Question

Figure P2.70 shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm² is attached to the top of the piston. The total mass of the piston and shaft is 25 kg. while the gas is slowly heated, the internal energy of the gas increases by 0.1 kJ, the potential energy of the piston-shaft combination increases by 0.2 kJ, and a force of 1334 N is exerted on the shaft as shown in the figure. The piston and cylinder are poor conductors, and friction between them is negligible. The local atmospheric pressure is 1 bar and g = 9.81 m/s². Determine, the heat transfer to the gas in kJ.

1 bar = 10⁵ Pa

Answer 1

Calculate the area of the piston.

\(A_{\mathrm{p}}=\frac{\pi D^{2}}{4}\)

Here, \(D\) is the diameter of the piston

$$ \begin{array}{l} \text { Substitute } D=10 \mathrm{~cm} \\ \begin{aligned} A_{\mathrm{p}} &=\frac{\pi(10 \mathrm{~cm})^{2}}{4} \\ &=78.54 \mathrm{~cm}^{2} \end{aligned} \end{array} $$

Cross sectional area of the shaft, \(A_{\mathrm{s}}=0.8 \mathrm{~cm}^{2}\)

Net area on which the atmospheric pressure is acting,

$$ \begin{aligned} A_{1} &=A_{\mathrm{p}}-A_{\mathrm{s}} \\ &=\left(78.54 \mathrm{~cm}^{2}\right)-\left(0.8 \mathrm{~cm}^{2}\right) \\ &=77.74 \mathrm{~cm}^{2} \end{aligned} $$

Free body diagram of the piston:

Force acting downwards on the piston due to the atmospheric pressure,

$$ F_{1}=p_{\mathrm{atm}} A_{1} $$

Substitute \(p_{\text {atm }}=1\) bar and \(A_{1}=77.74 \mathrm{~cm}^{2}\)

\(F_{1}=(1\) bar \()\left(77.74 \mathrm{~cm}^{2}\right)\left|\frac{100000 \mathrm{~N} / \mathrm{m}^{2}}{1 \mathrm{bar}} \| \frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right|\)

$$ =777.4 \mathrm{~N} $$

Force acting on the shaft,

$$ F=1334 \mathrm{~N} $$

Weight of the shaft and piston acting,

$$ W=M g $$

Here, \(M\) is the mass of the piston and shaft assembly

Substitute \(M=25 \mathrm{~kg}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

\(W=(25 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\)

$$ =245.25 \mathrm{~N} $$

Force acting upwards due to the pressure of the gas,

\(F_{2}=p_{\mathrm{gas}} A_{\mathrm{p}}\)

Balance the forces acting on the piston in the vertical direction,

$$ F_{2}=F+F_{1}+W $$

Substitute \(\quad F_{1}=777.4 \mathrm{~N}, F=1334 \mathrm{~N}, W=245.25 \mathrm{~N}\) and \(F_{2}=p_{\mathrm{gas}} A_{\mathrm{p}}\)

\(p_{\mathrm{gas}} A_{\mathrm{p}}=(1334 \mathrm{~N})+(777.4 \mathrm{~N})+(245.25 \mathrm{~N})\)

\(p_{\text {gas }}=\frac{2356.65 \mathrm{~N}}{A_{\mathrm{p}}}\)

Substitute \(A_{\mathrm{p}}=78.54 \mathrm{~cm}^{2}\)

\(p_{\mathrm{gas}}=\frac{2356.65 \mathrm{~N}}{78.54 \mathrm{~cm}^{2}}\left|\frac{1 \mathrm{~cm}^{2}}{10^{-4} \mathrm{~m}^{2}}\right|\)

$$ \begin{array}{l} =0.3000572 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2} \\ =0.3 \mathrm{MPa} \end{array} $$

(a)

Calculate the height through which the piston is displaced.

$$ \begin{array}{l} \Delta P E=M g \Delta h \\ \Delta h=\frac{\Delta P E}{M g} \end{array} $$

Here, \(M\) is the mass of the piston and shaft, \(g\) is the acceleration due to gravity, \(\Delta P E\) is the

increase in potential energy.

Substitute \(25 \mathrm{~kg}\) for \(M, 9.81 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 0.2 \mathrm{~kJ}\) for \(\Delta P E\)

\(\Delta h=\frac{0.2 \times 1000}{25(9.81)}\)

\(=0.8154 \mathrm{~m}\)

Calculate the work done by the shaft.

\(W_{s}=F \times \Delta h\)

Substitute \(1334 \mathrm{~N}\) for \(F\) and \(0.8154 \mathrm{~m}\) for \(\Delta h\).

$$ \begin{aligned} W_{s} &=1334 \times 0.815 \\ &=1.0872 \times 10^{3} \mathrm{~J} \\ &=1.0872 \mathrm{~kJ} \end{aligned} $$

Therefore, work done by the shaft is \(1.0872 \mathrm{~kJ}\)

(b)

Calculate the work done in displacing the atmosphere.

$$ W_{\text {atmasphere }}=F_{1} \Delta h $$

Substitute \(777.4 \mathrm{~N}\) for \(F_{1}\) and \(0.815 \mathrm{~m}\) for \(\Delta h\)

$$ \begin{aligned} W_{\text {atmasphere }} &=(777.4)(0.815) \\ &=633.581 \mathrm{~J} \\ &=0.633 \mathrm{~kJ} \end{aligned} $$

Therefore, the work done in displacing the atmosphere is \(0.633 \mathrm{~kJ}\)

(c)

Write the energy balance for the gas inside the cylinder,

$$ \Delta U+\Delta P E=Q-W $$

Here, \(\Delta P E \quad \Delta U, Q\) and \(W\) are the change in potential energy, internal energy, heat energy supplied and work done by the system respectively.

$$ Q=\Delta U+\Delta P E+W $$

Substitute \(\Delta P E=0.2 \mathrm{~kJ}, W=(1.087+0.633) \mathrm{kJ},\) and \(\Delta U=0.1 \mathrm{~kJ}\)

\(Q=(0.1 \mathrm{~kJ})+(1.087+0.633)+(0.2 \mathrm{~kJ})\)

$$ =2.02 \mathrm{~kJ} $$

Hence, the heat transfer to the gas is \(2.02 \mathrm{~kJ}\)

(d)

As it is shown, a total of \(2.02 \mathrm{~kJ}\) of the heat energy is supplied to the gas.

Out of this amount, \(0.2 \mathrm{~kJ}\) of energy is transferred to the potential energy of the piston and shaft

assembly, through the work done by the gas.

The remaining \(0.1 \mathrm{~kJ}\) resulted in the increase of the internal energy of the gas.