## Chapter 2

## 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.

## Step-by-Step

## 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