Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer:
For
- 5.556 lb/s
For
- 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T = ![20^{\circ}](https://tex.z-dn.net/?f=20%5E%7B%5Ccirc%7D)
Temperature rise, T' =
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air, ![C_{p} = 0.24 btu/lb ^{\circ}F](https://tex.z-dn.net/?f=C_%7Bp%7D%20%3D%200.24%20btu%2Flb%20%5E%7B%5Ccirc%7DF)
Now, for a rise of
in temeprature:
![\dot{m}C_{p}\Delta T = 96000](https://tex.z-dn.net/?f=%5Cdot%7Bm%7DC_%7Bp%7D%5CDelta%20T%20%3D%2096000)
![\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7B96000%7D%7BC_%7Bp%7D%5CDelta%20T%7D%20%3D%20%5Cfrac%7B96000%7D%7B0.24%5Ctimes%2020%7D%20%3D%2020000%20lb%2Fh%20%3D%20%5Cfrac%7B20000%7D%7B3600%7D%20%3D%205.556%20lb%2Fs)
Now, for
:
![\dot{m}C_{p}\Delta T = 96000](https://tex.z-dn.net/?f=%5Cdot%7Bm%7DC_%7Bp%7D%5CDelta%20T%20%3D%2096000)
![\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7B96000%7D%7BC_%7Bp%7D%5CDelta%20T%7D%20%3D%20%5Cfrac%7B96000%7D%7B0.24%5Ctimes%2015%7D%20%3D%2026666.667%20lb%2Fh%20%3D%20%5Cfrac%7B26666.667%7D%7B3600%7D%20%3D%207.4074%20lb%2Fs)
The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg
<h3>
How to calcultae the heat?</h3>
The Net Change in Enthalpy will be:
= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg
Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)
= 1/2 x ( 75 + 225) x (5 - 2)
W = 450 KJ
From the First Law of Thermodynamics, Q = U + W
So, Heat Transfer = Change in Internal Energy + Work Done
= 13331.618 + 450
Q = 13781.618 kJ/kg
Learn more about heat on:
brainly.com/question/13439286
#SP1
Answer:
the police officer cruise each streets precisely once and he enters and exit with the same gate.
Explanation:
NB: kindly check below for the attached picture.
The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.
The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.
Here in this question, the door is the vertex and the road is the edge.
The information needed to detemine a Euler circuit and a Hamilton circuit is;
"the police officer cruise each streets precisely once and he enters and exit with the same gate."
Check attachment for each type of circuit and the differences.