A pipe operates at steady-state, steady-flow conditions. R-134a enters the pipe at 120oF and 60 psia, and exits at 50 psia and a quality of 0.90. The volumetric flow rate of the R-134a entering the pipe is 1.75 ft3/s. Assume that there are no work interactions. (a) Determine the heat transfer rate to or from the pipe. (b) Using your equations, plot the rate of heat transfer for the same inlet conditions and outlet pressure, but for outlet qualities varying between 0.0 and 1.0.

Question

Given: R-134a; T1 = 120oF; P1 = 60 psia; P2 = 50 psia; x2 = 0.90; V?1 = 1.75 ft3/s
Steady-state, steady-flow; (Inlet = State 1; Outlet = State 2)
Assume: W?= 0 ; neglect changes in kinetic and potential energy

What will be an ideal response?

Answer 1

(a) For a single inlet, single-outlet, steady-state, steady-flow device:



With no power, and no changes in kinetic and potential energy:

Q?= m? (h2 ? h1)

The mass flow rate can be found from the volumetric flow rate of the inlet:

m? =V?1/v1

The specific volume of the inlet, and the enthalpy values of the inlet and outlet can be

found with on-line property program:

v1 = 0.9482 ft3/lbm

h1 = 124.88 Btu/lbm

h2 = 99.10 Btu/lbm

So, m? = (1.75 ft3/s) / (0.9482 ft3/lbm) = 1.846 lbm/s

Q? = (1.846 lbm/s)(99.10 – 124.88) Btu/lbm = -47.6 Btu/s

(b)