1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mnenie [13.5K]
3 years ago
5

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.

Engineering
1 answer:
Kruka [31]3 years ago
8 0

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

You might be interested in
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0
3 years ago
There is a dispute between the multiple parties storing financial transaction data on a blockchain over the validity of a transa
spayn [35]

Answer:

dgjkkkkkkkfdcvv

Explanation:

hjklllgfddsssssyjjjkkkk

p.s sorry

5 0
3 years ago
Multiple Choice
12345 [234]

Answer:https://global.oup.com/us/companion.websites/9780199385423/student/ch6/mcq/     just go here

Explanation:

6 0
3 years ago
A discrete MOSFET common-source amplifier has RG = 2 MΩ, gm = 5 mA/V, ro = 100 kΩ, RD = 20kΩ, Cgs = 3pF, and Cgd = 0.5pF. The am
Papessa [141]

Answer:

a) -36.36 V/V

b) 15.17 kHz

c) 1.6 GHz

Explanation:

See attached picture.

7 0
3 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0
3 years ago
Other questions:
  • A closed system undergoes a process in which work is done on the system and the heat transfer Q occurs only at temperature Tb. F
    8·1 answer
  • Plot the absorbance, A, versus the FeSCN2 concentration of the standard solutions (the values from your Pre-lab assignment). Fro
    7·1 answer
  • A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg lid. If the local atmospheric pressure
    9·2 answers
  • he Weather Channel reports that it is a hot, muggy day with an air temperature of 90????F, a 10 mph breeze out of the southwest,
    6·1 answer
  • When will the entropy value of the universe attained its maximum value?
    13·1 answer
  • For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari
    15·1 answer
  • Resistors of 150 Ω and 100 Ω are connected in parallel. What is their equivalent resistance?
    13·1 answer
  • ¿Cómo llevan a cabo el lavado ropa?​
    8·1 answer
  • (Blank) welding involves manual welding with equipment anomalously controls one or more of the windy conditions while (blank) we
    7·1 answer
  • I need the answer please
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!