Tugas Perpindahan Panas [PDF]

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NAMA



: MARUSAHA W SINAG



NIM



: 5143220023



M. KULIAH : PERPINDAHAN PANAS



SOAL... A section of a composite wall with the dimensins shown below has uniform temperatures of 200°C and 50°C over the left and right surfaces. Respectively if the thermal conductivitas of the wall materials are ka=70,kb= 60, kc= 40, kd= 20.determine the rate of heat transfer throug this section of the wall and the temperature at the interfaces. JAWAB : DIKETAHUI :Ka = 70 W/m°K Kb = 60 W/m°K Kc = 40 W/m°K Kd = 20 W/m°K



6cm



6



La = 2cm = 0.02m Lb=Lc = 2.5cm = 0.025 Ld = 4cm = 0.04m



6cm



A



B



D



2.5cm



4cm



A = 6cm x 6cm = 36 cm² = 0.036m² T1 = 200°C = 473°K T2 = 50°C = 323°K



2cm



PENYELESAIAN : Q..........?



C



JAWAB :



Q= R=



𝑇1−𝑇2 𝑅 𝐿



Rc =



𝑘 .𝐴



Ra =



𝐿𝑎



𝐿𝑐 𝑘𝑐 . 𝐴 0.025 𝑚



𝑎. 𝐴 0.02 𝑚



Ra = 70𝑊 𝑚°𝐾



𝑥 0.036𝑚²



Rc = 40𝑊 𝑚°𝐾



Rc =



𝑥 0.036𝑚²



0.025 1.44𝑊/°𝐾



0.02



Ra =



Rc = 0.01736°K/W



2.52𝑊/°𝐾



Ra = 0.007936°K/W Rb =



Rd =



𝐿𝑏



0.04 𝑚



𝑚°𝐾



0.025 𝑚



𝑚°𝐾



Rd =



𝑥 0.036𝑚²



0.025



Rb =



𝑘𝑑 . 𝐴



Rd = 20𝑊



𝑘𝑏 . 𝐴



Rb = 60𝑊



𝐿𝑑



𝑥 0.036𝑚²



0.04 0.72𝑊/°𝐾



Rd = 0.055555°K/W



2.16𝑊/°𝐾



Rb = 0.01157°K/W Karena Rb dan Rc paralel maka kita harus membuatnya menjadi seri 1 𝑅𝑝 1 𝑅𝑝 1 𝑅𝑝



=



1 0.01157°𝐾/𝑊 0.01736°𝐾 𝑊



= =



+



0.0002



+



1 0.01736°K/W



0.01157°𝐾 𝑊



0.0002



0.02893°𝐾/𝑊



Rp =



0.0002 0.0002 0.02893°𝐾/𝑊



Rp = 0.006913°K/W



Rt = Ra + Rp + Rd Rt = 0.007936°K/W + 0.006913°K/W + 0.055555°K/W Rt = 0.070398 °K/W



Q= Q= Q=



𝑇1−𝑇2 𝑅𝑡 473°𝐾−323°𝐾 0.070398 K/W 150 °𝐾 0.070398 °K/W



Q = 2130.7424 W