Let's solve the problem step by step! --- ## **Given Data** - Ratio of specific heats (\( k \)): **1.4** - Exit pressure (\( p_x \)): **14.7 lbf/in²** (atmospheric) - Exit temperature (\( T_x \)): **520°R** - Exit Mach number (\( M_x \)): **1.5** - Exit area (\( A_x \)): **3.3 in²** - Upstream of the shock, **isentropic flow**. --- # **(a) Stagnation Pressure (\( p_{x} \)), in lbf/in²** For an **isentropic flow** of an ideal gas: \[ \frac{p_{x}}{p_x} = \left[1 + \frac{k-1}{2} M_x^2 \right]^{\frac{k}{k-1}} \] Insert values: - \( k = 1.4 \) - \( M_x = 1.5 \) - \( p_x = 14.7 \) lbf/in² Calculate the bracket: \[ 1 + \frac{1.4-1}{2} \times (1.5)^2 = 1 + .2 \times 2.25 = 1 + .45 = 1.45 \] Now the exponent: \[ \frac{k}{k-1} = \frac{1.4}{.4} = 3.5 \] So, \[ \frac{p_{x}}{14.7} = (1.45)^{3.5} \] Calculate \( (1.45)^{3.5} \): \[ \ln(1.45) = .3716 \\ 3.5 \times .3716 = 1.3006 \\ e^{1.3006} \approx 3.67 \] So, \[ p_{x} = 14.7 \times 3.67 = 53.95 \text{ lbf/in}^2 \] --- ## **Final Answer (a):** \[ \boxed{p_{x} = 54. \ \text{lbf/in}^2} \] (Rounded to 1 decimal) --- # **(b) Stagnation Temperature (\( T_{x} \)), in °R** \[ \frac{T_{x}}{T_x} = 1 + \frac{k-1}{2} M_x^2 \] Insert values: \[ \frac{T_{x}}{520} = 1 + .2 \times (1.5)^2 = 1 + .2 \times 2.25 = 1 + .45 = 1.45 \] \[ T_{x} = 1.45 \times 520 = 754\ ^\circ\text{R} \] --- ## **Final Answer (b):** \[ \boxed{T_{x} = 754\ ^\circ\text{R}} \] --- # **(c) Mass Flow Rate (\( \dot{m} \)), in lb/s** First, use the **mass flow rate equation** at the exit: \[ \dot{m} = \rho_x V_x A_x \] Or, using Mach and isentropic relations: \[ \dot{m} = \frac{p_x A_x M_x}{\sqrt{R T_x}} \cdot \frac{1}{\sqrt{k}} \left[ 1 + \frac{k-1}{2} M_x^2 \right]^{-\frac{k+1}{2(k-1)}} \] ### **Step 1: Find R (gas constant for air in ft·lb/(lbm·°R) and convert units)** - \( R = 1716 \) ft·lb/(lbm·°R) But let's be careful with units. We need everything in consistent units. #### **Convert area to ft²:** \[ A_x = 3.3\ \text{in}^2 \times \left(\frac{1\ \text{ft}}{12\ \text{in}}\right)^2 = 3.3 \times \frac{1}{144} \approx .0229\ \text{ft}^2 \] #### **Convert pressure to lb/ft²:** \[ p_x = 14.7\ \text{lbf/in}^2 \times 144\ \text{in}^2/\text{ft}^2 = 2117\ \text{lbf/ft}^2 \] #### **Density at exit:** \[ \rho_x = \frac{p_x}{R T_x} \] \[ \rho_x = \frac{2117}{1716 \times 520} = \frac{2117}{892320} \approx .00237\ \text{lb/ft}^3 \] #### **Speed of sound at exit:** \[ a_x = \sqrt{k R T_x} = \sqrt{1.4 \times 1716 \times 520} \] \[ a_x = \sqrt{1.4 \times 893,280} = \sqrt{1,250,592} \approx 1,119\ \text{ft/s} \] #### **Velocity at exit:** \[ V_x = M_x \cdot a_x = 1.5 \times 1,119 = 1,678.5\ \text{ft/s} \] #### **Mass flow rate:** \[ \dot{m} = \rho_x V_x A_x = .00237 \times 1,678.5 \times .0229 \approx .0911\ \text{lb/s} \] --- ## **Final Answer (c):** \[ \boxed{\dot{m} = .091\ \text{lb/s}} \] --- # **Summary Table** | Quantity | Value | |--------------------------|----------------------| | Stagnation Pressure | \( 54. \) lbf/in² | | Stagnation Temperature | \( 754 \) °R | | Mass Flow Rate | \( .091 \) lb/s | --- If you need further clarifications or steps, let me know!