Question

Water enters the tubes of a cold plate at 70°F with an average velocity of 40 ft/min and leaves at 105°F. The diameter of the tubes is 0.25 in. Assuming 14 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation and the remaining 86 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. The properties of water at room temperature are rho = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm·°F.

Answer 1

The total amount of heat generated;Q' = 2067 Btu/h

Step-by-step explanation:

We are given;

Water entering temperature;T1 = 70°F

Water leaving temperature;T2 = 105°F

average velocity of water;V = 40 ft/min

Diameter of tube;D = 0.25 in = 0.25/12 ft = 0.02083 ft

Water density;ρ = 62.1 lbm/ft³

cp = 1.00 Btu/lbm·°F.

Now, the mass flow rate of the water is calculated from;

m' = ρAV

Where ρ is density, A is area and V is velocity

Area = πD²/4 = π*0.02083²/4 = 0.00034077555 ft²

m' = 62.1 * 0.00034077555 * 40

m' = 0.8465 lbm/min

Converting to lbm/hr = 0.8465 * 60 = 50.79 lbm/hr

From energy balance equation, we have;

E_in = E_out

So,

Q_in,w + m'h1 = m'h2

Q_in,w = m'h2 - m'h1

Q_in,w = m'(h2 - h1)

Now, m'(h2 - h1) can be written as;

m'cp(T2 - T1).

Thus ;

Q_in,w = m'cp(T2 - T1)

Plugging in the relevant values, we have;

Q_in,w = (50.79*1)(105 - 70)

Q_in,w = 1777.65 Btu/h

We are told that remaining 86 percent of heat generaged is removed by the cooling water. Thus;

The total amount of heat generated could be defined as;

Q' = Q_in,w/0.86

Q' = 1777.65/0.86

Q' = 2067 Btu/h