3. Liquid nitrogen at 90 K, 400 kPa flows into a probe used in a cryogenic survey. In the return line the nitrogen is then at 160 K, 400 kPa. Find the specific heat transfer to the nitrogen. If the return line has a cross-sectional area 100 times larger than that of the inlet line, what is the ratio of the return velocity to the inlet velocity?

Specific heat transfer = 236.16 kJ/kg

Ratio of return velocity to inlet velocity = 0.80

Explanation:

Given

Temperature of liquid nitrogen, T1 = 90 K

Pressure of liquid nitrogen, P1 = 400 kPa

Temperature of nitrogen, T2 = 160 K

Pressure of nitrogen, T2 = 400 kPa

A (e) = 100 A (i)

To solve, we use the formula

h (i) + 1/2v (i) ² + q = h (e) + 1/2v (e) ² + q

The mass flow is

m = m (i) = m (e)

m = (Av/V) i = (Av/V) e

Ratio of return velocity to inlet velocity is

v (e) / v (i) = A (i) / A (e) * V (e) / V (i)

v (e) / v (i) = 1/100 * V (e) / V (i)

From the saturated Nitrogen table, at 100 K, we have

h (i) = h (f) = - 73.2

v (i) = v (f) = 0.001452

From the saturated Nitrogen table again, at 160 K and 400 kPa

h (e) = 162.96 kJ/kg

v (e) = 0.11647 m³/kg

Substituting these in the formula, we have

v (e) / v (i) = 1/100 * 0.11647/0.001452

v (e) / v (i) = 1/100 * 80.2

v (e) / v (i) = 0.80

Energy equation is given by

q + h (i) = h (e)

q = h (e) - h (i)

Now, calculating specific heat transfer

q = 162.96 - - 73.2

q = 236.16 kJ/kg