Pressure variation in the environment illustrates how pressure varies with verti

Question

Pressure variation in the environment illustrates how pressure varies with vertical elevation due to the force required to support the weight of the fluid column above For a liquid region of moderate depth the fluid density (rho) and gravitational acceleration (g) are constant. So the pressure difference (dp) for each unit change in elevation (dz) is constant (dp/dz = -rho g = constant). However in Earth's atmosphere (with elevation changes of many kilometers) the air density is not constant (and g may change a small amount). Hence the pressure difference (dp) for each unit change n elevation (dz) varies with elevation (dp/dz = -rho g = function of z). A rough first estimate for the pressure distribution in the standard atmosphere is obtained based on simplifications of assuming the temperature is constant and the gravitational acceleration is constant. e.g. assume rho g = (p/RT_0, sea level) g_sea level = (p/p_0, seal level) rho_0, sea level) g_0, sea level solve p vs. z with dp/dz = -rho g = -(p/p_0) p_0 g_0 with rho_0 = 1.2256 kg/m^3, g_0 = 9.807 m/s^2, p_0 = 101300 Pa, phi = (rho_0 g_0/p_0) = 1.1865 times 10^-6 m^-1 When executed with small layer thickness the Euler approximation will match the integral. But variable temperature must be included to match the actual data. The approximate integral method is simple to modify to improve the validity of the estimate. See the 2 steps below

Explanation / Answer

The answer of the following questions are as follows

Data by integration: pressure p(z)=112 kPa

Data by local linear approximation: pressure p (z)= 168 kPa

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