I.TINH TOAN TRUYEN NHIET DO DAN NHIET HINH TRU

Cong thuc tinh dien tich trung binh logarit

Alm=¿(A 2−A1 )

ln (A 2 )−ln (A 1)

Cong thuc tinh nhiet tro:

Rdn=r 2−r 1Alm K

Cong thuc tinh ton that nang luong:

Q=T 2−T 1Rdn

Bai tap 1:

Ong thep co duong kinh trong 6cm va day 2cm co he so dan nhiet 43(W/Mc) duoc lam ong dan hoi nuoc voi nhiet do 115 C.Neu nhiet do ben ngoai ong la 90.Tinh ton that nang luong voi moi truong xung quanh.

Bai giai:

R1=0,03m

R2=0,05m R1=0,03m R2 R2

90C

1.dien tinh trung binh logarit

115C

115C

R

Alm=(A2−A1)

ln ( A2 )−ln (A1)= 2 πLR2−2πLR1

ln (2 πLR2 )−ln (2πLR 1 )

= 2∗3,14∗40∗0,05−2∗3,14∗40∗0,03ln (2∗3,14∗40∗0,05 )− ln (2∗3,14∗40∗0,03)

= 9,84m2.

2.Tinh toan nhiet tro:

Rd n=R2−R1A lmK

=0,05−0,039,84∗43

=0,0000473( CW

)

3.Tinh toan ton that nang luong:

Q=T 1−T 2Rdn

= 115−900,0000473

=528,541(w).

BAI TAP 2

1 lop cach nhiet day 2cm co he so dan nhiet 0,02(W/mC) bao quanh mot duong ong co duong kinh trong 4cm va day 2mm dai 5m co he so dan nhiet 16(W/ mC).Nhiet do ben trong 90C, nhiet do ben ngoai 35.Tinh nhiet do ben ngoai ong kim loai

BAI GIAI

1.ve so do R1=0,02m, R2=0,02+0,002=0,022m

R3=0,02m+ 0,002m+0,02m=0,042m

R2R1

1.Tinh dien tich trung binh logarit

Alm 1=A 2−A 1

ln (A 2 )− ln ( A1 )= 2πLR2−2 πlR1

ln (2πlR2 )−ln (2πLR 1 )

¿ 2∗3,14∗5∗0,022−2∗3,14∗5∗0,02ln (2∗3,14∗5∗0,022 )−ln (2∗3,14∗5∗0,02)

=¿ 0,6589m2.

Alm 2=A 3−A2

ln (A 3 )−ln (A2 ) =2πLR 3−2 πlR2

ln (2πlR3 )−ln (2πLR 2 )

= 2∗3,14∗5∗0,041−2∗3,14∗5∗0,022ln (2∗3,14∗5∗0,041 )− ln (2∗3,14∗5∗0,02 2)=0,958m2

2.Tinh toan nhiet tro

R1=R2−R1A lm1 K1

=0,022−0,020,6589∗16

=0,00019( CW

)

R2=R3−R2A lm2 K 2

=¿ 0,041−0,0220,958∗0,02

=0,991( CW )TONG NHIET TRO:

R=R1+R2=0,9912 (C/W)

3.NHIET LUONG TRUYEN QUA HE THONG

Q=T 1−T 3R

=90−350,9912

=55.48W

DO truyen nhiet on dinh nen nhiet luong truyen khong doi

Q=T 1−T 2R1

VAY T 2=T 1−QR1=¿ 90-55.48*0,00019=89,98C

Q= T 2−T 3R2

VAY T 2=T 3+QR 2=35+55.48∗0.991=89.98C