Answer:
T2 ( final temperature ) = 576.9 K
a) 853.4 kJ/kg
b) 1422.3 kJ / kg
Explanation:
given data :
pressure ( P1 ) = 90 kPa
Temperature ( T1 ) = 30°c + 273 = 303 k
P2 = 450 kPa
Determine final temperature for an Isentropic process
[tex]T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}[/tex] ----------- ( 1 )
T2 = 303 [tex]( \frac{450}{90})^{(1.667- 1)/1.667}[/tex] = 576.9K
Work done in a piston-cylinder device can be calculated using this formula
[tex]w_{in} = c_{v} ( T2 - T1 )[/tex] ------- ( 2 )
where : cv = 3.1156 kJ/kg.k for helium gas
T2 = 576.9K , T1 = 303 K
substitute given values Back to equation 2
[tex]w_{in}[/tex] = 853.4 kJ/kg
work done in a steady flow compressor can be calculated using this
[tex]w_{in} = c_{p} ( T2 - T1 )[/tex]
where : cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K
T2 = 576.9 k , T1 = 303 K
substitute values back to equation 3
[tex]w_{in}[/tex] = 1422.3 kJ / kg
1. The final temperature for this reversible, adiabatic process is 576.82 Kelvin.
2a. Assuming the process took place in a piston-cylinder device, the work done is equal to 853.11 kJ/kg.
2b. Assuming the process took place in a steady-flow compressor, the work done is equal to 1422.82 kJ/kg.
Given the following data:
Constant, k = 1.667
Constant pressure for helium gas, [tex]C_p[/tex] = 5.1926 kJ/kg.K
Constant volume for helium gas, [tex]C_v[/tex] = 3.1156 kJ/kg.k
First of all, we would determine the final temperature for this isentropic process.
For an isentropic process, the final temperature is given by the formula:
[tex]T_2 = T_1 (\frac{P_2}{P_1})^{\frac{k-1}{k}}\\\\T_2 = 303 \times (\frac{450}{90})^{\frac{1.667-1}{1.667}}\\\\T_2 = 303 \times (5)^{\frac{0.667}{1.667}}\\\\T_2 = 303 \times 5^{0.40}\\\\T_2 = 303 \times 1.9037\\\\T_2 = 576.82 \;K[/tex]
a. In a piston-cylinder device, work done is given by the formula:
[tex]W = C_v(T_2 -T_1)\\\\W = 3.1156(576.82 -303)\\\\W = 3.1156 \times 273.82\\\\W = 853.11[/tex]
Work done, W = 853.11 kJ/kg
b. In a steady-flow compressor, work done is given by the formula:
[tex]W = C_p(T_2 -T_1)\\\\W = 5.1926(576.82 -303)\\\\W = 5.1926 \times 273.82\\\\W = 1422.82[/tex]
Work done, W = 1422.82 kJ/kg
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