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Kazeer [188]
3 years ago
14

The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg·°C. Express this value in kJ/kg·K, J/g·°C, kcal/ kg·°C, and Bt

u/lbm·°F. Round the final answers to three decimal place
Engineering
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

a) C_v = 1.005 KJ/kgK

b) C_v = 1005.000 J/kgC

c) C_v = 0.240 kcal/kgC

d) C_v = 0.240 Btu/lbmF

Explanation:

Given:

- constant-pressure specific heat C_v = 1.005 KJ/kgC

Find C_v in units of:

a) kJ/kg·K

b) J/g·°C

c) kcal/ kg·°C

d) Btu/lbm·°F

Solution:

a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,

C_v = 1.005 KJ/kgK

b)

                           C_v = 1.005 KJ/kgC * ( 1000 J / KJ)

                           C_v = 1005.000 J/kgC

c)

                           C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)

                           C_v = 0.240 kcal/kgC

d)

C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)

C_v = 0.240 Btu/lbmF

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Answer:

a) V = 0.354

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Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

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Using Area, A = π/4 · d²

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E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

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Also final diameter d(f) = 19.9837 mm

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Poisson said that V = Е(elasticity)/Е(long)

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