Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
![L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m](https://tex.z-dn.net/?f=L_c%20%3D%20%5Cfrac%7BV%7D%7BA%7D%20%3D%20%5Cfrac%7BLA%7D%7B2A%7D%20%3D%20%5Cfrac%7B5%2A10%5E%7B-3%7D%7D%7B2%7D%20%3D%200.0025%20%5C%20m)
The Biot number is given as;
![B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952](https://tex.z-dn.net/?f=B_i%20%3D%20%5Cfrac%7Bh%20L_c%7D%7Bk%7D%5C%5C%5C%5CB_i%20%3D%20%5Cfrac%7B80%2A0.0025%7D%7B21%7D%20%5C%5C%5C%5CB_i%20%3D%200.00952)
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
![\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B%5Crho%20C_p%20V%7D%7BhA_s%7D%20%3D%20%5Cfrac%7B%5Crho%20C_p%20L_c%7D%7Bh%7D%5C%5C%5C%5C%5Ctau%20%3D%20%5Cfrac%7B8000%2A%20570%2A%200.0025%7D%7B80%7D%5C%5C%5C%5C%5Ctau%20%3D%20142.5%20s)
The time for the heating process is given as;
![t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7BV%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B3%20%5C%20m%7D%7B0.01%20%5C%20m%2Fs%7D%20%3D%20300%20s)
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
![T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C](https://tex.z-dn.net/?f=T%28t%29%20%3D%20T_%7B%20%5Cinfty%7D%20%2B%20%28T_i%20-T_%7B%5Cinfty%7D%29e%5E%7B-t%2F%20%5Ctau%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%2820%20-930%29e%5E%7B-300%2F%20142.5%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%28-110.85%29%5C%5C%5C%5CT_%7B%28t%29%7D%20%3D%20819.15%20%5C%20%5E0%20C)
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer:
e.Fire resistance,Inexpensive,Non-toxic.
Explanation:
Desirable hydraulic property of fluid as follows
1. Good chemical and environment stability
2. Low density
3. Ideal viscosity
4. Fire resistance
5. Better heat dissipation
6. Low flammability
7. Good lubrication capability
8. Low volatility
9. Foam resistance
10. Non-toxic
11. Inexpensive
12. Demulsibility
13. Incompressibility
So our option e is right.
Answer:
(a) T = W/2(1-tanθ) (b) 39.81°
Explanation:
(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:
Summation of moment in clockwise direction is equivalent to zero. Therefore,
T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0
T*l*(cosθ - sinθ) = W*(l/2)*cosθ
T = W*cosθ/2(cosθ - sinθ)
Dividing both the numerator and denominator by cosθ, we have:
T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)
(b) If T = 3W, then:
3W = W/2(1-tanθ),
Further simplification and rearrangement lead to:
1 - tanθ = 1/6
tanθ = 1 - (1/6) = 5/6
θ = tan^(-1) 5/6 = 39.81°
Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number (
) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
![Re=\frac{\rho v D}{\mu}](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B%5Crho%20v%20D%7D%7B%5Cmu%7D)
where
is the density of the fluid,
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
![v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7BRe%2A%5Cmu%7D%7B%5Crho%2AD%7D%20%3D%5Cfrac%7B2100%2A2.1x10%5E%7B-2%7DPa%2As%20%7D%7B855kg%2Fm%5E3%2A0.01m%7D%20%3D5.16m%2Fs)
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:
![D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm](https://tex.z-dn.net/?f=D%3D%5Cfrac%7BRe%2A%5Cmu%7D%7B%5Crho%2Av%7D%20%3D%5Cfrac%7B2100%2A1.5x10%5E%7B-2%7DPa%2As%7D%7B925kg%2Fm%5E3%2A5.16m%2Fs%7D%3D6.6%20mm)
Answer:
porosity = 0.07 or 7%
dry bulk density = 3.25g/cm3]
water content =
Explanation:
bulk density = dry Mass / volume of sample
dry mass = 0.490kg = 490g
volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3
density = 490/150.8 = 3.25g/cm3
porosity =
=
= 0.07 or 7%
water content =
= 7%