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

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

Step-by-step 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=(0.3*0.15)/(0.1165x10^(-6) ) =3.863x10^(5) ,Re<5x10^(5)`

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

`C_(fn) =(0.664)/(√(Ren) ) =\frac{0.664}{\sqrt{(v_(\alpha ) *0.075)/(v) } } =\frac{0.664}{\sqrt{(0.3*0.075)/(0.1165x10^(-6) ) } } =0.00151`

The local drag at the midpoint is equal:

`\tau =(C_(fn)PV_(\alpha )^(2) )/(2) =(0.0015*13557.5*0.3^(2) )/(2) =0.921N/m^(2)`

b) The average coefficient for the plate is:

`C_(D) =2*(0.664)/(√(ReL) ) =2*\frac{0.664}{\sqrt{(0.3*0.15)/(0.1165x10^(-6) ) } } =0.00214`

The total drag force is:

`F_(D) =(C_(D)Pv_(\alpha )^(2)A )/(2) =(0.0024*13557.5*0.3^(2)*0.15*0.3 )/(2) =0.059N`

c) The equation:

`Nv_(n) =(h-n)/(k) =0.332Re^(1/2) Pr^(1/3)`

at midpoint n = 0.075 m

`(h-0.075)/(8.5675) =0.332\sqrt{(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:

`(h*0.5)/(8.5675) =0.644((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`