Question

A swine nursery building is being designed to hold 300 pigs weighing 50 pounds each. Water vapor production in the building (from animal respiration and evaporation from wet surfaces) is expected to be 40.2 lbs H2O/hr. Outside air at 30 degrees F dry bulb and 50% RH is to be blown through a fan into the building. The air inside the building is to be at 65 degrees F dry bulb and 70% RH.

Find:
A) Volumetric flow rate (cfm) of air that the fan should move to remove moisture from the building, and B) The rate at which heat (BTU/hr) will be removed from the building in the ventilation air.

Answer 1

The volumetric flow rate (V) = 1110.648 cfm

The rate at which heat will be removed is 8929.887 Btu/hr

Step-by-step explanation:

Given that :

the mass of each pig = 50 lb

water vapor production
`M_w` = 40.2 lbs/hr

Dry bulb temperature of the outside air
`T \phi_1` = 30° F

Relative humidity of outside air ,
`\phi_1` = 50%

Dry bulb of inside air
`T \phi_2` = 65° F

Relative humidity of inside air
`\phi_2` = 70%

Obtaining the following properties of air;

At
`T \phi_1` = 30° F and
`\phi_1` = 50%

Enthalpy
`h_1` = 9.0429 Btu/lb

Humidity ratio
`\omega _1` = 0.00173 lb/lb

Specific volume
`v_1` = 12.37 ft³/lb

At
`T \phi_2` = 65° F and
`\phi_2` = 70%

`h_2` = 25.618 Btu/lb

`\omega_2` = 0.009192 lb/lb

`v_2` = 13.41 ft³/lb

The mass balance equation is given as:

`M_a \omega_1 + M \omega_2 = M_a \omega_2`

`M_a (0.00173)+ 40.2 = M_a (0.009192)`

`40.2 = M_a (0.009192) - M_a (0.00173) \\ \\ 40.2 = M_a(0.007462) \\ \\ M_a = (40.2)/(0.007462)`

`M_a =` 5387.291 Btu/hr

where:

`M_a` = the mass of the dry air

NOW;

Volumetric flow rate (V) =
`M_a *v_1`

V = 5387.291 × 12.37

V = 66.460.8496 ft³/hr

V = 1110.648 cfm

B)

The rate at which heat (BTU/hr) will be removed from the building in the ventilation air is determined as follows:

`Q = M_a(h_2-h_1)`

`Q =5387.291*(25.618-9.0429)`

Q = 8929.887 Btu/hr

Thus; the rate at which heat will be removed is 8929.887 Btu/hr