Question

A one-shell-pass and eight-tube-passes heat exchanger is used to heat glycerin $\left(c_{p}=0.60 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( from \)80^{\circ} \mathrm{F}\( to \)140^{\circ} \mathrm{F}$ by hot water $\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( that enters the thin-walled \)0.5$-in-diameter tubes at \(175^{\circ} \mathrm{F}\) and leaves at \(120^{\circ} \mathrm{F}\). The total length of the tubes in the heat exchanger is \(400 \mathrm{ft}\). The convection heat transfer coefficient is $4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\( on the glycerin (shell) side and \)50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$ on the water (tube) side. Determine the rate of heat transfer in the heat exchanger \((a)\) before any fouling occurs and \((b)\) after fouling with a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2}-\mathrm{F} / \mathrm{B}\) tu on the outer surfaces of the tubes.

Short answer

Based on the given heat exchanger with one-shell-pass and eight-tube-passes and the provided information, the rate of heat transfer before fouling is 11403.1 Btu/h and after fouling is 6473.7 Btu/h.

Step-by-step solution

Calculate the Heat Transfer Area

The total length of the tubes in the heat exchanger is 400ft. Each tube has a diameter of 0.5 inches. To find the total heat transfer area, we first need to convert the diameter to feet and then multiply it by the total length of the tubes. Diameter in feet: \((0.5\,\text{in})\cdot(\frac{1\,\text{ft}}{12\,\text{in}})= \frac{1}{24}\,\text{ft}\) Heat Transfer Area \(A_h\): \( 8 \cdot \pi \cdot (\frac{1}{24}\,\text{ft}) \cdot 400\,\text{ft} = 100\,\pi\,\mathrm{ft}^{2}\)

Calculate the Overall Heat Transfer Coefficient

Before any fouling occurs, we can use the arithmetic mean of the convection heat transfer coefficients on the glycerin and water sides to calculate the overall heat transfer coefficient. $$ \frac{1}{U} = \frac{1}{h_{g}}+\frac{1}{h_{w}} \Rightarrow U = \frac{2h_{g}h_{w}}{h_{g}+h_{w}} $$ Substituting the given values, we have: $$ U = \frac{2\cdot4\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}\cdot50\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}}{4\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}+50\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}} = 7.2727\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}} $$

Calculate the Rate of Heat Transfer before Fouling

The heat transfer rate can be calculated using the heat transfer equation: $$ q = U \cdot A_h \cdot \Delta T_{lm} $$ The Log Mean Temperature Difference (LMTD) is given by: $$ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln{\frac{\Delta T_1}{\Delta T_2}}} $$ Where \(\Delta T_1 = T_{h,in} - T_{c,in} = 175 - 80 = 95\,^{\circ}\mathrm{F}\) and \(\Delta T_2 = T_{h,out} - T_{c,out} = 120 - 140 = -20\,^{\circ}\mathrm{F}\). Plugging these values in LMTD formula gives us: $$ \Delta T_{lm} = \frac{95 - (-20)}{\ln{\frac{95}{-20}}} = 49.34\,^{\circ}\mathrm{F} $$ Now, substituting the known values in q: $$ q = (7.2727\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}) \cdot (100\pi\,\text{ft}^{2}) \cdot (49.34\,^{\circ}\mathrm{F}) = 11403.1\,\mathrm{Btu\,h^{-1}} $$

Calculate the Rate of Heat Transfer after Fouling

After fouling, we need to take into account the fouling factor (R) of \(0.002\,\mathrm{h\,ft^{-2}\,^{\circ}F\,Btu^{-1}}\). The overall heat transfer coefficient after fouling, \(U_{fouled}\), is given by: $$ \frac{1}{U_{fouled}} = \frac{1}{U} + R \Rightarrow U_{fouled} = \frac{1}{\frac{1}{U}+R} $$ Substituting the known values: $$ U_{fouled} = \frac{1}{\frac{1}{7.2727\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}} + 0.002\,\mathrm{h\,ft^{-2}\,^{\circ}F\,Btu^{-1}}} = 4.1667\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}} $$ Now, we can calculate the heat transfer rate after fouling: $$ q_{fouled} = U_{fouled} \cdot A_h \cdot \Delta T_{lm} = (4.1667\,\mathrm{Btu\,h^{-1}\,ft^{-2}\,^{\circ}F^{-1}}) \cdot (100\pi\,\mathrm{ft}^{2}) \cdot (49.34\,^{\circ}\mathrm{F}) = 6473.7\,\mathrm{Btu\,h^{-1}} $$ So, the rate of heat transfer in the heat exchanger (a) before any fouling occurs is \(11403.1\,\mathrm{Btu\,h^{-1}}\) and (b) after fouling is \(6473.7\,\mathrm{Btu\,h^{-1}}\).