Recent content by xray2golf

I got V1 to equal .831 When I plug these numbers into the equation for flow rate through the pipe I get very close numbers at both ends...41.8 m^3/sec and 41.5 m^3/sec. I assume the rounding could be the reason for the difference...the bottom line is the numbers are the same, as the should...

OK, I plugged Vin = .250Vout into the original Bernoulli's equation and ended up with V2 = 3.30. (25000) + (.5)*(1000)*(.250V2)^2 + (1000)*(9.8)*(0) = (15000) + (.5)*(1000)*(V2)^2 + (1000)*(9.8)(.5) V2 = 3.30 If this is correct then I will plug that answer back into the equation and...

Ok, by solving V_{in} = \frac{A_{out}}{A_{in}}V_{out} with the actual data, V_{in} = (12.6/50.3)V_{out} or V_{in} = .250V_[out] I assume I plug that into the original P_{}1 + Pgy + 1/2P V^{}2 = constant or into the longer version? (sorry, I am struggling with the Latex) I will...

Homework Statement Water moves through a constricted pipe in steady, ideal flow. At one point, where the pressure is 2.50*10^4 Pa, the diameter is 8.0 cm. At another point .5 Meters higher, the pressure is equal to 1.50*10^4 Pa and the diameter is 4.0cm. Find the speed of flow at the lower...