Question 12.3: Calculate the rise in temperature of water which is passed a......
Calculate the rise in temperature of water which is passed at 3.5 m/s through a smooth 25 mm diameter pipe, 6 m long. The water enters at 300 K and the tube wall may be assumed constant at 330 K. The following methods may be used:
(a) the simple Reynolds analogy (equation 12.139);
\frac{h}{C_{p}\rho u}=\left(\frac{h}{C_{p}\rho u_{s}}\right)\left(\frac{u_{s}}{u} \right) = \left({\frac{R}{\rho u_{s}^{2}}}\right)\left({\frac{u_{s}}{u}}\right) = \left({\frac{R}{\rho u^{2}}}\right)\left({\frac{u}{u_{s}}}\right) = 0.032R e^{-1/4} (12.139)
(b) the Taylor-Prandtl modification (equation 12.140);
\frac{h}{C_{p}\rho u}=\frac{0.032R e^{-1/4}}{1+(u_{b}/u)(u/u_{s})(P r-1)} = \frac{0.032R e^{-1/4}}{1+2.0R e^{-1/8}(P r-1)} (12.140)
(c) the universal velocity profile (equation 12.141);
S t=\frac{h}{C_{p}\rho u}=\frac{0.817(R/\rho u^{2})}{1+0.817\sqrt{(R/\rho u^{2})}\,5[(P r-1)+\ln(\frac{5}{6}P r+\frac{1}{6})]} = \frac{0.032R e^{-1/4}}{1+0.817R e^{-1/8}[(P r-1)+ln(\frac{5}{6}P r+\frac{1}{6})]} (12.141)
(d)N u=0.023R e^{0.8}P r^{0.33} (equation 9.64).
Taking the fluid properties at 310 K and assuming that fully developed flow exists, an approximate solution will be obtained neglecting the variation of properties with temperature.
R e={\frac{0.025\times3.5\times1000}{0.7\times10^{-3}}}=1.25\times10^{5}P r={\frac{4.18\times10^{3}\times0.7\times10^{-3}}{0.65}}=4.50
(a) Reynolds analogy
\frac{h}{C_{p}\rho u}=0.032R e^{-0.25} (equation 12.139)
\frac{h}{C_{p}\rho u}=\left(\frac{h}{C_{p}\rho u_{s}}\right)\left(\frac{u_{s}}{u} \right) = \left({\frac{R}{\rho u_{s}^{2}}}\right)\left({\frac{u_{s}}{u}}\right) = \left({\frac{R}{\rho u^{2}}}\right)\left({\frac{u}{u_{s}}}\right) = 0.032R e^{-1/4} (12.139)
h=[4.18\times1000\times1000\times3.5\times0.032(1.25\times10^{5})^{-0.25}]= 24,902 W/m² K or 24.9 kW/m² K
Heat transferred per unit time in length dL of pipe = h π 0.025 dL (330 – θ) kW, where θ is the temperature at a distance L m from the inlet.
Rate of increase of heat content of fluid = \left(\frac{\pi}{4}0.025^{2}\times3.5\times1000\times4.18\right)\,\mathrm{d}\theta\ \mathrm{kW}
The outlet temperature θ’ is then given by:
\int_{300}^{\theta^{\prime}}\frac{\mathrm{d}\theta}{(330-\theta)}=0.0109h\int_{0}^{6}\mathrm{d}Lwhere h is in kW/m² K.
∴ \log_{10}(330-\theta^{\prime})=\log_{10}30-{\biggl(}{\frac{0.0654h}{2.303}}{\biggr)}=1.477-0.0283h
In this case: h = 24.9 kW/m² K
∴ \log_{10}(330-\theta^{\prime})=(1.477-0.705)=0.772
and: \underline{\underline{\theta ^{\prime}=324.1\ K}}
(b) Taylor-Prandtl equation
\frac{h}{C_{p}\rho u}=0.032R e^{-1/4}[1+2.0R e^{-1/8}(P r-1)]^{-1} (equation 12.140)
\frac{h}{C_{p}\rho u}=\frac{0.032R e^{-1/4}}{1+(u_{b}/u)(u/u_{s})(P r-1)} = \frac{0.032R e^{-1/4}}{1+2.0R e^{-1/8}(P r-1)} (12.140)
∴ h={\frac{24.9}{(1+2.0\times3.5/4.34)}}
= 9.53 kW/m² K
and: \log_{10}(330-\theta^{\prime})=1.477-(0.0283\times9.53) = 1.207
(c) Universal velocity profile equation
= \frac{24.9}{1+(0.82/4.34)(3.5+2.303\times0.591)}
= 12.98 kW/m² K
S t=\frac{h}{C_{p}\rho u}=\frac{0.817(R/\rho u^{2})}{1+0.817\sqrt{(R/\rho u^{2})}\,5[(P r-1)+\ln(\frac{5}{6}P r+\frac{1}{6})]} = \frac{0.032R e^{-1/4}}{1+0.817R e^{-1/8}[(P r-1)+ln(\frac{5}{6}P r+\frac{1}{6})]} (12.141)
∴ \log_{10}(330-\theta^{\prime})=1.477-(0.0283\times12.98)=1.110
and: \underline{\underline{\theta ^{\prime}=317.1\ K}}
(d) N u=0.023R e^{0.8}P r^{0.33}
h={\frac{0.023\times0.65}{0.0250}}(1.25\times10^{5})^{0.8}(4.50)^{0.33} (equation 9.64)
= 0.596 x 1.195 x 10^{4} x 1.64
= 1.168 x 10^{4} W/m² K or 11.68 kW/m² K
and: \log_{10}(330-\theta^{\prime})=1.477-(0.0283\times11.68) = 1.147
Comparing the results:
It is seen that the simple Reynolds analogy is far from accurate in calculating heat transfer to a liquid.
| Method | h(kW/m² K) | θ'(K) |
| (a) | 24.9 | 324.1 |
| (b) | 9.5 | 313.9 |
| (c) | 13.0 | 317.1 |
| (d) | 11.7 | 316.0 |