ME 200 – Thermodynamics I – Spring 2020

Homework 14: Open System Mass Balance

Part (i): Nozzle

Part (ii): Garden Hose

Part (iii): Pressure Drop Device

Part (i): Nozzle

Given:

A nozzle at steady state using air as the working fluid has inlet conditions of 3 m/s, 110 kPa, and 60°C. The air leaves the nozzle at 15 m/s, 100 kPa, and 60°C.

In [1]:
# Given Inputs:
V_1 = 3              # inlet air velocity [m/s]
P_1 = 110            # inlet air pressure [kPa]
T_1 = 60 + 273.15    # inlet air temperature [K]
V_2 = 15             # exit air velocity [m/s]
P_2 = 100            # exit air pressure [kPa]
T_2 = 60. + 273.15   # exit air temperature [K]

Find:

the ratio of inlet to exit area of the nozzle

System Diagram:

The air inside the nozzle is chosen as the system. This system is open.

Assumptions:

1) open system, 2) steady flow, 3) air treated as an ideal gas

Basic Equations:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$$$\dot m=\rho VA$$$$Pv=RT$$

Solution:

First the conservation of mass is used to relate the mass flow entering and leaving the nozzle.

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$

Since the flow is steady, then $dm/dt = 0$. Also, since there is only one inlet and one outlet, then summations are replaced with single mass flow terms, one for the mass entering and one for the mass leaving the system. Thus, the conservation of mass equation simplifies to the following result showing that the mass flow rates into and out of the system are equal.

$$\dot m_{out}=\dot m_{in}$$

The mass flow rate can also be related to the fluid density, cross-sectional flow area, and velocity

$$\dot m=\rho AV$$

and when combined with conservation of mass gives

$$\rho _1 A_1V_1 = \rho _2A_2V_2 \rightarrow \dfrac{A_1V_1}{v_1} = \dfrac{A_2V_2}{v_2}$$

From the ideal gas law, specific volume is

$$v =\dfrac{RT}{P}$$

such that the mass balance can be rewritten as

$$\dfrac{A_1V_1P_1}{RT_1}=\dfrac{A_2V_2P_2}{RT_2}$$

This equation can then be rearranged to solve for the ratio of exit to inlet areas in terms of the inlet and outlet states.

$$\dfrac{A_2}{A_1}=\dfrac{P_1}{P_2}\dfrac{T_2}{T_1}\dfrac{V_1}{V_2}$$
In [3]:
A_ratio = P_1*T_2*V_1/(P_2*T_1*V_2)

print('A_1/A_2 = ',round(1/A_ratio,2))

A_1/A_2 =  4.55

Part(ii): Garden Hose

Given

A hose is filling a 5 gallon bucket. The hose inner diameter is 2.54 cm and is reduced to 1.27 cm using a nozzle. The water flows at a velocity of 2 m/s in the hose and a temperature of 20°C.

In [39]:
# Given Inputs:
Vol_bucket = 0.0189271  # bucket volume [m^3]
d_1 = 0.0254            # hose (nozzle inlet) diameter [m]
d_2 = 0.0127            # nozzle exit diameter [m]
V_1 = 2                 # nozzle inlet velocity [m/s]
T_1 = 20 + 273.15       # nozzle inlet temperature [K]

Find:

a) Volumetric flow rate and mass flow rate of water in the hose

b) Velocity at the exit of the nozzle

c) Time required to fill the bucket

System Diagram:

For parts a and b, we can consider the water in the nozzle that is connected to the hose as the system. For part c, the volume of the bucket can be considered as the system.

Assumptions:

1) open system, 2) steady flow, 2) water is a considered an incompressible substance

Basic Equations:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$$$\dot m=\rho VA$$

Solution:

Part a)

The flow is assumed to be steady, so the mass flow in the hose is the mass flow that both enters and leaves the nozzle. The mass flow can be related to the fluid density, cross-sectional flow area, and flow velocity according to

$$\dot m=\rho AV$$

The hose and its connection to the nozzle have a circular cross-section. The water is considered to be an incompressible liquid and its density can be determined from the saturated liquid specific volume at the water temperature from the SLVM table for water. Thus,

$$\dot m= \dfrac{\pi d_1^2}{4} \dfrac{V_1}{v_1}$$

where $ v_1 = v_f(20°C) = 0.0010018m^3/kg$

The volumetric flow rate is the product of the velocity and cross-sectional flow area. Since the density is constant for an incompressible, then volummetric flow rate is the same at the inlet and outlet.

$$\dot V = V_1A_1 = V_1\dfrac{\pi d_1^2}{4}$$
In [40]:
from math import pi

v_1 = 0.0010018                  # specific volume of water @20C [m^3/kg]

m_dot = pi*d_1**2/4*V_1/v_1      # mass flow rate [kg/s]
V_dot = pi*d_1**2/4*V_1          # volumetric flow rate [m^3/s]

print('m_dot = ',round(m_dot,3),'kg/s')
print('V_dot = ',round(V_dot,6),'m^3/s')

m_dot =  1.012 kg/s
V_dot =  0.001013 m^3/s

Part b)

In order to find the velocity at the exit of the nozzle, conservation of mass applied to the water in the nozzle is used to relate the entering and leaving mass flow rates.

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$

Assuming steady flow ($dm/dt = 0$) and relating the mass flow rate to the fluid density, cross-sectional flow area, and velocity.

$$\dot m_{out}=\dot m_{in} \rightarrow \rho _1 A_1V_1 = \rho _2A_2V_2$$

The inlet and outlet of the nozzle are circular and the density is replaced with the specific volume from part a.

$$\dfrac{\pi d_1^2}{4}\dfrac{V_1}{v_1}= \dfrac{\pi d_2^2}{4}\dfrac{V_2}{v_2}$$

Then, solving for the exit velocity in terms of the inlet velocity and hose diameters gives

$$V_2=V_1 \dfrac{d_1^2}{d_2^2}$$
In [41]:
V_2 = V_1*d_1**2/d_2**2

print('V_2 = ',round(V_2,3),'m/s')

V_2 =  8.0 m/s

Part c)

The time required to fill up the bucket can be determined using a transient mass balance with the bucket as the system.

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$

In this case, there is no outlet flow such that $\dot m_{out}=0$. Also, since the density is constant for an incompressible, then

$$\rho\dfrac{dV_{water}}{dt} = \rho \dot V \rightarrow \dfrac{dV_{water}}{dt} = \dot V$$

where $V_{water}$ is the volume of water in the bucket at any time. Integration from the start to the end of the filling process with a constant volumetric flow rate and solving for the filling time gives

$$t_{fill} = \dfrac{V_{bucket}}{\dot V}$$
In [42]:
t_fill = Vol_bucket/V_dot

print('t_fill = ',round(t_fill,1),'s')

t_fill =  18.7 s

Part(iii): Pressure Drop Device

Given:

Saturated liquid water at 50°C enters a device and leaves as a two phase mixture at 20 kPa and quality of 0.2. The cross-sectional areas of the inlet and outlet are equal. The system operates with a steady flow.

In [43]:
# Given Inputs:
T_1 = 50        # inlet temperature of water [C]
X_1 = 0.        # saturated liquid condition at inlet
P_2 = 20        # outlet pressure of the water [kPa]
x_2 = 0.2       # outlet quality of water

Find:

Ratio of inlet and outlet velocities

System Diagram:

Assumptions:

1) open system, 2) steady flow

Basic Equations:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$$$\dot m=\rho VA$$$$z_{mix}=xz_g+(1-x)z_f$$

Solution:

In order to find the ratio of inlet and outlet velocities, conservation of mass is used to relate the mass entering and leaving the device.

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$

With steady flow, $dm/dt=0$. Also, there is only one inlet and one outlet. As a result,

$$\dot m_1 = \dot m_2 \rightarrow \rho _1 A_1V_1=\rho _2A_2V_2$$

Replacing the density with the inverse of the specific volume and solving for the ratio of velocities assuming the same inlet eand exit areas results in the following.

$$\dfrac{V_2}{V_1}=\dfrac{v_2}{v_1}$$

where

$$v_1 = v_{f,50°C} = 0.0010121m^3/kg$$$$v_2 = x_2v_{g,20kPa} + (1-x_2)v_{f,20kPa}$$
In [44]:
v_1 = 0.0010121                     # saturated liquid specific volume of T_1=50C [m^3/kg]
v_g_20 = 7.6480                     # saturated water vapor specific volume P_2=20kPa [m^3/kg]
v_f_20 = 0.0010172                  # saturated liquid water specific volume at P_2=20kPa [m^3/kg]
v_2 = x_2*v_g_20 + (1-x_2)*v_f_20   # SLVM specific volume for state 2[m^3/kg]
V_ratio = v_2/v_1                   # ratio of exit to inlet velocity

print('V_2/V_1 = ',round(V_ratio))

V_2/V_1 =  1512
In [ ]: