Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
Answer:
The answer is A, B, C and D
Explanation:
(is that how it works?)
Answer:
Velocity
Explanation:
Velocity is an object's change in motion per unit time in a specified direction
Answer:
The maximum mass of water that could be produced by the chemical reaction is 10.1 g
Explanation:
The equation of reaction involves the combustion of 2 moles of hexane (C6H14) with 19 moles of oxygen (O2) to produce 12 moles of carbon dioxide (CO2) and 14 moles of water (H2O)
From the equation of reaction above,
2 moles of C6H14 (172 g) produced 14 moles of H2O (252 g)
6.9 g of C6H14 would produce (6.9×252/172) = 10.1 g of water (to 3 significant figures)
Also, from the equation of reaction,
19 moles of O2 (608 g) produced 14 moles of H2O (252 g)
17.3 g of O2 would produce (17.3×252/608) = 7.17 g of water (to 3 significant figures)
Maximum mass of water produced = 10.1 g