Let's break down the problem step by step: --- # **Step 1: List the Given Data** - **Location:** Detroit, Michigan - Latitude (\(\phi\)) = 42.3314° N - Longitude = 83.0458° W - **Date:** June 1 - **Time:** 12:00 noon EST - **Collector Tilt (\(\beta\))**: 30° - **Collector Azimuth (\(\alpha_w\))**: +20° (East of South) - **We need:** Instantaneous solar radiation (irradiance) on the collector. --- # **Step 2: Calculate Solar Declination (\(\delta\))** Solar declination for day \( n \) of the year: \[ \delta = 23.45^\circ \sin\left(\frac{360}{365}(n - 81)\right) \] June 1 is typically day 152. \[ \delta = 23.45^\circ \sin\left(\frac{360}{365}(152 - 81)\right) = 23.45^\circ \sin\left(\frac{360}{365} \times 71\right) = 23.45^\circ \sin(70.03^\circ) \] \[ \sin(70.03^\circ) \approx .940 \] \[ \delta \approx 23.45^\circ \times .940 \approx 22.^\circ \] --- # **Step 3: Calculate Solar Hour Angle (\(H\))** Solar noon is when the sun is highest. At local solar noon, \(H = ^\circ\). --- # **Step 4: Calculate Solar Zenith Angle (\(\theta_z\))** \[ \cos\theta_z = \sin\phi \sin\delta + \cos\phi \cos\delta \cos H \] Insert values (\(\phi = 42.33^\circ\), \(\delta = 22.^\circ\), \(H = ^\circ\)): \[ \sin\phi = \sin(42.33^\circ) \approx .674 \] \[ \sin\delta = \sin(22.^\circ) \approx .375 \] \[ \cos\phi = \cos(42.33^\circ) \approx .739 \] \[ \cos\delta = \cos(22.^\circ) \approx .927 \] \[ \cos H = \cos(^\circ) = 1 \] So, \[ \cos\theta_z = (.674)(.375) + (.739)(.927)(1) = .253 + .685 = .938 \] \[ \theta_z = \cos^{-1}(.938) \approx 20.3^\circ \] --- # **Step 5: Calculate Solar Azimuth Angle (\(\gamma_s\))** At solar noon in the northern hemisphere, the sun is due south, so \(\gamma_s = ^\circ\). --- # **Step 6: Angle of Incidence (\(\theta\)) on the Collector** \[ \cos\theta = \sin\delta \sin\phi \cos\beta + \sin\delta \cos\phi \sin\beta \cos\alpha_w + \cos\delta \cos\phi \cos\beta \cos H + \cos\delta \sin\phi \sin\beta \cos\alpha_w \cos H + \cos\delta \sin\beta \sin\alpha_w \sin H \] At solar noon (\(H = \)), \(\sin H = \), \(\cos H = 1\): - \(\beta = 30^\circ\) - \(\alpha_w = 20^\circ\) \[ \sin\beta = \sin(30^\circ) = .5 \] \[ \cos\beta = \cos(30^\circ) = .866 \] \[ \cos\alpha_w = \cos(20^\circ) = .940 \] \[ \sin\alpha_w = \sin(20^\circ) = .342 \] Plug in values (from previous steps): - \(\sin\delta = .375\) - \(\cos\delta = .927\) - \(\sin\phi = .674\) - \(\cos\phi = .739\) Now, calculate each term: **Term 1:** \[ \sin\delta \sin\phi \cos\beta = .375 \times .674 \times .866 \approx .219 \] **Term 2:** \[ \sin\delta \cos\phi \sin\beta \cos\alpha_w = .375 \times .739 \times .5 \times .94 \approx .130 \] **Term 3:** \[ \cos\delta \cos\phi \cos\beta \cos H = .927 \times .739 \times .866 \times 1 \approx .592 \] **Term 4:** \[ \cos\delta \sin\phi \sin\beta \cos\alpha_w \cos H = .927 \times .674 \times .5 \times .94 \times 1 \approx .293 \] **Term 5:** \[ \cos\delta \sin\beta \sin\alpha_w \sin H = .927 \times .5 \times .342 \times = \] Sum: \[ \cos\theta = .219 + .130 + .592 + .293 = 1.234 \] **Note:** This is greater than 1, which is not possible. Let's check the formula. Usually, the angle of incidence formula is: \[ \cos\theta = \sin\delta \sin\phi \cos\beta - \sin\delta \cos\phi \sin\beta \cos\alpha_w + \cos\delta \cos\phi \cos\beta \cos H + \cos\delta \sin\phi \sin\beta \cos\alpha_w \cos H + \cos\delta \sin\beta \sin\alpha_w \sin H \] Notice the second term should be **minus**, not plus. So: **Term 2:** \[ - .130 \] New sum: \[ \cos\theta = .219 - .130 + .592 + .293 = .974 \] \[ \theta = \cos^{-1}(.974) \approx 12.7^\circ \] --- # **Step 7: Extraterrestrial Solar Irradiance (\(I_\))** \[ I_{sc} = 1367 \, \text{W/m}^2 \] \[ I_ = I_{sc} \left[1 + .033 \cos\left(\frac{360n}{365}\right)\right] \] \[ I_ = 1367 \times \left[1 + .033 \cos\left(\frac{360 \times 152}{365}\right)\right] \] \[ \frac{360 \times 152}{365} = 149.86^\circ \] \[ \cos(149.86^\circ) \approx -.870 \] \[ I_ = 1367 \times [1 + .033 \times (-.870)] = 1367 \times (1 - .0287) = 1367 \times .9713 \approx 1328 \, \text{W/m}^2 \] --- # **Step 8: Air Mass Ratio (\(AM\))** \[ AM = \frac{1}{\cos \theta_z} \] \[ AM = \frac{1}{.938} \approx 1.066 \] --- # **Step 9: Estimate Instantaneous Solar Radiation \(I\) on the Collector** Assuming a clear sky, the global horizontal irradiance at noon is about 90% of extraterrestrial after atmospheric losses: \[ I_{GH} \approx I_ \times .9 \times \cos\theta_z \] \[ I_{GH} \approx 1328 \times .9 \times .938 \approx 112 \, \text{W/m}^2 \] On the tilted surface: \[ I_{tilted} = I_{GH} \frac{\cos\theta}{\cos\theta_z} \] \[ I_{tilted} = 112 \times \frac{.974}{.938} \approx 1163 \, \text{W/m}^2 \] --- # **Final Answer** ## **The instantaneous solar radiation at 12:00 noon EST on June 1 in Detroit, Michigan on the specified tilted solar collector is approximately:** \[ \boxed{116 \ \text{W/m}^2} \] *(rounded to three significant digits)* --- ### **Summary of Steps** 1. Determined declination angle for June 1. 2. Found solar hour angle at noon. 3. Calculated solar zenith angle. 4. Calculated angle of incidence on collector. 5. Found extraterrestrial irradiance. 6. Estimated atmospheric losses. 7. Adjusted horizontal irradiance to collector tilt. **Assumptions:** Clear sky, no shading, typical atmospheric conditions. Let me know if you need further clarification on any step!