Given Information:
Velocity of water flow = 2 m/s
Temperature of water = 17° C
Heat dissipation = 2500 W
Area of copper plate = 0.04 m²
Required Information:
Temperature of copper plate = ?
Answer:
° ![C](https://tex.z-dn.net/?f=C)
Explanation:
Each chip dissipates 25 W so 100 chips will dissipate 25*100 = 2500 W
Area of copper plate = 0.2*0.2 = 0.04 m²
According to the convection rate equation
![T_{p}= T_{w} + \frac{q}{hA}](https://tex.z-dn.net/?f=T_%7Bp%7D%3D%20T_%7Bw%7D%20%2B%20%5Cfrac%7Bq%7D%7BhA%7D)
Where Tp is the temperature of copper plate, Tw is the temperature of water, q is the the heat dissipation of chips, A is the area of copper plate and h is the convection coefficient
The convection coefficient is given by turbulent flow correlation
![h = Nu_{L}(k/L) =0.037Re_{L}^{4/5}P_{r}^{1/3}(k/L)](https://tex.z-dn.net/?f=h%20%3D%20Nu_%7BL%7D%28k%2FL%29%20%3D0.037Re_%7BL%7D%5E%7B4%2F5%7DP_%7Br%7D%5E%7B1%2F3%7D%28k%2FL%29)
Where Nu is Nusselt number, Re is Reynolds number, Pr = 5.2 is Prandtl number and k = 0.620 W/m.K
![Re_{L}= uL/v](https://tex.z-dn.net/?f=Re_%7BL%7D%3D%20uL%2Fv)
Where u = 2 m/s and L = 0.2 m and v = 0.96x10⁻⁶m²
/s
![Re_{L}= 2*0.2/0.96x10^{-6}](https://tex.z-dn.net/?f=Re_%7BL%7D%3D%202%2A0.2%2F0.96x10%5E%7B-6%7D)
![Re_{L}= 416666.66](https://tex.z-dn.net/?f=Re_%7BL%7D%3D%20416666.66)
![h = 0.037(416666.66)^{4/5}(5.2)^{1/3}(0.620/0.2)](https://tex.z-dn.net/?f=h%20%3D%200.037%28416666.66%29%5E%7B4%2F5%7D%285.2%29%5E%7B1%2F3%7D%280.620%2F0.2%29)
![W/m^{2}K](https://tex.z-dn.net/?f=W%2Fm%5E%7B2%7DK)
![T_{p}= 17 + \frac{2500}{(6223.89)0.04}](https://tex.z-dn.net/?f=T_%7Bp%7D%3D%2017%20%2B%20%5Cfrac%7B2500%7D%7B%286223.89%290.04%7D)
° ![C](https://tex.z-dn.net/?f=C)
Therefore, the temperature of the copper plate is 27° C