I need an enthalpy calculation formula to put in a macro datasheet and then this macro do it automaticlly to avoid do another extra calculation in the psychometric chart.

I saw this formula in this forum, but this was put it here two years ago:

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Here's a short BASIC program that calculates the enthalpy using dry bulb temperature and releative humidity.

10 REM ENTHALPY CALCULATION

20 REM Assumes standard atmospheric pressure (14.7 psi), see line 110

30 REM Dry-bulb temperature in degrees F (TEMP)

40 REM Relative humidity in percentage (RH)

50 REM Enthalpy in BTU/LB OF DRY AIR (H)

60 T = TEMP + 459.67

70 N = LN( T ) : REM Natural Logrithm

80 L = -10440.4 / T - 11.29465 - 0.02702235 * T + 1.289036E-005 * T ^ 2 - 2.478068E-009 * T ^ 3 + 6.545967 * N

90 S = LN-1( L ) : REM Inverse Natural Logrithm

100 P = RH / 100 * S

110 W = 0.62198 * P / ( 14.7 - P )

120 H = 0.24 * TEMP + W * ( 1061 + 0.444 * TEMP )

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But the problem is that

According to the formula, the results that I get are not acurate,

maybe

I'm putting something wrong when I calculate the "L" I check everything, But I don't get it.

Can somebody help me?

The air inlet that I have is 100°F and the humidity is 40%RH.

The enthalphy is sopposed to be in 42.9 BTU/LB according to the machine results, but I'm getting 24 BTU/Lb with this formula, what could be wrong????

Does anybody has another formula to calculate the enthalphy of the air to avoid using the psychometric char????

I saw this formula in this forum, but this was put it here two years ago:

---

Here's a short BASIC program that calculates the enthalpy using dry bulb temperature and releative humidity.

10 REM ENTHALPY CALCULATION

20 REM Assumes standard atmospheric pressure (14.7 psi), see line 110

30 REM Dry-bulb temperature in degrees F (TEMP)

40 REM Relative humidity in percentage (RH)

50 REM Enthalpy in BTU/LB OF DRY AIR (H)

60 T = TEMP + 459.67

70 N = LN( T ) : REM Natural Logrithm

80 L = -10440.4 / T - 11.29465 - 0.02702235 * T + 1.289036E-005 * T ^ 2 - 2.478068E-009 * T ^ 3 + 6.545967 * N

90 S = LN-1( L ) : REM Inverse Natural Logrithm

100 P = RH / 100 * S

110 W = 0.62198 * P / ( 14.7 - P )

120 H = 0.24 * TEMP + W * ( 1061 + 0.444 * TEMP )

---

But the problem is that

According to the formula, the results that I get are not acurate,

maybe

I'm putting something wrong when I calculate the "L" I check everything, But I don't get it.

Can somebody help me?

The air inlet that I have is 100°F and the humidity is 40%RH.

The enthalphy is sopposed to be in 42.9 BTU/LB according to the machine results, but I'm getting 24 BTU/Lb with this formula, what could be wrong????

Does anybody has another formula to calculate the enthalphy of the air to avoid using the psychometric char????

hi

Saturation vapour pressure, ps, in pascals:

ps = 610.78 *exp( t / ( t + 238.3 ) *17.2694 )

where t is the temperature in degrees Celsius

The Relative Humidity (RH) is the ratio of the actual water vapour pressure to the saturation water vapour pressure at the prevailing temperature.

RH = p/ps

RH is usually expressed as a percentage rather than as a fraction.

kg water vapour / kg dry air = 0.018 *p / ( 0.029 *(P - p ) )

= 0.62 *p / (P - p )

At room temperature P - p is nearly equal to P, which at ground level is close to 100,000 Pa, so, approximately:

kg water vapour / kg dry air = 0.62 *10-5 *p

The enthalpy of moist air, in kJ/kg, is therefore:

h = (1.007*t - 0.026) + g*(2501 + 1.84*t)

g is the water content in kg/kg of dry air

Saturation vapour pressure, ps, in pascals:

ps = 610.78 *exp( t / ( t + 238.3 ) *17.2694 )

where t is the temperature in degrees Celsius

The Relative Humidity (RH) is the ratio of the actual water vapour pressure to the saturation water vapour pressure at the prevailing temperature.

RH = p/ps

RH is usually expressed as a percentage rather than as a fraction.

kg water vapour / kg dry air = 0.018 *p / ( 0.029 *(P - p ) )

= 0.62 *p / (P - p )

At room temperature P - p is nearly equal to P, which at ground level is close to 100,000 Pa, so, approximately:

kg water vapour / kg dry air = 0.62 *10-5 *p

The enthalpy of moist air, in kJ/kg, is therefore:

h = (1.007*t - 0.026) + g*(2501 + 1.84*t)

g is the water content in kg/kg of dry air

Substituting ARI conditions: 80F DB/67F WB (26.67C/51%RH)in the formula:

h = (1.007*t - 0.026) + g*(2501 + 1.84*t)

g is Relative Humidity (0<RH<1)

t is dry bulb temperature in Celsius

h is enthalpy (kJ/kg)

h = (1.007*26.7 - 0.026) + (0.51)(2501 + 1.84*26.7) = 1327

The answer should be about 55.5 kJ/kg...

h = (1.007*t - 0.026) + g*(2501 + 1.84*t)

g is Relative Humidity (0<RH<1)

t is dry bulb temperature in Celsius

h is enthalpy (kJ/kg)

h = (1.007*26.7 - 0.026) + (0.51)(2501 + 1.84*26.7) = 1327

The answer should be about 55.5 kJ/kg...

The formula is correct, except for the 2501 constant, that should read 2.501 (the initial poster forgot the dot!). Anyway, for very low and very high values of enthalpy the formula gives inaccurate data.

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