JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II PROCESS HEAT TRANSFER Previous Question Paper (November/December - 2020 - R13)
Answer any Five Questions. All Questions Carry Equal Marks.
Main
A hollow sphere has an inside surface temperature 300°C and outside surface temperature of 30°C. Calculate the heat loss by conduction for an inside diameter of 5 cm and outside diameter of 15 cm. The thermal conductivity of material is 15 kcal/hr m °C.
Answer
For a hollow sphere, the steady-state heat conduction rate Q is determined by the formula:
Given parameters:
Inside temperature (Ti) = 300°C
Outside temperature (To) = 30°C
Inside radius (ri) = 2.5 cm = 0.025 m
Outside radius (ro) = 7.5 cm = 0.075 m
Thermal conductivity (k) = 15 kcal/hr m °C
Substituting the values into the formula:
The total heat loss by conduction through the hollow sphere is 1908.5 kcal/hr.
Calculate the rate of heat loss through a plane of homogeneous wall whose thickness is considerably smaller than its width and length. Assuming the wall to be made from steel (k = 40 W/m K), concrete (k = 1.2 W/m K), Diatomite brick (k = 0.12 W/m K). Wall thickness for all the materials is equal to 5 cm. The inner and outer surfaces are at 1000°C and 100°C respectively.
Answer
To calculate the rate of heat loss through a composite wall consisting of three layers in series, we use the thermal resistance concept. The total thermal resistance Rtotal is the sum of the individual resistances of each layer.
For a plane wall, the thermal resistance is given by the formula:
Given that the thickness L = 0.05 m for each layer, and assuming a unit area A = 1 m2, the resistances are:
- Resistance of steel (R1): 0.05 / (40 × 1) = 0.00125 K/W
- Resistance of concrete (R2): 0.05 / (1.2 × 1) = 0.04167 K/W
- Resistance of Diatomite brick (R3): 0.05 / (0.12 × 1) = 0.41667 K/W
The total resistance is:
The rate of heat transfer Q is calculated using the temperature difference ΔT = 1000°C - 100°C = 900 K:
Thus, the rate of heat loss per unit area through the composite wall is approximately 1958.27 W/m2.
A counter flow heat exchanger is used for cooling oil 2500 kg/hr of oil from 160°C temperature to 80°C. Cooling water enters the tubes at 35°C at a flow rate of 4000 kg/hr. The predicted clean heat transfer coefficient is 960 W/m2°C. The heat transfer area is 4.1 m2. The specific heat of oil is 0.72 cal/gm °C. What is the outlet temperature of oil?
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Discuss about various types of fluid flows in a heat exchanger.
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Air is heated in an air heater from 30 to 110°C, and the hot gases cool from 210 to 145°C. Calculate the LMTD between air and gas with two fluids in parallel and counter flows?
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Derive the relation between Graetz and Peclet numbers and show that NGz = π/NFo.
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Write a short note on “heat transfer to liquid metals”?
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Describe the mechanism of boiling with the help of q/A Vs ΔT curve. Indicate different regimes of boiling.
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Derive the total heat flux equation for combined heat transfer by conduction-convection and radiation.
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Compare the merits and demerits of different methods of feeding in multiple effect evaporators.
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