Question

An electric resistance heater is embedded in a long cylinder of diameter 30 mm. When water with a temperature of 25 oC and velocity of 1 m/s flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of 90 oC is 28 kW/m. When air, also at 25 oC, but with a velocity of 10 m/s is flowing, the power per unit length required to maintain the same surface temperature is 400 W/m.

Calculate and compare the convection coefficients for the flows of water and air.

Answer 1

`h_(w)` = 4.570 w /m². k

`h_(a)` = 65 w /m². k

Step-by-step explanation:

Flow is cross-wise over cylinder which is very long in the direction normal to flow.

Therefore, from analysis, the convection heat rate from the cylinder per unit length of the cylinder has the form.

Note: the air velocity is 10 times that of the water flow, yet
`h_(w)` = 70 k
`h_(a)`

These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases

Answer 2

Water flow,
`h_w` = 4572.9 w/m².k

Air flow,
`h_a` = 65.32 w/m²k

Flow of water is 70 times faster than flow of air, which means that the from of water was forced.

Step-by-step explanation:

Given:

q' = 28 kw/m = 28 x 10³ w/m

D = 30mm = 30 x 10⁻³ m

T∞ (Air) = 25°C

Ts = 90°C

Step 1: The convection heat rate from the cylinder per unit length of the cylinder has the form

q' = h(πD)(Ts - T∞)

h = heat transfer convection coefficient

h = q' ÷ [πD(Ts - T∞)]

⇒ Substitute for the given values

`h_w` = 28×10³ w/m ÷ [π × 0.030 m (90 - 25)°C]

`h_w` = 4572.9 w/m².k

Also for air
`h_a` ,
`q'_a` = 400 w/m

`h_a` = 400 ÷ [π × 0.030 m (90 - 25)°C]

`h_a` = 65.32 w/m²k

Note that the air velocity is 10 times the water flow, but
`h_w` is 70 times
`h_a`. This indicate that the values for convection coefficient are typical for forced convection heat transfer with liquid and gases.