Question #273544

Say & lb/sec of CO₂ gas are com .2 H pressed polytropically (pV¹² = C) from Pi =. 15 psia, ₁ 140°F to 1₂ = 440°F. As suming ideal gas action, find P2, W, Q, AS (a) as a nonflow process, (b) as a steady flow process where AP = 0, AK = 0.

1
Expert's answer
2021-11-30T17:44:43-0500

Solution;

Given;

P1=15psia=103.421kPaP_1=15psia=103.421kPa

T1=140°F=333.15KT_1=140°F=333.15K

T2=440°F=499.82KT_2=440°F=499.82K

m˙=1lb/s=0.4536kg/s\dot{m}=1lb/s=0.4536kg/s

For carbon dioxide;

R=0.1889kJ/kgKR=0.1889kJ/kgK

P1P2=(T1T2)nn1\frac{P_1}{P_2}=(\frac{T_1}{T_2})^{\frac{n}{n-1}}

103.421P2=(333.15499.82)1.20.2\frac{103.421}{P_2}=(\frac{333.15}{499.82})^{\frac{1.2}{0.2}}

P2=1179.09kPaP_2=1179.09kPa

Work;

W=mR(T2T1)n1W=\frac{mR(T_2-T_1)}{n-1}

W=0.4536×0.1889(499.82333.15)1.21W=\frac{0.4536×0.1889(499.82-333.15)}{1.2-1}

W=71.41kWW=71.41kW

Heat;

Q=W+ΔuQ=W+\Delta u

Δu=mcv(T2T1)\Delta u=mc_v(T_2-T_1)

0.2cv=R0.2c_v=R

cv=0.9445kJ/kgKc_v=0.9445kJ/kgK

Δu=0.4536×0.9445(499.82333.15)=71.41kW\Delta u=0.4536×0.9445(499.82-333.15)=71.41kW

Q=W+Δu=142.82kWQ=W+\Delta u=142.82kW






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