Given:
Temperature of water, =
=273 +(-6) =267 K
Temperature surrounding refrigerator, =
=273 + 21 =294 K
Specific heat given for water, = 4.19 KJ/kg/K
Specific heat given for ice, = 2.1 KJ/kg/K
Latent heat of fusion, = 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max =
= = 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max =
= 10.89
Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
is given as
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
A is given as
With initial grade, the elevation of PVC is
The station is given as
Low point is given as
The station of low point is given as
The elevation is given as
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
