Test report of pressure loss in the TGC ps-pack-cooling system

T.Takashita 26/Feb/2002

________________________________________________________

1. Test set up to measure the pressure loss

Test has been carried out at the similar scheme when the new aluminum cooling pipes of 4m each were introduced. The cooling test is descrived elsewhere .

picture of the test set up

picture of the differential pressure gauge

There I got a differential pressure air gauge and put it in parallel with the cooling pipe, where a bypass line was filled with air. In the middle of the bypass line, I put a differential pressure gauge so that the pressure difference between the inlet and outlet water could be measured through the air-pressure

. I have measured the pressure difference (or loss) by changing the water flow rate from 200 to 500ml/min. It is shown in the next plot.

2. results : plot depending on the water flow rate

According to the theory of liquid flow, the pressure loss (dP [kgf/m^2]) depends on the length (l [m]), diameter (d [m]), velocity (v [m/s]) and gravitational acceleration (g [m/s^2]) as follows;, where lamda is the coeffcient of friction depending on the Reynolds number and roughness of the pipe inner surface. Here I suppose the relative weight (gamma [kgf/m^3]) of the liquid to be 10^3. This means the density is 1g/cm^3. This equation can be rewritten as , where F is the flow rate [m^3/s]. In the result figure, the pressure loss ( difference) is ploted as a function of the flow rate (F). There seems to be such dependence, although it shows a pedestal even at the zero-flow rate. Since the diameter of the aluminum cooling pipe is 8.5mm, the flow velocity becomes 15cm/s, when the flow rate is 500ml/min. Here the result of dP=3.5kPa means, the lamda ( coefficient of friction) is equal to be 0.34. This means the water flow in the pipe is quite smooth and laminar (= Poiseuille's flow). One can calculate the kinematic viscosity (neu) as . It is calculated to be 6.8 x 10 ^-6 [m^2/s] which is rather large than that of well know, although in the same order.

Normaly it is around 1.2 x 10 ^-6 at 15 degree. The Reynolds number is equivalent to be 190.