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slava [35]
4 years ago
13

The heat transfer coefficient for hydrogen flowing over a sphere is to be determined by observing the temperature–time history o

f 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?
Physics
1 answer:
mr_godi [17]4 years ago
8 0

Answer:

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.

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v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}

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What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0
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