Question

A thermodynamic cycle ABCDE is piotted on T-s diagram. The relation between T and s can be expressed as
T = ($400+100\sin (2\pi s$)) k
if the mass of fluid is 0.1 kg. What is the heat supplied in the process $A\to C$ ?

• A. 0
  
• B. 231.83 kJ
• C. 23.183
• D. 400 kJ

Answer 1

The correct option is B 231.83 kJ

Heat supplied in the process $A\to C$ is the area under $A-B-C$ on $T-s$ diagram, so
$q_{AC}=\underset{1.0}{\overset{1.5}{\int }}Tds$
$=\underset{1.0}{\overset{1.5}{\int }}(400+100\sin 2\pi s)ds$
$=\left[\right(400s{)}_0^{1.5}-{\left(\frac{100}{2\pi }\cos 2\pi s\right)}_1^{1.5}]$
$=[400\times (1.5-1)-\frac{100}{2\pi }(\cos 3\pi -\cos 2\pi \left)\right]$
For 0.1 kg of working fluid, heat supplied $=0.1\times 231.83=23.183\text{kJ}$
$=[400\times 0.5-\frac{100}{2\pi }(-1-1\left)\right]=231.83\text{kJ/kg}$