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Heat Convection by Latif M. Jiji - solutions. 1.

1. Radiation Heat Transfer
2. Solution Manual For Convection Heat Transfer Latifah
3. Solution Manual For Convection Heat Transfer Latifi

PROBLEM 1.1 Heat is removed from a rectangular surface by convection to an ambient fluid at T. The heat transfer coefficient is h. Surface temperature is given by sT = 2/1 x A where A is constant. Determine the steady state heat transfer rate from the plate. (1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable.

(ii) Ambient temperature and heat transfer coefficient are uniform. (iii) Surface temperature varies along the rectangle. (2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient. (3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform.

This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer. (4) Plan Execution. (i) Assumptions.

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(1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature. (ii) Analysis.

Newton's law of cooling states that qs = h As (Ts - T ) (a) where As = surface area, m2 h = heat transfer coefficient, W/m2 -o C qs = rate of surface heat transfer by convection, W Ts = surface temperature, o C T = ambient temperature, o C Applying (a) to an infinitesimal area dAs dqs = h (Ts - T ) dAs (b) The next step is to express )(xTs in terms of distance x along the triangle. )(xTs is specified as sT = 2/1 x A (c) L x0 W x0 L W sdq dx. PROBLEM 1.1 (continued) The infinitesimal area dAs is given by dAs = W dx (d) where x = axial distance, m W = width, m Substituting (c) and into (b) dqs = h( 2/1 x A - T ) Wdx (e) Integration of (f) gives qs qs = sdq = L dxTAxhW 0 2/1 )( (f) Evaluating the integral in (f) LTALhWqs 2/1 2 Rewrite the above TALhWLqs 2/1 2 (g) Note that at x = L surface temperature )(LTs is given by (c) as 2/1 )( ALLTs (h) (h) into (g) TLThWLq ss )(2 (i) (iii) Checking. Dimensional check: According to (c) units of C are 2/1o C/m. Therefore units sq in (g) are W.

Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i) satisfies these limiting cases. (5) Comments. Integration is necessary because surface temperature is variable. The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.

PROBLEM 1.2 A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to an ambient fluid at T.

The heat transfer coefficient h varies along the surface according to h = C x1 2/ where C is constant and x is distance along the base measured from the apex. Determine the total heat transfer rate from the triangle.

Radiation Heat Transfer

(1) Observations. (i) Heat is removed from the surface by convection.

Therefore, Newton's law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform. (iii) Surface area and heat transfer coefficient vary along the triangle. (2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient. (3) Solution Plan.

Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface area and heat transfer coefficient are not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer. (4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid temperature. (ii) Analysis. Newton's law of cooling states that qs = h As (Ts - T ) (a) where As = surface area, m2 h = heat transfer coefficient, W/m2 -o C qs = rate of surface heat transfer by convection, W Ts = surface temperature, o C T = ambient temperature, o C Applying (a) to an infinitesimal area dAs dqs = h (Ts - T ) dAs (b) The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer coefficient h is given by h = C x1 2/ (c) The infinitesimal area dAs is given by W x L dx dqs. PROBLEM 1.2 (continued) dAs = y(x) dx (d) where x = distance along base of triangle, m y(x) = height of the element dAs, m Similarity of triangles give y(x) = W L x (e) where L = base of triangle, m W = height of triangle, m Substituting (c), (d) and (e) into (b) dqs = C x1 2/ (Ts - T ) W L x dx (f) Integration of (f) gives qs.

Keeping in mind that C, L, W, Ts and T are constants, (f) gives qs = sdq = )( TT L WC s L x x 0 2/1 dx (g) Evaluating the integral in (g) qs = 2 3 C W L1/2 (Ts - T ) (h) (iii) Checking. Dimensional check: According to (c) units of C are W/m3/2 -o C. Therefore units of qs in (h) are qs = C(W/m3/2 -o C) W(m) L1/2 (m1/2 ) (Ts - T )(o C) = W Limiting checks: If h = 0 (that is C = 0) then qs = 0. Similarly, if W = 0 or L = 0 or Ts = T then qs = 0. Equation (h) satisfies these limiting cases. (5) Comments.

Integration was necessary because both area and heat transfer coefficient vary with distance along the triangle. The same procedure can be followed if the ambient temperature or surface temperature is non-uniform. PROBLEM 1.3 A high intensity light bulb with surface heat flux sAq )/( is cooled by a fluid at T. Sketch the fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where h1.

Solution manual for heat convection 2nd ed latif m. Solution Manual For Heat Convection 2nd edition By Latif M. Jiji Access full file only for 12$ www.bookfi.in. PROBLEM 1.1 Heat is removed from a rectangular surface by convection to an ambient fluid at T. The heat transfer coefficient is h. Surface temperature is given by sT = 2/1 x A where A is constant. Determine the steady state heat transfer rate from the plate.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform.

(iii) Surface temperature varies along the rectangle. (2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient. (3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform. This means that the rate of heat transfer varies along the surface.

Solution Manual For Convection Heat Transfer Latifah

Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer. (4) Plan Execution. (i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.

(ii) Analysis. Newton's law of cooling states that qs = h As (Ts - T ) (a) where As = surface area, m2 h = heat transfer coefficient, W/m2 -o C qs = rate of surface heat transfer by convection, W Ts = surface temperature, o C T = ambient temperature, o C Applying (a) to an infinitesimal area dAs dqs = h (Ts - T ) dAs (b) The next step is to express )(xTs in terms of distance x along the triangle.

)(xTs is specified as sT = 2/1 x A (c) L x0 W x0 L W sdq dx Download Complete Solution Manual for Heat Convection 2nd ed by Latif M. Jiji Only for 12$. PROBLEM 1.1 (continued) The infinitesimal area dAs is given by dAs = W dx (d) where x = axial distance, m W = width, m Substituting (c) and into (b) dqs = h( 2/1 x A - T ) Wdx (e) Integration of (f) gives qs qs = sdq = L dxTAxhW 0 2/1 )( (f) Evaluating the integral in (f) LTALhWqs 2/1 2 Rewrite the above TALhWLqs 2/1 2 (g) Note that at x = L surface temperature )(LTs is given by (c) as 2/1 )( ALLTs (h) (h) into (g) TLThWLq ss )(2 (i) (iii) Checking. Dimensional check: According to (c) units of C are 2/1o C/m.

Therefore units sq in (g) are W. Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0.

Equation (i) satisfies these limiting cases. (5) Comments. Integration is necessary because surface temperature is variable.

The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform. Download Solution Manual for Heat Convection 2nd ed by Latif M. Jiji full version only 12 $. PROBLEM 1.2 A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to an ambient fluid at T.

The heat transfer coefficient h varies along the surface according to h = C x1 2/ where C is constant and x is distance along the base measured from the apex. Determine the total heat transfer rate from the triangle. (1) Observations. (i) Heat is removed from the surface by convection.

Therefore, Newton's law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform. (iii) Surface area and heat transfer coefficient vary along the triangle. (2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient. (3) Solution Plan.

Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface area and heat transfer coefficient are not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer. (4) Plan Execution. (i) Assumptions.

(1) Steady state, (2) negligible radiation and (3) uniform ambient fluid temperature. (ii) Analysis. Newton's law of cooling states that qs = h As (Ts - T ) (a) where As = surface area, m2 h = heat transfer coefficient, W/m2 -o C qs = rate of surface heat transfer by convection, W Ts = surface temperature, o C T = ambient temperature, o C Applying (a) to an infinitesimal area dAs dqs = h (Ts - T ) dAs (b) The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer coefficient h is given by h = C x1 2/ (c) The infinitesimal area dAs is given by W x L dx dqs.

Solution Manual For Convection Heat Transfer Latifi

PROBLEM 1.2 (continued) dAs = y(x) dx (d) where x = distance along base of triangle, m y(x) = height of the element dAs, m Similarity of triangles give y(x) = W L x (e) where L = base of triangle, m W = height of triangle, m Substituting (c), (d) and (e) into (b) dqs = C x1 2/ (Ts - T ) W L x dx (f) Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and T are constants, (f) gives qs = sdq = )( TT L WC s L x x 0 2/1 dx (g) Evaluating the integral in (g) qs = 2 3 C W L1/2 (Ts - T ) (h) (iii) Checking. Dimensional check: According to (c) units of C are W/m3/2 -o C. Therefore units of qs in (h) are qs = C(W/m3/2 -o C) W(m) L1/2 (m1/2 ) (Ts - T )(o C) = W Limiting checks: If h = 0 (that is C = 0) then qs = 0. Similarly, if W = 0 or L = 0 or Ts = T then qs = 0. Equation (h) satisfies these limiting cases.

(5) Comments. Integration was necessary because both area and heat transfer coefficient vary with distance along the triangle. The same procedure can be followed if the ambient temperature or surface temperature is non-uniform. PROBLEM 1.3 A high intensity light bulb with surface heat flux sAq )/( is cooled by a fluid at T. Sketch the fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where h1.