standard.html...
How did you get this number?
Density for air: 1.29 Kg/m3
I am unable to get your point.
Can you elaborate with numbers?
Thank you
How did you get this number? or any formula to calculate?
Thanks with Regards,
Sathish Rosario
I am clear with this concept.
Can you please explain with numbers.
Say, my gauge pressure is zero barg. Standard atmospheric pressure is 1 bar.
Hence, absolute pressure = 0+1 = 1 bar. right?
(Earlier I have mentioned 2 bar as absolute)
Can you explain the calculation steps?
My calculation...
@LittleInch
Even though the tank pressure is 11.5 barg, we are controlling the input pressure to test specimen by a regulator.
Test circuit:
Air source ---> Tank ---> Pressure gauge ---> Filter ---> Regulator ---> Pressure gauge (P1) ---> Test specimen ---> pressure gauge (P2) ---> Ball...
Our practical results:
Upstream pipe dia: 7 mm
Orifice dia (Test specimen - plain bore): 5 mm
Discharge tank capacity: 1000 liters.
Method: Charged the tank to 12.5 barg.
Since 1 bar = 1000 liters, 12.5 barg = 12, 500 liters of free air.
Set the upstream pressure to 6 barg.
Control the...
We are practically measuring volumetric flow as per ISO 6358-2
Discharge method.
Upstream pressure is controlled by a regulator.
Downstream pressure is controlled by a ball valve.
Upstream Pressure gauge is mounted between regular and test specimen.
Downstream Pressure gauge is mounted...
For case 2 upstream pressure is 6 bar. Downstream pressure is not controlled. It is fully opened.
So I am not sure how to tell the downstream pressure. Whether same 6 bar or zero bar
Hi..
Can we calculate volumetric flow for given data:
Case 1:
Orifice diameter, mm = 5
Inlet pressure, Bar = 6
Pressure drop, Bar = 1
Medium: Compressed air
Case 2:
Orifice diameter, mm = 5
Inlet pressure, Bar = 6
Pressure drop, Bar = 0
Medium: Compressed air
Thank you.