lets look at a column of water 100m high, with cross section of 1m^2
weight of water= density water*volume water*average height
= 1E3kg/m^3*9.8N/kg*100m^3*50m
= 4.9E8N
pressure= weight/area= 4.9E8P or 4.9E5kpa
= 490 kpa or in atmospheres,
= 490/101 atmospheres
check my math.
Calculate the water pressure at the bottom of the 100-m-high water tower
2 answers
well, let's say the density of water is 1000 kg/m^3.
Then the weight of 1 m^3 of water is 9810 Newtons
Then the weight of a stack of those 100 m high = 9,810,000 Newtons and that is the force down on a square meter of bottom above atmospheric (the "gage" pressure). A Newton per square meter of pressure is called a "Pascal".
To get the total pressure you must add the pressure of the air on the top of the stack (atmospheric pressure) which is about 10^5 or 100,000 Pascals
So the total pressure is about 9,910,000 or 10^7 Pascals
Then the weight of 1 m^3 of water is 9810 Newtons
Then the weight of a stack of those 100 m high = 9,810,000 Newtons and that is the force down on a square meter of bottom above atmospheric (the "gage" pressure). A Newton per square meter of pressure is called a "Pascal".
To get the total pressure you must add the pressure of the air on the top of the stack (atmospheric pressure) which is about 10^5 or 100,000 Pascals
So the total pressure is about 9,910,000 or 10^7 Pascals
0 answers