Thermo-hydraulic performance of solar air heater having winglet type roughness element

Abstract

An experimental study has been done to evaluate the thermo-hydraulic performance of solar air heater having roughness element over the absorber plate in the form of a winglet type vortex generator. LCT technique is used to measure the Nusselt number over the absorber surface for Reynolds number (Re) of 3000–22,000 and roughness parameters such as relative roughness pitch (P/e) of 5–12, angle of attack (α) ranges of 30°–75° and relative roughness width (W/w) of 3–7. The Nusselt number and friction factor with this roughness are compared with smooth surface for similar flow condition. The maximum enhancement in Nusselt number and friction factor are obtained to be 2.91 and 2.85 times that of smooth surface, respectively. The optimum values of the roughness parameters are evaluated based on the thermo-hydraulic performance parameter criterion. The optimum roughness parameters obtained as the relative roughness pitch (P/e) of 8, angle of attack (α) of 60° and relative roughness width (W/w) of 5. The maximum value of the thermo-hydraulic performance parameter (THPP) is observed to be 2.95.

Appendix AU: Uncertainty analysis

The error of all the measured values leads to achieving some amount of uncertainty in results for the collected data. The uncertainty of an evaluated parameter may be computed as [26]:

$$\delta f_{{\text{e}}} = \pm \sqrt {\left[ {\sum\limits_{{j = 1}}^{n} {\left( {\frac{{\partial f_{\text{e}} }}{{\partial x_{\text{j}} }}\delta x_{\text{j}} } \right)^{2} } } \right]}$$

(Au.0)

where, $\delta f_{\text{e}}$, total error encountered for the parameters,$\delta {x}_{j}$, each parameter error, $\frac{{\partial f_{\text{e}} }}{{\partial x_{j} }}$ known as sensitivity coefficient. The sampling procedure for the relevant data are shown below:

Thermo-physical properties of air are represented as:

$${Cp}_{\text{a}}=1006 {\left(\frac{{T}_{mf}}{293}\right)}^{0.0155}$$

$${\mu }_{\text{a}}=1.81 x {10}^{-5} {\left(\frac{{T}_{\text{mf}}}{293}\right)}^{0.735}$$

$${K}_{\text{a}}=0.0275 {\left(\frac{{T}_{\text{mf}}}{293}\right)}^{0.86}$$

ρ_a = 1.174 kg m⁻³.

Sl. no.	Measured data	Values
1	Absorber plate length/mm, Ld	1240 mm
2	Duct width/mm, W	160 mm
3	Duct height/mm, H	40 mm
4	Diameter of pipe/mm, dp	80 mm
5	Diameter of orifice meter/mm, do	40 mm
6	Orifice meter pressure head, Δho	11.3 cm of H2O
7	Pressure drop*(test section)/Pa, ΔPd	18 Pascal
8	Inlet air temperature/°C, Ti	25.25 °C
9	Outlet air temperature/°C, To	29.87 °C
10	Temperature rise, ΔT	4.62 °C
11	Mean air temperature/°C, Tmf	27.56 °C
12	Mean plate temperature/°C, Tlc	40.55 °C
13	Mass flow rate of air/ kg s−1, ṁ	0.0404 kg s−1
14	Heat transfer to air/W, Qconv	187.84 W
15	Heat transfer coefficient/ W m−2 K, h	70.61 W m−2 K
16	Friction factor, fr	0.01889
17	Nusselt number, Nu	172.03

A_U.1: Duct hydraulic diameter, D _h

$${D}_{\text{h}}= \frac{4WH}{2\left(W+H\right)}$$

where W = 160 mm, H = 40 mm.

D_h = 64 mm

$$\frac{{D}_{\text{h}}}{\partial W}= \frac{{D}_{\text{h}}}{W}- \frac{{D}_{\text{h}}}{\left(W+H\right)}= 1.28$$

$$\frac{{D}_{\text{h}}}{\partial H}= \frac{2W}{\left(W+H\right)}- \frac{2WH}{\left(W+H\right)}=0.08$$

Using Eq. (A_U.0);

$$\frac{{\delta D}_{\text{h}}}{{D}_{\text{h}}}= {\left[{\left(\frac{{\partial D}_{\text{h}}}{\partial W}\delta W\right)}^{2} + {\left(\frac{{\partial D}_{\text{h}}}{\partial H}\delta H\right)}^{2}\right]}^{1/2}$$

 = 0.1%

A_U.2: Duct area, A _duc

A_duc = WH = 6400 mm²

$$\frac{\partial {A}_{\text{duc}}}{\partial W}=H$$

 = 40 mm

$$\frac{\partial {A}_{\text{duc}}}{\partial W}=W$$

 = 160 mm.

Using Eq. (A_U.0);

$$\frac{\delta {A}_{\text{duc}}}{{A}_{\text{duc}}}={\left[{\left(\frac{\partial {A}_{\text{duc}}}{\partial W}\delta W\right)}^{2}+ {\left(\frac{\partial {A}_{\text{duc}}}{\partial H}\delta H\right)}^{2}\right]}^{1/2}$$

 = 0.1288%

A_U.3: Area of absorber plate, AP

A_P = WL_d.

$\frac{\partial {A}_{\text{p}}}{\partial W}={L}_{\text{d}}$ = 1240 mm.

$\frac{\partial {A}_{\text{p}}}{\partial {L}_{\text{d}}}=W$ = 160 mm.

Using Eq. (A_U.0);

$$\frac{\delta {A}_{\text{p}}}{{A}_{\text{p}}}={\left[{\left(\frac{\partial {A}_{\text{p}}}{\partial {L}_{\text{d}}}\delta {L}_{\text{d}}\right)}^{2}+ {\left(\frac{\partial {A}_{\text{p}}}{\partial W}\delta W\right)}^{2}\right]}^\frac{1}{2}$$

$$\frac{\delta {A}_{\text{p}}}{{A}_{\text{p}}}={\left[{\left(\frac{\delta {L}_{\text{d}}}{{L}_{\text{d}}}\right)}^{2}+ {\left(\frac{\delta W}{W}\right)}^{2}\right]}^\frac{1}{2}$$

$$\frac{\delta {A}_{\text{p}}}{{A}_{\text{p}}}= 0.000864= 0.0864 \%$$

A_U.4: Orifice throat area, A _ori

A_ori = $\left( {\frac{\pi }{4}D_{0}^{2} } \right)$ = 1256.6 mm².

Derivatives was written as:

$$\frac{{\partial A_{\text{ori}} }}{{\partial D_{\text{o}} }} = \frac{{2\pi D_{\text{o}} }}{4} = \frac{{\pi D_{\text{o}} }}{2}$$

Using Eq. (A_U.0);

$$\frac{{\delta A_{\text{ori}} }}{{A_{\text{ori}} }} = \left[ {\left( {\frac{{\partial A_{\text{ori}} }}{{\partial D_{\text{o}} }}\delta D_{\text{o}} } \right)^{2} } \right]^{0.5}$$

$$\frac{{\delta A_{\text{ori}} }}{{A_{\text{ori}} }} = \frac{{2 \times \delta D_{\text{o}} }}{{D_{\text{o}} }}$$

 = 0.25%

A_U.5: Mass flow rate,$\mathop {\varvec{m}}\limits^{*}$

$$\mathop {\mathop m\limits }\limits^{{}} = C_{\text{d}} A_{\text{ori}} \left[ {\frac{{2P_{\text{atm}} }}{{RT_{\text{o}} }}(\Delta P)_{\text{o}} } \right]^{0.5}$$

$\mathop m\limits^{*}$ = f_n [C_d, A_ori, P_atm, T_o, (∆P) _o].

Using Eq. (A_U.0);

$$\frac{{\delta \mathop m\limits^{ \bullet } }}{{\mathop m\limits^{ \bullet } }} = \left[ {\left( {\frac{{\delta C_{\text{d}} }}{{C_{\text{d}} }}} \right)^{2} + \left( {\frac{{\delta A_{\text{ori}} }}{{A_{\text{ori}} }}} \right)^{2} + \frac{1}{4}\left( {\frac{{\delta T_{\text{o}} }}{{T_{\text{o}} }}} \right)^{2} + \frac{1}{4}\left( {\frac{{\delta P_{\text{atm}} }}{{P_{\text{atm}} }}} \right)^{2} + \frac{1}{4}\left( {\frac{{\delta (\Delta P)_{\text{o}} }}{{\Delta P_{\text{o}} }}} \right)^{2} } \right]^{0.5}$$

As δC_d/C_d = 1.6%

δA_ori/A_ori = 0.25%

δ (∆P)_o/(∆P)_o = (0.001/1108.53).

δ $\mathop m\limits^{*}$/$\mathop m\limits^{*}$ = 0.01683 = 1.68%

A_U.6: Air density, ρ _a

$$\rho_{\text{a}} = \frac{{P_{\text{atm}} }}{{RT_{\text{mf}} }}\alpha \frac{{P_{\text{atm}} }}{{T_{\text{mf}}}}$$

Using Eq. (A_U.0);

$$\frac{{\delta \rho_{\text{a}} }}{{\rho_{\text{a}} }} = \left[ {\left( {\frac{{\delta P_{\text{atm}} }}{{P_{\text{atm}} }}} \right)^{2} + \left( {\frac{{\delta T_{\text{o}} }}{{T_{\text{o}} }}} \right)^{2} } \right]^{0.5}$$

Substituting the values of

$$\frac{{\delta P_{\text{atm}} }}{{P_{\text{atm}} }} = \frac{0.05}{{99}}$$

and

$$\frac{{\delta T_{\text{o}} }}{{T_{\text{o}} }} = \frac{0.25}{{29.87}}$$

$$\frac{{\delta \rho_{\text{a}} }}{{\rho_{\text{a}} }} = 8.38 \times 10^{ - 3} = 0.84\%_{{}}$$

A_U.7: Velocity of air, V _a

$${V}_{\text{a}}= \frac{\dot{m}}{{\rho }_{\text{a}}WH}$$

Using Eq. (A_U.0);

$$\frac{\delta {V}_{\text{a}}}{{V}_{\text{a}}}={\left[{\left(\frac{\delta \dot{m}}{\dot{m}}\right)}^{2}+ {\left(\frac{\delta {\rho }_{\text{a}}}{{\rho }_{\text{a}}}\right)}^{2}+{\left(\frac{\delta W}{W}\right)}^{2}+ {\left(\frac{\delta H}{H}\right)}^{2}\right]}^{1/2}$$

$$\frac{\delta {V}_{\text{a}}}{{V}_{\text{a}}}=0.01916=1.92 \%$$

A_U.8: Heat transfer, Q _conv

$$Q_{\text{conv}} = \mathop m\limits c_{\text{p}} (T_{\text{o}} - T_{\text{i}} ) = \mathop m\limits c_{\text{p}} \Delta T$$

Using Eq. (A_U.0);

$$\frac{{\delta Q_{\text{conv}} }}{{Q_{\text{conv}} }} = \left[ {\left( {\frac{\delta \mathop m\limits }{{\mathop m\limits }}} \right)^{2} + \left( {\frac{{\delta c_{\text{p}} }}{{c_{\text{p}} }}} \right)^{2} + \left( {\frac{\delta (\Delta T)}{{\Delta T}}} \right)^{2} } \right]^{0.5}$$

$$\frac{{\delta Q_{\text{conv}} }}{{Q_{\text{conv}} }} = 0.0566 = 5.6\%$$

A_U.9: Heat transfer coefficient, h

$$h= \frac{{Q}_{\text{u}}}{{A}_{\text{p}}\left({T}_{\text{lc}}-{T}_{\text{mf}}\right)}= \frac{{Q}_{\text{u}}}{{A}_{\text{p}}\left(\Delta {T}_{\text{m}}\right)}$$

Using Eq. (A_U.0);

$$\frac{\delta h}{h}={\left[{\left(\frac{\delta \dot{{Q}_{\text{u}}}}{{Q}_{\text{u}}}\right)}^{2}+ {\left(\frac{\delta {A}_{\text{p}}}{{A}_{\text{p}}}\right)}^{2}+{\left(\frac{\delta \Delta {T}_{\text{mf}}}{{\Delta T}_{\text{mf}}}\right)}^{2}\right]}^{1/2}$$

$$\frac{\delta h}{h}=0.06=6 \%$$

A_U.10: Nusselt number, Nu

Nusselt number is calculated by,

$$Nu= \frac{h{D}_{\text{h}}}{{k}_{\text{a}}}$$

Using Eq. (A_U.0);

$$\frac{\delta Nu}{Nu}={\left[{\left(\frac{\delta h}{h}\right)}^{2}+ {\left(\frac{\delta {D}_{\text{h}}}{{D}_{\text{h}}}\right)}^{2}+{\left(\frac{\delta {k}_{\text{a}}}{{k}_{\text{a}}}\right)}^{2}\right]}^{1/2}$$

As, δk_a = 0.00001 W/mK,

$$\frac{\delta Nu}{{Nu}} = 0.0599 = 5.9\%$$

A_U.11: Reynolds number, Re

Reynolds number evaluated by;

$$Re= \frac{{{\rho }_{\text{a}}V}_{\text{a}}{D}_{\text{h}}}{\mu }$$

Using Eq. (A_U.0);

$$\frac{\delta Re}{Re}={\left[{\left(\frac{\delta {V}_{\text{a}}}{{V}_{\text{a}}}\right)}^{2}+ {\left(\frac{\delta {\rho }_{\text{a}}}{{\rho }_{\text{a}}}\right)}^{2}+ {\left(\frac{\delta {D}_{\text{h}}}{{D}_{\text{h}}}\right)}^{2}+ {\left(\frac{\delta \mu }{\mu }\right)}^{2}\right]}^{1/2}$$

As δµ = 0.001 N-s/m²,

$$\frac{{\delta {\text{Re}} }}{{\text{Re}}} = 0.0212 = 2.12\%$$

A_U.12: Friction factor, f _r

The friction factor is written as;

$${f}_{\text{r}}= \frac{{2\left(\Delta P\right)}_{\text{d}}{D}_{\text{h}}}{4{\rho }_{\text{a}}{L}_{\text{d}}{V}_{\text{a}}^{2}}$$

Using Eq. (A_U.0);

$$\frac{\delta {f}_{\text{r}}}{{f}_{\text{r}}}={\left[{\left(\frac{\delta {V}_{\text{a}}}{{V}_{\text{a}}}\right)}^{2}+ {\left(\frac{\delta {\rho }_{\text{a}}}{{\rho }_{\text{a}}}\right)}^{2}+ {\left(\frac{\delta {D}_{\text{h}}}{{D}_{\text{h}}}\right)}^{2}+ {\left(\frac{\delta {L}_{\text{d}}}{{L}_{\text{d}}}\right)}^{2}+{\left(\frac{\delta {\left(\Delta P\right)}_{\text{d}}}{{\left(\Delta P\right)}_{\text{d}}}\right)}^{2}\right]}^{1/2}$$

$$\frac{{\delta f_{\text{r}} }}{{f_{\text{r}} }} = 0.0394 = 3.94\%$$

Sl. no	Parameters	Values	Uncertainty %
1	Mass flow of air, $\mathop m\limits^{*}$ kg/s	0.0053–0.0404	1.68–3.01
2	Heat transfer coefficient, h, W/m2 K	9.12–70.61	3.12–6.56
3	Nusselt number, Nu	22.02–172.03	3.12–6.00
4	Friction factor, fr	0.01888–0.04149	3.91–6.45