An open tank holds water 1.25m deep. If a small hole of cross section area 3cm^2 is made at the bottom of the tank, calculate the mass of water per second initially flowing out of the hole. (g=10m/s^2, density of water= 1000kg/m^3)
To calculate the mass of water per second flowing out of the hole, we can use Bernoulli's equation, which states that the pressure at a given point in a fluid is equal to the sum of the pressure, height, and kinetic energy at a reference point.
1. First, let's calculate the pressure at the bottom of the tank, where the hole is. The pressure at a certain depth in a fluid can be calculated using the formula:
P = ρgh
Where:
P is the pressure
ρ is the density of the fluid (given as 1000 kg/m^3)
g is the acceleration due to gravity (given as 10 m/s^2)
h is the depth of the fluid (given as 1.25m)
P = (1000 kg/m^3) * (10 m/s^2) * (1.25 m)
P = 12500 Pa
2. Next, let's calculate the mass flow rate of water through the hole. We can use the formula:
Mass flow rate (ṁ) = ρ * A * v
Where:
ṁ is the mass flow rate
ρ is the density of the fluid (given as 1000 kg/m^3)
A is the cross-sectional area of the hole (given as 3 cm^2, which is equal to 3 * 10^(-4) m^2)
v is the velocity of the water flowing out of the hole
Since the hole is at the bottom and the tank is open to the atmosphere, the velocity of water flowing out of the hole can be calculated using a Torricelli's Law:
v = √(2gh)
v = √(2 * 10 m/s^2 * 1.25 m)
v = √(25) m/s
v = 5 m/s
ṁ = (1000 kg/m^3) * (3 * 10^(-4) m^2) * 5 m/s
ṁ = 0.15 kg/s
Therefore, the mass of water flowing out of the hole initially is 0.15 kg/s.
To calculate the mass of water per second initially flowing out of the hole, we first need to find the volume of water flowing out per second and then multiply it by the density of water.
Here is the step-by-step explanation to find the answer:
Step 1: Find the height of the water column above the hole.
Given that the tank holds water 1.25m deep and the hole is at the bottom, the height of the water column above the hole is 1.25m.
Step 2: Find the velocity of water flowing out of the hole.
To find the velocity of water flowing out of the hole, we can use Torricelli's law, which states that the velocity of liquid flowing out of a small hole is given by:
v = √(2gh)
Where:
v = velocity of liquid flowing out
g = acceleration due to gravity (given as 10m/s^2)
h = height of the water column above the hole
Substituting the given values, we get:
v = √(2 * 10 * 1.25)
v = √25
v = 5m/s
Step 3: Find the volume of water flowing out per second.
The volume of water flowing out per second can be calculated using the equation:
V = A * v
Where:
V = volume of water flowing out per second
A = cross-sectional area of the hole (given as 3cm^2)
v = velocity of water flowing out
Converting the area from cm^2 to m^2:
A = 3 * 10^-4 m^2
Substituting the values, we get:
V = 3 * 10^-4 * 5
V = 1.5 * 10^-3 m^3/s
Step 4: Calculate the mass of water flowing out per second.
To calculate the mass, we multiply the volume (in cubic meters) by the density of water (given as 1000kg/m^3):
Mass (M) = density (ρ) * volume (V)
Substituting the values, we get:
M = 1000 * 1.5 * 10^-3
M = 1.5kg/s
Therefore, the mass of water per second initially flowing out of the hole is 1.5kg/s.
Wouldn't Bernoulli's equation reduced to
rho*g(h)=1/2 rho v^2
h=1.25, rho divides out, but it is given, g is given.
V^2=gh=1.25*10
V= 3.53m/s
massrate=V*rho*area=3.53m/s*1000kg/m^2*.03^2
= 1.06kg/sec