Oil enters a counterflow heat exchanger at 600 K with a mass flow rate of 10 kg/s and exits at 200 K. A separate stream of liqui
d water enters at 20°C, 5 bar. Each stream experiences no significant change in pressure. Stray heat transfer with the surroundings of the heat exchanger and kinetic and potential energy effects can be ignored. The specific heat of the oil is constant, c = 2 kJ/kg · K. If the designer wants to ensure no water vapor is present in the exiting water stream, what is the minimum mass flow rate for the water, in kg/s?
1 answer:
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Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:
isobaric expansion = 281.09 Btu
isothermal compression= 72 Btu
Explanation:
The first law of thermodynamics is:
where:
Q=heat transferred
W= work
U=internal energy
P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.
In a isobaric process heat transferred is:
For an isothermal process (T2-T1 = 0) so
From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.
For the first process
we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)
we use this information in the appropriate unit to find the pressure in thermodynamic tables.
T1= 176°C
v1= 1.49 m^3/kg
P=1.37 bar
in the second state we have
P=1.37 bar =137000Pa
with thee properties we check in the thermodynamic tables
T2= 255°C
we usually find Cp on tables for water but from the Mayer relation we have:
Cp for water vapor is: 33.12 J/mol*K
R=8.314 J/mol*K
Cv= 41.434 J/mol*K
replacing in the equation for Q
296569J =281.09 Btu