# Hot and cold air interaction in a sealed environment

What would happen if air heated to 1200 degree's Fahrenheit in a sealed environment was injected with colder air? No specific pressure defined yet just a general idea of potential interaction between the two.

Edit
My apologies, I meant within a sealed environment like an engine cylinder. I know that the hot air would naturally expand and push a piston but how would the injection of cold air alter this, if at all?
For example purposes we can assume the cylinder is iron while the piston is aluminum which is common design today.

• Maybe this question should be moved to Physics SE. – kaliaden Apr 21 '13 at 12:30
• @kaliaden: A note: While this is on topic on Phys.SE, it is not off topic here (gas laws/thermodynamics is fine here) -- migrations of such posts are only done if the OP requests or if they don't get answers in a few days :) – ManishEarth Apr 21 '13 at 12:47

In order to work out this problem, several parameter values are required to define the initial state, preferably

• amount $n_1$, volume $V_1$, or mass $m_1$ of the hot air in the cylinder
• initial temperature $T_1$ of the hot air
• initial pressure $p_1$ of the hot air
• amount $n_2$, volume $V_2$, or mass $m_2$ of the injected cold air
• initial temperature $T_2$ of the cold air
• initial pressure $p_2$ of the cold air

Unfortunately, only the initial temperature of the hot air is defined as $T_1 = 1200\ ^\circ\mathrm F$, which is $T_1 \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$ in proper units (the use of the unit degree Fahrenheit is deprecated). This leaves us with five unknown variables, which is certainly too many for a vivid analytical description.

Nevertheless, we may investigate this problem by guessing reasonable parameter values as follows. By way of example, we arbitrarily assume that equal masses of hot and cold air $(m_1 = m_2 = 1\ \mathrm{kg})$ are mixed. The assumed temperature of the cold air sample is equal to room temperature, i.e. $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$, whereas the temperature of the hot air sample is given as $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$. The assumed pressures of the air samples is equal to normal atmospheric pressure, i.e. $p_1 = p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. These conditions define the initial state and also allow to determine other interesting properties using tabulated reference values for air, e.g. density $\rho$, Volume $V$, specific internal energy $u$, and internal energy $U$.

1. Hot air sample

• temperature $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$
• pressure $p_1 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
• mass $m_1 = 1\ \mathrm{kg}$
• density $\rho_1 = 0.3827\ \mathrm{kg/m^3}$
• volume $V_1 = \frac{m_1}{\rho_1} = 2.613\ \mathrm{m^3}$
• specific internal energy $u_1 = 819.4\ \mathrm{kJ/kg}$
• internal energy $U_1 = u_1 \cdot m_1 = 819.4\ \mathrm{kJ}$

2. Cold air sample

• temperature $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$
• pressure $p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
• mass $m_2 = 1\ \mathrm{kg}$
• density $\rho_2 = 1.2046\ \mathrm{kg/m^3}$
• volume $V_2 = \frac{m_2}{\rho_2} = 0.830\ \mathrm{m^3}$
• specific internal energy $u_2 = 335.3\ \mathrm{kJ/kg}$
• internal energy $U_2 = u_2 \cdot m_2 = 335.3\ \mathrm{kJ}$

3. Mixture at constant volume

The hot air sample is confined by a piston in a cylinder. The cold air sample is added, thus allowing the two air samples to mix. The total mass $m_3$ is given by \begin{align} m_3 &= m_1 + m_2 \\ &= 1\ \mathrm{kg} + 1\ \mathrm{kg} \\ &= 2\ \mathrm{kg} \end{align}

For simplicity’s sake, we ignore any work $W$ performed by or on the system (e.g. the work that is required to inject the cold air into the cylinder). Furthermore, we assume that the process is adiabatic, i.e. no heat $Q$ is transferred into or out of the system. Therefore, in accordance with the first law of thermodynamics, the total internal energy $U_3$ is given by \begin{align} U_3 &= U_1 + U_2 \\ &= 819.4\ \mathrm{kJ} + 335.3\ \mathrm{kJ} \\ &= 1154.7\ \mathrm{kJ} \end{align} and the new specific internal energy $u_3$ is \begin{align} u_3 &= \frac{U_3}{m_3} \\ &= \frac{1154.7\ \mathrm{kJ}}{2\ \mathrm{kg}} \\ &= 577.3\ \mathrm{kJ/kg} \end{align}

At constant volume $V$, i.e. \begin{align} V_3 &= V_1 \\ &= 2.613\ \mathrm{m^3} \end{align} the new density is \begin{align} \rho_3 &= \frac{m_3}{V_3} \\ &= \frac{2\ \mathrm{kg}}{2.613\ \mathrm{m^3}} \\ &= 0.7654\ \mathrm{kg/m^3} \end{align}

The new temperature $T_3$ and pressure $p_3$ can be determined from the density $\rho_3$ and the specific enthalpy $h_3$ using tabulated reference values: \begin{align} T_3 &= 658\ ^\circ\mathrm F \approx 348\ ^\circ\mathrm C \approx 621\ \mathrm K \\ p_3 &= 1.37\ \mathrm{bar} = 0.137\ \mathrm{MPa} \end{align}

4. Mixture at constant pressure (ignoring work)

Allowing the system to expand to normal atmospheric pressure by moving the piston, the new pressure $p_4$ is $$p_4 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$

We know that the mass $m$ of the air in the closed system does not change, i.e. \begin{align} m_4 &= m_3 \\ &= 2\ \mathrm{kg} \end{align} and we still (incorrectly) assume that no work $W$ is performed by or on the system and that no heat $Q$ is transferred into or out of the system. Therefore, \begin{align} U_4 &= U_3 \\ &= 1154.7\ \mathrm{kJ} \end{align} and thus \begin{align} u_4 &= u_3 \\ &= 577.3\ \mathrm{kJ/kg} \end{align}

The new density $\rho_4$ can be determined from the pressure $p_4$ and the specific internal energy $u_4$ using tabulated reference values: $$\rho_4 = 0.5681\ \mathrm{kg/m^3}$$

Thus, the new volume $V_4$ is \begin{align} V_4 &= \frac{m_4}{\rho_4} \\ &= \frac{2\ \mathrm{kg}}{0.5681\ \mathrm{kg/m^3}} \\ &= 3.521\ \mathrm{m^3} \end{align}

The volume change $\Delta V$ during expansion is \begin{align} \Delta V &= V_4 - V_3 \\ &= 3.521\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\ &= 0.908\ \mathrm{m^3} \end{align}

The corresponding pressure-volume work $W$ can be calculated assuming an external pressure of $p_\mathrm{ex} = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. \begin{align} W &= p_\mathrm{ex} \cdot \Delta V \\ &= 101\,325\ \mathrm{Pa} \times 0.908\ \mathrm{m^3} \\ &= 92.0\ \mathrm{kJ} \end{align}

The amount of energy transferred to the surroundings as work $W$ significantly reduces the internal energy $U$ of the system. We ignored this fact when we assumed that $W_4 = 0$ and $U_4 = U_3$. We can eliminate this error by calculating a new state as follows.

5. Mixture at constant pressure (including work)

If the internal energy is reduced (due to energy transfer to the surroundings as work) to $$U_5 = 1087.8\ \mathrm{kJ}$$ and thus the specific internal energy is reduced to \begin{align} u_5 &= \frac{U_5}{m_5} \\ &= \frac{1087.8\ \mathrm{kJ}}{2\ \mathrm{kg}} \\ &= 543.9\ \mathrm{kJ/kg} \end{align} the new density $\rho_5$ after expansion to normal atmospheric pressure $$p_5 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$ can be determined from the pressure $p_5$ and the specific internal energy $u_4$ using tabulated reference values as $$\rho_5 = 0.6111\ \mathrm{kg/m^3}$$

Thus, the new volume $V_5$ is \begin{align} V_5 &= \frac{m_5}{\rho_5} \\ &= \frac{2\ \mathrm{kg}}{0.6111\ \mathrm{kg/m^3}} \\ &= 3.273\ \mathrm{m^3} \end{align} the volume change $\Delta V$ during expansion is \begin{align} \Delta V &= V_5 - V_3 \\ &= 3.273\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\ &= 0.660\ \mathrm{m^3} \end{align} and the corresponding pressure-volume work $W_5$ is \begin{align} W_5 &= p_\mathrm{ex} \cdot \Delta V \\ &= 101\,325\ \mathrm{Pa} \times 0.660\ \mathrm{m^3} \\ &= 66.9\ \mathrm{kJ} \end{align}

We find that \begin{align} U_5 &= U_3 - W_5 \\ 1087.8\ \mathrm{kJ} &= 1154.7\ \mathrm{kJ} - 66.9\ \mathrm{kJ} \\ \end{align} This result is in agreement with the first law of thermodynamics, which states that the change in the internal energy $U$ of a closed system is equal to the amount of heat $Q$ supplied to the system (which is $Q = 0$ in this case), minus the amount of work $W$ done by the system on its surroundings.

$$\small \begin{array}{llllll} \hline \text{Quantity} & \text{Symbol} & \text{Unit} & \text{State 1} & \text{State 2} & \text{State 3} & \text{State 4} & \text{State 5} \\ \hline \text{Temperature} & T & ^\circ\mathrm F & 1200 & 68 & 658 & 658 & 580 \\ & & ^\circ\mathrm C & 649 & 20 & 348 & 348 & 304 \\ & & \mathrm K & 922 & 293 & 621 & 621 & 577 \\ \text{Pressure} & p & \mathrm{bar} & 1.01325 & 1.01325 & 1.37 & 1.01325 & 1.01325 \\ & & \mathrm{MPa} & 0.101325 & 0.101325 & 0.137 & 0.101325 & 0.101325 \\ \text{Mass} & m & \mathrm{kg} & 1 & 1 & 2 & 2 & 2 \\ \text{Density} & \rho & \mathrm{kg/m^3} & 0.3827 & 1.2046 & 0.7654 & 0.5681 & 0.6111 \\ \text{Volume} & V & \mathrm{m^3} & 2.613 & 0.830 & 2.613 & 3.521 & 3.273 \\ \text{Specific internal energy} & u & \mathrm{kJ/kg} & 819.4 & 335.3 & 577.3 & 577.3 & 543.9 \\ \text{Internal energy} & U & \mathrm{kJ} & 819.4 & 335.3 & 1154.7 & 1154.7 & 1087.8 \\ \text{Work} & W & \mathrm{kJ} & \text{n/a} & \text{n/a} & 0 & 0 & 66.9 \\ \hline \end{array}$$

If the volume is constant (sealed container that is not a rubber ballon, for example), the overall temperature would cool. And the net pressure would be a weighted average of the two pressures if the volume was constant. Without actual amounts it cannot be determined if the pressure would be higher or lower than the original.

If the container were a rubber balloon, it is not possible to determine the final volume (even as bigger or smaller) without information about relative amounts and the temperature of the 'colder air'.

• So this is a closed system energy balance then with the heat transfer rate between the hot and cold air is a function of internal energy change only? An eventual equilibrium temperature may be calculated by $Q_\text{lost by hot air} = Q_\text{gained by cold air}$? – dearN Mar 21 '13 at 17:19