Mechanical Engineering Department, Faculty of Engineering, Kafrelsheikh University, Kafrelsheikh, Egypt.
Computers and Control Systems Engineering Department, Faculty of Engineering, Mansoura University, 35516, Mansoura, Egypt.
Environ Sci Pollut Res Int. 2022 Dec;29(60):90632-90655. doi: 10.1007/s11356-022-21850-2. Epub 2022 Jul 23.
This research work intends to enhance the stepped double-slope solar still performance through an experimental assessment of combining linen wicks and cobalt oxide nanoparticles to the stepped double-slope solar still to improve the water evaporation and water production. The results illustrated that the cotton wicks and cobalt oxide (CoO) nanofluid with 1wt% increased the hourly freshwater output (HP) and instantaneous thermal efficiency (ITE). On the other hand, this study compares four machine learning methods to create a prediction model of tubular solar still performance. The methods developed and compared are support vector regressor (SVR), decision tree regressor, neural network, and deep neural network based on experimental data. This problem is a multi-output prediction problem which is HP and ITE. The prediction performance for the SVR was the lowest, with 70 (ml/m h) mean absolute error (MAE) for HP and 4.5% for ITE. Decision tree regressor has a better prediction for HP with 33 (ml/m h) MAE and almost the same MAE for ITE. Neural network has a better prediction for HP with 28 (ml/m h) MAE and a bit worse prediction for ITE with 5.7%. The best model used the deep neural network with 1.94 (ml/m h) MAE for HP and 0.67% MAE for ITE.
本研究工作旨在通过实验评估，将亚麻纤维芯和氧化钴纳米粒子结合到阶梯式双斜坡太阳能蒸馏器中，以提高水蒸发和产水性能，从而提高阶梯式双斜坡太阳能蒸馏器的性能。结果表明，棉芯和 1wt%的氧化钴（CoO）纳米流体提高了每小时淡水产量（HP）和瞬时热效率（ITE）。另一方面，本研究比较了四种机器学习方法，以创建管状太阳能蒸馏器性能的预测模型。开发和比较的方法是支持向量回归器（SVR）、决策树回归器、神经网络和基于实验数据的深度神经网络。这个问题是一个多输出预测问题，即 HP 和 ITE。SVR 的预测性能最低，HP 的平均绝对误差（MAE）为 70（ml/m h），ITE 的 MAE 为 4.5%。决策树回归器对 HP 的预测更好，MAE 为 33（ml/m h），与 ITE 的 MAE 几乎相同。神经网络对 HP 的预测更好，MAE 为 28（ml/m h），对 ITE 的预测稍差，MAE 为 5.7%。使用深度神经网络的最佳模型，HP 的 MAE 为 1.94（ml/m h），ITE 的 MAE 为 0.67%。
