A 1.20m-high, 1m-diameter cylindrical water tank has the top open to the atmosph
ID: 1815009 • Letter: A
Question
A 1.20m-high, 1m-diameter cylindrical water tank has the top open to the atmosphere and is initially filled with water. Next, the discharge plug near the bottom of the tank is pulled out, and a water jet with diameter of 1cm streams out. The mean velocity of the jet is given by v = (2gh)^1/2 where h is the height of the water in the tank (measured from the bottom) and g is the gravitational acceleration. Determine how long it will take for the water level in the tank to drop to 0.6 m from the bottom. Take g = 9.81m/s^2Explanation / Answer
64klx , you should be able to understand thethermodynamics home work problem you are asking about by thefollowing steps height of the cylinder H=120m diameter of the cylinder D=1m volume of the cylinder V1 =/4*D2H =/4*1*120 =94.24m^3 volume of the cylinder when h=0.6m V2 =/4*D2*h =/4*1*0.6 =0.4712m^3 volume of water leaving from the cylinder V=V1-V2 =94.24-0.4712 =93.76m^3 =94.24-0.4712 =93.76m^3 jet diameter d=1cm =0.01m area of jet a=/4 *d^2 =/4*0.01^2 =0.0000785m^2 when water level in the tank h=0.6m mean velocity of the jet v=2gh^1/2 =2*9.81*0.6^1/2 =3.431m/s volume rate of the water Q2 =a*v =0.0000785*3.431 =0.000269m^3/s when H=120m the volume rate of the waterQ1=a*2gH =0.0000785*(2*9.81*120) =0.00381m^3/s volume rate of water discharged from the cylinderQ=Q1-Q2 =0.00381-0.000269 =0.00354m^3/s =0.00381-0.000269 =0.00354m^3/s time required t=V/Q =93.76/0.00354 =26488.36s =7.3578hr I hope this helps best of luck with rest of your courseworkRelated Questions
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