Answer to Question #279973 in Mechanical Engineering for Kiran

Question #279973

Find the values of Cp and ᵞ for a gas whose gas



constant is 0.324 kJ/kg K and Cv = 0 84 kJ/kg K.



If 2.25 kg of this gas with an initial volume of



1.15 m3 undergoes a process during which its



pressure remains constant at 7 bar and its



temperature becomes 280‘C at the end of the



process, calculate: (a) the change in internal



energy, (b) the change in enthalpy, (c) the



transferred heat, and (d) the work done.

1
Expert's answer
2021-12-24T10:25:02-0500

Solution;

Given;

R=0.324kJ/kgKR=0.324kJ/kgK

Cv=0.84kJ/kgKC_v=0.84kJ/kgK

m=2.25kgm=2.25kg

V1=1.15m3V_1=1.15m^3

P1=P2=7barP_1=P_2=7bar

T2=28°C=301KT_2=28°C=301K

Value of CpC_p can be calculated as;

Cp=Cv+RC_p=C_v+R

Cp=0.84+0.324=1.164kJ/kgKC_p=0.84+0.324=1.164kJ/kgK

Value for γ\gamma can be calculated as;

γ=CpCv=1.1640.84=1.386\gamma=\frac{C_p}{C_v}=\frac{1.164}{0.84}=1.386

(a)change in internal energy;

dU=mCv(T1T2)dU=mC_v(T_1-T_2)

From the relation;

PV=mRTPV=mRT

V2=2.25×324×3017×105=0.3134m3V_2=\frac{2.25×324×301}{7×10^5}=0.3134m^3

For ideal gases;

V1T1=V2T2\frac{V_1}{T_1}=\frac{V_2}{T_2}

T1=1.150.3134×301=1104.5KT_1=\frac{1.15}{0.3134}×301=1104.5K

dU=2.25×840×(1104.5301)dU=2.25×840×(1104.5-301)

dU=1518.615kJdU=1518.615kJ

(b)change in enthalpy;

dH=mCp(T1T2)dH=mC_p(T_1-T_2)

dH=2.25×1164×(1104.5301)dH=2.25×1164×(1104.5-301)

dH=2104.366kJdH=2104.366kJ

(c)Heat transferred;

For isobaric process;

dH=dQ=2104.366kJdH=dQ=2104.366kJ

(d)Work done;

W=pΔVW=p\Delta V

W=7×105(1.150.3134)W=7×10^5(1.15-0.3134)

W=585.620kJW=585.620kJ


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment