The stagnation point of a high-speed missile at 934 Â°R and 7.8 atm, respectively

1 atm = 2116 lb/ft2 = 1.01 Ã— 10^5 N/m^2. Calculate cp, cv, e, and h for a. The stagnation point of a high-speed missile at 934 Â°R and 7.8 atm, respectively. b. Air at standard sea level conditions.

The Correct Answer and Explanation is:

To calculate the thermodynamic properties cpc_pcp, cvc_vcv, eee, and hhh (specific heat at constant pressure, specific heat at constant volume, internal energy, and enthalpy) for the given scenarios, we need to make use of specific formulas and relationships from thermodynamics, particularly for an ideal gas.

1. High-Speed Missile at Stagnation Point:

Given:

Temperature (T) = 934 °R
Pressure (P) = 7.8 atm
R (specific gas constant for air) ≈ 287 J/(kg·K)
Gas constant R = 287 J/(kg·K)
Ideal gas behavior is assumed for air.

a. cpc_pcp and cvc_vcv:

For air, the specific heat ratio (γ\gammaγ) is approximately 1.4.
The relationship between cpc_pcp and cvc_vcv is given by the formula: γ=cpcv\gamma = \frac{c_p}{c_v}γ=cvcp and also: cp−cv=Rc_p – c_v = Rcp−cv=R Using these, we can calculate cpc_pcp and cvc_vcv. First, find cvc_vcv and cpc_pcp by solving: cp=γ⋅cvc_p = \gamma \cdot c_vcp=γ⋅cv and cp−cv=Rc_p – c_v = Rcp−cv=R Substituting: cp=1.4⋅cvc_p = 1.4 \cdot c_vcp=1.4⋅cv 1.4⋅cv−cv=2871.4 \cdot c_v – c_v = 2871.4⋅cv−cv=287 0.4⋅cv=2870.4 \cdot c_v = 2870.4⋅cv=287 cv=2870.4=717.5 J/kg\cdotpKc_v = \frac{287}{0.4} = 717.5 \, \text{J/kg·K}cv=0.4287=717.5J/kg\cdotpK Then, calculate cpc_pcp: cp=1.4⋅717.5=1004.5 J/kg\cdotpKc_p = 1.4 \cdot 717.5 = 1004.5 \, \text{J/kg·K}cp=1.4⋅717.5=1004.5J/kg\cdotpK

b. Enthalpy (h) and Internal Energy (e):

Enthalpy is given by: h=cp⋅Th = c_p \cdot Th=cp⋅T Substituting cp=1004.5c_p = 1004.5cp=1004.5 J/kg·K and T=934T = 934T=934 °R (convert to Kelvin by subtracting 459.67): T=934−459.67=474.33 KT = 934 – 459.67 = 474.33 \, \text{K}T=934−459.67=474.33K h=1004.5⋅474.33=476,922.4 J/kgh = 1004.5 \cdot 474.33 = 476,922.4 \, \text{J/kg}h=1004.5⋅474.33=476,922.4J/kg
Internal energy is given by: e=cv⋅Te = c_v \cdot Te=cv⋅T e=717.5⋅474.33=340,649.7 J/kge = 717.5 \cdot 474.33 = 340,649.7 \, \text{J/kg}e=717.5⋅474.33=340,649.7J/kg

2. Standard Sea Level Conditions:

At standard sea level conditions:

Pressure (P) = 1 atm
Temperature (T) = 288.15 K (standard temperature)
R = 287 J/(kg·K)

Using the same process, you can calculate cpc_pcp, cvc_vcv, hhh, and eee for these conditions as we did for the missile case.

a. Specific Heats:

Using γ=1.4\gamma = 1.4γ=1.4 for air, we find: cv=287/0.4=717.5 J/kg\cdotpKc_v = 287 / 0.4 = 717.5 \, \text{J/kg·K}cv=287/0.4=717.5J/kg\cdotpK cp=1.4⋅717.5=1004.5 J/kg\cdotpKc_p = 1.4 \cdot 717.5 = 1004.5 \, \text{J/kg·K}cp=1.4⋅717.5=1004.5J/kg\cdotpK

b. Enthalpy and Internal Energy:

Enthalpy: h=1004.5⋅288.15=289,858.6 J/kgh = 1004.5 \cdot 288.15 = 289,858.6 \, \text{J/kg}h=1004.5⋅288.15=289,858.6J/kg
Internal energy: e=717.5⋅288.15=206,624.3 J/kge = 717.5 \cdot 288.15 = 206,624.3 \, \text{J/kg}e=717.5⋅288.15=206,624.3J/kg

Summary of Results:

For the High-Speed Missile:

cp=1004.5 J/kg\cdotpKc_p = 1004.5 \, \text{J/kg·K}cp=1004.5J/kg\cdotpK
cv=717.5 J/kg\cdotpKc_v = 717.5 \, \text{J/kg·K}cv=717.5J/kg\cdotpK
h=476,922.4 J/kgh = 476,922.4 \, \text{J/kg}h=476,922.4J/kg
e=340,649.7 J/kge = 340,649.7 \, \text{J/kg}e=340,649.7J/kg

For Standard Sea-Level Conditions:

cp=1004.5 J/kg\cdotpKc_p = 1004.5 \, \text{J/kg·K}cp=1004.5J/kg\cdotpK
cv=717.5 J/kg\cdotpKc_v = 717.5 \, \text{J/kg·K}cv=717.5J/kg\cdotpK
h=289,858.6 J/kgh = 289,858.6 \, \text{J/kg}h=289,858.6J/kg
e=206,624.3 J/kge = 206,624.3 \, \text{J/kg}e=206,624.3J/kg