Answer:
6.4 m/s
Explanation:
From the equation of continuity
A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively
where r1 and r2 are radius of inlet and outlet respectively
v1 is given as 1.6 m/s
Therefore
Answer:
The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.
<u>Explanation:</u>
To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is
, we know ρ=density of fluid=1000
,
g=acceleration due to gravity=9.81 , h=height of the pool=2 m and l=length of the pool=8 m.
hydrostatic force on each wall= = 313920 N.
<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>
At 1 m from the ground this hydrostatic force is acted on each wall.
The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =
=
= 1255680 N. This is 4 times 313920 N.
Answer:
T2 ( final temperature ) = 576.9 K
a) 853.4 kJ/kg
b) 1422.3 kJ / kg
Explanation:
given data :
pressure ( P1 ) = 90 kPa
Temperature ( T1 ) = 30°c + 273 = 303 k
P2 = 450 kPa
Determine final temperature for an Isentropic process
----------- ( 1 )
T2 = 303 = 576.9K
Work done in a piston-cylinder device can be calculated using this formula
------- ( 2 )
where : cv = 3.1156 kJ/kg.k for helium gas
T2 = 576.9K , T1 = 303 K
substitute given values Back to equation 2
= 853.4 kJ/kg
work done in a steady flow compressor can be calculated using this
where : cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K
T2 = 576.9 k , T1 = 303 K
substitute values back to equation 3
= 1422.3 kJ / kg
