Question

Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters at a temperature and exits at 284 C. The pipe is smooth and its length is 10 m. temperature is 25 ° C. Since the smear coefficient of the pipe surface is given as 0.8; a-) Indoor and outdoor convection coefficients (W / m2K), b-) Heat loss from the pipe to the environment (W), c-) The surface temperature of the pipe (C), d-) Calculate the required fan control (W) and interpret the results.

Answer 1

a)
`h_c = 0.1599 W/m^2-K`

b)
`H_(loss) = 5.02 W`

c)
`T_s = 302 K`

d)
`\dot{Q} = 25.125 W`

Step-by-step explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature,
`T_a = 25^0 C = 298 K`

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

`\dot{ m} = 4.5 tons/hr\\\dot{m} = (4.5*1000)/(3600) = 1.25 kg/sec`

Rate of heat transfer,

`\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W`

a) To calculate the convection coefficient relationship for heat transfer by convection:

`\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K`

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe,
`k_c = 0.8`

`\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K`

b) Heat loss from the pipe to the environment:

`H_(loss) = h_c A(T_s - T_a)\\H_(loss) = 0.1599 * 7.855( 302 - 298)\\H_(loss) = 5.02 W`

d) The required fan control power is 25.125 W as calculated earlier above