Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 k
1 answer:
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
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Answer:
Explanation:
The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.
Given:
T1=170C
To convert to Kelvin
= 17+273 =290K
T1 = 290K
Pressure (P)= 95KPa
Specific heat ratio = CP/CV= K
WhereK=1.005/0.718
K = 1.4
The final temperature can be calculated using the formula below.
T2 = CP/CV × T1
=. K × T1
T2 = 1.4 × 290
Answer:
|W|=169.28 KJ/kg
ΔS = -0.544 KJ/Kg.K
Explanation:
Given that
T= 100°F
We know that
1 °F = 255.92 K
100°F = 310 .92 K
We know that work for isothermal process
Lets take mass is 1 kg.
So work per unit mass
We know that for air R=0.287KJ/kg.K
W= - 169.28 KJ/kg
Negative sign indicates compression
|W|=169.28 KJ/kg
We know that change in entropy at constant volume
ΔS = -0.544 KJ/Kg.K