An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches
are SL = ST = 24 mm. There are 3 rows of elements in the flow direction (NL = 3) and 4 elements per row (NT = 4). Atmospheric air with an upstream velocity of 12 m/s and a temperature of 25°C moves in cross-flow over the elements, which have a diameter of 12 mm, a length of 250 mm, and are maintained at a surface temperature of 350°C.
(A) Determine the total heat transfer to the air and the temperature of the air leaving the duct heater.
(B) Determine the pressure drop across the element bank and the fan power requirement.
(C) Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results.
(A) Determine the total heat transfer to the air and the temperature of the air leaving the duct heater.
(B) Determine the pressure drop across the element bank and the fan power requirement.
(C) Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results.
1 answer:
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Answer:
isobaric expansion = 281.09 Btu
isothermal compression= 72 Btu
Explanation:
The first law of thermodynamics is:
where:
Q=heat transferred
W= work
U=internal energy
P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.
In a isobaric process heat transferred is:
For an isothermal process (T2-T1 = 0) so
From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.
For the first process
we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)
we use this information in the appropriate unit to find the pressure in thermodynamic tables.
T1= 176°C
v1= 1.49 m^3/kg
P=1.37 bar
in the second state we have
P=1.37 bar =137000Pa
with thee properties we check in the thermodynamic tables
T2= 255°C
we usually find Cp on tables for water but from the Mayer relation we have:
Cp for water vapor is: 33.12 J/mol*K
R=8.314 J/mol*K
Cv= 41.434 J/mol*K
replacing in the equation for Q
296569J =281.09 Btu
Given:
outer radius, R' = 10 m
inner diameter, d = 2 m
inner radius, R = = 1 m
surface temperature, T' =
Thermal conductivity of soil, K = 0.52 W/mK
Solution:
To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :
Q =
500 =
T = 68.86 +20 =
Therefore, outside surface temperature is