Respuesta :
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:
[tex]Re=\frac{0.3*0.15}{0.1165x10^{-6} } =3.863x10^{5} ,Re<5x10^{5}[/tex]
The flow is laminar. The drag coefficient at the midpoint is:
[tex]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[/tex]
The local drag at the midpoint is equal:
[tex]\tau =\frac{C_{fn}PV_{\alpha }^{2} }{2} =\frac{0.0015*13557.5*0.3^{2} }{2} =0.921N/m^{2}[/tex]
b) The average coefficient for the plate is:
[tex]C_{D} =2*\frac{0.664}{\sqrt{ReL} } =2*\frac{0.664}{\sqrt{\frac{0.3*0.15}{0.1165x10^{-6} } } } =0.00214[/tex]
The total drag force is:
[tex]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[/tex]
c) The equation:
[tex]Nv_{n} =\frac{h-n}{k} =0.332Re^{1/2} Pr^{1/3}[/tex]
at midpoint n = 0.075 m
[tex]\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[/tex]
The local heat transfer is:
[tex]Q=h(T_{plate} -T_{\alpha } )=4937.6(70-15)=27158.6W/m^{2}[/tex]
d) The average heat coefficient is:
[tex]h_{v} =2hL=0.15\\[/tex]
The average Nussele:
[tex]\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[/tex]
[tex]Q_{T} =Ah(T_{plate} -T_{\alpha } )=0.15*0.3*3982.8*(70-5)=17282.5W[/tex]
