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What is RSHF, GSHF, ERSHF, BPF?

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Define RSHF, ERSHF. Explain how to draw RSHF, ERSHF and GSHF lines on the psychrometric chart

Subject:- Refrigeration and Air Conditioning

Topic:- Design of air conditioning systems

Difficulty:- Low

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RSHF

It is defined as the ratio of room sensible heat to room total heat. The supply air having conditions given by any point on this line will satisfy requirements of the room with quantity of air supplied different for different points

$RSHF=\frac{RSH}{RSH+RLH}$

GSHF

It is defined as ratio of total sensible heat to the grand total heat which the cooling coil is required to handle after the outside fresh air and recirculated air mixing has taken place.

GSHF=$\frac{(Room \ sensible \ heat \ + \ Outside \ air \ sensible \ heat)}{((RSH \ +ouside \ air \ sensible \ heat)+(RLH+ \ outside \ air \ latent \ heat))}$ The BPF fraction of outside air adds SH load and LH load to the room. However BPF fraction of return air has no effect as it is already at room condition. Thus,

Effective room sensible heat load (ERSH) =RSH+BPF x outside air SH

Effective room latent heat load (ERLH) =RLH+BPF x outside air LH

Hence, ERSHF=$\frac{ERSH}{(ERSH+ERLH)}$ BPF: Given:

$t_d1=30℃,t_d2=20℃,RH_1=55%,p_b=1.0132 bar=1.0132×100 kPa$

$ω_1-ω_2=0.004 \frac{(kg \ of \ water \ vapour)}{(kg \ of \ dry \ air)}$

To find:

$RH_2$

$DPT_2$ or $t_dp2$

We know,

Humidity ratio =

ω=0.622×$\frac{p_v}{(p_b-p_v )}$

∴$ω_1-ω_2=0.622[\frac{p_v1}{(p_b-p_v1 )}-\frac{p_v2}{(p_b-p_v2 )]}]$

∴0.004=0.622$[\frac{p_v1}{(1.0132×100-p_v1 )}-\frac{p_v2}{(1.0132×100-p_v2 )]}]$……..(1)

Also,

$RH_1= \frac{p_v1}{p_vs1}$ ……(2)

Now,$p_{vs1}=p_(sat_{t_d1})$=0.04242 bar (from steam table)

Substituting in (2) we get,

$p_v1=0.023331 bar=0.023331×100$ kPa

Substituting in (1) we get,

$p_v2=1.70725$ kPa

Now, $RH_2= \frac{p_v2}{p_vs2}$

Substituting,

$p_v2$ and $p_vs2$=$p_{sat_{t_{d2}}}$=2.337 kPa (from steam table)

We get, RH_2=73.05%

We know, $DPT_2$ or $t_dp2=T_{sat{p_v2}}$

From steam table using interpolation we find T_sat at pressure of 0.0170725 bar,

$\frac{0.02-0.0170725}{0.02-0.015}=\frac{17.51-T_sat}{17.51-13.04}$

∴$DPT_2$ or $t_dp2$=$T_{sat_{p_v2 }}$=14.893℃