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

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