enters at 30c C with a mass flow rate of 2.09 kg/ s. If the effectiveness of the heat exchanger is 0.8, the LMTD ( in c C)

Transcription

1 CHAPTER 7 HEAT TRANSFER YEAR 0 ONE MARK MCQ 7. For an opaque surface, the absorptivity ( α ), transmissivity ( τ ) and reflectivity ( ρ ) are related by the equation : (A) α+ ρ τ (B) ρ+ α+ τ 0 (C) α+ ρ (D) α+ ρ 0 MCQ 7. Which one of the following configurations has the highest fin effectiveness? (A) Thin, closely spaced fins (B) Thin, widely spaced fins (C) Thick, widely spaced fins (D) Thick, closely spaced fins YEAR 0 TWO MARKS MCQ 7.3 Consider two infinitely long thin concentric tubes of circular cross section as shown in the figure. If D and D are the diameters of the inner and outer tubes respectively, then the view factor F is give by (A) D b D l (B) zero (C) D b D l (D) D D b l MCQ 7.4 Water ( c p 4.8 kj/ kgk) at 80c C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/ s. Air ( c p kj/ kgk) enters at 30c C with a mass flow rate of.09 kg/ s. If the effectiveness of the heat exchanger is 0.8, the LMTD ( in c C) is (A) 40 (B) 0 (C) 0 (D) 5 Published by: NODIA and COMPANY ISBN:

2 PAGE 304 HEAT TRANSFER CHAP 7 YEAR 0 ONE MARK MCQ 7.5 MCQ 7.6 In a condenser of a power plant, the steam condenses at a temperatures of 60c C. The cooling water enters at 30c C and leaves at 45c C. The logarithmic mean temperature difference (LMTD) of the condenser is (A) 6.c C (B).6cC (C) 30c C (D) 37.5cC A pipe of 5 mm outer diameter carries steam. The heat transfer coefficient between the cylinder and surroundings is 5 W/ m K. It is proposed to reduce the heat loss from the pipe by adding insulation having a thermal conductivity of 0.05 W/ mk. Which one of the following statements is TRUE? (A) The outer radius of the pipe is equal to the critical radius. (B) The outer radius of the pipe is less than the critical radius. (C) Adding the insulation will reduce the heat loss. (D) Adding the insulation will increases the heat loss. YEAR 0 TWO MARKS MCQ 7.7 MCQ 7.8 A spherical steel ball of mm diameter is initially at 000 K. It is slowly cooled in surrounding of 300 K. The heat transfer coefficient between the steel ball and the surrounding is 5 W/ m K. The thermal conductivity of steel is 0 W/ mk. The temperature difference between the centre and the surface of the steel ball is (A) large because conduction resistance is far higher than the convective resistance. (B) large because conduction resistance is far less than the convective resistance. (C) small because conduction resistance is far higher than the convective resistance. (D) small because conduction resistance is far less than the convective resistance. The ratios of the laminar hydrodynamic boundary layer thickness to thermal boundary layer thickness of flows of two fluids P and Q on a flat plate are / and respectively. The Reynolds number based on the plate length for both the flows is 0 4. The Prandtl and Nusselt numbers for P are /8 and 35 respectively. The Prandtl and Nusselt numbers for Q are respectively (A) 8 and 40 (B) 8 and 70 (C) 4 and 40 (D) 4 and 35 Published by: NODIA and COMPANY ISBN:

3 CHAP 7 HEAT TRANSFER PAGE 305 YEAR 00 TWO MARKS MCQ 7.9 A fin has 5 mm diameter and 00 mm length. The thermal conductivity of fin material is 400 Wm K. One end of the fin is maintained at 30c C and its remaining surface is exposed to ambient air at 30c C. If the convective heat transfer coefficient is 40 Wm K, the heat loss (in W) from the fin is (A) 0.08 (B) 5.0 (C) 7.0 (D) 7.8 YEAR 009 ONE MARK MCQ 7.0 A coolant fluid at 30c C flows over a heated flat plate maintained at constant temperature of 00c C. The boundary layer temperature distribution at a given location on the plate may be approximated as T exp( y) where y (in m) is the distance normal to the plate and T is in c C. If thermal conductivity of the fluid is.0 W/ mk, the local convective heat transfer coefficient (in W/ m K) at that location will be (A) 0. (B) (C) 5 (D) 0 YEAR 009 TWO MARKS MCQ 7. MCQ 7. In a parallel flow heat exchanger operating under steady state, the heat capacity rates (product of specific heat at constant pressure and mass flow rate) of the hot and cold fluid are equal. The hot fluid, flowing at kg/ s with cp 4 kj/kg K, enters the heat exchanger at 0c C while the cold fluid has an inlet temperature of 5c C. The overall heat transfer coefficient for the heat exchanger is estimated to be kw/m K and the corresponding heat transfer surface area is 5m. Neglect heat transfer between the heat exchanger and the ambient. The heat exchanger is characterized by the following relations: ε exp( NTU) The exit temperature (in c C) for the cold fluid is (A) 45 (B) 55 (C) 65 (D) 75 Consider steady-state conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides. Published by: NODIA and COMPANY ISBN:

4 PAGE 306 HEAT TRANSFER CHAP 7 Given : hi 0 W/m K, ho 50 W/m K;, T3,i 0cC; T3,o cc, k 0 W/mK; k 50 W/mK; L 0.30 m and L 0.5 m. Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in c C), of the two walls will be (A) 0.50 (B).75 (C) 3.75 (D) 4.50 Common Data For Q. 3 and Q.4 Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate ) is a black surface and is the warmer one being maintained at 77cC, the lower plate (plate ) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 7c C. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady-state conditions to exits. Stefan-Boltzmann constant is given as 5.67 # 0 8 W/m K 4 MCQ 7.3 The irradiation (in kw/ m ) for the plate (plate ) is (A).5 (B) 3.6 (C) 7.0 (D) 9.5 MCQ 7.4 If plate is also diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kw/m ) between plate and plate is (A) 7.0 (B) 9.5 (C) 3.0 (D) 3.7 YEAR 008 ONE MARK MCQ 7.5 For flow of fluid over a heated plate, the following fluid properties are known Viscosity 0.00Pa-s; Specific heat at constant pressure kj/ kg. K; Thermal conductivity W/ m K Published by: NODIA and COMPANY ISBN:

5 CHAP 7 HEAT TRANSFER PAGE 307 The hydrodynamic boundary layer thickness at a specified location on the plate is mm. The thermal boundary layer thickness at the same location is (A) 0.00 mm (B) 0.0 mm (C) mm (D) 000 mm YEAR 008 TWO MARKS MCQ 7.6 MCQ 7.7 The logarithmic mean temperature difference (LMTD) of a counter flow heat exchanger is 0c C. The cold fluid enters at 0c C and the hot fluid enters at 00c C. Mass flow rate of the cold fluid is twice 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 (A) is 40c C (B) is 60cC (C) is 80c C (D) cannot be determined For the three-dimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer coefficient of 0 W/ m K. The ambient temperature is 30c C. Heat is uniformly generated inside the object at the rate of 00 W/ m 3. Assuming the face PQRS to be at uniform temperature, its steady state temperature is (A) 0c C (B) 0cC (C) 30c C (D) 40cC MCQ 7.8 A hollow enclosure is formed between two infinitely long concentric cylinders of radii m and m, respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface-) and the outer surface of the smaller cylinder (surface-). The radiating surfaces are diffuse and the medium in the enclosure is non-participating. The fraction of the thermal radiation leaving the larger surface and striking itself is Published by: NODIA and COMPANY ISBN:

6 PAGE 308 HEAT TRANSFER CHAP 7 (A) 0.5 (B) 0.5 (C) 0.75 (D) MCQ 7.9 Steady two-dimensional heat conduction takes place in the body shown in the figure below. The normal temperature gradients over surfaces P and Q can be considered to be uniform. The temperature gradient T/ x at surface Q is equal to 0 K/ m. Surfaces P and Q are maintained at constant temperature as shown in the figure, while the remaining part of the boundary is insulated. The body has a constant thermal conductivity of 0. W/ mk. The values of T and T at surface P are x y (A) T 0 K/ m, T 0 K/ m x y (B) T 0 K/ m, T 0 K/ m x y (C) T 0 K/ m, T 0 K/ m x y (D) T 0 K/ m, T 0 K/ m x y YEAR 007 TWO MARKS MCQ 7.0 The temperature distribution within the thermal boundary layer over a heated isothermal flat plate is given by Published by: NODIA and COMPANY ISBN:

7 CHAP 7 HEAT TRANSFER PAGE 309 T T y y T T 3 3 w b w δ l 3 t bδ l, t where T w and T 3 are the temperature of plate and free stream respectively, and y is the normal distance measured from the plate. The local Nusselt number based on the thermal boundary layer thickness δ t is given by (A).33 (B).50 (C).0 (D) 4.64 MCQ 7. MCQ 7. In a counter flow heat exchanger, hot fluid enters at 60c C and cold fluid leaves at 30c C. Mass flow rate of the fluid is kg/ s and that of the cold fluid is kg/ s. Specific heat of the hot fluid is 0 kj/ kgk and that of the cold fluid is 5 kj/ kgk. The Log Mean Temperature Difference (LMTD) for the heat exchanger in c C is (A) 5 (B) 30 (C) 35 (D) 45 The average heat transfer co-efficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assume the plate temperature to be uniform at any instant of time and radiation heat exchange with the surroundings negligible. The ambient temperature is 5c C, the plat has a total surface area of 0m. and a mass of 4kg. The specific heat of the plate material is.5 kj/ kgk. The convective heat transfer co-efficient in W/m K, at the instant when the plate temperature is 5c C and the change in plate temperature with time dt/ dt 0.0 K/ s, is (A) 00 (B) 0 (C) 5 (D) 0 MCQ 7.3 MCQ 7.4 Common Data For Q. 3 and Q.4 Consider steady one-dimensional heat flow in a plate of 0 mm thickness with a uniform heat generation of 80 MW/ m 3. The left and right faces are kept at constant temperatures of 60c C and 0c C respectively. The plate has a constant thermal conductivity of 00 WmK. / The location of maximum temperature within the plate from its left face is (A) 5 mm (B) 0 mm (C) 5 mm (D) 0 mm The maximum temperature within the plate in c C is (A) 60 (B) 65 (C) 00 (D) 50 Published by: NODIA and COMPANY ISBN:

8 PAGE 30 HEAT TRANSFER CHAP 7 YEAR 006 ONE MARK MCQ 7.5 In a composite slab, the temperature at the interface (T inter ) between two material is equal to the average of the temperature at the two ends. Assuming steady one-dimensional heat conduction, which of the following statements is true about the respective thermal conductivities? (A) k k (B) k k (C) k 3k (D) k k YEAR 006 TWO MARKS MCQ 7.6 A 00 W electric bulb was switched on in a 5. m# 3m# 3m size thermally insulated room having a temperature of 0c C. The room temperature at the end of 4 hours will be (A) 3c C (B) 34cC (C) 450c C (D) 470cC MCQ 7.7 A thin layer of water in a field is formed after a farmer has watered it. The ambient air conditions are : temperature 0c C and relative humidity 5%. An extract of steam tables is given below. Temp( cc) Saturation Pressure (kpa) Neglecting the heat transfer between the water and the ground, the water temperature in the field after phase equilibrium is reached equals (A) 0. 3c C (B) 0.3cC (C) 4.5cC (D) 4.5cC MCQ 7.8 With an increase in the thickness of insulation around a circular pipe, heat loss to surrounding due to (A) convection increase, while that the due to conduction decreases Published by: NODIA and COMPANY ISBN:

9 CHAP 7 HEAT TRANSFER PAGE 3 (B) convection decrease, while that due to conduction increases (C) convection and conduction decreases (D) convection and conduction increases YEAR 005 ONE MARK MCQ 7.9 MCQ 7.30 In a case of one dimensional heat conduction in a medium with constant properties, T is the temperature at position x, at time t. Then T is t proportional to (A) T (B) T x x (C) T x t (D) T x The following figure was generated from experimental data relating spectral black body emissive power to wavelength at three temperature T, T and T ( T > T > T) 3 3. The conclusion is that the measurements are (A) correct because the maxima in E bλ show the correct trend (B) correct because Planck s law is satisfied (C) wrong because the Stefan Boltzmann law is not satisfied (D) wrong because Wien s displacement law is not satisfied YEAR 005 TWO MARKS MCQ 7.3 Heat flows through a composite slab, as shown below. The depth of the slab is m. The k values are in W/ mk. The overall thermal resistance in K/ W is Published by: NODIA and COMPANY ISBN:

10 PAGE 3 HEAT TRANSFER CHAP 7 (A) 7. (B).9 (C) 8.6 (D) 39. MCQ 7.3 MCQ 7.33 A small copper ball of 5mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/ mk, its density 9000 kg/ m 3 and its specific heat 385 JkgK. / If the heat transfer coefficient is 50 W/ m K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s, (A) 8.7 (B) 3.9 (C) 7.3 (D) 7.7 A solid cylinder (surface ) is located at the centre of a hollow sphere (surface ). The diameter of the sphere is m, while the cylinder has a diameter and length of 0.5 m each. The radiation configuration factor F is (A) (B) 0.65 (C) 0.75 (D) MCQ 7.34 Hot oil is cooled from 80 to 50c C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to 40c C. The designer uses a LMTD value of 6c C. The type of heat exchange is (A) parallel flow (B) double pipe (C) counter flow (D) cross flow Common Data For Q. 35 and Q.36 An uninsulated air conditioning duct of rectangular cross section m# 0. 5m, carrying air at 0c C with a velocity of 0 m/ s, is exposed to an ambient of 30c C. Neglect the effect of duct construction material. For air in the range of 0 30cC, data are as follows; thermal conductivity 0.05 W / mk ; viscosity 8 μpas, Prandtl number 0.73; density. kg/m 3. The laminar flow Nusselt number is 3.4 for constant wall temperature conditions and for turbulent flow, Nu 0.03Re Pr Published by: NODIA and COMPANY ISBN:

11 CHAP 7 HEAT TRANSFER PAGE 33 MCQ 7.35 MCQ 7.36 The Reynolds number for the flow is (A) 444 (B) 890 (C) 4.44 # 0 5 (D) 5.33 # 0 5 The heat transfer per meter length of the duct, in watts is (A) 3.8 (B) 5.3 (C) 89 (D) 769 YEAR 004 ONE MARK MCQ 7.37 One dimensional unsteady state heat transfer equation for a sphere with heat generation at the rate of q can be written as (A) r r r T q T b r l + (B) r T q T k α t r r b r l+ k α t (C) T q T + (D) q ( rt) T r k α + t r k α t YEAR 004 TWO MARKS MCQ 7.38 A stainless steel tube ^k s 9 W/ mkh of cm ID and 5cm OD is insulated with 3cm thick asbestos ^k a 0. W/ mkh. If the temperature difference between the innermost and outermost surfaces is 600c C, the heat transfer rate per unit length is (A) 0.94 W/m (B) 9.44 W/m (C) W/m (D) W/m MCQ 7.39 MCQ 7.40 A spherical thermocouple junction of diameter mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 WmK /. Thermo-physical properties of thermocouple material are k 0 W/mK, c 400 J/kg K and ρ 8500 kg/m 3. If the thermocouple initially at 30c C is placed in a hot stream of 300c C, then time taken by the bead to reach 98c C, is (A).35 s (B) 4.9 s (C) 4.7 s (D) 9.4 s In a condenser, water enters at 30c C and flows at the rate 500 kg/ hr. The condensing steam is at a temperature of 0c C and cooling water leaves the condenser at 80c C. Specific heat of water is 4.87 kj/ kgk. If the overall heat transfer coefficient is 000 W/m K, then heat transfer area is (A) m (B) 7.07 m (C) 70.7 m (D) 4.4 m Published by: NODIA and COMPANY ISBN:

12 PAGE 34 HEAT TRANSFER CHAP 7 YEAR 003 ONE MARK MCQ 7.4 A plate having 0 cm area each side is hanging in the middle of a room of 00 m total surface area. The plate temperature and emissivity are respectively 800 K and 0.6. The temperature and emissivity values for the surfaces of the room are 300 K and 0.3 respectively. Boltzmann s constant 8 4 σ 5.67 # 0 W/m K. The total heat loss from the two surfaces of the plate is (A) 3.66 W (B) 7.3 W (C) 7.87 W (D) 3.66 MW YEAR 003 TWO MARKS MCQ 7.4 MCQ 7.43 In a counter flow heat exchanger, for the hot fluid the heat capacity kj/kgk, mass flow rate 5 kg/s, inlet temperature 50cC, outlet temperature 00cC. For the cold fluid, heat capacity 4kJ/kgK, mass flow rate 0 kg/s, inlet temperature 0cC. Neglecting heat transfer to the surroundings, the outlet temperature of the cold fluid in c C is (A) 7.5 (B) 3.5 (C) 45.5 (D) 70.0 Consider a laminar boundary layer over a heated flat plate. The free stream velocity is U 3. At some distance x from the leading edge the velocity boundary layer thickness is δ v and the thermal boundary layer thickness is δ T. If the Prandtl number is greater than, then (A) δ > δ (B) δ > δ v T / (C) δ. δ + ( U3x) (D) δ. δ + x / v T T v v T Common Data For Q. 44 and Q.45 Heat is being transferred by convection from water at 48c C to a glass plate whose surface that is exposed to the water is at 40c C. The thermal conductivity of water is 0.6 WmK / and the thermal conductivity of glass is. W/ mk. The spatial gradient of temperature in the water at the waterglass interface is dt/ dy # 0 K/m. 4 Published by: NODIA and COMPANY ISBN:

13 CHAP 7 HEAT TRANSFER PAGE 35 MCQ 7.44 MCQ 7.45 The value of the temperature gradient in the glass at the water-glass interface in K/m is (A) # 0 4 (B) 0.0 (C) 0.5 # 0 4 (D) # 0 4 The heat transfer coefficient h in W/m K is (A) 0.0 (B) 4.8 (C) 6 (D) 750 YEAR 00 ONE MARK MCQ 7.46 For the same inlet and outlet temperatures of hot and cold fluids, the Log mean Temperature Difference (LMTD) is (A) greater for parallel flow heat exchanger than for counter flow heat exchanger (B) greater for counter flow heat exchanger than for parallel flow heat exchanger (C) same for both parallel and counter flow heat exchangers (D) dependent on the properties of the fluids. YEAR 00 ONE MARK MCQ 7.47 For the circular tube of equal length and diameter shown below, the view factor F 3 is 07.. The view factor F in this case will be (A) 07. (B) 0. (C) 079. (D) 083. MCQ 7.48 In descending order of magnitude, the thermal conductivity of (a) pure iron, (b) liquid water, (c) saturated water vapour and (d) aluminum can be arranged as (A) abcd (B) bcad (C) dabc (D) dcba ********** Published by: NODIA and COMPANY ISBN:

14 PAGE 36 HEAT TRANSFER CHAP 7 SOLUTION SOL 7. Option (C) is correct. The sum of the absorbed, reflected and transmitted radiation be equal to α+ ρ+ τ α Absorpivity, ρ Reflectivity, τ Transmissivity For an opaque surfaces such as solids and liquids τ 0, Thus, α+ ρ SOL 7. Option (A) is correct. The performance of the fins is judged on the basis of the enhancement in heat transfer area relative to the no fin case. The fin effectiveness ε fin Heat transfer rate from the fin of base area Heat transfer rate from the surface area When determining the rate of heat transfer from a finned surface, we must consider the unfinned portion of the surface as well as the fins and number of fins. Thin and closed spaced fin configuration, the unfinned portion of surface is reduced and number of fins is increased. Hence the fin effectiveness will be maximum for thin and closely spaced fins. SOL 7.3 Option (D) is correct. According to the reciprocity relation. AF AF Which yields F A F DL D A # π π DL # bd l F 0 since no radiation leaving surface and strikes F, since all radiation leaving surface and strikes The view factor F is determined by applying summation rule to surface, F + F Thus F F D D b l SOL 7.4 Option (C) is correct. Given : 80cC, 30cC, mo h 0.5 kg/ sec, mo c.09 kg/ sec., ε 0.8 t h t c Published by: NODIA and COMPANY ISBN:

15 CHAP 7 HEAT TRANSFER PAGE 37 Capacity rate for hot fluid C h 4.8 # kj/ Ksec. C c # kj/ K sec. So, C h C c Qo ( th th) Ch Effectiveness ε Q o ( t t ) C max h c c th or, 80 t h 40 t h 40cC From energy balance, Ch( th th) Cc( tc tc) t c 30 t c 70cC Now LMTD θ m θ θ θ...(i) θ θ th tc cC θ th tc cC θ θ...(ii) So LMTD is undefined Let θ x & θ xθ θ Put in equation (i), so ( ) θ m lim xθ θ θ x lim x" x θ x" x θ It is a 0 : 0D form, applying L-Hospital rule θ( 0) θ m lim lim xθ x" x" x θ m θ θ From equation (ii) θ t t cC m θ h c Published by: NODIA and COMPANY ISBN:

16 PAGE 38 HEAT TRANSFER CHAP 7 SOL 7.5 Option (B) is correct. SOL 7.6 Given : t t 60 C, 30cC, 45cC h h c From diagram, we have θ t t cC t c h c And θ th tc cC Now LMTD, θ m t c θ θ θ 30 5 bθ l b 30 5 l.6cc Option (C) is correct. Given : d 0 5 mm 0.05 m, r m, h 5 W/ m K, k 0.05 W/ mk Hence, Critical radius of insulation for the pipe is given by, r c k m h 5 r c < r 0 or r 0 > r c...(i) So, from equation (i) option a and b is incorrect. The critical radius is less than the outer radius of the pipe and adding the insulation will not increase the heat loss. Hence the correct statement is adding the insulation will reduce the heat loss. Published by: NODIA and COMPANY ISBN:

17 CHAP 7 HEAT TRANSFER PAGE 39 SOL 7.7 Option (D) is correct. 3 Given : D mm # 0 m, h 5 W/ m K, k 0 W/ mk For spherical ball, 0 m 6 0 # 3 # l volume πr 3 D surface area 4πR 6 The non-dimensional factor ( hl/ k ) is called Biot Number. It gives an indication of the ratio of internal (conduction) resistance to the surface (convection) resistance. A small value of Bi implies that the system has a small conduction resistance i.e., relatively small temperature gradient or the existence of a practically uniform temperature within the system. Biot Number, Bi hl 5# # k 0 Since, Value of Biot Number is very less. Hence, conduction resistance is much less than convection resistance. SOL 7.8 Option (A) is correct. H Given : δ bδ l and δh b δ l Here, And Th P Published by: NODIA and COMPANY ISBN: Th Q δ H "Thickness of laminar hydrodynamic boundary layer δ Th "Thickness of thermal boundary layer 4 ( Re) P ( Re) Q 0 ( Pr) P 8 ( Nu) P 35 For thermal boundary layer prandtl Number is given by, (For fluid Q) 3 ( Pr ) / H Q δ bδ l Th Q 3 ( Pr) Q () 8 For laminar boundary layer on flat plate, relation between Reynolds Number, Prandtl Number and Nusselt Number is given by, hl / / 3 Nu ( Re) ( Pr) k Since, Reynolds Number is same for both P and Q. 3 / ( Nu) P ( Pr) P So, ( Nu) 3 / Q ( Pr) Q 3 / 3 / ( Pr) Q () 8 ( Nu) Q 3 / # ( Nu) P 3 / # ( 35) ( Pr) P (/) 8 40 / # 35

18 PAGE 30 HEAT TRANSFER CHAP 7 SOL 7.9 SOL 7.0 Option (B) is correct. Given, d 5 mm m, l 00 mm 0. m, k 400 W/ mk T 0 30cC, T a 30cC, h 40 W/ m K Heat loss by the fin is given by, Q fin mkac( T0 Ta) tanh( ml)...(i) Perimeter p d 4 Cross sectional Area Ac π π 4 d d p (ii) Ac And m h p k ba l 40 c 400 # From equation(i), Q fin π # # ( )( 30 30) tanh( 80 0.) 4 # # # # 400 #. 96 # 0 # 00 # tanh( ) 7. 0 # W Option (B) is correct. Given : T 30cC, T 00cC, k.0 W/ mk, T exp( y)...(i) Under steady state conditions, Heat transfer by conduction Heat transfer by convection ka dt haδt A " Area of plate dy ka d ( ) haδt dy e y Solving above equation, we get ka( 70e y ) haδt At the surface of plate, y 0 Hence 70kA haδt h 70kA 70k AΔT ΔT 70 # ( 00 30) W/ m K SOL 7. Option (B) is correct. Given : Co h Co c, mh o kg/sec, c ph 4 kj/ kg K, t h 0cC, t c 5cC U kw/ m K, A 5m Published by: NODIA and COMPANY ISBN:

19 CHAP 7 HEAT TRANSFER PAGE 3 The figure shown below is for parallel flow. C o 4 / h mc o h ph kj sk The heat exchanger is characterized by the following relation, exp( NTU) ε..(i) For parallel flow heat exchanger effectiveness is given by exp[ NTU( + C)] ε...(ii) + C Comparing equation (i) and equation (ii), we get capacity ratio C Cc Cmin...(iii) Ch Cmax Applying energy balance for a parallel flow C ( t t ) C ( t t ) h h h c c c Cc th th From equation(iii) Ch tc tc th th tc tc Number of transfer units is given by, NTU UA # 5 5. Cmin 4 exp(. 5) Effectiveness, ε # Maximum possible heat transfer is, Q max C ( t t ) min h c 4 # 6 ( ) ( kw But Actual Heat transfer is, Q a εq max # kw And Q a C ( t t ) c c c 60 4( t 5) c t c cC Published by: NODIA and COMPANY ISBN:

20 PAGE 3 HEAT TRANSFER CHAP 7 SOL 7. Option (C) is correct. The equivalent resistance diagram for the given system is, R eq # L L ha i ka ka ha 0 A L L h k k h R eq i Published by: NODIA and COMPANY ISBN: m K/ W Q Heat flux, q ΔT Q A AR ΔT eq / R Under steady state condition, q T3i T o 3 k( T T) k( T T) hi( T3 i T)...(i) AReq L L T i T o 0 ( ) W/ m...(ii) AReq T3 i T 0 T From equation(i) hi ( 0 T). 5 0 T & T cC Again from equation(i), k( T T) q L 50 0 (. 75 T) T & T 3.75cC Alternative : Under steady state conditions, Heat flow from I to interface wall Heat flow from interface wall to O 0

21 CHAP 7 HEAT TRANSFER PAGE 33 ( T3, i T) ( T T,o) 3 L + L + ha i ka ka ha 0 T3, i T T T, o 3 L + L + h k k h i o ( 0 T) T ( ) ( 0 T) T ( 0 T). 86( T + ). 86T T cC 86 SOL 7.3 Option (D) is correct. 8 4 Given : σ b 5.67 # 0 W/ m K, T (7 + 73) K 500 K T ( ) K 000 K Let, α " The absorptivity of the gray surface E " The radiant energy of black surface E " The radiant energy of gray surface Now, Plate emits radiant energy E which strikes the plate. From it a part α E absorbed by the plate and the remainder ( E αe) is reflected back to the plate. On reaching plate, all the part of this energy is absorbed by the plate, because the absorptivity of plate is equal to one (it is a black surface). Irradiation denotes the total radiant energy incident upon a surface per unit time per unit area. Energy leaving from the plate is, E E + ( α) E...(i) Published by: NODIA and COMPANY ISBN:

22 PAGE 34 HEAT TRANSFER CHAP 7 Hence, E is the energy emitted by plate. 4 E εσ b T 0.7 # 5.67 # 0 # (500) # # 0 # 65 # 0 And fraction of energy reflected from surface is, ( α)e ( ασ ) T E 4 εσ b T W/ m # 0 8 ( 0. 7) #( 000) W/ m Now, Total energy incident upon plate is, E E + ( α) E W/ m 9.49 kw/ m, 9.5 kw/ m SOL 7.4 Option (D) is correct. Given : ε 08., ε 07. As both the plates are gray, the net heat flow from plate to plate per unit time is given by, Q εε σb ( T 4 T 4 ) σ ε + ε b( T T ) εε ε ε. [( ) ( ) ] # 5 67 # # # W/ m kw/ m SOL 7.5 Option (C) is correct. Given : μ 0.00 Pa s, c p kj/ kg K, k W/ mk The prandtl Number is given by, 3 μc Pr p # # 0 k And δ δ t Given, δ m δ / () 3 δ t hydrodynamic bondary layer thickness ( Pr) Thermal boundary layer thickness / δ δ t mm Hence, thermal boundary layer thickness at same location is mm. 3 SOL 7.6 Option (C) is correct. The T L curve shows the counter flow. Published by: NODIA and COMPANY ISBN:

23 CHAP 7 HEAT TRANSFER PAGE 35 Given : θm 0cC, t c 0cC, t h 00cC mo c m mc o o h & mo h...(i) cph c ph cpc & cpc...(ii) Energy balance for counter flow is, Heat lost by hot fluid Heat gain by cold fluid mc o ( t t ) mc o ( t t ) h ph h h c c ph pc c pc c c ( th th) mo c ( t c t c) mo h h ( t t ) ( t t ) t h c c t t t h c h c θ θ...(iii) And θ m θ θ...(iv) θ bθ l Substituting the equation (iii) in equation (iv), we get undetermined form. Let θ x, & θ θx...(v) θ Substitute θ in equation(iv), θ m lim x θ θ θ ( ) lim x...(vi) x " θ x x " x b θ l 0 : 0D form, So we apply L-Hospital rule, θ( 0) θ m lim lim x θ x" x" x θ m θ θ From equation(iii) Now we have to find exit temperature of cold fluid ( t c ), So, θ t t m θ h c t c t h θ m cC Published by: NODIA and COMPANY ISBN:

24 PAGE 36 HEAT TRANSFER CHAP 7 SOL 7.7 Option (D) is correct. 3 Given : h 0 W/ m K, T i 30cC, q g 00 W/ m Five faces of the object are insulated, So no heat transfer or heat generation by these five faces. Only sixth face (PQRS) interacts with the surrounding and generates heat. Hence, Heat generated throughout the volume Q Rate of heat Generated # Volume of object 00 #( # # ) 400 W And heat transfer by convection is given by Q ha( T T ) f #( # )( T f 30) T f cC i SOL 7.8 Option (B) is correct. Given : D m, D m Hence, the small cylindrical surface (surface ) cannot see itself and the radiation emitted by this surface strikes on the enclosing surface. From the conservation principal (summation rule). For surface, F + F F 0 F...(i) From the reciprocity theorem AF AF F A DL A π π. DL D D 05 and from the conservation principal, for surface, we have F + F F F So, the fraction of the thermal radiation leaves the larger surface and striking itself is F 05.. SOL 7.9 Option (D) is correct. Given : b T l 0 K/ m, ( T) P ( T) Q, ( k) P () k Q 0. W/ mk x Q Direction of heat flow is always normal to surface of constant temperature. So, for surface P, T 0 x Because, Q ka( T/ x) and T is the temperature difference for a short perpendicular distance dx. Let width of both the bodies are unity. From the law of energy conservation, Published by: NODIA and COMPANY ISBN:

25 CHAP 7 HEAT TRANSFER PAGE 37 Heat rate at P Heat rate at Q 0. T # # cy m 0. T # # bx l P Because for P heat flow in y direction and for Q heat flow in x direction T cy m 0. # # 0 0 K/ m 0. P Q SOL 7.0 SOL 7. Option (B) is correct. The region beyond the thermal entrance region in which the dimensionless temperature profile expressed as T Tw b T 3 T l remains unchanged is called w thermally fully developed region. Nusselt Number is given by, N u hl T k c y l m...(i) at yl 0 Here, T T Tw y and yl T3 Tw t y y So, N u y 3 3 l; bδ l t bδ le y ( y ) t y 3 3 : l l D yl 0 yl 0 y 3 3 ; bδ le. t 3 5 yl 0 Option (B) is correct. The counter flow arrangement of the fluid shown below : Given: for hot fluid : t h 60cC, mo h kg/sec, c h 0 kj/ kg K And for cold fluid : t c 30cC, mo c kg/sec, c c 5 kj/ kg K Heat capacity of Hot fluid, C h mc o h h # 0 0 kj/ k. sec And heat capacity of cold fluid, C c mc o c c # 5 0 kj/ k sec By energy balance for the counter flow mc o ( t t ) mc o ( t t ) h h h h c c c c C ( t t ) C ( t t ) C C h h h t c c c t t t h c h c Published by: NODIA and COMPANY ISBN: h c

26 PAGE 38 HEAT TRANSFER CHAP 7 θ θ LMTD, θ m θ θ...(i) θ bθ l Let, θ x θ is equal to θ θ and θ m is undetermined θ x θ Substituting θ in equation (i), we get, θ m lim x θ θ θ ( ) lim x x " () x x " () x 0 b form 0 l, So we apply L-hospital rule, θ m lim θ x # lim x θ x " x " θ m θ θ & θ t t cC h c SOL 7. Option (D) is correct. Given : T 5 cc (73 + 5) 98 K, A 0. m, m 4 kg, c.5 kj/ kg K h?, T 5cC K Temperature Gradient, dt 0.0 K/ s dt Here negative sign shows that plate temperature decreases with the time. From the given condition, Heat transfer by convection to the plate Rate of change of internal energy ha( T T) mc dt dt h 3 mc dt AT ( T) # 4#. 5# 0 ( 00. ) dt 0. ( ) # 0 W/ m K SOL 7.3 Option (C) is correct. Let the location of maximum temperature occurs at the distance x from the left face. We know that steady state heat flow equation in one dimension Published by: NODIA and COMPANY ISBN:

27 CHAP 7 HEAT TRANSFER PAGE 39 with a uniform heat generation is given by, T q g (i) x k Here q g Heat generated per unit volume and per unit time, 6 Given : q g 80 MW/ m 80 # 0 W/ m, k 00 W/ mk Substituting the value of q g and k in equation (i), we get 6 T # 0 x 00 T # 0 x Integrating the above equation, T + 4# 0 5 # x + c 0...(ii) x Again integrating, we get 5 T x # # + cx + c 0...(iii) Applying boundary conditions on equation (iii), we get () At x 0, T 60cC 60 + c 0 c 60...(iv) () At x 0 mm 0.00 m, T 0cC 5 (. 0 00) # 0 # + c ( 60) # + 0 c c 60 0 c 0. 00c (v) To obtain the location of maximum temperature, applying maxima-minima principle and put dt 0 in equation (ii), we get dx x # 0 x + ( 000) 0 c # 0 5# 0 3 m 5mm 4# 0 SOL 7.4 Option (B) is correct. From the previous part of the question, at x 5mm temperature is maximum. 3 So, put x 5mm 5# 0 m in equation(iii), we get 3 5 ( 5 ) T 4 0 # # # + ( 000) 5 0 ( 60) 0 # # T + 5 # 0 # Published by: NODIA and COMPANY ISBN:

28 PAGE 330 HEAT TRANSFER CHAP 7 T & T 65cC SOL 7.5 Option (D) is correct. Given : T T T inter + Heat transfer will be same for both the ends ka ( T Tinter) ka ( Tinter T) So, Q b b Q ka dt dx There is no variation in the horizontal direction. Therefore, we consider portion of equal depth and height of the slab, since it is representative of the entire wall. So, A A and T So, we get k T T T + ; b le k inter T + T k T T + : T D T T T : D k T T T + : D k [ T T ] k [ T T] k k SOL 7.6 Option (D) is correct. Given : P 00 W, ν.5 # 3 # 3.5 m 3, T i 0cC Now Heat generated by the bulb in 4 hours, Q 00 # 4 # 60 # MJ...(i) Volume of the room remains constant. Heat dissipated, Q mcvdt ρνcv( Tf Ti) m ρv Where, T f Final temperature of room ρ Density of air. kg/ m 3 c v of air 0.77 kj/ kg K Substitute the value of Q from equation (i), we get #. 5 # # 0 ( Tf 0) #. 5 # 0. 77( Tf 0) ( Tf 0) T f cC - 470cC SOL 7.7 Option (C) is correct. Given : Relation humidity 5% at temperature 0cC Relative humidity, Published by: NODIA and COMPANY ISBN:

29 CHAP 7 HEAT TRANSFER PAGE 33 Actual mass of water vapour in a given volume of moist air φ mass of water vapour in the same volume of saturated air at same temperature & pressure φ mv pv (i) ms ps Where, p v Partial pressure of vapor at 0cC From given table at T 0cC, p s.34 kpa From equation (i), p v 005. # ps 005. # kpa Phase equilibrium means, ps pv The temperature at which p v becomes saturated pressure can be found by interpolation of values from table, for ps 00. to ps ( 5) T 5 + ; ( ) E # cC SOL 7.8 Option (B) is correct. The variation of heat transfer with the outer radius of the insulation r, when r r < cr The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as Q o T T T T Rins Rconv. r ar k + πlk h( πr L) The value of r at which Q o reaches a maximum is determined from the requirement that dq o 0. By solving this we get, dr r cr, pipe k...(i) h Published by: NODIA and COMPANY ISBN:

30 PAGE 33 HEAT TRANSFER CHAP 7 From equation (i), we easily see that by increasing the thickness of insulation, the value of thermal conductivity increases and heat loss by the conduction also increases. But by increasing the thickness of insulation, the convection heat transfer co-efficient decreases and heat loss by the convection also decreases. These both cases are limited for the critical thickness of insulation. SOL 7.9 SOL 7.30 Option (D) is correct. The general heat equation in cartesian co-ordinates, T T T + + T x y z α t For one dimensional heat conduction, T T ρcp T α k Thermal Diffusitivity x α t k t ρcp For constant properties of medium, T \ T t x Option (D) is correct. Given : T> T> T3 From, Wien s displacement law, λ maxt mk Cns ο tant λ max \ T If T increase, then λ m decrease. But according the figure, when T increases, then λ m also increases. So, the Wien s law is not satisfied. SOL 7.3 Option (C) is correct. Assumptions : Published by: NODIA and COMPANY ISBN:

31 CHAP 7 HEAT TRANSFER PAGE 333 () Heat transfer is steady since there is no indication of change with time. () Heat transfer can be approximated as being one-dimensional since it is predominantly in the x -direction. (3) Thermal conductivities are constant. (4) Heat transfer by radiation is negligible. Analysis : There is no variation in the horizontal direction. Therefore, we consider a m deep and m high portion of the slab, since it representative of the entire wall. Assuming any cross-section of the slab normal to the x direction to be isothermal, the thermal resistance network for the slab is shown in the figure. R L K/ W ka 00. ( # ) R L K/ W ka 00. #( # 05. ) R L K/ W ka #( # 05. ) Resistance R and R 3 are in parallel. So the equivalent resistance R eq will be R eq R + R 3 R R3 R + eq RR 3 R RR 3 eq 5# K/ W R + R Resistance R and R eq are in series. So total Resistance will be R R + R eq K/ W SOL 7.3 Option (C) is correct. Given : D 5 mm m, T i 500 K, T a 300 K, k 400 W/ mk, ρ 9000 kg/ m 3, c 385 JkgK /, h 50 W/ m K, Given that lumped analysis is assumed to be valid. So, T Ta exp hat T T c ρνc m exp ht c ρlc m...(i) i a Published by: NODIA and COMPANY ISBN:

32 PAGE 334 HEAT TRANSFER CHAP l ν Volume of ball πr 3 l ν A Surface Area 4πR A l R D m On substituting the value of l and other parameters in equation. (i), T 300 exp 50 # t c 9000 # m t T # e On differentiating the above equation w.r.t. t, dt t 00 ( ) e dt # # Rate of fall of temperature of the ball at the beginning of cooling is (at beginning t 0) dt b dt l 00 #( ) # 7.36 K/sec t 0 Negative sign shows fall of temperature. SOL 7.33 Option (C ) is correct. Given : d m, d 0.5 m, L 0.5 m The cylinder surface cannot see itself and the radiation emitted by this surface falls on the enclosing sphere. So, from the conservation principle (summation rule) for surface, F + F From the reciprocity theorem, For sphere, F F F 0 AF AF F A F A # + F Published by: NODIA and COMPANY ISBN: A...(ii) A F F...(iii) From equation (ii) and (iii), we get

33 CHAP 7 HEAT TRANSFER PAGE 335 F A rl π rl A πd d # # SOL 7.34 Option (D) is correct. The figure shown below are of parallel flow and counter flow respectively. For parallel flow, 80cC, 50cC, 30cC, 40cC t h t h t c ( th tc) ( th tc) θ mp θ θ θ th tc b θ l b t h t l c Where, θ mp denotes the LMTD for parallel flow. ( 80 30) ( 50 40) θ mp cC 50 () 5 b 0 l For counter flow arrangement 80cC, 50cC, 40cC, 30cC t h t h t c Where, θ mc denotes the LMTD for counter flow. θ mc θ θ θ bθ l t c t c ( th tc) ( th tc) th tc b t t l h c ( 80 30) ( 50 40) cC 50 () 5 b 0 l Now for defining the type of flow, we use the correction factor. θ m Fθmc Fθmp...(i) Where F correction factor, which depends on the geometry of the heat exchanger and the inlet and outlet temperatures of the of the hot and cold streams. Published by: NODIA and COMPANY ISBN:

34 PAGE 336 HEAT TRANSFER CHAP 7 F <, for cross flow and F, for counter and parallel flow So, From equation (i), m F θ < θmc m and also F θ > θmp So, cross flow in better for this problem. SOL 7.35 Option (C) is correct. Given : A duct of rectangular cross section. For which sides are a m and b 0.5 m 30cC, 0cC, V 0 m/ sec, k 0.05 W/ mk T T Viscosity 8 μpas, Pr 0.73, ρ. kg/ m 3, Nu 0.03 Re Pr Hence, For a rectangular conduit of sides a and b, Hydraulic diameter, 4 p A D H Where, A is the flow cross sectional area and p the wetted perimeter D H 4ab ab ( a+ b) ( a+ b) # # m ( 0. 5) ρvd Reynolds Number, Re H μ. # 0 # # # 0 SOL 7.36 Option (D) is correct. From the first part of the question, Re 444. # 0 5 Which is greater than 3# 0 5. So, flow is turbulent flow Therefore, Nu Re Pr hl ( 0. 73) k # # # 3954 # h # L k DH L m ^ h Nu # 5.64 W/ m K 666 hl k Total Area, A ( a+ b) L ( ) L 3L Heat transfer by convection is given by, Q ha( T T) Published by: NODIA and COMPANY ISBN:

35 CHAP 7 HEAT TRANSFER PAGE # 3 L #[( ) ( )] Heat transfer per meter length of the duct is given by Q L # # 769. W W SOL 7.37 Option (B) is correct. The one dimensional time dependent heat conduction equation can be written more compactly as a simple equation, r r r n T q ρc n : + r D T...(i) k k t Where, n 0, For rectangular coordinates n, For cylindrical coordinates n, For spherical coordinates Further, while using rectangular coordinates it is customary to replace the r -variable by the x -variable. For sphere, substitute r in equation (i) r r r T q ρc : + r D T k k t r r r T q : + r D T α k thermal diffusivity k α t ρc SOL 7.38 Option (C) is correct. Let Length of the tube l Given : r d / cm cm, r 5 cm.5 cm Radius of asbestos surface, r 3 k s 9 W/ mk, k a 0. W/ mk d cm And T T 600cC From the given diagram heat is transferred from r to r and from r to r 3. So Equivalent thermal resistance, Σ R r kl r r 3 π a k + s πkl ar k For hollow cylinder Rt a Published by: NODIA and COMPANY ISBN: loge( r/ r) πkl

36 PAGE 338 HEAT TRANSFER CHAP 7 Σ R# l r k r r π a k + πk ar k s a # # b l+ # # b l mk/ W...(i) Heat transfer per unit length, Q T T W/ m ( ΣR# l) SOL 7.39 Option (B) is correct. Given : h 400 W/ m K, k 0 W/ mk, c 400 JkgK /, ρ 8500 kg/ m 3 T i 30cC, D mm, T a 300cC, T 98cC Biot Number, B i hl..(i) k 4 3 And l Volume R 3 π πd 3 6 Surface Area 4πR πd D # m 6 6 # 4 From equation (i), we have Bi hl 400 #. 76 # k 0 Bi < 0. The value of Biot Number is less than one. So the lumped parameter solution for transient conduction can be conveniently stated as T T hat a e c Ti T c ρν m e ht cρ cl m ν a A l exp 400t 300 b # 400 #. 76 # 0 l 70 e t 70 e t Take natural logarithm both sides, we get b 70 l t " t 490. sec SOL 7.40 Option (A) is correct. Given : t c 30cC, dm m o 500 kg/ hr 500 kg/sec kg/ sec dt 3600 th th 0cC, t c t c 80cC, c w 4.87 kj/ kg K, U 000 W/ m K. Figure for condensation is given below : Published by: NODIA and COMPANY ISBN:

37 CHAP 7 HEAT TRANSFER PAGE 339 Hence, θ th tc cC And θ th tc cC So, Log mean temperature difference (LMTD) is, θ m θ θ θ 90 ^ cC _ θ i 40h Energy transferred is given by, Q mc o wδt UAθm A mcw o ΔT # # 000 # m U θ m 000 # SOL 7.4 Option (B) is correct. Given, for plate : A 0 cm 0 (0 ) m # 0 3 m, T 800 K, ε 0.6 For Room : A 00 m, T 300 K, ε 0.3 and σ 5.67 # 0 W/ m K 8 4 Total heat loss from one surface of the plate is given by, ( Q ) Eb Eb ( ε) ( ε ) + + A ε A F A ε Published by: NODIA and COMPANY ISBN: If small body is enclosed by a large enclosure, then F and from Stefan s

38 PAGE 340 HEAT TRANSFER CHAP 7 Boltzman law E ( Q ) b 4 σt. So we get σ( T 4 T 4 ) ε + + ε A ε A A ε # 0 [( 800) ( 300) ] # # # W Q is the heat loss by one surface of the plate. So, heat loss from the two surfaces is given by, Q net # Q # W SOL 7.4 Option (B) is correct. In counter flow, hot fluid enters at the point and exits at the point or cold fluid enter at the point and exit at the point. Given : for hot fluid, c h kj/ kg K, m h 5 kg/sec, t h 50cC, t h 00cC and for cold fluid, c c 4 kj/ kg K, m c 0 kg/ sec, t c 0cC, t c? From the energy balance, Heat transferred by the hot fluid Heat gain by the cold fluid mc o ( t t ) mc o ( t t ) h h h h c c c c # # 0 ( 50 00) 0 # 4 # 0 ( tc 0) 0 4 # # 0 ( tc 0 ) t c cC 4 Hence, outlet temperature of the cold fluid, t c 3.5cC SOL 7.43 Option (A) is correct. The non-dimensional Prandtl Number for thermal boundary layer is, δv ( Pr ) δ 3 / (i) When Pr δv δt T Published by: NODIA and COMPANY ISBN:

39 CHAP 7 HEAT TRANSFER PAGE 34 (ii) When Pr > δ v > δ T (iii) When Pr < δ v < δ T So for Pr >, δ > δ v T SOL 7.44 Option (C) is correct. Given for water : And for glass : T g T w 48cC, k w 0.6 W/ mk 40cC, k g. W/ mk Spatial gradient dt 4 cdy m # 0 K/ m w Heat transfer takes place between the water and glass interface by the conduction and convection. Heat flux would be same for water and glass interface. So, applying the conduction equation for water and glass interface. k dt w cdy m k dt Q ka dt g cdy m q dx k dt A A dx w dt cdy m g g kw dt k cdy m g w 06.. # # 0 K/ m SOL 7.45 SOL 7.46 Option (D) is correct. From the equation of convection, Heat flux, q ht [ w Tg]...(i) Where, h Heat transfer coefficient First find q, q k dt k dt w c g dy m cdy m 06. # W/ m w g Now from equation (i), q h W/ m K T T Option (C) is correct. w g Given : (A) For counter flow t LMTD, θ mc θ θ θ θ t, t t h C h C Published by: NODIA and COMPANY ISBN:

40 PAGE 34 HEAT TRANSFER CHAP 7 ( th tc) ( th tc) θ mc th tc : t h t D C (B) For parallel flow given : t θ mp LMTD, θ mp θ θ θ bθ l ( th tc) ( th tc) th tc : t t D h C ( th th) ( th th) ( th th) th th th th : th t D : h th t D h t, t t h C From equation (i) and (ii), we get θ mc θmp h C ( th th) ( th th) th th : t t D h h ( th th) th th : t t D h h...(i)...(ii) SOL 7.47 Option (D) is correct. Given : F Applying summation rule : F + F + F3 The flat surface cannot see itself. So, F 0 This gives, F F F SOL 7.48 Option (C) is correct. S. No. Materials Thermal Conductivity ( W/ m K). Aluminum 37. Pure Iron Liquid Water Saturated Water Vapour 0.06 ********** Published by: NODIA and COMPANY ISBN:

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