Respuesta :
Answer:
a) The flow rate of the air is 0.0104 kg/s
b) The fraction of the temperature is 23.91%
Explanation:
a) Given:
N = Number of PCBs = 8
Q = heat dissipated = 10 W
W = power supplied = -25 W
ΔT = rise temperature = 10°C
The total amount of heat dissipated is equal to:
[tex]Q_{T} =N*Q=8*10=80W[/tex]
The expression of conservation of energy is:
[tex]E_{in} =E_{out} \\Q_{T} +m_{in} h_{in} =m_{out} h_{out} +W\\m_{in}=m_{out} m_{air},(mass-balance)\\Q_{T}=m_{air}(h_{out} -h_{in})+W\\h=CpT\\Q_{T}=m_{air}Cp(T_{out} -T_{in})+W[/tex]
Replacing:
[tex]80=m_{air} *1.005x10^{3} *10+(-25)\\m_{air} =0.0104kg/s[/tex]
b) The amount of heat is equal:
[tex]Q_{fan} =m_{air} Cp*delta-T\\25=0.0104*1.005x10^{3} *delta-T\\delta-T=2.391C[/tex]
The fraction of the temperature is:
[tex]f=\frac{2.391}{10} *100=23.91[/tex]%
