Question

A heat exchanger for heating liquid mercury is under development. The exchanger is visualized as a 15cm-long and 0.3m-wide flat plate. The plate is maintained at 70 °C and the mercury flows parallel to the short side at 15 °C with a velocity of 0.3 m/s. a) Find the local drag (in N/m2 ) at the midpoint of the plate. b) Find the total drag force (in N) on the plate. c) Find the local heat transfer rate (in W/m2 ) at the midpoint of the plate. d) Find the total heat transfer rate (in W) on the plate.

Answer 1

Answer:

a) The local drag at the midpoint is 0.921 N/m²

b) The drag force is 0.059 N

c) The local heat transfer is 27158.6 W/m²

d) The total heat transfer is 17282.5 W

Explanation:

Given:

ρ = 13557.5 kg/m³

v = 0.1165x10⁻⁶m²/s

Pr = 0.026

k = 8.5675 W/m K

vα = 0.3 m/s

Tα = 15°C

a) The Reynold´s number is:

$Re=\frac{0.3*0.15}{0.1165x10^{-6} } =3.863x10^{5} ,Re$

The flow is laminar. The drag coefficient at the midpoint is:

$C_{fn} =\frac{0.664}{\sqrt{Ren} } =\frac{0.664}{\sqrt{\frac{v_{\alpha } *0.075}{v} } } =\frac{0.664}{\sqrt{\frac{0.3*0.075}{0.1165x10^{-6} } } } =0.00151$

The local drag at the midpoint is equal:

$\tau =\frac{C_{fn}PV_{\alpha }^{2} }{2} =\frac{0.0015*13557.5*0.3^{2} }{2} =0.921N/m^{2}$

b) The average coefficient for the plate is:

$C_{D} =2*\frac{0.664}{\sqrt{ReL} } =2*\frac{0.664}{\sqrt{\frac{0.3*0.15}{0.1165x10^{-6} } } } =0.00214$

The total drag force is:

$F_{D} =\frac{C_{D}Pv_{\alpha }^{2}A }{2} =\frac{0.0024*13557.5*0.3^{2}*0.15*0.3 }{2} =0.059N$

c) The equation:

$Nv_{n} =\frac{h-n}{k} =0.332Re^{1/2} Pr^{1/3}$

at midpoint n = 0.075 m

$\frac{h-0.075}{8.5675} =0.332\sqrt{\frac{0.3*0.075}{0.1165x10^{-6} } } *(0.026)^{1/3} \\h=4937.6W/m^{2} K$

The local heat transfer is:

$Q=h(T_{plate} -T_{\alpha } )=4937.6(70-15)=27158.6W/m^{2}$

d) The average heat coefficient is:

$h_{v} =2hL=0.15$

The average Nussele:

$\frac{h*0.5}{8.5675} =0.644(\frac{0.3*0.15}{0.1165x10^{-6} } )^{1/2} *(0.016)^{1/3} \\h=6982.8W/m^{2} K$

$Q_{T} =Ah(T_{plate} -T_{\alpha } )=0.15*0.3*3982.8*(70-5)=17282.5W$