Question #204301

A steam enters a boiler at 3.42 bar, specific volume of 0.567 m3/kg, specific internal energy of

1892kJ/kg at velocity of 5 m/s. At rate 3 kg/s, th steam exit with 755 kN/m2, specific volume

of 0.892 m3/kg,specific internal energy of 2014 kJ/kg and velocity of 35 m/s. The boiler's exit

is 2m higher than the entrance. Calculate the heat transfer in kW and state either it is into or

out of the system.


1
Expert's answer
2021-06-07T19:16:14-0400

P=mtgFS1ρ2Q2m2v2hp1ρ1Q1m1v1=P=\frac{\frac mt\cdot g\cdot \frac FS\cdot \frac{1}{\rho_2}\cdot \frac{Q_2}{m_2}\cdot v_2\cdot h}{p\cdot \frac{1}{\rho _1}\cdot \frac{Q_1}{m_1}\cdot v_1}=

=39.87551030.89220141033523.421050.56718921035=1.52 kW=\frac{3\cdot 9.8\cdot 755\cdot 10^3\cdot 0.892\cdot 2014\cdot 10^3\cdot 35\cdot 2}{3.42\cdot 10^5\cdot 0.567\cdot 1892\cdot 10^3\cdot 5}=1.52~\text{kW} out of the system.


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