Chapter 6, Problem 6.12P

Interpretation Introduction

Interpretation:

The time taken to lower the surface of oil by 2 ft in the given tank is to be calculated.

Concept Introduction:

The general energy balance equation according to the first law of thermodynamics written for a control volume is:

  $\frac{\delta Q}{dt}-\frac{\delta W}{dt}={\displaystyle {\iint }_{C.S}\left(e+\frac{P}{\rho }\right)\rho \left(v\cdot n\right)dA}+\frac{\partial }{\partial t}{\displaystyle {\iiint }_{C.V}e\rho dV}+\frac{\delta W_{\mu }}{dt}$   ...... (1)

The general continuity equation to be used for a controlled volume is:

  ${\displaystyle {\iint }_{C.S}\rho \left(v\cdot n\right)dA}+\frac{\partial }{\partial t}{\displaystyle {\iiint }_{C.V}\rho dV}=0$   ...... (2)

Here, C.S is the control surface over which the integral dA is taken, C.V is the control volume over which the integral dV is taken, $\rho$ is the density of the fluid, $v$ is the velocity vector, $n$ is the direction of the vector $v$ , $\delta Q/dt$ is the rate of heat added (positive) or removed (negative) from the system, $\delta W/dt$ is the rate of work done by (positive) or on (negative) the system, and $\delta W_{\mu }/dt$ is the rate of work done to overcome the viscous effect at the control surface. The product $v\cdot n$ is scalar defined as:

  $v\cdot n=\left|v\right|\left|n\right|\cos \theta$