Question

A shell and tube heat exchanger must be designed to heat 2.5 kg/s water from 15 to 85 degrees celsius. The heating is to be accomplished by passing hot engine oil, which is available at 160 degrees celsius, through the shell side of the exchanger. The oil is known to provide an average convection coefficient of h0 = 400 W/mK on the outside of the tubes. Ten tubes pass the water through the shell. Each tube is thin-walled, of diameter D = 25 mm, and makes eight passes through the shell.
If the oil leaves the exchanger at 100 degrees celsius, what is its flow rate? How long must the tubes be to accomplish the desired heating?

Answer 1

Answer:

a) $\dot m_{o} = 6.105\,\frac{kg}{s}$, b) $L_{tube,total} = 29.144\,m$

Explanation:

a) The heat transfer rate needed to heat water is:

$\dot Q_{in} = \dot m_{w} \cdot c_{p,w}\cdot \Delta T_{w}$

$\dot Q_{in} = (2.5\,\frac{kg}{s} )\cdot (4186\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (85^{\textdegree} C-15^{\textdegree} C)$

$\dot Q_{in} = 732550\,W$

The mass flow rate of oil is:

$\dot m_{o} = -\frac{\dot Q_{in}}{c_{p,o}\cdot \Delta T_{o}}$

$\dot m_{o} = - \frac{732550\,W}{\left(2000\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot \left (100^{\textdegree}C-160^{\textdegree}C \right)}$

$\dot m_{o} = 6.105\,\frac{kg}{s}$

b) The formula for outer convection is:

$\dot Q_{in} = h_{o}\cdot A_{s} \cdot (\overline T_{o}-\overline T_{w})$

Mean temperatures are, respectively:

$\overline T_{o} = \frac{160^{\textdegree}C + 100^{\textdegree}C}{2}$

$\overline T_{o} = 130^{\textdegree}C$

$\overline T_{w} = \frac{15^{\textdegree}C + 85^{\textdegree}C}{2}$

$\overline T_{w} = 50^{\textdegree}C$

The required area is cleared in the convection equation:

$A_{s} = \frac{\dot Q_{in}}{h_{o}\cdot \left(\overline T_{o}-\overline T_{w}\right)}$

$A_{s} = \frac{732550\,W}{\left(400\,\frac{W}{m^{2}\cdot ^{\textdegree} C} \right)\cdot (130^{\textdegree}C-50^{\textdegree}C)}$

$A_{s} = 22.892\,m^{2}$

Last result is also given by this expression, whose length is cleared:

$A_{s} = \pi \cdot D \cdot n_{tubes} \cdot n_{pass}\cdot L_{tube,unit}$

$L_{tube,unit} = \frac{A_{s}}{\pi \cdot D \cdot n_{tubes}\cdot n_{pass}}$

$L_{tube,unit} = \frac{22.892\,m^{2}}{\pi \cdot (0.025\,m)\cdot (10)\cdot (8)}$

$L_{tube,unit} \approx 3.643\,m$

Total length of each tube is:

$L_{tube, total} = 8 \cdot (3.643\,m)$

$L_{tube,total} = 29.144\,m$