Question 3.SP.2: An open tank contains water 1.40 m deep covered by a 2-m-th......

– From inside cover of book: \gamma_{w}=9.81 \mathrm{kN} / \mathrm{m}^{3} .

Sec. 2.3: \quad \gamma_{o}=0.855(9.81)=8.39 \mathrm{kN} / \mathrm{m}^{3}

Eq. (3.4) for interface: p_{i}=\gamma_{o} h_{o}=(8.39) 2=16.78 \mathrm{kN} / \mathrm{m}^{2}=16.78 \mathrm{kPa}

Eq. (3.5) for water equivalent of oil:

h_{o e}=\frac{p_{i}}{\gamma_{w}}=\frac{16.78 \mathrm{kN} / \mathrm{m}^{2}}{9.81 \mathrm{kN} / \mathrm{m}^{3}}=1.710 \mathrm{~m} \text { of water }

So

h_{w e}=h_{w}+h_{o e}=1.40+1.710=3.11 \mathrm{~m} \text { of water }

Solution 2

From Eq. (3.4) for bottom of tank:

p_{b}=\gamma_{o} h_{o}+\gamma_{w} h_{w}=(8.39) 2+9.81(1.4)=30.51 \mathrm{kN} / \mathrm{m}^{2}=30.51 \mathrm{kPa}

Eq. (3.5) for total water equivalent:

h_{\mathrm{we}}=\frac{p_{b}}{\gamma_{w}}=\frac{30.51 \mathrm{kN} / \mathrm{m}^{2}}{9.81 \mathrm{kN} / \mathrm{m}^{3}}=3.11 \mathrm{~m} \text { of water }

  p=\gamma h          (3.4)

 h=\frac{p}{\gamma}      (3.5)