A computer cooled by a fan contains eight printed circuitboards or \"PCB\'s\". E

Question

A computer cooled by a fan contains eight printed circuitboards or "PCB's". Each PCB dissipates 10 watts of power andmeasures 12 cm by 18 cm. The cooling air is supplied by a 25 wattfan mounted at the air inlet. The temperature rise of air as itflows through the case of the computer is not to exceed 10 degreesC. Determine the flow rate of air that the fan needs to deliver inkg per second, and determine the percentage of the temperature riseof the air that is due to the heat generated by the fan and it'smotor. Please show all steps. This is a lifesaver problem. Thank you. A computer cooled by a fan contains eight printed circuitboards or "PCB's". Each PCB dissipates 10 watts of power andmeasures 12 cm by 18 cm. The cooling air is supplied by a 25 wattfan mounted at the air inlet. The temperature rise of air as itflows through the case of the computer is not to exceed 10 degreesC. Determine the flow rate of air that the fan needs to deliver inkg per second, and determine the percentage of the temperature riseof the air that is due to the heat generated by the fan and it'smotor. Please show all steps. This is a lifesaver problem. Thank you.

Explanation / Answer

Write the energy balance for the system: EIN - EOUT = ESYS =0 Therefore EIN = EOUT QIN + WIN + mh1 =mh2 (since ke = pe = 0) QIN + WIN = mcp(T2- T1) The fan power is 25W and the 8 PCBs transfer a total of 80W ofheat to air, the mass flow is then: QIN + WIN = mcp(Te- Ti) rearrange equation for m                   =[(8*10)W +25W]/ (1005 J/kgC)(10C) = 0.0104kg/s b. The fraction of temp. rise from heat generated by the fanand motor can be determined by: Q = mcpT T = Q/mcp       = 25W/(0.0104kg/s)(1005J/kgC)       = 2.4C f = 2.4C/10C = 24% Don't forget to rate!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.