Question

我正在尝试创建一个简单的程序来模拟供暖通风应用中的可变风扇。下面的脚本仅模仿基于风扇启动和启动的管道压力读数（以英寸水柱表示）。基本上，风扇会启动，管道压力将为零。当电动机加速时，风管压力将每10秒缓慢上升，每10秒将产生.1“ WC的值。当风扇的风管压力设定值为1.5时，程序将生成一个大约1.3-1.7的随机数，以模仿风扇在设定点附近徘徊。希望这是有道理的！

我觉得这可以写得更好一些，但形式最简单：

import time
from numpy.random import seed
from numpy.random import randint

# seed random number generator
seed(1)

#generate random number to mimic fan hovering at setpoint
def ductRandStatic():
    value = randint(13, 17, 20)
    return value * .1

ductStaticStart = 0
ductStaticEnd = 1.5

#mimic fan ramping up to setpoint slowly
def startFan():
    static = ductStaticStart + .1
    time.sleep(10)
    if static < ductStaticEnd:
        static = ductRandStatic()
    else:
        static = static
    print(static)


while True:
    startFan()

由于某种原因，该程序仅打印我认为随机数生成器...我希望它能每10秒打印一次模拟风管压力的值... [1.5 1.6 1.6 1.5 1.4 1.4 1.4 1.6 1.3 1.3 1.4 1.6 1.3 1.5 1.3 1.3 1.4 1.6 1.4 1.5]

Answer 1

我认为您想摆脱startFan()函数并将其大部分代码移入主循环，以便static可以记住其值：

# initialize to 0.1
static = 0.1

# mimic fan ramping up to setpoint slowly
while True:
    if static < 1.5:
        static = ductRandStatic()
    print(static)
    time.sleep(10)

另外，您对randint()的调用是错误的-应该有两个整数，但是要传递三个整数。

Answer 2

您的仿真有两个阶段：启动（一系列固定值）和维持（您的随机变化）。这样的代码：

from time import sleep
from numpy.random import randint


def ductRandStatic():
    return randint(13, 17) * 0.1


delay = 10

for i in range(16):
    pressure = i * 0.1
    print(pressure)
    sleep(delay)

while True:
    print(ductRandStatic() )
    sleep(delay)

输出：

0.0
0.1
0.2
0.30000000000000004
0.4
0.5
0.6000000000000001
0.7000000000000001
0.8
0.9
1.0
1.1
1.2000000000000002
1.3
1.4000000000000001
1.5
1.5
1.6
1.4000000000000001
1.4000000000000001
1.4000000000000001
1.6
1.4000000000000001
1.4000000000000001
1.6
1.4000000000000001
1.3
1.4000000000000001
1.3
1.3
1.5
1.3
^CTraceback (most recent call last):
  File "so.py", line 18, in <module>
    sleep(delay)
KeyboardInterrupt

风扇模拟程序HVAC

2 个答案: