# Problem 07 - Variation of Pressure

**Problem**

If the pressure in the tank of oil (sp gr 0.80) is 415 kPa, what is the equivalent head: (a) in meters of oil, (b) in meters of water, and (c) in centimeters of mercury?

**Solution**

Part (a) head in meters of oil

$h_{oil} = \dfrac{p}{\gamma_{oil}}$

$h_{oil} = \dfrac{p}{\gamma_{oil}}$

$h_{oil} = \dfrac{p}{s_{oil} \, \gamma_{water}}$

$h_{oil} = \dfrac{415}{0.80(9.81)}$

$h_{oil} = 52.88 \,\text{m}$ *answer*

Part (b) head in meters of water

$h_{water} = s_{oil} \, h_{oil}$

$h_{water} = 0.80(52.88)$

$h_{water} = 42.30 \,\text{m}$ *answer*

Part (c) head in centimeters of mercury

$h_{mercury} = \dfrac{s_{oil}}{s_{mercury}} h_{oil}$

$h_{mercury} = \dfrac{0.80}{13.6} (52.88)$

$h_{mercury} = 3.11 \,\text{m}$

$h_{mercury} = 311 \,\text{cm}$ *answer*

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