Problem 7-30:Hot engine oil at 150°C is flowing in parallel over a flat plate at a velocity of 2 m/s. Surfacetemperature of the 0.5-m-long flat plate is constant at 50°C. Determine (a) the local convection heattransfer coefficient at 0.2 m from the leading edge and the average convection heat transfercoefficient, and (b) repeat part (a) using the Churchill and Ozoe (1973) relation.Problem 7-76:In a geothermal power plant, the used geothermal water at 80°C enters a 15-cm-diameter and 400-m-long uninsulated pipe at a rate of 8.5 kg/s and leaves at 70°C before being reinjected back to theground. Windy air at 15°C flows normal to the pipe. Disregarding radiation, determine the averagewind velocity in km/h.Problem 7-112:Air is to be heated by passing it over a bank of 3-m-long tubes inside which steam is condensing at100°C. Air approaches the tube bank in the normal direction at 20°C and 1 atm with a mean velocityof 5.2 m/s. The outer diameter of the tubes is 1.6 cm, and the tubes are staggered with longitudinaland transverse pitches of S₂ = ST=4 cm. There are 20 rows in the flow direction with 10 tubes ineach row. Determine (a) the rate of heat transfer, (b) the pressure drop across the tube bank, and (c)the rate of condensation of steam inside the tubes. Evaluate the air properties at an assumed meantemperature of 35°C and 1 atm. Is this a good assumption?Problem 7-113:Repeat Prob. 7-112 for in-line arrangement with S,= ST = 6 cm.

Disclaimer: “Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Introduction When an item is either hotter or colder than the surrounding fluid, convective heat transfer happens. Never a solid, only a flowing liquid or gas experiences convection. Newton's Law of Cooling is the fundamental formula for calculating the speed of convectional heat transfer. Given data: temperature of engine oilT=150°Cflow velocity of oilV=2 m/stemperature of the plateTs=50°C Find out: (a). The local convection heat transfer coefficient at 0.2m from the leading edge and average convective heat transfer coefficient. (b). The local heat transfer coefficient at 0.2m from the leading edge and average convective heat transfer coefficient by using the Churchill and Ozoe relation.(a). Calculation of the local convective heat transfer coefficient at 0.2m from the leading edge and average convective heat transfer coefficient. For the calculation of the local convection heat transfer coefficient the following things we have to consider, The film temperature(T)f: (T)f=T+Ts2Tf=film temperatureT=engine oil temperatureT=150°CTs=plate surface temperatureTs=50°C(T)f=T+Ts2(T)f=150+502(T)f=100°C The oil property at the 100oC is mentioned below, ν=kinematic viscosity of oilν=2.046×10-5 m2/sk=conductivity of oilk=0.1367 W/m KPr=Prandlt numberPr=279.1 The Reynold number(Re) Re=V×lxνRe=Reynold numberV=velocity of oilV=2 m/sv=2.046×10-5 m2/slx=distance from the leading edgelx=0.2mRe=V×lxνRe=2×0.22.046×10-5 Re=1.955×104 The obtained Reynold number is less than, 5×105 hence the flow of oil will be laminar flow.The local convective heat transfer coefficient(h) For the laminar flow local Nusselt number (Nu)x Nux=0.332Re0.5×Pr1/3Nux=local Nusselt numberRe=1.955×104Pr=279.1(Nu)x=0.332(Re)0.5×(Pr)1/3(Nu)x=0.332(1.955×104)0.5×(279.1)1/3(Nu)x=303.36(Nu)x=h×lxk303.36=h×0.2m0.1367 W/m Kh=303.36×0.1367 W/mK0.2 mh=207.35 w/m2K Hence the local convective heat transfer coefficient is 207.35 W/m2K. The average convective heat transfer coefficient (h¯) The average Nusselt number Nu¯ Nu¯=0.664×Re¯0.5×Pr1/3Re¯=Reynold number at the average length of plateRe¯=V×lavRe¯=2×0.52.046×104Re¯=4.887586×104Nu¯=0.664×Re¯0.5×Pr1/3Nu¯=0.664×4.887586×1040.5×279.11/3Nu¯=959.33Nu¯=h¯×lak959.33=h¯×0.5m0.1367 W/m Kh¯=262.28 W/m2K Hence the average convective heat transfer coefficient is 262.28 W/m2K.(b). Calculation of the local convective heat transfer coefficient at 0.2m from the leading edge and average convective heat transfer coefficient by using the Churchill and Ozoe relation. The local convective heat transfer coefficient(h) The local Nusselt nunberNu Nu=0.3387×Re0.5×Pr1/31+0.0468Pr2/31/4Nu=0.3387×1.955×1040.5×279.11/31+0.0468279.12/31/4Nu=309.251Nu=h×lxkk=0.1367 W/m Klx=0.2 m309.251=h×0.20.1367 W/m Kh=211.373 W/m2K Hence the convective heat transfer coefficient is 211.373 W/m2K.The average convective heat transfer coefficient (h¯) For the entire plate, the average convective heat transfer coefficient can be obtained as follows, h¯=1L∫0Lhxdxh¯=1L∫0L0.3387×Re0.5×Pr1/31+0.0468Pr2/31/4×klxdlRe=V×lxvh¯=1L∫0L0.3387×V×lxv0.5×Pr1/31+0.0468Pr2/31/4×klxdlh¯=1L0.3387×Vv0.5×Pr1/3×k1+0.0468Pr2/31/4×∫0Ll0.5×dllxh¯=1L0.3387×Vv0.5×Pr1/3×k1+0.0468Pr2/31/4×∫0Llx-1/2dlh¯=1L0.3387×Vv0.5×Pr1/3×k1+0.0468Pr2/31/4×2lxh¯=1L0.3387×22.046×10-50.5×279.11/3×0.13671+0.0468Pr2/31/4×2lxlx=lala=average lenght of platela=0.5h¯=10.50.3387×22.046×10-50.5×279.11/3×0.13671+0.0468279.12/31/4×20.5h¯=267.536 W/m2 K Hence the average convective heat transfer coefficient is 267.536 W/m2K. Results: The local convective heat transfer coefficient (h)=207.35 w/m2K The average convective heat transfer coefficient h¯=262.28 W/m2K By using the Churchill and Ozoe relation The local convective heat transfer coefficient (h)=211.373 W/m2K The average convective heat transfer coefficient h¯=267.536 W/m2 K