Imagine a house with two bathrooms/showers: one downstairs, the other upstairs. The volume flow rate of the shower head downstairs (2 meters above the ground level) is 4*10-4 m3/s. The cross sectional area of the shower head is 30 mm2. What is the volume flow rate of an identical shower head installed in the bathroom upstairs (5 meters above the ground level) under identical conditions?
Density of water: 1000 kg/m3
Atmospheric pressure=101300 Pa g=9.8 m/s2 1 Liter = 1000 cm3
To find the volume flow rate of the shower head upstairs, we need to use the principle of conservation of mass.
The volume flow rate is calculated by multiplying the cross-sectional area of the shower head by the velocity of the water flowing through it. Since the cross-sectional area of both shower heads is the same, we can focus on determining the velocity of the water upstairs.
To find the velocity, we first need to calculate the pressure difference between the two locations. The pressure difference is caused by the difference in height between the two shower heads.
Using the equation for pressure difference due to height change:
ΔP = ρ * g * Δh
Where:
ΔP is the pressure difference
ρ is the density of water (given as 1000 kg/m^3)
g is the acceleration due to gravity (given as 9.8 m/s^2)
Δh is the height difference between the two shower heads (5 - 2 = 3 meters)
Plugging in the values, we get:
ΔP = 1000 kg/m^3 * 9.8 m/s^2 * 3 m
ΔP = 29400 Pa
Now, we can use Bernoulli's equation to find the velocity of the water upstairs. Bernoulli's equation relates pressure, velocity, and height:
P₁ + 1/2 * ρ * v₁^2 + ρ * g * h₁ = P₂ + 1/2 * ρ * v₂^2 + ρ * g * h₂
Where:
P₁ and P₂ are the pressures at the two locations (atmospheric pressure at both locations)
v₁ and v₂ are the velocities at the two locations (v₁ is given, we need to find v₂)
h₁ and h₂ are the heights at the two locations (2 and 5 meters)
Since atmospheric pressure is the same at both locations and velocity is zero at both locations, the equation simplifies to:
1/2 * ρ * v₁^2 + ρ * g * h₁ = 1/2 * ρ * v₂^2 + ρ * g * h₂
Plugging in the values:
1/2 * 1000 kg/m^3 * (4*10^-4 m^3/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * 2 m = 1/2 * 1000 kg/m^3 * v₂^2 + 1000 kg/m^3 * 9.8 m/s^2 * 5 m
Now, we can solve for v₂ by rearranging the equation:
1/2 * (4*10^-4 m^3/s)^2 + 2 * 9.8 m + 0 = 1/2 * v₂^2 + 5 * 9.8 m
Simplifying further:
1.6 * 10^-7 m^6/s^2 + 19.6 m + 0 = 0.5 * v₂^2 + 49 m
Rearranging again:
0.5 * v₂^2 = 49 m - 1.6 * 10^-7 m^6/s^2 - 19.6 m
0.5 * v₂^2 = 29.4 m
Finally, solving for v₂:
v₂ = √(29.4 m / 0.5)
v₂ ≈ 7.66 m/s
Therefore, the volume flow rate of the shower head upstairs is equal to the cross-sectional area (30 mm^2 = 30 * 10^-6 m^2) multiplied by the velocity (7.66 m/s):
Volume flow rate upstairs = 30 * 10^-6 m^2 * 7.66 m/s
Volume flow rate upstairs ≈ 0.23 * 10^-3 m^3/s