The heat transfer coefficient for hydrogen flowing over a sphere is to be determined by observing the temperature–time history of a sphere fabricated from pure copper. The sphere, which is 20.0 mm in diameter, is at 70°C before it is inserted into the gas stream having a temperature of 27°C. A thermocouple on the outer surface of the sphere indicates 50°C 97 s after the sphere is inserted into the hydrogen. Find

a) What is the value of the specific heat of the copper at the average temperature of the process, in J/kg·K?

b) What is the value of the lumped thermal capacitance of the sphere, in J/K?

c) Solve for the thermal time constant, in sec.

d) What is the value of the heat transfer coefficient, in W/m2· K?

e) What is the value of the Biot number?

h = 74.7177 W/m^2 K    

Explanation:

Given:-

- The diameter of the sphere, Ds = 20.0 mm

- The initial Temperature of sphere, Ti = 70°C

- The temperature of gas stream, Ts = 27°C

- The thermocouple reading after 97s, Tr = 50°C

Solution:-

- We will use Table A-1, to extract the following properties of copper at Ti = 70°C ( 343 K ):

                         Density ρ = 8933 kg/m^3

                         Specific Heat capacity cp = 389 J / kg.K

                         Thermal conductivity, k = 388 W/m.K

- The time-temperature history is given by the equations:

                        θ(t) / θi = exp ( - t / Rt*Ct )

Where,

                        Rt = 1 / h*As , Ct = ρ*V*cp , As = πDs^2 / 4 ,  θ = T - T∞

                        V =  πDs^3 / 6.

- We will use the above relationships and given data to calculate:

                       θ(t) = T( at 97s) - T∞ = Tr - Ts

                             = 50 - 27 = 23°C

                        θi = Ti - T∞ = Ti - Ts

                            = 70 - 27 = 43°C

- Then we have:

                       θ(t) / θi = 23 / 43 = 0.53488372

                       θ(t) / θi = exp ( - t / τ )

                       0.53488372 = exp ( - t / τ )

Solve for τ:

                       τ = - 97 / Ln ( 0.53488372)

                       τ = 155.025 s

- Then solve for h:

                       τ = ρ*V*cp / h*As

                       h = ρ*V*cp / τ*As

                       h = ( 8933 )*(0.02/6)*(389) / ( 155.025)

                       h = 74.7177 W/m^2 K    

- Verifying the use of spatial isothermal assumption:

                      Lc = Ds / 6 = 20 / 6 = 0.003333

                      Bi = h*Lc / k = (74.7177)*(0.00333) / 388 = 0.00064

Hence, Bi < 0.1 so, spatial isothermal assumption is valid.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.