Question #122897
The specific heat at constant pressure of 1kg fluid undergoing a MF fluid constant pressure process is given by
Cp(2.5+40÷T+20)kg/kg⁰C
Where T is in ⁰C.
The pressure during the process is maintained at 2 bar and volume changes from 1m³ to 1.8m³ and temperature changes from 50⁰C to 450⁰C. Determine
(i)Heat added
(ii)Work done
(iii)Change in internal Energy
(iv)Change in Enthalpy
1
Expert's answer
2020-06-21T20:13:40-0400


(i) Heat added 


Q=mCPdtQ = m C_Pdt


Q=m×Cp(2.5+40÷T+20)×dtQ=m \times Cp(2.5+40÷T+20)\times dt


Q=1×4.187(2.5+40÷400+20)×400=37.85kJQ= 1 \times 4.187(2.5+40÷400+20)\times 400=37.85kJ


(ii) Work done  


W=p(V2V1)=200000(1.81)=160kJW = p (V_2-V_1)=200000(1.8-1)=160 kJ


(iii) Change in internal Energy:


U=WQ=16037.85=122.15kJU = W - Q =160-37.85 =122.15 kJ



(iv) Change in Enthalpy:


dH=dQ=Q=37.85kJdH = dQ =Q = 37.85 kJ






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