# Laval No Flow Problem (Step-by-Step Solution) We are given: - **Tank absolute pressure**: \( P_ = 400\,\text{kPa} \) - **Tank temperature**: \( T_ = 70^\circ C = 343.15\,\text{K} \) - **Throat diameter**: \( d_t = 20\,\text{mm} = .02\,\text{m} \) - **Exit diameter**: \( d_e = 50\,\text{mm} = .05\,\text{m} \) - **Gas constant for air**: \( R = 286.9\,\text{J/(kg}\cdot\text{K)} \) - **Specific heat ratio for air**: \( k = 1.40 \) - **Pipe absolute pressure (for part 2)**: \( P_b = 150\,\text{kPa} \) We need to find: 1. **The absolute pressure within the connected pipe that will just choke the nozzle while keeping isentropic subsonic flow in the divergent portion.** 2. **The mass flow rate when the pipe pressure is \(150\,\text{kPa}\).** --- ## 1. **Find the critical (choking) pressure in the pipe** A Laval nozzle chokes at its throat when the Mach number at the throat \( M_t = 1 \). For isentropic flow, the pressure ratio for choked flow is: \[ \frac{P^*}{P_} = \left( \frac{2}{k+1} \right)^{\frac{k}{k-1}} \] Where: - \( P^* \) = critical (throat) pressure - \( P_ \) = stagnation (tank) pressure **Calculate:** \[ \frac{2}{k+1} = \frac{2}{2.4} = .8333 \] \[ \frac{k}{k-1} = \frac{1.4}{.4} = 3.5 \] \[ \left( .8333 \right)^{3.5} \approx .528 \] \[ P^* = .528 \times 400\,\text{kPa} = 211\,\text{kPa} \] ### **Choking Condition** - If the pipe pressure \( P_b \leq 211\,\text{kPa} \), the nozzle will choke. - To ensure subsonic flow in the divergent (exit) section, **the back pressure must not be low enough to cause supersonic flow at the exit.** - **To maintain subsonic flow at the exit, \( P_b \) must be just below the critical pressure, but above the pressure that would cause the exit Mach number to be 1.** #### **Find the pressure ratio for Mach 1 at the exit:** The Mach number at the exit is given by the area-Mach relation: \[ \frac{A_e}{A_t} = \frac{M_e}{M_t} \left[ \frac{1 + \frac{k-1}{2} M_t^2}{1 + \frac{k-1}{2} M_e^2} \right]^{\frac{k+1}{2(k-1)}} \] For \( M_t = 1 \): \[ \frac{A_e}{A_t} = M_e \left[ \frac{1 + \frac{k-1}{2}}{1 + \frac{k-1}{2} M_e^2} \right]^{\frac{k+1}{2(k-1)}} \] But to **avoid supersonic flow at the exit**, \( M_e < 1 \). - Therefore, the **maximum back pressure for subsonic flow in the exit** is when \( P_b = P^* = 211\,\text{kPa} \). **Final answer (part 1):** > The absolute pressure in the pipe must be **no lower than \( 211\,\text{kPa} \)** to ensure the nozzle chokes and the flow remains subsonic in the divergent section. --- ## 2. **Find the Mass Flow Rate for \( P_b = 150\,\text{kPa} \)** Now, the back pressure is **below the critical pressure**, so the nozzle is choked at the throat. ### **Choked Mass Flow Rate Formula (Isentropic):** \[ \dot{m} = C A_t P_ \sqrt{ \frac{k}{R T_} } \left( \frac{2}{k+1} \right)^{\frac{k+1}{2(k-1)} } \] Where: - \( C \) = discharge coefficient (assume \( C = 1 \) for ideal nozzle) - \( A_t \) = throat area #### **Calculate throat area:** \[ A_t = \frac{\pi}{4} d_t^2 = \frac{\pi}{4} (.02)^2 = 3.142 \times 10^{-4} \,\text{m}^2 \] #### **Calculate mass flow rate:** First, compute the dimensionless factor: \[ \sqrt{ \frac{k}{R T_} } = \sqrt{ \frac{1.4}{286.9 \times 343.15} } \] \[ 286.9 \times 343.15 = 98479 \] \[ \frac{1.4}{98479} = 1.422 \times 10^{-5} \] \[ \sqrt{1.422 \times 10^{-5}} = .003775 \] Now, the pressure ratio power: \[ \left( \frac{2}{k+1} \right)^{\frac{k+1}{2(k-1)} } \] \[ \frac{2}{2.4} = .8333 \] \[ \frac{k+1}{2(k-1)} = \frac{2.4}{.8} = 3 \] \[ (.8333)^3 = .5787 \] Now, plug in all values: \[ \dot{m} = 1 \times 3.142 \times 10^{-4} \times 400000 \times .003775 \times .5787 \] Breakdown: - \( 3.142 \times 10^{-4} \times 400000 = 125.68 \) - \( 125.68 \times .003775 = .4745 \) - \( .4745 \times .5787 = .2746 \) **Final mass flow rate:** \[ \dot{m} \approx .275\,\text{kg/s} \] --- # **Summary of Answers** ### 1. **Minimum pipe absolute pressure for choked, subsonic-divergent flow:** \[ \boxed{211\,\text{ kPa}} \] ### 2. **Mass flow rate for pipe pressure \(150\,\text{kPa}\):** \[ \boxed{.275\,\text{ kg/s}} \]