fundamentals of heat and mass transfer solutions manual phần 10 doc
... of heat transfer to tube without ash deposit, (b) Rate of heat transfer with an ash deposit of diameter D d = 0.06 m, (c) Effect of deposit diameter and convection coefficient on heat rate and contributions ... specification of nearly optimum α a ( λ ) for absorber. PROBLEM 13 .109 KNOWN: Dimensions and emissivity of double pane window. Thickness of air gap....
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... value of the heat rate. PROBLEM 1.39 KNOWN: Power consumption, diameter, and inlet and discharge temperatures of a hair dryer. FIND: (a) Volumetric flow rate and discharge velocity of heated ... KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas. FIND: Daily cost of...
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... Temperature and volume of hot water heater. Nature of heater insulating material. Ambient air temperature and convection coefficient. Unit cost of electric power. FIND: Heater dimensions and insulation ... K q= ln 1.04 1 2 2m 1.4 W/m K 1.04m 2m 10 W/m K 400 300 K + 11 1 0.962 m 2 1.4 W/m K 1.04m 10 W/m K 100 K 100 K q= 2.23 10 K/W + 15.30 10 K/W 4.32 10 K/W + 29.43...
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fundamentals of heat and mass transfer solutions manual phần 4 docx
... 120 s) and T(0.15 m, 120s) using the finite-element software FEHT for a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your ... independent of time. Hence, () ( ) p1 p p o 11 2 T T 1 2Fo Bi Fo Fo T T Bi FoT + ∞ =−−⋅+ ++⋅ .(5) End Nodal Point 10: p1 p 10 10 abc c p TT x qqq A c 2t ρ + − ∆ ++=⋅ ⋅ ∆ ( ) (...
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fundamentals of heat and mass transfer solutions manual phần 6 docx
... pharmaceutical. Heat transfer coefficient at outer surface and dimensions of coil. FIND: (a) Expressions for T c (t) and T h,o (t), (b) Plots of T c (t) and T h,o (t) for prescribed conditions. Effect of ... 8.34 KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length, diameter and surface temperature. FIND: (a) Outlet water tempe...
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fundamentals of heat and mass transfer solutions manual phần 3 pdf
... guesses. (2) Note that the rate of heat transfer by convection to the top surface of the rod must balance the rate of heat transfer by conduction to the sides and bottom of the rod. NOTE TO INSTRUCTOR: ... ( )( ) q50 W/mK[1/2120.5 5100 120.6 4100 121.2 9100 123.8 9100 134.5 7100 1/2147.1 4100 ] ′ =⋅−+−+− +−+−+− q6711 W/m. ′ = < COMMENTS: For nodes a through d, there is...
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fundamentals of heat and mass transfer solutions manual phần 5 ppsx
... emissivity of Nichrome wire. Electrical current. Temperature of air flow and surroundings. Velocity of air flow. (a) Surface and centerline temperatures of the wire, (b) Effect of flow velocity and ... stream velocity and temperature of air coolant. (a) Average pin convection coefficient, (b) Pin heat transfer rate, (c) Total heat rate, (d) Effect of velocity an...
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fundamentals of heat and mass transfer solutions manual phần 7 pot
... emissivity, and power dissipation of cylindrical heater. Temperature of ambient air and surroundings. FIND: Steady-state temperature of heater and time required to come within 10 ° C of this temperature. SCHEMATIC: ASSUMPTIONS: ... Summary of the calculations above and for water and ethylene glycol: Fluid Ra D ( ) 2 D hW/mK ⋅ q(W) Air 6.750 × 10 4 10. 6 1.55 < W...
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fundamentals of heat and mass transfer solutions manual phần 8 ppsx
... −°=° < The rate of heat transfer to the air is () pc,o c,i q mc T T 0.03kg/s 100 8J/ kg K 62.8 C 1900 W =−=×⋅×°= and the rate of condensation is 4 cond 6 fg q 1900W m 8.70 10 kg /s h 2.183 10 J /kg − == ... (a) Overall heat transfer coefficient, U, for the HXer, outlet temperature of cooling water, T c,o , and condensation rate of the steam h m ; and (b) Com...
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fundamentals of heat and mass transfer solutions manual phần 9 ppt
... ) 4 824 4 s 2 ss h sr 0.95.6710W/mK273K 1T G I90.2W/msr sr εσ ρ πππ − ×⋅ − ====⋅ ( ) ( ) 26258 sd q46.290.2W/msr510m2.2210sr1.5110W −−− − =+⋅×××=× 86232 d G1.5110W/210m7.5710W/m. −−− =××=× < (b) ... allowable value of T s = 1800 K, it follows that 84 84 84 o 0.391 5.67 10 (2400) 0.8 5.67 10 (300) 2 0.46 5.67 10 (1800) h (1800 300) −− − ×× +×× −××× = − 525 2 o 7.335 10 3.674 10...
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