Question

After an initial test run John determines that his cooling system generates 45 W of heat loss. Calculate the amount of heat loss (H2), in W, that Mike expects his pump to do if its fan speed were 3.5 times greater and the coolant density was 9.5 times smaller.

Answer 1

Given:

heat generated by John's cooling system,  $H = \rho A v^{3}$  = 45 W    (1)

If ρ, A, and v corresponds to John's cooling system then let $\rho_{1}, A_{1}, v_{1}$ be the variables for Mike's system then:

$\rho  = 9.5\rho_{1}$

$\rho_{1}  = \frac{\rho}{9.5}$

$v_{1} =3.5 v$

Formula use:

Heat generated, $H = \rho A v^{3}$

where,

$\rho$ = density

A = area

v = velocity

Solution:

for Mike's cooling system:

$H_{2}$ = $v_{1}^{3}{1}A_{1}\rho_{1}$

⇒ $H_{2}$ = $(3.5v)^{3}$ × A × $\frac{\rho}{9.5}$

$H_{2}$ = 4.513$v^{3}$ A  $\rho$

Using eqn (1) in the above eqn, we get:

$H_{2}$ = 4.513 × 45 = 203.09 W

Answer 2

We have that The Heat loss H2 is given as

• H2=1347.5w

From the question we are told

• After an initial test run John determines that his cooling system generates 45 W of heat loss.
• Calculate the amount of heat loss (H2), in W, that Mike expects his pump to do if its fan speed were 3.5 times greater and the coolant density was 9.5 times smaller.

heat loss (H2)

Generally the equation for the   is mathematically given as

H=PAV^3

Therefore

$\frac{110}{h_2}=\frac{P_1}{p1/3.5}*(\frac{V_1}{3.5*v_1})^3$

H_2=1347.5w

Therefore

The Heat loss H2 is given as

• H2=1347.5w

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