Answer to Question #309314 in Mechanical Engineering for Engineering

Question

The specific heat at constant pressure of 1kg fluid undergoing a non –flow constant pressure

process is given by:

𝐶𝑝 = [2.5 +

40

𝑇+20]kJ/kg-℃ where T is in degrees centigrade. The pressure during the process

is maintained at 2 bar and volume changes from 1m3

to 1.8 m3

and temperature changes from

50℃ to 450℃. Determine (a) heat added (b) work done (c) change in internal energy (d) change

in enthalpy

Answer 1

Solution;

(a)Heat added;

$Q=\int_{T_1}^{T_2}C_pdT$

$Q=\int_{50}^{450}(2.5-\frac {40}{T+20})dT$

$Q=[2.5T+40ln(T+20)]_{50}^{450}$

$Q=1076.17kJ/kg$

(b)Work done;

$W=\int_{V_1}^{V_2}pdV=p(V_2-V_1)$

$W=2(1.8-1)×10^5$

$W=160kJ/kg$

(c) Change in internal energy;

$\Delta u =Q-W=1076.17-160=916.17kJ/kg$

(d)Change in enthalpy;

$\Delta h=Q=1076.17kJ/kg$