Concept explainers
The common thermal polymerase chain reaction (PCR) process requires the cycling of reagents through three distinct temperatures for denaturation (90 − 94°C), annealing (50 − 55°C), and extension (72°C). In continuous-flow PCR reactors, the temperatures of the three thermal zones are maintained as fixed while the reagents are cycled continuously through these zones. These temperature variations induce significant variations in the fluid density, which under appropriate conditions can be used to generate fluid motion. The figure depicts a thermosiphon-based PCR device. The closed loop is filled with PCR reagents. The plan of the loop is inclined at an angle α with respect to the vertical. The loop is surrounded by three heaters and coolers that maintain different temperatures.
- (a) Explain why the fluid automatically circulates in the closed loop in the counterclockwise direction.
- (b) What is the effect of the angle α on the fluid velocity?
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Fox and McDonald's Introduction to Fluid Mechanics
Additional Engineering Textbook Solutions
Web Development and Design Foundations with HTML5 (8th Edition)
Introduction To Programming Using Visual Basic (11th Edition)
Starting Out with Java: From Control Structures through Objects (7th Edition) (What's New in Computer Science)
Starting Out with Python (4th Edition)
Mechanics of Materials (10th Edition)
Degarmo's Materials And Processes In Manufacturing
- An unpressurized cylindrical tank with a 100-foot diameter holds a 40-foot column of water. What is total force acting against the bottom of the tank?arrow_forward7. In the following problems check to see if the set S is a vector subspace of the corresponding R. If it is not, explain why not. If it is, then find a basis and the dimension. (a) S = (b) S = {[],+,"} X1 x12x2 = x3 CR³ {[1], 4+4 = 1} CR³ X2arrow_forwardAAA Show laplace transform on 1; (+) to L (y(+)) : SY(s) = x (0) Y(s) = £ [lx (+)] = 5 x(+) · est de 2 -St L [ y (^) ] = So KG) et de D 2 D D AA Y(A) → Y(s) Ŷ (+) → s Y(s) -yarrow_forward
- 1) In each of the following scenarios, based on the plane of impact (shown with an (n, t)) and the motion of mass 1, draw the direction of motion of mass 2 after the impact. Note that in all scenarios, mass 2 is initially at rest. What can you say about the nature of the motion of mass 2 regardless of the scenario? m1 15 <+ m2 2) y "L χ m1 m2 m1 בז m2 Farrow_forward8. In the following check to see if the set S is a vector subspace of the corresponding Rn. If it is not, explain why not. If it is, then find a basis and the dimension. X1 (a) S = X2 {[2], n ≤ n } c X1 X2 CR² X1 (b) S X2 = X3 X4 x1 + x2 x3 = 0arrow_forward2) Suppose that two unequal masses m₁ and m₂ are moving with initial velocities V₁ and V₂, respectively. The masses hit each other and have a coefficient of restitution e. After the impact, mass 1 and 2 head to their respective gaps at angles a and ẞ, respectively. Derive expressions for each of the angles in terms of the initial velocities and the coefficient of restitution. m1 m2 8 m1 ↑ บา m2 ñ Вarrow_forward
- The fallowing question is from a reeds book on applied heat i am studying. Although the answer is provided, im struggling to understand the whole answer and the formulas and the steps theyre using. Also where some ov the values such as Hg and Hf come from in part i for example. Please explain step per step in detail thanks In an NH, refrigerator, the ammonia leaves the evaporatorand enters the cornpressor as dry saturated vapour at 2.68 bar,it leaves the compressor and enters the condenser at 8.57 bar with50" of superheat. it is condensed at constant pressure and leavesthe condenser as saturated liquid. If the rate of flow of the refrigerantthrough the circuit is 0.45 kglmin calculate (i) the compressorpower, (ii) the heat rejected to the condenser cooling water in kJ/s,an (iii) the refrigerating effect in kJ/s. From tables page 12, NH,:2.68 bar, hg= 1430.58.57 bar, hf = 275.1 h supht 50" = 1597.2Mass flow of refrigerant--- - - 0.0075 kgls 60Enthalpy gain per kg of refrigerant in…arrow_forwardstate the formulas for calculating work done by gasarrow_forwardExercises Find the solution of the following Differential Equations 1) y" + y = 3x² 3) "+2y+3y=27x 5) y"+y=6sin(x) 7) y"+4y+4y = 18 cosh(x) 9) (4)-5y"+4y = 10 cos(x) 11) y"+y=x²+x 13) y"-2y+y=e* 15) y+2y"-y'-2y=1-4x³ 2) y"+2y' + y = x² 4) "+y=-30 sin(4x) 6) y"+4y+3y=sin(x)+2 cos(x) 8) y"-2y+2y= 2e* cos(x) 10) y+y-2y=3e* 12) y"-y=e* 14) y"+y+y=x+4x³ +12x² 16) y"-2y+2y=2e* cos(x)arrow_forward
- The state of stress at a point is σ = -4.00 kpsi, σy = 16.00 kpsi, σ = -14.00 kpsi, Try = 11.00 kpsi, Tyz = 8.000 kpsi, and T = -14.00 kpsi. Determine the principal stresses. The principal normal stress σ₁ is determined to be [ The principal normal stress σ2 is determined to be [ The principal normal stress σ3 is determined to be kpsi. kpsi. The principal shear stress 71/2 is determined to be [ The principal shear stress 7½ is determined to be [ The principal shear stress T₁/, is determined to be [ kpsi. kpsi. kpsi. kpsi.arrow_forwardRepeat Problem 28, except using a shaft that is rotatingand transmitting a torque of 150 N * m from the left bearing to the middle of the shaft. Also, there is a profile keyseat at the middle under the load. (I want to understand this problem)arrow_forwardProb 2. The material distorts into the dashed position shown. Determine the average normal strains &x, Ey and the shear strain Yxy at A, and the average normal strain along line BE. 50 mm B 200 mm 15 mm 30 mm D ΕΙ 50 mm x A 150 mm Farrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning