#### GATE Practice Questions - Heat Transfer

# GATE Practice Questions - Heat Transfer

GATE Practice Questions for Heat Transfer is a highly searched term by GATE aspirants on the web, so we have decided to share GATE Practice questions on important topics in GATE syllabus with the students. In this post, we will share the questions on the topic of Heat Transfer in GATE syllabus. Feel free to share your queries and doubts on the shared questions in the comment area or email us at info@mindvis.in.

**Q1) **A composite hollow sphere with steady internal heating is made of 2 layers of materials of equal thicknesses with thermal conductivities in the ratio of 1:2 for inner to outer layers. Ratio of inside to outside diameters is 0.8. What is ratio of temperature drops across the inner and outer layers?

- 0.4
- 1.6
- 2.8
- 2.5

**Sol) **For spherical composite shell,

Let 'r' be the radius at the junction and 't' be the thickness of each layer.

Now

and

From Fourier's law of heat conduction,

Now,

As

**Q2) **A large concrete slab 1 m thick has one-dimensional temperature distribution . Where, T is temperature and x is distance from one face towards other face of slab. If the Slab material has thermal diffusivity of ,what is the rate of change of temperature at the other face of the slab?

**Sol) **Temperature distribution along one face to the other of a slab:

Thermal diffusivity: =2x10^{-3}m^{2}/h, Length of slab: L=1 m

To be determined: rate of change of temperature at other face i.e,

From 1D general heat conduction equation,

**Q3)** The temperature on the two surfaces of 25mm thick steel plate (K=480 W/m^{o}C) having a uniform volumetric generation of 30x10^{6} W/m^{3} are 190^{0}Cand 130^{0}C.Neglect end effects. The value of maximum temperature is ___________^{0}C.

**Sol) **K_{steel}=480 W/m-^{0}C, L_{plate}=25 mm,

t_{w1}=190^{0}C, t_{w2}=130^{0}C, q_{g}=30x10^{6} W/m^{3}

To be determined: Maximum temperature of the plate i.e, t_{max}

Now the temperature distribution of a plate with different temperatures at both end is:

, putting the values:

t(x)=190+5412.5x-312500x^{2}, for maximum value of temperature find ‘x’ from

dt/dx=0 5412.5-625000x=0 x=0.00866 m. Now putting the value in t(x)

t_{max}=213.44^{0}C

**Q4)** A thermocouple measures the temperature of a fluid having the property of spherical junction as specific heat= 400 J/kg-K, diameter of junction =3mm. The heat transfer coefficient is 40 W/m^{2}-K. The junction is initially kept at 30^{0}Cand it is immersed into the fluid temperature maintained at 360^{0}C. Find the temperature of junction after 8 sec.

**Sol) ** Here check whether Lumped heat analysis is applicable or not, i.e. Bi < 0.1

and for sphere

Apply . Solve for T and get T =

**Q5) **A thin flat plate 2m x 2m is hanging freely in air. The temperature of the surrounding is 25^{o}C. Solar radiation is falling on one side of the plate at the rate of 500 W/m^{2}. what should be the convective heat transfer coefficient in W//m^{2o}C, if the temperature of the plate is to remain constant at 30^{o}C?

- 25
- 100
- 50
- 200

**Sol) **Here solar radiation falls on one side but heat transfer takes place from both sides.

**So, ,after solving h=50 W/m ^{2}-^{0}C.**

**Q6) **Two fluids A and B exchange heat in a countercurrent heat exchanger. Fluid A enters at 420^{o}Cand has a mass flow rate of 1 kg/s. Fluid B enters at 20^{o}Cand also has a mass flow rate 1 kg/s. Effectiveness of heat exchanger is 75%. (Specific heat of fluid A= 1 kJ/kg-K and that of fluid B= 4 kJ/kg-K). The exit temperature of fluid B is

**Sol) ** From effectiveness,

We know, heat gain by cold fluid = heat lost by hot fluid i.e,

m_{B}c_{pB}(T_{B2}-T_{B1}) = **m**_{A}**c**_{pA}**(****T**_{A1}**-****T**_{A2}**), **putting the values: T_{B2}=95^{0}C.

**Q7)** A commercial aeroplane is modelled as a flat plate which is 1.5 *m *wide and 8 *m *long in size. It is maintained at 20^{o}C. the aeroplane is flying at a speed of 800 *km/h* in air at 0^{o}C and 60 cm of *Hg *pressure The properties of air at average temperature 10^{o}C, K=2.511x10^{-2}W/mK, =14.16x10^{-6}m^{2}/s, Pr= 0.705.

For laminar flow; Nu= 0.064Re^{1/2}Pr^{1/3}, for turbulent flow; Nu= 0.064Re^{0.8}Pr^{0.33 }Find the heat transfer coefficient in W/m^{2}-K.

- 253
- 303
- 375
- 408

**Sol) C**heck whether laminar or turbulent: i.e, '

Using the relation for turbulent flow Nu= 96661.33, and we know Nu=hL/K. Putting the values and solving we get, h = 303 W/m^{2}K

**Q8)** Configuration of a furnace can be approximated as an equilateral triangular duct which is sufficiently long that the end effects are negligible. The hot wall is maintained at T_{1}=1000 K. The cold wall is at T_{2}=500 K. The third wall is reradiating zone for which Q_{3}=0.Then the temperature (in K) of third wall is

- 605 K
- 928 K
- 853 K
- 492 K

**Sol) ** By symmetry,

Heat transfer from radiating surface,

Now

**Q9)** The LMTD of counter flow heat exchanger is 20^{o}C. The cold fluid enters at 20^{o}Cand the hot fluid enters at 100^{o}C. Mass flow rate of the cold fluid is twice than that of the hot fluid. Specific heat at constant pressure of the hot fluid is twice that of the cold fluid. The exit temperature of the cold fluid is

- Cannot be determined

**Sol) ** From energy conservation

Q_{h}=Q_{c} _{ }m_{h}c_{h}(T_{h1}-T_{h2})=m_{c}c_{c}(T_{c2}-T_{c1})

T_{h1}-T_{h2}=T_{c2}-T_{c1} _{ }T_{h1}-T_{c2}=T_{h2}-T_{c1}

As T_{1}=T_{2}, So LMTD==20^{0}C.

Now T_{h1}-T_{c2}=20^{0}C T_{c2}=80^{0}C.

**Q10)** Which non-dimensional number relates the thermal boundary layer and hydrodynamic boundary layer?

- Rayleigh number
- Prandtl number
- Grashof number
- Nusselt number

**Sol) ****The relation between thermal and hydrodynamic boundary layer is: **