List a consistent set of assumptions required to employ each of the following equations. Possible assumptions include:
Closed system, open system, steady state, adiabatic, no work, negligible changes in potential energy, negligible changes in kinetic energy, isothermal, constant pressure, incompressible, ideal gas, constant specific heats. Show how the simplified equation is developed from a basic equation.
(a) $\Delta u = c\Delta T$
(b) $\Delta h = c\Delta T$
(c) $\Delta h = c_{p}\Delta T$
(d) $\Delta h = \Delta u + v\Delta p$
(a) $\Delta u = c\Delta T$
$\underline{Assumptions}$: incompressible, constant specific heat
This relation comes from $\Delta u=\int {cdT}$ which is the integration of a basic equation for evaluating internal energy changes of an incompressible substance ($du=cdT$). You should know that this is not a relation for an ideal gas because there is no differentiation between $c_p$ and $c_v$ for an incompressible. The integration result follows from the assumption of constant $c$.
(b) $\Delta h = c\Delta T$
$\underline{Assumptions}$: incompressible, constant pressure, constant specific heat
This comes from integration of the basic equation for evaluating enthalpy changes of an incompressible: $dh = du + vdP$. Then for constant pressure, $dP = 0$ and so $\Delta h =\Delta u = \int {cdT}$. The final integration result then follows from the assumption of constant $c$.
(c) $\Delta h = c_{p}\Delta T$
$\underline{Assumptions}$: ideal gas, constant specific heat
This comes directly from integration of the basic equation for evaluating the enthalpy change of an ideal gas $\Delta h=\int {c_p dT}$ with assumption of constant $c_p$.
(d) $\Delta h = \Delta u + v\Delta P$
$\underline{Assumptions}$: incompressible
This comes from integration of $dh = d(u+Pv)$ with constant $v$. Constant $v$ is the assumption of an incompressible leading to the basic equation: $dh = du + vdP$.
Compare the changes in specific internal energy and enthalpy for liquid water as determined using the following approaches:
a) Saturated liquid values
b) Constant specific heat evaluated at the average temperature
c) Constant specific heat evaluated at the initial temperature
The given pressure information can be used in checking whether the water is a liquid at the specified temperatures. Since the saturation temperature at 100 kPa is 99.61°C for water, then water at both 20°C and 60°C is a compressed liquid.
a) In the absence of compressed liquid data, the change in internal energy can be determined as a difference in two saturated liquid values at the two specific temperatures: $\Delta u = u_{f,60 °C} - u_{f,20°C}$
For an incompressible ($v$ = constant), the enthalpy change is $\Delta h = \Delta u + v\Delta P$. Since the pressure is constant, then $\Delta P = 0$ and $\Delta h = \Delta u = u_{f,60 °C} - u_{f,20°C}$
u_f_60 = 251.16 # specific internal energy of sat liq water @ 60°C [kJ/kg]
u_f_20 = 83.912 # specific internal energy of sat liq water @ 20°C [kJ/kg]
Delta_u_a = u_f_60 - u_f_20 # change in specific internal energy [kJ/kg]
Delta_h_a = Delta_u_a # change in specific enthalpy [kJ/kg]
print('Delta_u_a = ',round(Delta_u_a,2),'(kJ/kg)')
print('Delta_h_a = ',round(Delta_h_a,2),'(kJ/kg)')
b) Using constant specific heats at $T = \bar T = 313.15 K$ requires linear interpolation between 300 and 325 K for the specific heat of water. The tables use the symbol $c_p$ for incompressible substances, but for an incompressible $c_p = c_v = c$ and so we'll use the symbol $c$. For a constant specific heat:
$$\Delta u=\int {cdT} = \bar c \Delta T$$Additionally, since the pressure is constant, then the enthalpy change for an incompressible is:
$$\Delta h = \Delta u + v\Delta P = \Delta u = \bar c \Delta T$$c_Tbar = 4.181 # specific heat at average temeprature of 313.15 K [kJ/(kg-K)]
Delta_T = 40 # temperature change [K]
delta_u_b = c_Tbar*Delta_T # change in specific internal energy with spec heat at average temp [kJ/kg]
delta_h_b = delta_u_b # change in specific enthalpy with spec heat at average temp [kJ/kg]
print('delta_u_b = ',round(delta_u_b,2),'(kJ/kg)')
print('delta_h_b = ',round(delta_h_b,2),'(kJ/kg)')
c) Using constant specific heats at $T = T_i = 293.15 K$ requires interpolation between 275 K and 300 K.
c_Ti = 4.188 # specific heat at T = T_i = 293.15 K [kJ/(kg-K)]
delta_u_c = c_Ti*Delta_T # change in specific internal energy with spec heat at initial temp [kJ/kg]
delta_h_c = delta_u_c # change in specific enthalpy with spec heat at initial temp [kJ/kg]
print('delta_u_c = ',round(delta_u_c,2),'(kJ/kg)')
print('delta_h_c = ',round(delta_h_c,2),'(kJ/kg)')
The three approaches give almost the same results!
Consider a constant volume process involving heat addition to a closed system consisting of an ideal gas with no work or changes in kinetic or potential energy.
Is the required heat transfer for raising the temperature from 295 to 305 K the same as the heat transfer required from 345 to 355 K? If not, then what assumption would be needed for them to be the same? Explain.
For a closed system, the governing equation is:
$$\Delta E = \Delta U + \Delta KE + \Delta PE = Q - W$$For the case of no work and no changes in kinetic and potential energy, the above equation reduces to:
$$Q = \Delta U = m\Delta u$$For an ideal gas, $\Delta u = \int_{T_{1}}^{T_{2}} c_{v}dT$. Thus,
$$Q = m\int_{T_{1}}^{T_{2}} c_{v}dT $$In general, the specific heat of a gas will vary with the temperature. Therefore, the heat transfer required to raise the temperature from 345 K to 355 K is not generally equal to the heat transfer required to raise the temperature from 295 K to 305 K due to variable specific heats.
The only way for the heat transfer to be the same would be if the specific heat were constant. If we assume constant specific heat, then we would need to assume the same specific heat for both calculations. Then, for both cases
$$Q_1 = Q_2 = mc_v(20K)$$When assuming constant specific heat, it is generally best to determine the value at the average temperature for the process. Although the average temperatures are different for the two cases, the temperatures are relatively close and the assumption of a single specific heat is very reasonable.
A fixed mass of an ideal gas is heated from 40 °C to 60 °C at a constant pressure of (a) 100 kPa and (b) 300 kPa.
For which case do you think the energy required will be greater? Use thermodynamic relations to explain.
The closed system energy balance for a constant volume process with negligible changes in kinetic and potential energy is developed as
$$\Delta U + \Delta KE + \Delta PE = Q - W \rightarrow \Delta U = Q - W = Q - P \Delta V$$Since $H = U + PV$, then this result can be rewritten as
$$Q = \Delta U + P \Delta V = \Delta (U+PV) = \Delta H = m\Delta h$$Futhermore, the change in specific enthalpy for an ideal gas is only a function of temperature and expressed as
$$\Delta h = \int_{T_{1}}^{T_{2}} c_{p} dT$$So,
$$Q = m\int_{T_{1}}^{T_{2}} c_{p} dT$$Since the specific heat of an ideal gas only depends on temperature and not pressure
$$c_{p,100 kPa, T} = c_{p,300 kPa, T} = c_{p, T}$$and the required heat transfer is the same for both pressures since the initial and final temperatures are the same.
A fixed mass of an ideal gas is heated from 40 to 60°C at a constant volume of (a) 1 $m^3$ and (b) 3 $m^3$
For which case do you think the energy required will be greater? Use thermodynamic relations to explain.
From Part (iii), the energy balance for a constant volume heat addition with negligible changes in kinetic and potential and an ideal gas reduces to
$$Q = m\Delta u = m\int_{T_{1}}^{T_{2}} c_{v}dT $$For a gas obeying the ideal gas model, the specific heat depends only on temperature. Therefore,
$$c_{v,1 m^3, T} = c_{v,3 m^3, T} = c_{v, T}$$and the required heat transfer is the same for both volumes since the initial and final temperatures are the same.
A fixed mass of an ideal gas is heated from 40 to 60°C (a) at constant volume and (b) at constant pressure.
For which case do you think the energy required will be greater? Use thermodynamic relations to explain.
From Part (iii), the heat transfer for an ideal in a closed constant volume system with no changes in kinetic or potential energy is
$$Q = m\Delta u = m\int_{T_{1}}^{T_{2}} c_{v}dT $$From Part (iv), the heat transfer for an ideal in a closed system undergoing a constant pressure process with no changes in kinetic or potential energy is
$$Q = \Delta U + P \Delta V = \Delta (U+PV) = \Delta H = m\Delta h = m\int_{T_{1}}^{T_{2}} c_{p} dT$$The constant pressure heating involves both changes in internal energy and boundary work, whereas the constant volume process involves just changes in internal energy. Since the beginning and ending temperatures are the same for both processes, then the constant pressure process requires more heat transfer. This can also be explained by the fact that $c_p > c_v$. In fact, $c_{p} = c_{v} + R$.
# Given Inputs
T_1 = 300 # initial temperature [K]
T_2 = 400 # final temperature [K]
Compare the change in internal energy using 3 different approaches:
a) Ideal gas tables for air
b) Constant specific heat for air evaluated at the average temperature
c) Constant specific heat for air evaluated at the initial temperature
# Solution
# Values of internal energy from ideal gas tables
u_400 = 286.1 # specific internal energy at 400 K [kJ/kg]
u_300 = 214.1 # specific internal energy at 300 K [kJ/kg]
# Values of specific heat at average temperature of 350 K and initial temperature of 300 K
c_v_350 = 0.721 # specific heat at constant volume at 350 K [kJ/(kg-K)]
c_v_300 = 0.718 # specific heat at constant volume at 300 K [kJ/(kg-K)]
# Changes in internal energy for 3 cases
delta_u_a = u_400 - u_300 # change in specific internal energy, ideal gas tables [kJ/kg]
delta_u_b = c_v_350*(T_2-T_1) # change in specific internal energy, spec heat at T_avg [kJ/kg]
delta_u_c = c_v_300*(T_2-T_1) # change in specific internal energy, spec heat at T_i [kJ/kg]
a) Ideal gas tables for air
$$\Delta u = u_{400 K} - u_{300 K}$$print('delta_u_a = ',round(delta_u_a,2),'(kJ/kg)')
b) Constant specific heat for air evaluated at the average temperature
$$\Delta u = c_{p, 350 K}\Delta T$$print('delta_u_b = ',round(delta_u_b,2),'(kJ/kg)')
c) Constant specific heat for air evaluated at the initial temperature
$$\Delta u = c_{p, 300 K}\Delta T$$print('delta_u_c = ',round(delta_u_c,2),'(kJ/kg)')
The three approaches give almost the same results!