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## Q. 2.36

The properties at state 1 of a reversible non-flow isobaric process are: $p_1$ = 1.5 bar, $ν_1$ = 0.3 m³/kg, $t_1$ = 20°C and at state 2 are: $t_2$ = 250°C, $ν_2$ = 0.60 m³/kg. The specific heat of the fluid is given by:

$c_p=\left[1.5+\frac{80}{t+50}\right] \mathrm{kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$

Where t is in °C.

Calculate (a) heat added, (b) work done, (c) change in internal energy, and (d) change in enthalpy per kg of fluid.

## Verified Solution

(a)

\begin{aligned}Q &=\int_{t_1}^{t_2} c_p d t=\int_{20}^{250}\left[1.5+\frac{80}{t+50}\right] d t \\&=|1.5 t+80 \ln (t+50)|_{20}^{250}=1.5(250-20)+80 \ln \frac{300}{70} \\&=461.423 \mathrm{~kJ} / \mathrm{kg}\end{aligned}

(b) $\quad w=\int_{v_1}^{v_2} p \cdot d V=p\left(v_2-v_1\right)=1.5 \times 10^2(0.60-0.30)=45 \mathrm{~kJ} / \mathrm{kg}$

(c)        dU = Q – W = 461.423 – 45 = 416.423 kJ/kg