A 10-m-long countercurrent-flow heat exchanger is being used to heat a liquid food from 20 to 808C. The heating medium is oil, w
hich enters the heat exchanger at 1508C and exits at 608C. The specific heat of the liquid food is 3.9 kJ/(kg K). The overall heat-transfer coefficient based on the inside area is 1000 W/ (m2 K). The inner diameter of the inside pipe is 7 cm. a. Estimate the flow rate of the liquid food. b. Determine the flow rate of the liquid food if the heat exchanger is operated in a concurrent-flow mode for the same conditions of temperatures at the inlet and exit from the heat exchanger.
1 answer:
Answer:
mlf=0.5038kg/s
Explanation:
a. Please kindly check attachment for the step by step solution
b. B) in concurrent flow heat exchanger same exit temperatures for both fluids cannot be obtained. Since tho<tco
Case ii) only liquid food is considered tci=20 &tco=80
Cp=3.9kj/kgk
&If Q is same then mlf=0.5038kg/s this is same as case a
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Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=<u>34.077MW</u>
