Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exits at the same pressure as saturated liquid. The steam mass flow rate is 1.5 kg/min. A separate stream of air with a mass flow rate of 100 kg/min enters at 30 degree C and exits at 60 degree C. The ideal gas model with c rho = 1.005 kJ/kg K can be assumed for air. Kinetic and potential energy effects are negligible. Determine (a) the quality of the entering steam and (b) the rate of heat transfer between the heat exchanger and its surroundings, in kW.

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.