Question

Density (kg/m) Enthalpy (kJ) Thermal conductivity (W/m-K) Specific heat capacity (kJ/kg K) Mass (kg) Problem 4 A barometer at the top of the mountain reads 85 kPa while one in the valley reads 101 kPa. Estimate the difference in elevation between the two locations in meters and feet. Assume an average air density of 1.2 kg/m'. Approximate answer: 4,000 fi.

Answer 1

Problem 4

Given : P₁ = 101 kPa

P₂ = 85 kPa

air density = 1.2 kg/m³

Formula used : $\dpi{100} \small \Delta$ P = ($\dpi{100} \small \Delta$h) * (density of air) * (g)

where

$\dpi{100} \small \Delta$P = pressure difference

$\dpi{100} \small \Delta$P = P₁ - P₂

$\dpi{100} \small \Delta$P = 101 kPa - 85 kPa

$\dpi{100} \small \Delta$P = 16 kPa

$\dpi{100} \small \Delta$P = 16000 Pa

$\dpi{100} \small \Delta$h = difference in elevation between two points

g = gravitational constant of Earth = 10 m/s²

Substituting the values,

16000 Pa = ($\dpi{100} \small \Delta$h) * (1.2 kg/m³) * (10 m/s²)

$\dpi{100} \small \Delta$h = (16000 Pa) / [(1.2 kg/m³) * (10 m/s²)]

$\dpi{100} \small \Delta$h = 1333.33 m

$\dpi{100} \small \Delta$h = 1333.33 m * (3.28 ft / 1 m)

$\dpi{100} \small \Delta$h = 4373.33 ft