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
Answer: hello some aspects of your question is missing below is the missing information
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m.
answer:
≈ 22.5 mm
Explanation:
Given data:
Inner diameter = 1.5 m
pressure = 5 MPa
factor of safety = 1.5
<u>Calculate the required minimum wall thickness</u>
maximum-shear-stress theory ( σ allow ) = σγ / FS
= 250(10)^6 / 1.5 = 166.67 (10^6) Pa
given that |σ| = σ allow
3.75 (10^6) / t = 166.67 (10^6)
∴ t ( wall thickness ) = 0.0225 m ≈ 22.5 mm
Answer:
7.07%
Explanation:
Thermal efficiency can be by definition seen as the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.
We will determine the thermal efficiency of the given problem at the attached file.
Answer:
Simple G Code Example Mill - This is a very simple G code example for beginner level ... Run the program on your cnc machine (Safety first, keep a professional around). ... This G code program example don't use Tool radius compensation ...
Explanation: