Question

a long hollow pipe has an inner dia pf 10cm and outer dia of 20cm.the inner surface is kept at 200 degrees c and the outer surface at 50 degree c the thermal conductivity is 0.12.how much heat is lost per minute for a portion of pipe 20m long find the temperature at a distance x=7.5cm from the centre of the pipe

Answer 1

Answer:

Rate of the heat transfer is $16.8 \ kW$

Step-by-step explanation:

Given:

Length of the cylindrical pipe $(\mathrm{I})=10 \mathrm{~cm}$

The inner radius of the cylindrical pipe $\left(r_{1}\right)=10 \mathrm{~cm}$

The outer radius of the cylindrical pipe $\left(r_{2}\right)=20 \mathrm{~cm}$

Temperature difference ${T}=200-50=150^{0} C$

Thermal conductivity $(k)=150 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$

Then,

$\begin{aligned}&\text { Rate of heat transfer }(Q)=\frac{2 \pi L k \Delta T}{\ln \left(\frac{r_{2}}{r_{1}}\right)} \\&=\frac{2 \times 3.14 \times 0.1 \times 150 \times 0.20}{\ln \left(\frac{20}{10}\right)} \\&=16.8 \mathrm{~kW}\end{aligned}$

Answer 2

Answer:

The heat loss per minute is 1.69 J/minute

Explanation:

Given data :

Length of the pipe(L) = 20 m

Thermal Conductivity (k) = 0.12

Inner Radius (r1) = 20 cm = 0.2 m

Outer Radius (r2) = 10 cm = 0.1 m

Difference between temperatures (T) = 200 - 50 = 150 ° C

Formula to be used :

Q = -k (A/L) T

Area (A) = 3.14 [ (0.2)^2 - (0.1)^1] = 0.0314 m^2

Obtaining Results :

Q = -(0. 12 × 0.0314 × 150 ) /20

Q = -0.02826 J s^-1

Heat Loss in one minute = 0.02826 × 60 = 1.69 J / minute.

Concept :

• The quantity of heat that is transmitted through a material per unit of time, often measured in watts, is known as the rate of heat flow (joules per second).
• The term "heat flow" is redundant (i.e., a pleonasm) because heat is the flow of thermal energy caused by thermal non-equilibrium.

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