Answer:
7.615 kW
Explanation:
Solution in pen paper form in the attachment section
The answer is true because if the effect is neglected, the saturation id region is considered true
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13

sat vapor
m=

b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
a)6.8 KPa
b)0.264 gallon
c)47.84 Pa.s
Explanation:
We know that
1 lbf= 4.48 N
1 ft =0.30 m
a)
Given that
P= 1 psi
psi is called pound force per square inch.
We know that 1 psi = 6.8 KPa.
b)
Given that
Volume = 1 liter
We know that 1000 liter = 1 cubic meter.
1 liter =0.264 gallon.
c)
