Answer:
Force per unit plate area is 0.1344 ![N/m^{2}](https://tex.z-dn.net/?f=N%2Fm%5E%7B2%7D)
Solution:
As per the question:
The spacing between each wall and the plate, d = 10 mm = 0.01 m
Absolute viscosity of the liquid, ![\mu =1.92\times 10^{- 3} Pa-s](https://tex.z-dn.net/?f=%5Cmu%20%3D1.92%5Ctimes%2010%5E%7B-%203%7D%20Pa-s)
Speed, v = 35 mm/s = 0.035 m/s
Now,
Suppose the drag force that exist between each wall and plate is F and F' respectively:
Net Drag Force = F' + F''
![F = \tau A](https://tex.z-dn.net/?f=F%20%3D%20%5Ctau%20A)
where
= shear stress
A = Cross - sectional Area
Therefore,
Net Drag Force, F = ![(\tau ' +\tau '')A](https://tex.z-dn.net/?f=%28%5Ctau%20%27%20%2B%5Ctau%20%27%27%29A)
![\frac{F}{A} = \tau ' +\tau ''](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Ctau%20%27%20%2B%5Ctau%20%27%27)
Also
F = ![\frac{\mu v}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmu%20v%7D%7Bd%7D)
where
= dynamic coefficient of viscosity
Pressure, P = ![\frac{F}{A}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D)
Therefore,
![\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Cfrac%7B%5Cmu%20v%7D%7Bd%7D%20%2B%20%5Cfrac%7B%5Cmu%20v%7D%7Bd%7D%20%3D%202%5Cfrac%7B%5Cmu%20v%7D%7Bd%7D)
![\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%202%5Cfrac%7B1.92%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%200.035%7D%7B0.010%7D%20%3D%200.01344%20N%2Fm%5E%7B2%7D)
Dot 3 is mostly used in a lot of v4 and v6
Except the Table of Contents
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
![dQ = m \times c_p \times (T_2 -T_1)](https://tex.z-dn.net/?f=dQ%20%3D%20m%20%5Ctimes%20c_p%20%5Ctimes%20%28T_2%20-T_1%29)
For ideal gas,
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.