Question

The weight density of fluid in a container varies with depth as given under $\gamma =0.2h^2+0.1h+8.2$, where h is in m and $\gamma$ is in $kN/m^3$.
The pressure (gauge in kPa) at a depth 15m will be :

• A. $359.25$
• B. $423.2$
• C. $461.3$
• D. $820.5$

Answer 1

The correct option is A $359.25$

$dp=\gamma dh$

$${\int }_0^{p}dp={\int }_{h=0}^{15m}(0.2h^2+0.1h+8.2)dh$$

$$p={[\frac{0.2h^3}{3}+\frac{0.1h^2}{2}+8.2h]}_0^{15}$$

$=\frac{0.2}{3}\times (15{)}^3+\frac{0.1}{2}\times (15{)}^2+8.2\times 1.5$

$=359.25kPa$