To figure this out, I've tried to solve for $\Delta T$ for $PV=n_{cold}RT_{cold}$ and $PV=n_{warm}RT_{warm}$ which assumes that $PV$ is equal. This gave me the equation: $$\Delta T = \frac{T_{cold}^2 \Delta m R}{PV m_m - T_{cold} \Delta m R}$$ Where:
- $\Delta m$ is the target mass (weight of the balloon) of air to displace in grams
- $m_m$ is the molar mass of air in g/mol
- $P$ is the pressure at the current elevation in Pa
- $V$ is the volume of the balloon in m³
However, looking at Why is a hot air balloon "stiff"?, I've come to realise that $P$ is not the same for each side of the balloon.
So my question is, am I correct that I'm wrong? If so, what can I do to correct this?
To be fair, I was considering that there was a support structure inside the balloon maintaining its shape, so maybe my assumption that $P$ is the same is valid?
The following is how I arrived at the formula if it is of interest:
\begin{array}{rl} \rm define & PV = n_c R T_c & \tt (1a) \\ \rm define & PV = n_w R T_w & \tt (1b) \\ \rm define & n_c = {m_c \over m_m} & \tt (2a) \\ \rm define & n_w = {m_w \over m_m} & \tt (2b) \\ \tt (2a)\Rightarrow(1a) & PV = {m_c \over m_m} R T_c & \tt (3a) \\ \tt (2b)\Rightarrow(2b) & PV = {m_w \over m_m} R T_w & \tt (3b) \\ & \Delta T = T_w - T_c & \tt (4a) \\ \rm isolate\ \it T_w\ \tt(4a) & T_w = \Delta T + T_c & \tt (4b) \\ \tt (4b)\Rightarrow(3b) & PV = {m_w \over m_m} R (\Delta T + T_c) & \tt (5) \\ \tt (5)\Rightarrow(3a) & {m_w \over m_m} R (\Delta T + T_c) = {m_c \over m_m} R T_c & \\ & m_w (\Delta T + T_c) = m_c T_c & \tt (6) \\ \rm isolate\ \it \Delta T\ \tt(6) & m_w \Delta T + m_w T_c = m_c T_c \\ & m_w \Delta T = m_c T_c - m_w \\ & \Delta T = \frac{m_c T_c - m_w T_c} {m_w} \\ & \Delta T = T_c \frac{m_c - m_w} {m_w} & \tt (7) \\ \rm define & \Delta m = m_c - m_w & \tt (8) \\ \tt (8)\Rightarrow(7) & \Delta T = T_c \frac{\Delta m} {m_w} & \tt (9) \\ \rm isolate\ \it m_w\ \tt (9) & m_w = T_c \frac{\Delta m}{\Delta T} & \tt (10) \\ \tt (10)\Rightarrow(3b) & PV = {T_c \frac{\Delta m}{\Delta T} \over m_m} R T_w & \\ & PV = \frac{T_c \Delta m R T_w}{\Delta T m_m} & \tt (11) \\ \tt(4b)\Rightarrow(11) & PV = \frac{T_c \Delta m R (\Delta T + T_c)}{\Delta T m_m} & \tt (12) \\ \rm isolate\ \it \Delta T\ \tt(12) & PV = \frac{T_c \Delta m R \Delta T + T_c \Delta m R T_c}{\Delta T m_m} \\ & PV = \frac{T_c \Delta m R \Delta T}{\Delta T m_m} + \frac{T_c^2 \Delta m R}{\Delta T m_m} \\ & PV = \frac{T_c \Delta m R}{m_m} + \frac{T_c^2 \Delta m R}{\Delta T m_m} \\ & PV - \frac{T_c \Delta m R}{m_m} = \frac{T_c^2 \Delta m R}{\Delta T m_m} \\ & \Delta T = \frac{T_c^2 \Delta m R}{\left(PV - \frac{T_c \Delta m R}{m_m}\right)m_m} \\ & \Delta T = \frac{T_c^2 \Delta m R}{\frac{PV m_m - T_c \Delta m R}{m_m}m_m} \\ & \Delta T = \frac{T_c^2 \Delta m R}{PV m_m - T_c \Delta m R} \\ \end{array}
Note that this is entirely based on Archimedes principle and is why I don't have any force equations listed, meaning that I'm assuming neutrally buoyant if I displace the weight of the balloon. That's a valid approach, right?
