Answer:
hand tracing
Explanation:
as a programmer when we pretend computer in the debugging process by the step of each statement in recording
then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method
so while recording value of each variable each step is hand tracing
Answer:
Explanation:
The python code to generate this is quite simple to run.
i hope you understand everything written here, you can as well try out other problems to understand better.
First to begin, we import the package;
Code:
import pandas as pd
import matplotlib.pyplot as plt
name = input('Enter name of the file: ')
op = input('Enter name of output file: ')
df = pd.read_csv(name)
df['Date'] = pd.to_datetime(df["Date"].apply(str))
plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")
plt.legend(loc="upper right")
plt.xticks(rotation=20)
plt.savefig(op)
plt.show()
This should generate the data(plot) as seen in the uploaded screenshot.
thanks i hope this helps!!!
Answer:
![Q_{cv} = -1007.86kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20-1007.86kJ)
Explanation:
Our values are,
State 1
![V=3m^3\\P_1=1bar\\T_1 = 295K](https://tex.z-dn.net/?f=V%3D3m%5E3%5C%5CP_1%3D1bar%5C%5CT_1%20%3D%20295K)
We know moreover for the tables A-15 that
![u_1 = 210.49kJ/kg\\h_i = 295.17kJkg](https://tex.z-dn.net/?f=u_1%20%3D%20210.49kJ%2Fkg%5C%5Ch_i%20%3D%20295.17kJkg)
State 2
![P_2 =6bar\\T_2 = 296K\\T_f = 320K](https://tex.z-dn.net/?f=P_2%20%3D6bar%5C%5CT_2%20%3D%20296K%5C%5CT_f%20%3D%20320K)
For tables we know at T=320K
![u_2 = 228.42kJ/kg](https://tex.z-dn.net/?f=u_2%20%3D%20228.42kJ%2Fkg)
We need to use the ideal gas equation to estimate the mass, so
![m_1 = \frac{p_1V}{RT_1}](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7Bp_1V%7D%7BRT_1%7D)
![m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7B1bar%2A100kPa%2F1bar%283m%5E3%29%7D%7B0.287kJ%2Fkg.K%28295k%29%7D)
![m_1 = 3.54kg](https://tex.z-dn.net/?f=m_1%20%3D%203.54kg)
Using now for the final mass:
![m_2 = \frac{p_2V}{RT_2}](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7Bp_2V%7D%7BRT_2%7D)
![m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7B1bar%2A100kPa%2F6bar%283m%5E3%29%7D%7B0.287kJ%2Fkg.K%28320k%29%7D)
![m_2 = 19.59kg](https://tex.z-dn.net/?f=m_2%20%3D%2019.59kg)
We only need to apply a energy balance equation:
![Q_{cv}+m_ih_i = m_2u_2-m_1u_1](https://tex.z-dn.net/?f=Q_%7Bcv%7D%2Bm_ih_i%20%3D%20m_2u_2-m_1u_1)
![Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3Dm_2u_2-m1_u_1-%28m_2-m_1%29h_i)
![Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20%2819.59%29%28228.42%29-%283.54%29%28210.49%29-%2819.59-3.54%29%28295.17%29)
![Q_{cv} = -1007.86kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20-1007.86kJ)
The negative value indidicates heat ransfer from the system