Answer:
m = 0.4005 kg/s
Q_out = 45.1 KJ/s
Explanation:
Given
Pipe inlet diameter D1 = 16 cm
Steam inlet pressure P1 = 2 Mpa
Steam inlet temperature T1 = 300 °C
Pipe outlet diameter D2 = 14 cm
Steam inlet velocity V1 = 2.5 m/s
Steam outlet pressure P2 = 1.8 MPa
Steam outlet temperature T2 = 250 °C
Required
Determine
(a) The mass flow rate of steam.
(b) The rate of heat transfer.
Assumptions
Kinetic and potential energy changes are negligible.
This is a steady flow process.
There is no work interaction.
Solution
Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C
vl = 0.12551 m^3/Kg
h1 = 3024.2 KJ/Kg
The mass flow rate of steam could be defined as the following
m = 1/v1*A1*V1
m = 0.4005 kg/s
Part b We take the pipe as our system.The energy balance could be defined as the following
E_in -E_out =ΔE_sys = 0
E_in = E_out
mh_1 = Q_out + mh_2
Q_out = m(h_1-h_2)
From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C h2= 2911.7 KJ/Kg The heat transfer could be defined as the following
Q_out = m(h_1-h_2)
Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s
