Oil (SG = 0.88) flows in an inclined pipe at a rate of 5 f t 3 s ? 1 as shown in fig .If the...

Question:

Oil (SG = 0.88) flows in an inclined pipe at a rate of 5 {eq}ft^3s^-1 {/eq} as shown in fig .If the differential reading in the mercury ( SG = 13.6 ) manometer is 3 ft, Calculate the power that the pump supplies to the oil if head losses are negligible. Use {eq}{\gamma}_{H_20} {/eq} = 62 .4 lb {eq}ft^-3 {/eq} where neccesary .

Principle of Manometry:

The pressure difference between the two different points at the different elevation is changes directly with the change of elevation between the points. This principles can be applied between different fluid layer provided the fluids are immiscible to each other.

Answer and Explanation:


Given data

  • The specific gravity of oil is {eq}S_{oil} = 0.88 {/eq}
  • The specific gravity of mercury is {eq}S_{Hg} = 13.6 {/eq}
  • The volume flow rate through pipe is {eq}Q = 5\;{\rm{ft}}^{\rm{3}} {\rm{/s}} {/eq}
  • The manometric height of mercury column is {eq}H = 3\;{\rm{ft}} {/eq}
  • The inlet diameter of pipe is {eq}d_1 = 12\;{\rm{in}} {/eq}
  • The exit diameter of pipe is {eq}d_2 = 6\;{\rm{in}} {/eq}
  • The weight density of water {eq}\gamma _{H_2 O} = 62.4\;{\rm{lb/ft}}^{\rm{3}} {/eq}


The expression for pressure difference between pump inlet and outlet is

{eq}\begin{align*} P_1 + \rho _{oil} gh + 3\rho _{Hg} g - \rho _{oil} g\left( {3 + 2h} \right) &= P_2 \\ \dfrac{{P_1 }}{{\rho _{oil} g}} + h + 3\dfrac{{\rho _{Hg} g}}{{\rho _{oil} g}} - \left( {3 + 2h} \right) &= \dfrac{{P_2 }}{{\rho _{oil} g}} \\ \dfrac{{P_1 }}{{\gamma _{oil} }} + h + 3 \times \dfrac{{S_{Hg} }}{{S_{oil} }} - \left( {3 + 2h} \right) &= \dfrac{{P_2 }}{{\gamma _{oil} }} \\ \dfrac{{P_1 }}{{\gamma _{oil} }} + 3 \times \dfrac{{13.6}}{{0.88}} - 3 - h &= \dfrac{{P_2 }}{{\gamma _{oil} }} \\ \end{align*} {/eq}

{eq}\begin{align*} \dfrac{{P_1 }}{{\gamma _{oil} }} - \dfrac{{P_2 }}{{\gamma _{oil} }} &= 3 + h - 3 \times \dfrac{{13.6}}{{0.88}} \\ \dfrac{{P_1 }}{{\gamma _{oil} }} - \dfrac{{P_2 }}{{\gamma _{oil} }} &= h - 43.3636\;{\rm{ft}} \\ \end{align*} {/eq}

The velocity of oil at inlet is

{eq}V_1 = \dfrac{{4Q}}{{\pi d_1^2 }} {/eq}

Substitute the value in above expression

{eq}\begin{align*} V_1 &= \dfrac{{4 \times 5\;{\rm{ft}}^{\rm{3}} {\rm{/s}}}}{{\pi \left( {12/12\;{\rm{ft}}} \right)^2 }} \\ &= \dfrac{{20\;{\rm{ft}}^{\rm{3}} {\rm{/s}}}}{{\pi \left( {12/12} \right)^2 \;{\rm{ft}}^{\rm{2}} }} \\ &= 6.37\;{\rm{ft/s}} \\ \end{align*} {/eq}

The velocity of oil at outlet is

{eq}V_2 = \dfrac{{4Q}}{{\pi d_2^2 }} {/eq}

Substitute the value in above expression

{eq}\begin{align*} V_2 &= \dfrac{{4 \times 5\;{\rm{ft}}^{\rm{3}} {\rm{/s}}}}{{\pi \left( {6/12\;{\rm{ft}}} \right)^2 }} \\ &= \dfrac{{20\;{\rm{ft}}^{\rm{3}} {\rm{/s}}}}{{\pi \left( {1/2} \right)^2 \;{\rm{ft}}^{\rm{2}} }} \\ &= 25.5\;{\rm{ft/s}} \\ \end{align*} {/eq}

The expression for energy equation is

{eq}\begin{align*} \dfrac{{P_1 }}{{\gamma _{oil} }} + \dfrac{{V_1^2 }}{{2g}} + Z_1 + h_p &= \dfrac{{P_2 }}{{\gamma _{oil} }} + \dfrac{{V_2^2 }}{{2g}} + Z_2 \\ \dfrac{{P_1 }}{{\gamma _{oil} }} - \dfrac{{P_2 }}{{\gamma _{oil} }} + h_p &= \dfrac{{V_2^2 }}{{2g}} - \dfrac{{V_1^2 }}{{2g}} + Z_2 - Z_1 \\ \end{align*} {/eq}

Substitute the value in above expression

{eq}\begin{align*} h - 43.3636\;{\rm{ft}} + h_p &= \dfrac{{\left( {25.5\;{\rm{ft/s}}} \right)^2 }}{{2 \times 32.2\;{\rm{ft/s}}^{\rm{2}} }} - \dfrac{{\left( {6.37\;{\rm{ft/s}}} \right)^2 }}{{2 \times 32.2\;{\rm{ft/s}}^{\rm{2}} }} + h \\ h_p &= 43.3636\;{\rm{ft}} + 10.09\;{\rm{ft}} - 0.63\;{\rm{ft}} \\ h_p &= 52.8236\;{\rm{ft}} \\ \end{align*} {/eq}

The expression for power developed by pump is

{eq}\begin{align*} P &= \rho _{oil} gQh_p \\ &= S_{oil} \rho _w gQh_p \\ &= S_{oil} \gamma _{H_2 O} Qh_p \\ \end{align*} {/eq}

Substitute the value in above expression

{eq}\begin{align*} P &= 0.88 \times 62.4\;{\rm{lb/ft}}^{\rm{3}} \times 5\;{\rm{ft}}^{\rm{3}} {\rm{/s}} \times 52.8236\;{\rm{ft}} \\ &= 14503.2476\;{\rm{lb}} \cdot {\rm{ft/s}} \\ \end{align*} {/eq}

Hence power of the pump supplied to the oil is {eq}14503.2476\;{\rm{lb}} \cdot {\rm{ft/s}} {/eq}.


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