OK so, I understand everything until the part where you\'re actually calculating

Question

OK so, I understand everything until the part where you're actually calculating the values for P2 and P3. I plugged all the values in my calculator and I got a number in the 1000's. I went to wolframalpha to confirm my calculation and it gave me the same number I got.

How is

40 (kPa) -9.8*1 (kPa) -999*10^2*ln(6/5) (kPa) = 12.0 kPa?

Was there a conversion that I missed?

Water flows around the vertical two-dimensional bend with circular streamlines and constant velocity as shown in Fig. P3. 7. If the pressure is 40 kPa at point (1), determine the pressures at points (2) and (3). Assume that the velocity profile is uniform as indicated.

Explanation / Answer

When u r aking every thing in Kilo pascal in the line extreme below, u should not take 999,but u should take 0.999 becz u have already considered it in kilo pascal,and the density of water is 999 kg/m^3,

to understand the units just multiply the units u must get Kpa = K N/m^2

In with......statement ,u get 40 Kpa =40 000pa,,,and every thing others in pascal,so u subtract it and finally obtain11986.08..which u r writing as 12 Kpa

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