DOC PREVIEW
CSU MECH 344 - HW7 - Solutions

This preview shows page 1-2-3 out of 8 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

HW7 Solutions.pdfsm7-117sm7-117.pdfHW7 SolutionsPROBLEM 7.17 KNOWN: Temperature, pressure and Reynolds number for air flow over a flat plate of uniform surface temperature. FIND: (a) Rate of heat transfer from the plate, (b) Rate of heat transfer if air velocity is doubled and pressure is increased to 10 atm. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface temperature, (3) Negligible radiation, (4) Rexc=×5105. PROPERTIES: Table A-4, Air (Tf = 348K, 1 atm): k = 0.0299 W/m⋅K, Pr = 0.70. ANALYSIS: (a) The heat rate is ()( )Lsqh wL T T.∞=× − Since the flow is laminar over the entire plate for ReL = 4 × 104, it follows that ()()L1/2 1/31/2 1/3LLhLNu 0.664 Re Pr 0.664 40,000 0.70 117.9.k== = = Hence hkL W/m K0.2m W/m KL2==⋅=⋅117 9 117 90029917 6.... and ()()2Wq 17.6 0.1m 0.2m 100 50 C 17.6 W.mK=×−=⋅o < (b) With p2 = 10 p1, it follows that ρ2 = 10 ρ1 and ν2 = ν1/10. Hence 5L,2 L,121uL uLRe 2 10 20 Re 8 10νν∞∞⎛⎞ ⎛⎞==× ==×⎜⎟ ⎜⎟⎝⎠ ⎝⎠ and mixed boundary layer conditions exist on the plate. Hence ()()()LL4/51/34/5 1/3 5LLhLNu 0.037 Re 871 Pr 0.037 8 10 871 0.70kNu 961.⎡⎤== − = ×× −⎢⎥⎣⎦= Hence, h W/m K0.2m W/m KL2=⋅=⋅961002991436.. ()()2Wq 143.6 0.1m 0.2m 100 50 C 143.6 W.mK=×−=⋅o < COMMENTS: Note that, in calculating ReL,2, ideal gas behavior has been assumed. It has also been assumed that k, μ and Pr are independent of pressure over the range considered.PROBLEM 7.24 KNOWN: Plate dimensions and initial temperature. Velocity and temperature of air in parallel flow over plates. FIND: Initial rate of heat transfer from plate. Rate of change of plate temperature. SCHEMATIC: ASSUMPTIONS: (1) Negligible radiation, (2) Negligible effect of conveyor velocity on boundary layer development, (3) Plates are isothermal, (4) Negligible heat transfer from sides of plate, (5) 5x,cRe 5 10 ,=× (6) Constant properties. PROPERTIES: Table A-1, AISI 1010 steel (573K): kp = 49.2 W/m⋅K, c = 549 J/kg⋅K, ρ = 7832 kg/m3. Table A-4, Air (p = 1 atm, Tf = 433K): ν = 30.4 × 10-6 m2/s, k = 0.0361 W/m⋅K, Pr = 0.688. ANALYSIS: The initial rate of heat transfer from a plate is () ()2si iq2hATT 2hLTT∞∞=−=− With 62 5LRe u L/ 10m/s 1m/30.4 10 m /s 3.29 10 ,ν−∞==× × =× flow is laminar over the entire surface and ()()L1/21/31/2 1/3 5LNu 0.664Re Pr 0.664 3.29 10 0.688 336==× = () ()L2h k / L Nu 0.0361W / m K /1m 336 12.1W / m K== ⋅=⋅ Hence, ()( )22q 2 12.1W / m K 1m 300 20 C 6780W=× ⋅ − °= < Performing an energy balance at an instant of time for a control surface about the plate, out stEE,−=&& we obtain, ()22iidTLc h2L T Tdtρδ∞=− − ()()23i2 12.1W / m K 300 20 CdT0.26 C/sdt7832 kg / m 0.006m 549J / kg K⋅−°=− =− °×× ⋅ < COMMENTS: (1) With ()4pBi h / 2 / k 7.4 10 ,δ−==× use of the lumped capacitance method is appropriate. (2) Despite the large plate temperature and the small convection coefficient, if adjoining plates are in close proximity, radiation exchange with the surroundings will be small and the assumption of negligible radiation is justifiable.PROBLEM 7.92 KNOWN: Geometry, surface temperature, and air flow conditions associated with a tube bank. FIND: Rate of heat transfer per unit length. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation effects and incompressible flow, (3) Gas properties are approximately those of air. PROPERTIES: Table A-4, Air (300K, 1 atm): Pr = 0.707; Table A-4, Air (700K, 1 atm): ν = 68.1 × 10-6 m2/s, k = 0.0524 W/m⋅K, Pr = 0.695, ρ = 0.498 kg/m3, cp = 1075 J/kg⋅K. ANALYSIS: The rate of heat transfer per unit length of tubes is ()()()( )[]si somsi soTT TTqhND T hND .nT T/T Tππ−−−′=Δ=−−ll With maxTmax D,max-6 2TVDS 20 10 m/s 0.01 mV V 5 m/s 10 m/s, Re 1468.SD 1068.1 10 m / sν×=== == =−× Tables 7.5 and 7.6 give C1 = 0.27, m = 0.63 and C2 = 0.97. Hence from the Zukauskas correlation, ()()( )( )D1/40.63 0.36 1/ 4m0.362D,max sNu C C Re Pr Pr/ Pr 0.26 1468 0.695 0.695/0.7071== DD2kNu 22.4 h Nu 0.0524 W/m K 22.4/0.01 m 117 W/m K.D=== ⋅× = ⋅ Hence, ()()()( )2so si3TTpDNh 0.01 m 500 117 W/m KT T T T exp 400K exp VN S c0.498 kg/m 5 m/s 50 0.02 m 1075J / kg Kππρ××× ⋅−=− − =− −⋅⎛⎞⎛⎞⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠ soT T 201.3K−=− and the heat rate is ()()()()( )[]2400 201.3 Kq 117 W/m K 500 0.01 m 532 kW/mn 400 / 201.3π−+′=⋅ =−−−l < COMMENTS: (1) There is a significant decrease in the gas temperature as it passes through the tube bank. Hence, the heat rate would have been substantially overestimated (- 768 kW) if the inlet temperature difference had been used in lieu of the log-mean temperature difference. (2) The negative sign implies heat transfer to the water. (3) If the temperature of the water increases substantially, the assumption of uniform Ts becomes poor. The extent to which the water temperature increases depends on the water flow rate.PROBLEM 7.105 KNOWN: Exit diameter of plasma generator and radius of jet impingement surface. Temperature and velocity of plasma jet. Temperature of impingement surface. Droplet deposition rate. FIND: Rate of heat transfer to substrate due to convection and release of latent heat. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation, (3) Negligible sensible energy change due to cooling of droplets to Ts. ANALYSIS: The total heat rate to the substrate is due to convection from the jet and release of the latent heat of fusion due to solidification, q = qconv + qlat. With Re = VeD/ν = (400 m/s)0.01 m/5.6 × 10-3 m2/s = 714, Ar = D2/4r2 =0.04, and H/D = 10, F1 = 2Re1/2(1 + 0.005 Re0.55)1/2 = 58.2 and G = 1/2 1/2 1/2rr r2A (1 2.2A )/ 1 0.2(H/D 6)A 0.193⎡⎤−+−=⎣⎦, the correlation for a single round nozzle (Section 7.7) yields ()()0.42 0.421Nu GF Pr 0.193 58.2 0.60 9.07== = () ()2h Nu k D 9.07 0.671W m K 0.01m 609W m K== ⋅ = ⋅ Hence, () ()( )22se sq hA T T 609W m K 0.025m 10,000 300 K 11,600Wπ=−= ⋅× −= < The release of latent heat is ()()226lat s p sfq A m h 0.025m 0.02kg s m 3.577 10 J kg 140Wπ′′== ⋅×=& < COMMENTS: (1) The large plasma temperature renders heat transfer due to droplet deposition negligible compared to convection from the plasma. (2) Note that Re = 714 is outside the range of applicability of the


View Full Document

CSU MECH 344 - HW7 - Solutions

Download HW7 - Solutions
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view HW7 - Solutions and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view HW7 - Solutions 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?