A cylindrical container is 150mm in diameter and weights 2.25n when empty

Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?

bonufazy [111]

Answer:

Free convection:

When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.

Force convection:

When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.

Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.

We know that convective heat transfer given as

q = h A ΔT

h=Heat transfer coefficient

A= Surface area

ΔT = Temperature difference

5 0

2 years ago

A flat plate 1.5 m long and 1.0 m wide is towed in water at 20 o C in the direction of its length at a speed of 15 cm/s. Determi

beks73 [17]

Answer:

15.8

0.0944

Explanation:

L = 1.5

B = 1.0

Speed of water = 15cm

Temperature = 20⁰C

At 20⁰C

Specific weight = 9790

Kinematic viscosity v = 1.00x10^-4m²/s

Dynamic viscosity u = 1.00x10^-3

Density p = 998kg/m²

Reynolds number

= 0.15x1.5/1.00x10^-4

= 225000

S = 5

5x1.5/225000^1/2

= 0.0158

= 15.8mm

Resistance on one side of plate

F = 0.664x1x1.0x10^-3x0.15x225000^1/2

= 0.04724N

Total resistance

= 2N

= 2x0.04724

= 0.0944N

3 0

2 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$