A steady-state, steady flow thermodynamic system receives 100 lbm/min of a fluid at 30 psia and 200F and discharges it from a point 80 ft above the entrance section at 150 psia and 600F. The fluid enters with a velocity of 7,200 fpm and leaves with a velocity of 2,400 fpm. During this process, there are supplied 25,000 Btu/hr of heat from an external source, and the increase in enthalpy is 2.0 Btu/lb. Determine the work done in horsepower.

To solve this problem, we need to use the first law of thermodynamics for a steady-state, steady-flow process. This law can be expressed as:

Q - W = ΔH + ΔKE + ΔPE

where: Q is the heat transfer, W is the work done, ΔH is the change in enthalpy, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy.

Given in the problem are: Q = 25,000 Btu/hr, ΔH = 2.0 Btu/lb, mass flow rate = 100 lbm/min, inlet pressure = 30 psia, outlet pressure = 150 psia, inlet temperature = 200F, outlet temperature = 600F, inlet velocity = 7,200 fpm, outlet velocity = 2,400 fpm, and the height difference = 80 ft.

First, we need to convert all the given values to consistent units.

The heat transfer Q is given in Btu/hr, which can be converted to Btu/min by dividing by 60, giving Q = 416.67 Btu/min.

The velocities are given in feet per minute (fpm), which can be converted to feet per second (fps) by dividing by 60, giving inlet velocity = 120 fps and outlet velocity = 40 fps.

The height difference is given in feet, which can be converted to Btu/lb by multiplying by the conversion factor 0.0000361273 Btu/lb-ft, giving ΔPE = 2.89 Btu/lb.

Next, we can calculate the change in kinetic energy ΔKE. The kinetic energy of a fluid is given by KE = 0.5 * v^2, where v is the velocity. Therefore, ΔKE = 0.5 * (outlet velocity^2 - inlet velocity^2) = 0.5 * (40^2 - 120^2) = -12,000 ft^2/s^2. This can be converted to Btu/lb by multiplying by the conversion factor 0.00001427643 Btu/lb-ft^2/s^2, giving ΔKE = -0.171 Btu/lb.

Finally, we can substitute all these values into the first law of thermodynamics to find the work done W:

416.67 Btu/min - W = 2.0 Btu/lb * 100 lbm/min + (-0.171 Btu/lb * 100 lbm/min) + 2.89 Btu/lb * 100 lbm/min

Solving this equation for W gives W = 416.67 - 200 + 17.1 - 289 = -89.33 Btu/min.

This is the work done by the system, but the problem asks for the work done in horsepower. We can convert Btu/min to horsepower by multiplying by the conversion factor 0.0235808 hp/Btu/min, giving W = -89.33 * 0.0235808 = -2.11 hp.

Therefore, the work done is -2.11 horsepower. The negative sign indicates that work is done on the system.