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
= ![\frac{T_{2}}{T_{2} - T_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B2%7D%20-%20T_%7B1%7D%7D)
=
= 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max
= ![\frac{T_{1}}{T_{2} - T_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%20-%20T_%7B1%7D%7D)
= 10.89
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
![\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1} = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}](https://tex.z-dn.net/?f=%5Cfrac%7BP1%7D%7B%5Crho%20%7D%20%2B%20%5Cfrac%7Bv_%7B2%7D%20%7D%7B2g%7D%20%2Bz_%7B1%7D%20%20%3D%20%5Cfrac%7BP2%7D%7B%5Crho%20%7D%20%2B%20%5Cfrac%7Bv2%5E%7B2%7D%20%7D%7B2g%7D%20%2B%20z_%7B2%7D%20%2B%20f%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B2g%7D)
thus
![\frac{P1-P2}{\rho } = f\frac{l}{D} \frac{v^{2} }{2g}](https://tex.z-dn.net/?f=%5Cfrac%7BP1-P2%7D%7B%5Crho%20%7D%20%20%3D%20f%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B2g%7D)
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
I dont know is your papers main idea stated clearly?