The radiator of a steam system has a volume of 20 l and is filled with superheated

Δs = 0.133 KJ/.K

Explanation:

Given that

V= 20 L = 0.02 m³

At P₁= 200KPa and T₁=150 C :

v₁=0.93 m³/kg

s₁=7.2 KJ/kg.K

At  T₂=40 C :

v₁ = v₂

vf= 0.001 m³/kg

vg = 19.51 m³/kg

v₂ = vf + x(vg-vf)

0.93 = 0.001 + x(19.51 - 0.001)

x=0.047

s₂ = sf + x(sg-sf)

s₂ = 0.57 + 0.047(7.638) KJ/kg.K

s₂ = 0.93 KJ/kg.K

Δs= s₁ - s₂

Δs= 7.2 - 0.93 =6.2 KJ/kg.K

m=V/v₁

m=0.02/0.93 = 0.021 kg

Δs = 0.021 x 6.2 = 0.133 KJ/.K

Δs = 0.133 KJ/.K

(2PL + yAL²) / (2EA)

Explanation:

For any position x along the bar measured from the top, the stress is:

σ = F(x) / A

where F(x) is the force at position x,

and A is the cross sectional area.

The stress also equals:

σ = E ε(x)

σ = E dδ / dx

where E is the modulus of elasticity,

dx is a short length of bar,

and dδ is the deflection of that short length.

Therefore:

E dδ / dx = F(x) / A

E dδ / dx = (P + yAx) / A

dδ = 1/(EA) (P + yAx) dx

The total deflection is:

δ = ∫ dδ

δ = ∫₀ᴸ 1/(EA) (P + yAx) dx

δ = 1/(EA) ∫₀ᴸ (P + yAx) dx

δ = 1/(EA) (Px + ½yAx²) |₀ᴸ

δ = 1/(EA) (PL + ½yAL²)

δ = (2PL + yAL²) / (2EA)