Explanation:
Note: Refer the diagram below
Obtaining data from property tables
State 1:
State 2:
State 3:
State 4:
Throttling process
(a)
Magnitude of compressor power input
(b)
Refrigerator capacity
(c)
Cop:
Air enters a counterflow heat exchanger operating at steady state at 27 C, 0.3 MPa and exits at 12 C. Refrigerant 134a enters at
0.4 MPa, a quality of 0.3, and a mass flow rate of 35 kg/h. Refrigerant exits at 10 C. Stray heat transfer is negligible and there is no significant change in pressure for either stream.(a) For the Refrigerant 134a stream, determine the rate of heat transfer, in kJ/h.(b) For each of the streams, evaluate the change in flow exergy rate, in kJ/h, and interpret its value and sign.Let T0 = 22 C, p0 = 0.1 MPa, and ignore the effects of motion and gravity.
1 answer:
Answer:
A) 337.21 kj/h
b) -224.823 kj/h
Explanation:
Steady state temperature = 27⁰C
Pressure = 0.3 MPa
exit temperature = 12⁰c
Refrigerant 134a :
entering pressure = 0.4 MPa
mass flow rate = 35 kg/h
quality = 0.3
inlet temp = 8.93⁰c
A) Determine the rate of heat transfer in kJ/h for Refrigerant 134a stream
calculate the specific enthalpy of refrigerant at the inlet
hm = hf + x ( hg - hf )
at 0.4 MPa : hf = 64 kj/kg , hg = 256 kj/kg
x = 0.3 ( quality )
hence : hm = 64 + 0.3 ( 256 - 64 ) = 64.3 + 57.6 = 121.9 kj/kg
next calculate the specific enthalpy at outlet
Tout = 10⁰c , T sat = 8.93⁰c
since the Tsat < Tout the refrigerant is super heated and from the super heated refrigerant table at P = 0.4 MPa, hout = 257 kj/kg
Heat gained by refrigerant ( Qin)
Qin = mref ( hout - hin )
= 35 ( 257 - 112) = 5075 kj/h
To determine the rate of heat transfer we have to apply the equation below
heat gained by refrigerant = heat gained by air
Qin = Qout
5075 kj/h = mair ( hout - hin ) ------------ 1
considering ideal conditions
at T = 300 k , hout = 300.19 kj/kg ( specific enthalpy )
at T = 285 k , hin = 285.14 kj/kg ( specific enthalpy )
back to equation 1
5075 kj/h = mair ( 300.19 - 285.14 )
mair ( rate of energy transfer ) = 5075 / 15.05 = 337.21 kj/h
B) evaluating the change in flow energy rate in kJ/h
attached below is the detailed solution
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Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re =
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re =
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s