Question

5. You are working to determine a correlation for the Nu number for cooling of a steam turbine blade. The design is to pass a cooling fluid (either steam or air) through internal passages within the turbine blade. You do careful measurements in which you are able to calculate the average convection coefficient h as a function of the average velocity of the cooling fluid. The data are presented below for Air and for Water Vapor (Steam). For both experiments, the average gas temperature was controlled at Too450 K, therefore, you may base your thermal properties on this temperature value as follows Property @ 450 K Steam 1.01 3.11E-05 2.99E-02 Air 0.686 Kinematic viscosity (v) (m/s) 3.24E-05 Conductivity (W/mK) 3.73E-02 The characteristic length (Lc) for your turbine blade is 10 cm (0.1 m) (i) Calculate the correlation coefficients (C, m and n) for a function of the form Nu CRemPrn Hint: First, pair the water and steam data for approximately matching Reynolds numbers By taking the ratio of each of these data pairs, you can cancel out C and Re", which allows you to calculate n. You can then average overall all data set pairs. Second, for the same Pr number, plot In(Nu) as a function of In(Re). The slope of a linear fit to the data is the average value ofm. Do this for your two different Pr numbers and average to two values Third, using all of your data points, plot Nu vs. Re" P. The slope of this plot is the average value of C (ii) Provide a plot Mu vs. Rem Prn showing all 12 data points. Use different symbols for the steam and air data, but put them all on the same plot. Air Data Velocit m/s 9.10 13.73 h (W/m2 36.0 43.1

Answer 1

Given

Lc = 0.1m

The formulas used in the answer are

where gamma in the denominator is the viscosity

Nusselt number

$Nu = \frac{hLc}{k}$

From the above we see that n = 0.107

In the above table, m = (mair + mwater)/2

Since the plots are nearly linear, The slope is calculated by (ylast - yfirst) / (xlast - xfirst), where xfirst and yfirst are the intital x and y values of the graph and xlast and ylast are the ini x and y values. Here mair = (5.202-4.578)/(11.633-10.244) = 0.449.

We get average value of m = 0.444

The values of m and n used in the above table are 0.444 and 0.107 respectively. From the above we see that C = 1.218

ii)