ME 200 – Thermodynamics I – Spring 2020¶
Homework 32: Reversible Steady Flow Processes¶
Part (i): Compressor¶
Part (ii): Pump¶
#Given Inputs:
P_1 = 100 # compressor inlet pressure [kPa]
T_1 = 300 # compressor inlet temperature [K]
P_2 = 1000 # compressor outlet pressure [kPa]
R = 0.287 # ideal gas constant for air [kJ/kgK]
Find:¶
a) The specific work (kJ/kg)
b) The exit temperature (K)
c) The heat transfer (kJ/kg)
Depict the process on a T-s diagram.
Assumptions:¶
1) open system, 2) steady state, steady flow (SSSF) 3) air is an ideal gas, 4) negligible changes in kinetic and potential energy, 5) internally reversible
Basic Equations:¶
$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$$$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$$$w_{int,rev,flow}=-\int v dP + (V_1^2-V_2^2)/2+g(z_1-z_2)$$$$PV=mRT$$Solution:¶
A mass balance is applied to the flow stream with the assumption of steady flow ($dm/dt=0$).
$$\dot m_{2} = \dot m_{1} = \dot m$$An energy balance is applied to the ystem with the assumption of SSSF ($dE/dt=0$) and negligible changes in potential and kinetic energy. The simplified energy balance becomes:
$$\dot Q_{12}= \dot m (h_2-h_1)-\dot W_{12}$$$$ q_{12}= (h_2-h_1)- w_{12}$$The specific work is found by simplifying the equation for internally reversible SSSF work by applying the assumption of negligible kinetic and potential energy changes.
$$w_{int,rev,flow}=-\int v dP$$Since the undergoes a polytropic process as it flows through the compressor, then $Pv^{n}=constant$ where $n=1.1$. Then substituting the expression
$$v=v_1\dfrac{P_1^{1/n}}{P^{1/n}}$$into the reversible work equation, integrating between the inlet and outlet conditions, and then simplying leads to
$$w_{12}=\dfrac{-n}{n-1}(P_2v_2-P_1v_1)$$Also, the ideal gas law can be used to further this result as
$$w_{12}=\dfrac{-n}{n-1}R(T_2-T_1)$$The values for $v_2$ and $v_1$ can be found using the following equations:
$$v_1=\dfrac{RT_1}{P_1}$$$$v_2=v_1\dfrac{P_1^{1/n}}{P_2^{1/n}}$$The temperature of the air leaving the compressor can be found using the ideal gas law.
$$T_2 = \dfrac{P_2v_2}{R}$$The following code evaluates these resulting equations to determine the exit temperature, specific work, and specific heat transfer.
v_1 = R*T_1/P_1 # specific volume of air at state 1 [m^3/kg]
h_1 = 300.1 # specific enthalpy of air at state 1 [kJ/kg]
v_2 = v_1*(P_1/P_2)**(1/1.1) # specific volume of air at state 2 [m^3/kg]
T_2 = P_2*v_2/R # temperature of air at state 2 [K]
n = 1.1
w_12 = -n/(n-1)*R*(T_2-T_1) # specific compressor work [kJ/kg]
h_2 = 370.7 # specific enthalpy of air at state 2 [kJ/kg]
q_12 = h_2-h_1 + w_12 # specific heat transfer for compressor [kJ/kg]
print('v_1 = ',round(v_1,5),' m^3/kg')
print('v_2 = ',round(v_2,5),' m^3/kg')
print('T_2 = ',round(T_2,1),' K')
print('w_12 = ',round(w_12,1),' kJ/kg')
print('q_12 = ',round(q_12,1),' kJ/kg')
#Given Inputs:
T_1 = 20+273 # pump inlet temperature [K]
P_1 = 100 # pump inlet pressure [kPa]
T_2 = 18+273 # pump outlet temperature [K]
P_2 = 600 # pump outlet pressure [kPa]
System Diagram:¶
Assumptions:¶
1) open system, 2) steady state, steady flow (SSSF), 3) water is incompressible, 4) constant specific heat, 5) internally reversible
Basic Equations:¶
$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$$$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$$$w_{int,rev,flow}=\int ^2_1vdP+(V^1_1-V^2_2)/2+g(z_1-z_2)$$$$dh=CdT+vdP$$Solution:¶
A mass balance is applied to the flow stream with the assumption of steady flow ($dm/dt=0$).
$$\dot m_{2} = \dot m_{1} = \dot m$$An energy balance is applied to the system with the assumption of SSSF ($dE/dt=0$), along with no change in potential and kinetic energy. The simplified energy balance becomes:
$$\dot Q_{12}= \dot m (h_2-h_1)-\dot W_{12}$$$$ q_{12}= (h_2-h_1)- w_12$$The enthalpy change can be determined by integraing the following relation for an incompressible fluid:
$$dh=du+vdP$$for the assumption of constant specific heat to get the following result.
$$\int_1^2 dh=\int_1^2 CdT+\int_1^2 vdP$$$$h_2-h_1=C(T_2-T_1)+v(P_2-P_1)$$The energy balance can then be further simplified:
$$ q_{12}= C(T_2-T_1)+v(P_2-P_1)- w_{12}$$The specific work is found by simplifying the equation for internally reversible SSSF work with the assumptions of negligible kinetic and potential energy changes.
$$w_{int,rev,flow}=-\int v dP$$The equation can then be integrated for constant specific volume (incompressible assumption):
$$w_{int,rev,flow}=v(P_2-P_1)$$The value for the specific volume of incompressible water can be taken from the SLVM table.
$$v=v_f(T)$$Combining the equation for energy conservation and the equation for work leads to:
$$ q_{12}= C(T_2-T_1)+v(P_2-P_1)- v(P_2-P_1)$$or
$$ q_{12}= C(T_2-T_1)$$v_f = 0.0010018 # specific volume of water at state 1 [m^3/kg]
w_12 = -v_f*(P_2-P_1) # specific pump work [kJ/kg]
C = 4.18 # specific heat of water [kJ/kgK]
q_12 = C*(T_2-T_1) # specific heat transfer [kJ/kg]
print('w_12 = ',round(w_12,3),'kJ/kg')
print('q_12 = ',round(q_12,3),'kJ/kg')