Question

A perfectly insulated, concentric tube countercurrent heat exchanger is used to cool lubricating oil using water as a coolant. Oil enters the outer annulus at a mass flow rate of \(2\ \mathrm{kg\,s^{-1}}\) with a temperature of \(100^{\circ}\mathrm{C}\) and leaves at \(40^{\circ}\mathrm{C}\). Water enters the inner tube at \(1\ \mathrm{kg\,s^{-1}}\) with a temperature of \(20^{\circ}\mathrm{C}\) and leaves at \(80^{\circ}\mathrm{C}\). Specific heats: \(c_{p,\text{oil}}=2089\ \mathrm{J\,kg^{-1}K^{-1}}\), \(c_{p,\text{water}}=4178\ \mathrm{J\,kg^{-1}K^{-1}}\). No phase change. Under steady state, the number of transfer units (NTU) is ________ (in integer).

Hint:

When \(C_h=C_c\) (i.e., \(C_r=1\)), the countercurrent effectiveness simplifies to \(\varepsilon=\mathrm{NTU}/(1+\mathrm{NTU})\).
Always compute \(\varepsilon=Q/Q_{\max}\) first; it avoids using the overall 64ac0493b52af67589bd410c coefficient explicitly.

Answer 1

Correct Answer: 3

Step 1: Capacity rates and heat duty
\[ C_h = \dot m_h c_{p,h} = 2\times 2089 = 4178\ \mathrm{W\,K^{-1}},\qquad C_c = \dot m_c c_{p,c} = 1\times 4178 = 4178\ \mathrm{W\,K^{-1}}. \] Thus \(C_{\min}=C_{\max}=4178\ \Rightarrow\ C_r=C_{\min}/C_{\max}=1\). Actual 64ac0493b52af67589bd410c: \[ Q = C_h\,(T_{h,in}-T_{h,out}) = 4178(100-40)=4178\times 60. \] Step 2: Effectiveness
\[ Q_{\max}=C_{\min}(T_{h,in}-T_{c,in})=4178(100-20)=4178\times 80, \] \[ \varepsilon=\frac{Q}{Q_{\max}}=\frac{4178\times 60}{4178\times 80}=\frac{60}{80}=0.75. \] Step 3: NTU from countercurrent relation (for \(C_r=1\))
For a countercurrent exchanger with \(C_r=1\), \[ \varepsilon=\frac{\mathrm{NTU}}{1+\mathrm{NTU}}. \] Hence \[ 0.75=\frac{\mathrm{NTU}}{1+\mathrm{NTU}} \ \Rightarrow\ \mathrm{NTU} = 3. \]