equation. If these two assumes are not true then you lus +KE Hw#9 Fall Thermo.do

Question

equation. If these two assumes are not true then you lus +KE Hw#9 Fall Thermo.docx- We now have enough skills to solve the problem introduced the first week of the semester. We can size the fan needed to cool the Plant Growth Unit designed to operate on the space shuttle (See pgba WebFlier" uploaded into Blackboard). The following is the relevant information 1.260 watts of total system electrical power input 2. Cooling air inlet temperature 25°C . Cooling air outlet temperature 40°C 4.4” x 18" cooling duct cross section s. Heat exchanger causes a "back pressure" of 5 inches- of-water (1.24 kPa gauge) Use the following steps to determine the design specification for the cooling fan (flow rating and power consumption Assume the cooling air is an ideal gas: a) First consider the entire Plant Growth Unit as a WELIL INSULATED open system. Apply the open system energy balance to determine the required volumetric air flow rate (in Liters/min) at the cooling inlet conditions. Use an average air Cp from Table A.5, page 760 b) With the solution to part "a," now determine the minimum power rating required for the cooling fan (in watts). For this step, take as your system only the cabin air outside of the air inlet, the cooling air inlet, cooling duct, and fan. This smaller system, which excludes the heat exchanger, may be solved with the mechanical energy balance equation with the P equal to the heat exchanger's "back pressure" and the (velocity) determined by the cooling duct's cross- sectional area. Here you will assume that the change in pressure causes a negligible change in the air density

Explanation / Answer

given

electrical power input P = 260 W

inlet temp, Ti = 25 C = 298.15 K

outlet temp, To = 40 C = 313.15 K

cross section, A = 4"*10" = 72 in^2 = 0.0464515 m^2

backpressure, p = 1.24*10^3 Pa

a. for insulated system

let volumetric flow rate of air = V, mass flow rate = m'

then heat flow per unit time intor air = m'*cp*(To - Ti)

now, m' = V*rho ( rho is density of air = 1.225 kg/m^3)

cp is specific heat of air = 1000 J/kg

so, V*rho*cp(To - Ti) = P

V*1.225*1000*(313.15 - 298.15) = 260

V = 0.0141 m^3/s = 14.14965 lit/s = 848.97959 lit/min

b. for the system excluding the heat exchanger

as the speed of the flow does not change ( cross section remains constant)

let power of the fan be P'

then

Power is used to counter the back pressure

P' = p*V

P' = 1.24*10^3*0.0141 = 17.484 W

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