(i) A cylindrical pipe of length L has inner and outer radii as a and b respectively. The inner surface of the pipe is at a temperature `T_(1)` and the outer surface is at a lower temperature of `T_(2)`. Calculate the radial heat current if conductivity of the mate- rial is K. (ii) A cylindrical pipe of length L has two layers of material of conduc- tivity `K_(1)` and `K_(2)`. (see figure). If the inner wall of the cylinder is maintained at `T_(1)` and outer surface is at `T_(2) (lt T_(1))`, calcu- late the radial rate of heat flow.

(i) A cylindrical pipe of length L has inner and outer radii as a and

1.	(i) A cylindrical pipe of length L has inner and outer radii as a and b respectively. The inner surface of the pipe is at a temperature `T_(1)` and the outer surface is at a lower temperature of `T_(2)`. Calculate the radial heat current if conductivity of the mate- rial is K. (ii) A cylindrical pipe of length L has two layers of material of conduc- tivity `K_(1)` and `K_(2)`. (see figure). If the inner wall of the cylinder is maintained at `T_(1)` and outer surface is at `T_(2) (lt T_(1))`, calcu- late the radial rate of heat flow.
Answer» Correct Answer - (i) `((T_(1)-T_(2))2piKL)/(ln((b)/(c)))` (ii) `(T_(1)-T_(2))/((1)/(2piK_(1)L)ln((r^(2))/(r^(1)))+(1)/(2piK_(2)L)ln((r_(3))/(r_(2))))`

Discussion

No Comment Found

Related InterviewSolutions

Consider two rods of same length and different specific heats (`S_1` and `S_2`), conductivities `K_1` and `K_2` and area of cross section (`A_1` and `A_2`) and both having temperature `T_1` and `T_2` at their ends. If the rate of heat loss due to conduction is equal thenA. `K_(1)=K_(2)`B. `K_(1)S_(1)=K_(2)S_(2)`C. `(K_(1))/(A_(1)S_(1))=(K_(2))/(A_(2)S_(2))`D. `K_(1)A_(1)=K_(2)A_(2)`
Two identical rectangular rods of metal are welded as shown in figure `(1)` and `20 J` of heat flows through the rods in `1 min`. How long would it take for `20 J` heat to flow through the rods if they are welded as shown in figure `(2)`
Two identical square rods of metal are welded end to end as shown in figure (a). Assume that `10` cal of heat flows through thr rod in `2min`. Now the rods are welded as shown in figure (b). The time it would take for `10` cal to flow through the rods now, is A. `0.75 min`B. `0.5 min`C. `1.5 min`D. `1 min`
A lake surface is exposed to an atmosphere where the temperature is `lt 0^(@)C`. If the thickness of the ice layer formed on the surface grows form `2cm` to `4cm` in `1` hour. The atmospheric temperature, `T_(a)` will be- (Thermal conductivity of ice `K = 4 xx 10^(-3) cal//cm//s//.^(@)C`, density of ice `= 0.9 gm//ice.` Latent heat of fustion of ice `= 80 cal//m`. Neglect the change of density during the state change. Assume that the water below the ice has `0^(@)` temperature every where)A. `-20^(@)C`B. `0^(@)C`C. `-30^(@)C`D. `-15^(@)C`
A copper slab is 2 mm thick. It is protected by a 2 mm layer of stainless steel on both sides. The temperature on one side of this composite slab is `400^(@)C` and `200^(@)C` on the other side. Value of thermal conductivities are- `k_(cu)=400 Wm^(-1)K^(-1) ` and `k_(s)=16 Wm^(-1)K^(-1)` (a) Just by knowing that thermal conductivity of steel is much less than copper, find (approximately) the temperature of the copper slab. (b) Plot the variation of temperature across the thickness of the composite wall.
monatomic ideal gas is contained in a rigid container of volume V with walls of totsl inner surface area A, thickness x and thermal condctivity K. The gas is at an initial temperature `t_(0)` and pressure `P_(0)` . Find tjr pressure of the gas as a function of time if the temperature of the surrounding air is `T_(s)` . All temperature are in absolute scale.
A and B are two sphere made of same material. Radius of A is double that of B and initially they are at same temperature (T). Both of them are kept far apart in a room at temperature `T_(0)(lt T)`. Calculate the ratio of initial rate of cooling (i.e. rate of fall of temperature) of sphere A and B if (a) the spheres are solid, (b) the spheres are hollow made of thin sheets of same thickness
The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is `1.4 "kilowatt"//m^2`. The average earth-sun distance is `1.5 xx 10^11` meters. The mass lost by the sun per day is.A. `4.4 xx 10^(9) kg`B. `7.6 xx 10^(14) kg`C. `3.8 xx 10^(12) kg`D. `3.8 xx 10^(14) kg`
Heat received by the Earth due to solar radiations is `1.35 KWm^(-2)`. It is also known that the temperature of the Earth’s crust increases `1^(@)C` for every 30 m of depth. The average thermal conductivity of the Earth’s crust is `K = 0.75 J (msK)^(-1)` and radius of the Earth is `R = 6400 km`. (i) Calculate rate of heat loss by the Earth’s core due to conduction. (ii) Assuming that the Earth is a perfect block body estimate the temperature of its surface.
The container `A` is constantly maintained at `100^(@)C` and insulated container `B` in the contains ice at `0^(@)C` Different rods are used to connect them For a rod made of copper, it takes 30 mintes for the ice to melt and for a rod of steel of same cross-section taken in different experiment it takes 60 minutes for ice to melt. When these rods are simultaneously connected in parallel, the ice melts is .

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment