Question

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450K, 350kPa, and a velocity of 3 m/s. At the exit, the temperature is 300K and the velocity is 460 m/s. Using the ideal gas model for air with constant cp =1.011 kJ/kg.K,
determine (a) the area at the inlet, in m2, and (b) the heat transfer between the nozzle and its surroundings, in kW. Specify whether the heat transfer is to or from the air.

Answer 1

Answer:

(a) $A_1=0.283m^2$

(b) $Q=-105.5kW$: From the air to the surroundings.

Explanation:

Hello,

(a) In this case, we can compute the area at the entrance by firstly computing the inlet volumetric flow:

$V_1=\frac{mRT_1}{P_1M}= \frac{2.3\frac{kg}{s} *8.314\frac{kPa*m^3}{kmol* K}*450K}{350kPa*28.97\frac{kg}{kmol} } =0.849\frac{m^3}{s}$

Then, with the velocity, we compute the area:

$A_1=\frac{V_1}{v_1}=\frac{0.849\frac{m^3}{s} }{3\frac{m}{s} } =0.283m^2$

(b) In this case, via the following energy balance for the nozzle:

$Q-W=H_2-H_1+\frac{1}{2} mV_2^2-\frac{1}{2} mV_1^2$

We can easily compute the change in the enthalpy by using the given Cp and neglecting the work (no done work):

$\Delta H=H_2-H_1=mCp\Delta T=2.3kg/s*1.011\frac{kJ}{kg*K}*(300K-450K)\\ \\\Delta H=-348.795kW$

Finally, the heat turns out:

$Q=-348.795kW+\frac{1}{2}*2.3\frac{kg}{s}*[(460\frac{m}{s})^2 -(3\frac{m}{s})^2 ]\\\\Q=-348.795kW+243329.65W*\frac{1kW}{1000W}\\ \\Q=-105.5kW$

Such sign, means the heat is being transferred from the air to the surroundings.

Regards.