
The two broad categories of display cases are _______and _______.

The two broad categories of the display cases.
Answer to Problem 1RQ
The two broad categories of display cases are open and closed.
Explanation of Solution
The display case is a cabinet made up of a transparent material. It is used to display objects for viewing. It can be seen in restaurants, stores, or maybe in a house.
Display cases are also used in retail stores to preserve food and other things, which are displayed to the customers for sale. The customers can select items from the section of the store and choose any of them to purchase. Some of the display cases are open which allows the customer to reach in without opening a door. The closed display cases are more energy efficient as compared to open display cases. These display cases are available in different temperature ranges such as low, medium, and high-temperature range. An open display case maintains temperature conditions. This is because the refrigerated air is heavy and it gets settled at the bottom. This is achieved by designing and maintaining air patterns, which keep the refrigerated air from leaving the case.
Conclusion:
The two broad categories of display cases are open and closed. Both of them have their own characteristics as explained earlier.
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Chapter 26 Solutions
Refrigeration and Air Conditioning Technology (MindTap Course List)
- Experiment تكنولوجيا السيارات - Internal Forced convenction Heat transfer Air Flow through Rectangular Duct. objective: Study the convection heat transfer of air flow through rectangular duct. Valve Th Top Dead Centre Exhaust Valve Class CP. N; ~ RIVavg Ti K 2.11 Te To 18.8 21.3 45.8 Nath Ne Pre Calculations:. Q = m cp (Te-Ti) m: Varg Ac Acca*b Q=hexp As (Ts-Tm) 2 2.61 18.5 20.846.3 Tm = Te-Ti = 25 AS-PL = (a+b)*2*L Nu exp= Re-Vavy D heep Dh k 2ab a+b Nu Dh the- (TS-Tm) Ts. Tmy Name / Nu exp Naxe بب ارتدان العشريarrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findarrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of thearrow_forward
- Procedure: 1- Cartesian system, 2D3D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D ∑Fx=0 ∑Fy=0 ∑Fz=0 ∑Mx=0 ∑My=0 ΣMz=0 2D ΣFx=0 ΣFy=0 ΣMz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of thearrow_forwardProcedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of thearrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forward
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- The triangular plate, having a 90∘∘ angle at AA, supports the load PP = 370 lblb as shown in (Figure 1).arrow_forwardDesign a 4-bar linkage to carry the body in Figure 1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis with the free choice values z = 1.075, q= 210°, ß2 = −27° for left side and s = 1.24, y= 74°, ½ = − 40° for right side. φ 1.236 P2 147.5° 210° 2.138 P1 Figure 1 Xarrow_forwardDesign a 4-bar linkage to carry the body in Figure 1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis with the free choice values z = 1.075, q= 210°, B₂ = −27° for left side and s = 1.24, y= 74°, ½ = − 40° for right side. 1.236 P2 147.5° 210° P1 Figure 1 2.138 Xarrow_forward
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