A hot toluene stream, which has mass flow rate of 8.0 kg/min, is cooled by cooling water in a cocurrent heat exchanger; its temp
erature drops from 90o C to 60o C after passing through the heat exchanger. The cooling water stream has a mass flow rate of 10 kg/min and an inlet temperature of 20o C. Given that the specific heats of toluene and water are 0.41 and 1.00 kcal/(kg.o C), respectively, determine (a, 3%) the heat duty of the heat exchanger, (b,3%) the temperature of the cooling water leaving the heat exchanger, (c, 3%) the log mean temperature difference (or "driving force") for the heat exchange process, and (d, 3%) the UA value of the heat exchanger. Assume a negligible heat loss to the surroundings.
1 answer:
Complete Question
The complete question is shown on the first question
Answer:
a) The duty of the heat exchanger is given as 6.8658 KJ /sec
b) The temperature of the water leaving the exchanger is TOUT = 29.84 ⁰C
c) The log mean difference is given as TZ = 47.317 ⁰ C
d) the UA value is UA = 145.10
Explanation:
The explanation is uploaded on the first and second ,third and fourth image
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<h3>Answer:</h3>
134 atm
<h3>Explanation:</h3>- Based on the pressure law, the pressure of a gas varies directly proportionally to the absolute temperature at a constant volume.
- Therefore; we are going to use the equation;
In this case;
Initial pressure, P1 = 144 atm
Initial temperature, T1 (48°C) = 321 K
Final temperature, T2 (25°C) = 298 K
We need to find the final pressure,
Therefore;
P2 = (P1/T1)T2
= (144/321)× 298 K
= 133.68 atm
= 134 atm
Therefore, the new pressure will be 134 atm.