Answer:
a) [tex]\dot m_{o} = 6.105\,\frac{kg}{s}[/tex], b) [tex]L_{tube,total} = 29.144\,m[/tex]
Explanation:
a) The heat transfer rate needed to heat water is:
[tex]\dot Q_{in} = \dot m_{w} \cdot c_{p,w}\cdot \Delta T_{w}[/tex]
[tex]\dot Q_{in} = (2.5\,\frac{kg}{s} )\cdot (4186\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (85^{\textdegree} C-15^{\textdegree} C)[/tex]
[tex]\dot Q_{in} = 732550\,W[/tex]
The mass flow rate of oil is:
[tex]\dot m_{o} = -\frac{\dot Q_{in}}{c_{p,o}\cdot \Delta T_{o}}[/tex]
[tex]\dot m_{o} = - \frac{732550\,W}{\left(2000\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot \left (100^{\textdegree}C-160^{\textdegree}C \right)}[/tex]
[tex]\dot m_{o} = 6.105\,\frac{kg}{s}[/tex]
b) The formula for outer convection is:
[tex]\dot Q_{in} = h_{o}\cdot A_{s} \cdot (\overline T_{o}-\overline T_{w})[/tex]
Mean temperatures are, respectively:
[tex]\overline T_{o} = \frac{160^{\textdegree}C + 100^{\textdegree}C}{2}[/tex]
[tex]\overline T_{o} = 130^{\textdegree}C[/tex]
[tex]\overline T_{w} = \frac{15^{\textdegree}C + 85^{\textdegree}C}{2}[/tex]
[tex]\overline T_{w} = 50^{\textdegree}C[/tex]
The required area is cleared in the convection equation:
[tex]A_{s} = \frac{\dot Q_{in}}{h_{o}\cdot \left(\overline T_{o}-\overline T_{w}\right)}[/tex]
[tex]A_{s} = \frac{732550\,W}{\left(400\,\frac{W}{m^{2}\cdot ^{\textdegree} C} \right)\cdot (130^{\textdegree}C-50^{\textdegree}C)}[/tex]
[tex]A_{s} = 22.892\,m^{2}[/tex]
Last result is also given by this expression, whose length is cleared:
[tex]A_{s} = \pi \cdot D \cdot n_{tubes} \cdot n_{pass}\cdot L_{tube,unit}[/tex]
[tex]L_{tube,unit} = \frac{A_{s}}{\pi \cdot D \cdot n_{tubes}\cdot n_{pass}}[/tex]
[tex]L_{tube,unit} = \frac{22.892\,m^{2}}{\pi \cdot (0.025\,m)\cdot (10)\cdot (8)}[/tex]
[tex]L_{tube,unit} \approx 3.643\,m[/tex]
Total length of each tube is:
[tex]L_{tube, total} = 8 \cdot (3.643\,m)[/tex]
[tex]L_{tube,total} = 29.144\,m[/tex]
