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1) The document provides calculations to determine the diameter of a shell and number of tubes needed for heat transfer between air and oil streams. 2) Given parameters of the air and oil streams, the heat transfer rate is calculated to be 958,950 J/s. 3) Using the heat transfer rate and properties of the tube, the required total heat transfer surface area is calculated to be 35.25 m^2. 4) Based on the total area and tube dimensions, the diameter of the shell is calculated to be 0.31 m and the required number of tubes is 401.
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1. **Time Slots Allocation**: The core principle of TDM is to assign distinct time slots to each signal. During each time slot, the respective signal is transmitted, and then the process repeats cyclically. For example, if there are four signals to be transmitted, the TDM cycle will divide time into four slots, each assigned to one signal.
2. **Synchronization**: Synchronization is crucial in TDM systems to ensure that the signals are correctly aligned with their respective time slots. Both the transmitter and receiver must be synchronized to avoid any overlap or loss of data. This synchronization is typically maintained by a clock signal that ensures time slots are accurately aligned.
3. **Frame Structure**: TDM data is organized into frames, where each frame consists of a set of time slots. Each frame is repeated at regular intervals, ensuring continuous transmission of data streams. The frame structure helps in managing the data streams and maintaining the synchronization between the transmitter and receiver.
4. **Multiplexer and Demultiplexer**: At the transmitting end, a multiplexer combines multiple input signals into a single composite signal by assigning each signal to a specific time slot. At the receiving end, a demultiplexer separates the composite signal back into individual signals based on their respective time slots.
### Types of TDM
1. **Synchronous TDM**: In synchronous TDM, time slots are pre-assigned to each signal, regardless of whether the signal has data to transmit or not. This can lead to inefficiencies if some time slots remain empty due to the absence of data.
2. **Asynchronous TDM (or Statistical TDM)**: Asynchronous TDM addresses the inefficiencies of synchronous TDM by allocating time slots dynamically based on the presence of data. Time slots are assigned only when there is data to transmit, which optimizes the use of the communication channel.
### Applications of TDM
- **Telecommunications**: TDM is extensively used in telecommunication systems, such as in T1 and E1 lines, where multiple telephone calls are transmitted over a single line by assigning each call to a specific time slot.
- **Digital Audio and Video Broadcasting**: TDM is used in broadcasting systems to transmit multiple audio or video streams over a single channel, ensuring efficient use of bandwidth.
- **Computer Networks**: TDM is used in network protocols and systems to manage the transmission of data from multiple sources over a single network medium.
### Advantages of TDM
- **Efficient Use of Bandwidth**: TDM all
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- 1. FAKHRI BURHANUDDIN TEKNIK MESIN 41313120068 PERPINDAHAN PANAS-QUIZ 10 /diketahui : Aliran air Tci = 200C = 293K Tco = 360C=309K Cpc = 4176,61 J/kg-K Aliran oli mh = 4,5 kg/s Thi = 1600C=433K Th0=1000C=373K Cph =2131 J/kg-K Dimensi Tube Diameter : 3/4in(OD = 19 mm;0,019m ID = 16 mm;0,016m) Panjang = 4,8m Jumlah lintasan 1 pass α=450 CL = 1 CTP = 0,93 PR = 1,25 dalam keadaan clean Uc=325 W/m2K os = 20% = 1,2
- 2. Ditanya = berapa diameter shell dan jumlah tube yang diperlukan? Dijawab : - Balans energi pada aliran di sisi shell )( hohihphh TTcmQ = (4,5kg/s)( 2131 J/kg-K)(433K-373K) =958950 J/s - Laju Aliran massa di sisi tube (Qh=Qc) )( cicopc c c TTc Q m = 958950 J/s 4176,61 J/kg-K ( 309K-293K) =14,3 kg/s - Beda temperatur rata-rata logaritmatik CFmcSTm TFT ,, 2 1 21 , ln T T TT T CFm ....(1) ciho TTT 1 =373K- 293K=80K cohi TTT 2 =433K-309K=124K 2 1 21 , ln T T TT T CFm = = , = 100,46 - Mencari uf dalam keadaan foul 2,1 f c c f U U A A OS )( cicopccc TTcmQ
- 3. Uf = Uc/1,2 =(325 W/m2K)/1,2 =271 W/m2K - Luas permukaan perpindahan panas total STmf tot TU Q A , = / / ( , ) = 35,25m2 -menentukan diameter shell 5,025,0 637,0 L dPRA CTP CL D oo s =0,637 (1/0,93)0,5((35,25m2)(1,25)2(0,019m)/4,8m)0,5 =0,637 (1.04)(0,47m) =0,310m -menentukan jumlah tube dalam shell 22 2 785,0 o s t dPR D CL CTP N =0,785 ((0,93/1))((0,310m)2/(1,25)2(0,019m)2) =0,785(0.93)(549,58) =401buah