Answer:
1. You have the courage to help without expecting a reward.
2. Because actions are more eloquent than words. Actions are far more valuable and counted than words, and that's how she inspired me.
3. Doing simple things that can make someone grateful and happy without knowing that someone is inspired and motivated by your good deeds, and also doing some interesting things By.
Explanation:
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:

It is necessary to get the Reynold's number:

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

The overall heat transfer coefficient:

Here

Substituting values:

Answer:
I=0.3636
Explanation:
See the attached picture for explanation.
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity


Temperature at 28 mm distance after t time = = 50 degree C
we know that

![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)

from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)

solving fot t we get
t = 2244.3 sec
Answer:
a) C_v = 1.005 KJ/kgK
b) C_v = 1005.000 J/kgC
c) C_v = 0.240 kcal/kgC
d) C_v = 0.240 Btu/lbmF
Explanation:
Given:
- constant-pressure specific heat C_v = 1.005 KJ/kgC
Find C_v in units of:
a) kJ/kg·K
b) J/g·°C
c) kcal/ kg·°C
d) Btu/lbm·°F
Solution:
a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,
C_v = 1.005 KJ/kgK
b)
C_v = 1.005 KJ/kgC * ( 1000 J / KJ)
C_v = 1005.000 J/kgC
c)
C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)
C_v = 0.240 kcal/kgC
d)
C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)
C_v = 0.240 Btu/lbmF