Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
<u>Determine the heat transfer rate and exit temperature of fluid</u> <u>B</u>
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
Answer:
The heat transfer is 29.75 kJ
Explanation:
The process is a polytropic expansion process
General polytropic expansion process is given by PV^n = constant
Comparing PV^n = constant with PV^1.2 = constant
n = 1.2
(V2/V1)^n = P1/P2
(V2/0.02)^1.2 = 8/2
V2/0.02 = 4^(1/1.2)
V2 = 0.02 × 3.2 = 0.064 m^3
W = (P2V2 - P1V1)/1-n
P1 = 8 bar = 8×100 = 800 kPa
P2 = 2 bar = 2×100 = 200 kPa
V1 = 0.02 m^3
V2 = 0.064 m^3
1 - n = 1 - 1.2 = -0.2
W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ
∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ
Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ
Jude be done. Die he ehe ehehevehe Ed
Answer:
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