g A heat exchanger is designed to is to heat 2,500 kg/h of water from 15 to 80 °C by engine oil. The configuration of the heat e
xchanger is in shell-and tube and the water is flowing through the tubes. The oil makes a single shell pass, entering at 160 °C and leaving at 90 °C, with an averaged heat transfer coefficient from the oil to the outer wall of the tubes equals 400 W/m2·K. The water flows through 11 brass tubes of 30 mm diameter with each tube making four passes through the shell. Assuming fully developed flow for the water inside the tubes, determine the tube length per pass to achieve these specified output temperatures. You may assume the tube wall thickness is negligible in the determination of overall heat transfer coefficient. (
1 answer:
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Answer:
- branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
- branch 2: i = 21.466∠63.435°; pf = 0.447 leading
- total: i = 31.693∠10.392° leading; pf = 0.984 leading
Explanation:
To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.
The inductive reactance is ...
The capacitive reactance is ...
<u>Branch 1</u>
The impedance of branch 1 is ...
Z1 = 8 +j4.99984 Ω
so the current is ...
I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°
The power factor is cos(-32.005°) ≈ 0.848 (lagging)
<u>Branch 2</u>
The impedance of branch 2 is ...
Z2 = 5 -j10 Ω
so the current is ...
I2 = 240/(5 +j10) ≈ 21.466∠63.435°
The power factor is cos(63.436°) ≈ 0.447 (leading)
<u>Total current</u>
The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.
It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)
It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°
The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)
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The phasor diagram of the currents is attached.
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<em>Additional comment</em>
Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.
Answer:
L = 475.718
T = 240.89 ft
M = 23.0195
LC = 472.728
R = 1225 ft
Explanation:
See the attached file for the calculation.
