Answer:
B. The thickness of the heated region near the plate is increasing.
Explanation:
First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.
From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.
Answer:
thickness1=1.4m
thickness2=2.2m
convection coefficient=0.33W/m^2K
Explanation:
you must use this equation to calculate the thickness:
L=K(T2-T1)/Q
L=thickness
T=temperature
Q=heat
L1=0.04*(0--350)/10=1.4m
L2=0.1(220-0)/10=2.2m
Then use this equation to calculate the convective coefficient
H=Q/(T2-T1)
H=10/(250-220)=0.33W/m^2K
Answer: Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.
Explanation: (I hope this helped!! ^^)
Answer:
Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;
1. Agentes de arrastre de aire (AEA) o
2. Materiales poliméricos súper absorbentes
Explanation:
La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes
Ejemplos de agentes de arrastre de aire son;
Sulfonatos alcalinos
Acidos de resinas sulfonadas
Sales de ácidos grasos
Ejemplos de materiales poliméricos superabsorbentes son;
SAP0.26CT
SAP0.39PT.
Answer:
>>pounds=13.2
>>kilos=pounds/2.2
Explanation:
Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.
