# Problem 8.42

Atmospheric air enters a 10m long, 150mm diameter uninsulated heating duct at $60^\circ\mathrm{C}$ and 0.04 kg/s. The duct surface temperature is approximately constant at $T_{s} = 15^\circ\mathrm{C}$.

(a) What are the outlet air temperature, the heat rate q, and the pressure drop $\Delta p$ for these conditions?

(b) To illustrate the tradeoff between heat transfer rate and pressure drop considerations, calculate $q$ and $\Delta p$ for diameters in the range from 0.1 to 0.2 m. In your analysis, maintain the total surface area, $A_{s} = \pi DL$, at the value computed for part (a). Plot $q,\Delta p$, and $L$ as a function of the duct diameter.  


```python

from IPython.display import clear_output

import schemdraw as schem
import schemdraw.elements as e

import matplotlib.pyplot as plt
import numpy as np
import math
import scipy.constants as sc

import sympy as sym

from Libraries import thermodynamics as thermo
from Libraries import HT_internal_convection as intconv


```


```python
fluid = thermo.Fluid('air',60.,"C")
L = 10 #m
D = 0.15 #m
mdot = 0.04 #kg/s

A = np.pi*(0.075**2)
bulk = mdot/(fluid.rho*A)

Re = (4*mdot)/(np.pi*D*fluid.mu)
print('The Reynolds number is %.2f'%Re)



pipe = intconv.PipeFlow(D, L, rho=fluid.rho, nu=fluid.nu, Um = bulk)
print(pipe.Re)



```

    The Reynolds number is 16836.53
    10


# Problem 8.62

Water at a flow rate of $\dot{m}$ = 0.215 kg/s is cooled from $70^\circ\mathrm{C}$ to $30^\circ\mathrm{C}$ by passing it through a thin-walled tube of diameter $D$ = 50 mm and maintaining a coolant at $T_{\infty}$ = $15^\circ\mathrm{C}$ in cross flow over the tube. 

(a) What is the required tube length if the coolant is air and its velocity is $V$ = 20 m/s?

(b) What is the tuve length if the coolant is water and $V$ = 2 m/s?


```python
T_m = (70+30)/2
T_infty = 15 #C
T_mi = 70 #C
T_mo = 30 #C
T_f = (T_infty + T_m)/2 #C
mdot = 0.215 #kg/s
D = 0.05 #m
water = thermo.Fluid('water',T_m,"C")
air_c = thermo.Fluid('air',T_f,"C")
water_c = thermo.Fluid('water',T_f,"C")

q =mdot*water.Cp*(70-30)
#Finding log mean temperature, initial Reynolds number, and friction factor
T_lm = -(T_mi - T_mo)/np.log((T_infty-T_mo)/(T_infty-T_mi))
Re_i = (4*mdot)/(np.pi*D*water.mu)
f = (0.79*np.log(Re_i))**-2

#Gnielinski
Nu_i = ((f/8)*(Re_i-1000)*water.Pr)/(1+12.7*(((f/8)**.5))*((water.Pr**(2/3))-1))

h_i = Nu_i*(water.k/D)
#Reynolds number at outlet
Re_o_1 = (20*D)/air_c.nu

#Seban, constant T_s
Nu_o_1 = 5+0.025*(Re_o_1*water.Pr)**0.8
h_o_1 = Nu_o_1*water.k/D

Q = ((1/h_i)+(1/h_o_1))**-1

L_1 = q/(Q*np.pi*D*T_lm)
print("The required length if the coolant is air at 20 m/s is %.2f m."%L_1)

Re_o_2 = (2*D)/(water_c.nu)
Nu_o_2 = 5+0.025*(Re_o_2*water.Pr)**0.8
h_o_2 = Nu_o_2*water.k/D

Q_2 = ((1/h_i)+(1/h_o_2))**-1
L_2 = q/(Q_2*np.pi*D*T_lm)
print("The required length if the coolant is water at 2 m/s is %.2f m."%L_2)
```

    30.78621092446306
    The required length if the coolant is air at 20 m/s is 15.13 m.
    The required length if the coolant is water at 2 m/s is 14.58 m.



```python

```

0a4153ade23c46e9472c6aeac1cd0a4e9f9c1046
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Jupyter Notebook
Pettersen_HW6.ipynb
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