### Given the resolution of the problem 5.102 below, develop a python program that shows the found temperature distribution PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q = 10³ W/m³, occurs when the element is inserted into the core while the surfaces experience convection (To,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to Fo aAt 5x10 m²/sx0.3s 4x² t=pAt=0.3p(s). =0.375 (0.002m)² Fuel element- x=5x10 m²/s k-30W/m-K Coolant Too-250°C h-1100 W/m²-K رجا T(x,0)=250°C 9-10°W/m³ at +>0 Coolant Too,h L-10mm k+ax=2mm ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q=0, initially; at t> 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal network of Example 5.8, the same finite-difference equations may be used. Interior nodes, m=1, 2, 3, 4 TP+1 = Fo TP m-1 +TP 4(x) + m+1 2 -2 Fo)T (1) Midplane node, m = 0 Same as Eq. (1), but with TP Surface node, m = 5 T}+1=2 Fo|T{ +BiTot m-1 =TP 2k m+1' +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: hAx 1100W/m² K×(0.002m) Bi= k 30W/m-K 1/2 Fo =0.466 (1+Bi) = 0.0733 Substituting numerical values with q = 108W/m³, the nodal equations become 18-0.375 2+10 W/m³ (0.002m)²/30W/mK]+(1-2x0.375)T TP+1=0.375 2TP +13.33 +0.25 T 3.33] -0.25 TP TP+1 = 0.375 [T+T+13.33] .33] +0.25 T=0.375[T+T+13.33] +0.25 T T¹-0.375 T+T+13.33+0.25 T = 13.33 2 TP+1=0.375[T+T+13.33] +0.25 T TP+1=2×0.375 T +0.0733×250+- +(1-2x0.375-2×0.0733×0.375) TP+1 =0.750 [T+24.99] +0.195 T The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. (2) р t(s) To T₁ T2 T3 ΤΑ T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 255.00 255.00 255.00 250 254.99 2 0.6 260.00 260.00 260.00 260.00 260.00 259.72 3 0.9 265.00 265.00 4 1.2 270.00 270.00 265.00 265.00 264.89 264.39 270.00 269.96 269.74 268.97 5 1.5 275.00 275.00 274.98 274.89 274.53 273.50 ΔΙΣ =0.466 Fo(Ax)² (0.002m)² α = 0.373s. -6. 2 5x10 m/s Continued..... The desired temperature distribution T(x, 1.5s), corresponds to p = 5. COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the coolant during the first 1.5s time period. (4) (5) (7)

Experiment: A cooling tower uses forced air and column packing to cool downward-flowing water. Inlet water temperature and water flow rate are varied to investigate effects on outlet water temperature, outlet air temperature, and outlet air humidity. The system is first observed operating with ambient room temperature water. A heat load is then applied to the water tank, and the system response is observed. This is to simulate a power plant starting up and placing a cooling load on the cooling water supply. The aim is to compare the system response with and without the load. Data from the Experiment and the make-up water mass flow rate are both shown in the following tables below. For the load cases, determine the net rate of water evaporation from the cooling water to the air using the equation for air flow rate. Compare this with the rate at which make-up water enters the system. For the load cases, determine the rate of work supplied by the pump and compare it to the pump power…

Question 2 The composite wall of an oven consists of three materials, two of which are of known thermal conductivity, ka = 20 W/m•K and kc = 50 W/m•K, and known thickness, LA = 0.30 m and Lc = 0.15 m. The third material, B, which is sandwiched between materials A and C, is of known thickness, LB = 0.15 m, but unknown thermal conductivity ke. Under steady-state operating conditions, measurements reveal an outer surface temperature of T0= 20°C, an inner surface temperature of T; = 600° C, and an oven air temperature of T= 800C. The inside convection coefficient h is known to be 25 W/m2 •K. a) Sketch the equivalent thermal resistance of the circuit b) What is the value of kg? Ts0 Ts, i S,0 kA kc kB Air To h LA LB Lc

A Zamboni is a multi-step mechanical ice rink resurfacer. The first step is shaving the ice surfacewith an auger conveyor, pictured below. In an experimental conveyor design, a stream of warmwater is piped through the conveyor tube. This design can be modeled as an array of annular fins,illustrated on the right. Assume steady state.In this auger conveyor, you are asked to calculate the amount of heat needed to maintain the airand ice temperature in the chamber at T ¥ = 0°C, with convection coefficient h = 25 W/m2 ·K. Theconveyor inside wall tube temperature is T b = 40°C. The annular fin array dimensions are:r 1 = 2 cm, r 2 = 10 cm, t = 2 mm, and S = 8 cm. There are N = 16 fins, with the first and last finsattached to the ends. The auger conveyor assembly is made of stainless steel (ksteel = 15 W/m·K).(a) Using the annular fin efficiency plot on the next page, approximate the fin efficiency of a singleannular fin,h f .Answer: ____________________________ [%]Continued on the back side…