An electric resistance heater is embedded in a long cylinder of diameter 30 mm. when water with a temperature of 25 c 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 c is 28 kw/m.

What is known: Long, 30mm diameter cylinder with entrenched electrical heater; power to maintain a specified surface temperature for water and air flows.

What to look for: Convection coefficients for the water and air flow convection processes, h sub w and h sub a, correspondingly.

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

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

q’ =h (πD) (Tsub s – T sub infinite)

and solving for the heat transfer convection coefficient is

h = q’ / (πD) (Tsub s – T sub infinite)

Solution: substitute the numerical values for the water and air:

Water

h sub W = 2.8 x 10^3 W/m / π x 0.030m (90-25) degress Celsius

= 4,570 W/m^2 K <

Air

h sub a = 400W/m/ π x 0.030m (90-25) = 65 W/m^2 K <

Other info: the air velocity is 10 times that of the water flow, yet h sub w = 70 x h sub a

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .